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UBC Theses and Dissertations

The relationship between volume conservation and a volume algorithm for a rectangular parallelepiped Feghali, Issa Nehme 1979

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THE RELATIONSHIP BETWEEN VOLUME CONSERV&TION AND ft VOLUHE ALGORITHM FOR A RECTANGULAR PARALLELEPIPED by ISSA NEHME FEGHALI B.Sc. , C a l i f o r n i a S t a t e P o l y t e c h n i c U n i v e r s i t y , M.S., Baylor U n i v e r s i t y , 1972 1971 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF . THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF EDUCATION i n THE FACULTY OF GRADUATE STUDIES (Mathematics Education) We accept t h i s t h e s i s as conforming to the r e q u i r e d standard THE UNIVERSITY OF BRITISH COLUMBIA October 1979 @ Issa N. F e g h a l i , 1979 -In p r e s e n t i n g t h i s t h e s i s i n p a r t i a l f u l f i l m e n t o f the r e q u i r e m e n t s f o r an advanced degree at the U n i v e r s i t y o f B r i t i s h Co l umb i a , I ag r ee tha t the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r r e f e r e n c e and s t u d y . I f u r t h e r ag ree t h a t p e r m i s s i o n f o r e x t e n s i v e c o p y i n g o f t h i s t h e s i s f o r s c h o l a r l y pu rpo se s may be g r a n t e d by the Head o f my Department o r by h i s r e p r e s e n t a t i v e s . I t i s u n d e r s t o o d t h a t c o p y i n g o r p u b l i c a t i o n o f t h i s t h e s i s f o r f i n a n c i a l g a i n s h a l l not be a l l o w e d w i t h o u t my w r i t t e n p e r m i s s i o n . Depar tment o f M a t h w n a M r s E d u c a t i o n The U n i v e r s i t y o f B r i t i s h Co l umb i a 2075 Wesbrook Place Vancouver, Canada V6T 1WS Date D e c 2 4 , 1979 ABSTRACT Chairman: Dr. Douglas Owens T h i s study was designed t o i n v e s t i g a t e the r e l a t i o n s h i p between the l e v e l of c o n s e r v a t i o n of d i s p l a c e d volume and the degree t o which s i x t h grade c h i l d r e n l e a r n the volume a l g o r i t h m of a c u b o i d , "Volume = Length X Width X Height (V = L x W x H ) f " a t the knowledge and comprehension l e v e l s . The problem i s a consequence of an apparent d i s c r e p a n c y between present s c h o o l programs and P i a g e t * s theory c o n c e r n i n g the grade l e v e l at which t h i s a l g o r i t h m i s i n t r o d u c e d . While some s c h o o l programs i n t r o d u c e the a l g o r i t h m as e a r l y as grade 4, P i a g e t (1960) c l a i m s that i t i s not u n t i l the formal o p e r a t i o n a l stage t h a t c h i l d r e n understand how they can f i n d volume by m u l t i p l y i n g the boundary measures. Very few c h i l d r e n i n grade 4 are expected t o e x h i b i t formal o p e r a t i o n s . In such a predicament t h e r e seems t o be a need f o r r e s e a r c h i n order to j u s t i f y our present s c h o o l c u r r i c u l u m or to suggest m o d i f i c a t i o n s . Subjects of t h r e e suburban s c h o o l s i n B r i t i s h Columbia were c l a s s i f i e d as nonconservers (N = 57), p a r t i a l c o n s ervers (N = 16) and conservers (N = 32) using a judgement-based t e s t of volume c o n s e r v a t i o n . The s u b j e c t s were then d i v i d e d i n t o two e x p e r i m e n t a l groups and one c o n t r o l group by randomizing each i i i c o n s e r v a t i o n group ac r o s s the three treatments. One of the experimental groups (N = 36) was taught the volume a l g o r i t h m using an approach (Volume Treatment) which resembles t h a t of s c h o o l programs used i n North America. A c t i v i t i e s of t h i s treatment i n c l u d e d comparison, o r d e r i n g , and f i n d i n g the volume of cuboids by counting cubes and l a t e r by using the a l g o r i t h m "V = L x W x H." The other experimental group (N = 39) was taught the a l g o r i t h m using a method t h a t emphasized m u l t i p l i c a t i o n s k i l l s ( M u l t i p l i c a t i o n Treatment). T h i s treatment i n c l u d e d t r a i n i n g on compensating f a c t o r s with r e s p e c t to v a r i a t i o n s i n other f a c t o r s and was supplemented by a b r i e f d i s c u s s i o n of the volume a l g o r i t h m . The c o n t r o l group (N = 30) was taught a u n i t on numeration systems. Four d i f f e r e n t t e s t s were used: Volume Conservation (11 i t e m s ) , Volume achievement (27 i t e m s ) , M u l t i p l i c a t i o n Achievement (20 items) and the computation s e c t i o n (45 items) o f the S t a n f o r d Achievement T e s t . The p r e t e s t s were: Volume C o n s e r v a t i o n , Volume Achievement, and Computation. The p o s t t e s t s and r e t e n t i o n t e s t s were: Volume C o n s e r v a t i o n , Volume Achievement, and M u l t i p l i c a t i o n Achievement. Data from the p o s t t e s t s and r e t e n t i o n t e s t s were analyzed s e p a r a t e l y u s i n g a 3 X 3 f u l l y c r o s s e d two-way a n a l y s i s of c o v a r i a n c e . S u b j e c t s i n the volume treatment showed they were a b l e to apply the volume al g o r i t h m t o computation and comprehension q u e s t i o n s r e g a r d l e s s of t h e i r c o n s e r v a t i o n l e v e l . On the p o s t t e s t and r e t e n t i o n t e s t , s u b j e c t s o f t h i s group showed a 65 per c e n t performance l e v e l . For the grade 6 students i n the study, c o n s e r v a t i o n l e v e l was not a s i g n i f i c a n t f a c t o r i n l e a r n i n g the volume a l g o r i t h m a t the - computation and comprehension l e v e l s . On the p o s t t e s t , s u b j e c t s of the m u l t i p l i c a t i o n treatment performed s i g n i f i c a n t l y (F = 10,33, p < 0.01) b e t t e r than those i n the other groups on the M u l t i p l i c a t i o n achievement T e s t . Subjects of the volume treatment d i d s i g n i f i c a n t l y (F = 12.24, p < 0.01) b e t t e r than those i n the other groups on the Volume Achievement P o s t t e s t . I t seems a p p r o p r i a t e , t h e r e f o r e , to teach the volume a l g o r i t h m of a cuboid u s i n g a method t h a t i n c l u d e s students* a c t i v e involvement i n manipulating p h y s i c a l o b j e c t s . There was, g e n e r a l l y , an improvement of the students* c o n s e r v a t i o n l e v e l s r e g a r d l e s s of t h e i r volume achievement s c o r e s or treatments. The t r a n s i t i o n from a lower to a higher l e v e l of c o n s e r v a t i o n was found a) independent of treatments between the p r e t e s t and each of the p o s t t e s t ( X 2 = 0.93, df = 2) and r e t e n t i o n t e s t ( X 2 = 0.97, df = 2) and b) independent of volume achievement scores between the p r e t e s t and each of the p o s t t e s t ( b i s e r i a l r = 0.13) and r e t e n t i o n t e s t ( b i s e r i a l r = 0. 09) . In an addendum to the C o n s e r v a t i o n T e s t students were asked t o w r i t e reasons f o r t h e i r judgements i n items i n v o l v i n g equal and unequal volumes. Those w r i t t e n reasons were more e x p l i c i t on the items of unequal volumes than of equal volumes. V TABLE OF CONTENTS Page ABSTRACT . i i LIST OF TABLES ,. i x ACKNOWLEDGEMENTS x i i i CHAPTER I THE PROBLEM , 1 D e f i n i t i o n of Terms ................................ 3 Statement of the Problem 5 J u s t i f i c a t i o n of the Problem ....................... 7 I m p l i c a t i o n s of P i a g e t ' s C o g n i t i v e Theory 7 V a l i d a t i o n c f some of Pi a g e t ' s F i n d i n g s ......... 10 Need f o r Research ............................... 11 CHAPTER I I REVIEW OF RELATED LITERATURE ................. 13 Summary of P i a g e t ' s Theory 13 Pi a g e t ' s S t u d i e s of Volume ...................... 19 Rea c t i o n s to P i a g e t ' s Theory and Experiments ....... 23 T r a i n i n g Experiments ............................... 25 D i s c u s s i o n of T r a i n i n g Experiments other than Volume ....................................... 25 D i s c u s s i o n c f Experiments I n v o l v i n g Volume Conservation Tasks ........................... 30 Summary and I m p l i c a t i o n s of L i t e r a t u r e Reviewed .... 34 v i Age of Subjects ................................. 34 Nature of Volume Conservation T e s t s ............. 35 Choice of Treatments ............................ 39 C o n c l u s i o n 43 CHAPTER I I I PROCEDURES 45 Sub j e c t s . . 4 5 D e s c r i p t i o n of the Treatments ...................... 46 Volume Treatment 46 M u l t i p l i c a t i o n Treatment 47 C o n t r o l Treatment ............................... 48 D e s c r i p t i o n o f Tests ............................... 49 I n s t r u c t i o n and T e s t i n g Procedures ................. 55 I n s t r u c t o r s ............. ........................ 55 Schedule of I n s t r u c t i o n and T e s t i n g ............. 55 Design of the Study ................................ 57 Hypotheses .........................................58 S t a t i s t i c a l Analyses 60 CHAPTER IV RESULTS 62 Test s R e l i a b i l i t i e s and Item A n a l y s i s 62 Volume Achievement T e s t ......................... 63 Volume Conservation Test ........................ 63 M u l t i p l i c a t i o n Achievement Te s t ................. 63 P r e l i m i n a r y A n a l y s i s ............................... 64 C o v a r i a t e s ...................................... 64 Assumptions of A n a l y s i s of Covariance ............ 65 I n s t r u c t o r E f f e c t ...66 A n a l y s i s of Covariance ............................. 68 v i i Hypotheses 1-3 ., 68 Hypothesis 4 73 C o r r e l a t i o n Study 74 Hypotheses 5 and 7 .............................. 74 Hypotheses 9 and 12 ............................. 77 Hypothesis 10 78 Te s t s o f Independence 79 Hypotheses 6 and 8 79 Hypothesis 11 81 Post Hoc Q u a l i t a t i v e Analyses ....................... 82 T r a n s i t i o n between C o n s e r v a t i o n L e v e l s .......... 82 Students' Reasons f o r t h e i r Responses on Question 11 86 Co n s i s t e n c y between L e v e l s of Co n s e r v a t i o n and Responses t o Question 12 of the Volume Cons e r v a t i o n Test ............................ 90 Students' Reasons f o r t h e i r Responses on Question 12 92 CHAPTER V SUMMARY, CONCLUSIONS AND RECOMMENDATIONS ...... 97 Review of the Problem .............................. 97 F i n d i n g s and C o n c l u s i o n s ...........................99 Summary of C o n c l u s i o n s and D i s c u s s i o n 102 Importance of Conservation L e v e l s ............102 E f f e c t of Treatments .........................103 T r a n s i t i o n between Con s e r v a t i o n L e v e l s .......105 E f f e c t o f Mathematics Achievement .............106 E f f e c t of Sex .106 v i i i L i m i t a t i o n s o f the Study - .................107 I m p l i c a t i o n s f o r E d u c a t i o n a l P r a c t i c e ..............109 Eecommendations f o r Future Research ...110 REFERENCES ....114 APPENDIX A D e s c r i p t i o n of the I n s t r u c t i o n a l U n i t s ......123 APPENDIX B Item S t a t i s t i c s 191 APPENDIX C Raw Data ..195 APPENDIX D Unadjusted D e s c r i p t i v e S t a t i s t i c s ...........199 APPENDIX E Tests .201 i x LIST OF TABLES Table Page 3.1 Experimental Design ................................. 57 4.1 F Values f o r E n t e r i n g of C o v a r i a t e s 65 4.2 A n a l y s i s of Covariance - I n s t r u c t o r s E f f e c t ......... 68 4.3 A n a l y s i s of Covariance of Volume Achievement P o s t t e s t Scores ................................................ .71 4.4 A n a l y s i s of Covariance of Volume Achievement Re t e n t i o n Scores 71 4.5 Means, Standard D e v i a t i o n s , and Group S i z e s of Volume Achievement P r e t e s t Scores f o r Treatments by Con s e r v a t i o n L e v e l s ................................... 71 4.6 Adjusted Means, Standard D e v i a t i o n s , and Group S i z e s of Volume Achievement P o s t t e s t Scores f o r Treatments by Con s e r v a t i o n L e v e l s 72 4.7 Adjusted Means, Standard D e v i a t i o n s , and Group S i z e s of Volume Achievement Hetention Test Scores f o r Treatments by Conservation L e v e l s ..................... 72 4.8 A n a l y s i s of Covariance of M u l t i p l i c a t i o n P o s t t e s t S c o r e s . . . 73 4.9 A n a l y s i s of Covariance of M u l t i p l i c a t i o n Retention Scores ................................................ 74 X 4.10 Adjusted Means, Standard D e v i a t i o n s , and Group S i z e s of M u l t i p l i c a t i o n Achievement P o s t t e s t and Be t e n t i o n Test Scores f o r Treatments 74 4.11 B i s e r i a l C o r r e l a t i o n C o e f f i c i e n t s Between Volume Achievement Scores on the P o s t t e s t and B e t e n t i o n T e s t and T r a n s i t i o n to a Higher or Lower L e v e l of Co n s e r v a t i o n .......................................... 76 4.12 C e l l S i z e s of Co n s e r v a t i o n L e v e l s X Treatments on the P r e t e s t , P o s t t e s t and B e t e n t i o n Test .................. 76 4.13 Number of Su b j e c t s Whose Cons e r v a t i o n L e v e l s Changed/ Did Not Change betwween the P r e t e s t and each of the Post and B e t e n t i o n T e s t ............................... 77 4.14 Pearson's Product Moment C o r r e l a t i o n C o e f f i e n t s and S i g n i f i c a n c e L e v e l s Between Volume Achievement Scores and SAT Scores on the P o s t t e s t and the B e t e n t i o n T e s t . 78 4.15 P o i n t - B i s e r i a l C o r r e l a t i o n C o e f f i e n t s and Sign ...... 78 4.16 Contingency Table: T r a n s i t i o n Op, Down, o r S t a y i n g the Same Versus Treatments Between P r e t e s t - P o s t t e s t and P r e t e s t - R e t e n t i o n Test ................................ 80 4.17 Contingency Table: T r a n s i t i o n Op Versus Treatments on the P o s t t e s t and the B e t e n t i o n Test ................... 80 4.18 Contingency Table: T r a n s i t i o n Down Versus Treatments on the P o s t t e s t and the B e t e n t i o n Test ................ 81 4.19 Contingency Tahle: P r e t e s t C o n s e r v a t i o n L e v e l Versus Sex 82 4.20 Contingency Table: P r e t e s t - P o s t t e s t T r a n s i t i o n Among Con s e r v a t i o n L e v e l s by Treatments ..................... 85 4.21 Contingency Table: P r e t e s t - R e t e n t i o n Test T r a n s i t i o n Among Co n s e r v a t i o n L e v e l s by Treatments ............... 85 4.22 Contingency Table: P o s t t e s t - R e t e n t i o n T e s t T r a n s i t i o n Among Conservation L e v e l s by Treatments ............... 86 4.23 Number of Students with Respect to t h e i r Reasons f o r t h e i r Responses on Question 11 of the Volume Cons e r v a t i o n P r e t e s t .................................. 88 4.24 Number of Students with Respect to t h e i r Reasons f o r t h e i r Responses on Question 11 of the Volume Co n s e r v a t i o n P o s t t e s t .................................89 4.25 Number of Students with Respect to t h e i r Reasons f o r t h e i r Responses on Question 11 of the Volume Cons e r v a t i o n R e t e n t i o n Test ............................ 89 4.26 Contingency Table: C o r r e c t or I n c o r r e c t Response on Item 12 of the Volume Con s e r v a t i o n P r e t e s t , P o s t t e s t and R e t e n t i o n T e s t Versus C o n s e r v a t i o n L e v e l s ......... 92 4.27 Number of Students with Respect to t h e i r Reasons f o r t h e i r Responses on Question 12 of the Volume Cons e r v a t i o n P r e t e s t . • 95 4.28 Number of Students with Respect to t h e i r Reasons f o r t h e i r Responses on Question 12 of the Volume Con s e r v a t i o n P o s t t e s t ................................. 95 4.29 Number of Students with Respect to t h e i r Reasons f o r t h e i r Responses on Question 12 of the Volume Co n s e r v a t i o n R e t e n t i o n Test ........................... 96 B1 Volume Achievement P r e t e s t Item S t a t i s t i c s ............ 191 B2 Volume C o n s e r v a t i o n P r e t e s t Item S t a t i s t i c s 191 B3 Volume achievement P o s t t e s t Item S t a t i s t i c s ,...192 B4 Volume Con s e r v a t i o n P o s t t e s t Item S t a t i s t i c s ..........192 B5 M u l t i p l i c a t i o n P o s t t e s t Item S t a t i s t i c s 193 B6 Volume achievement B e t e n t i o n Test Item S t a t i s t i c s 193 B7 Volume Con s e r v a t i o n B e t e n t i o n T e s t Item S t a t i s t i c s .,..194 B8 M u l t i p l i c a t i o n B e t e n t i o n T e s t Item S t a t i s t i c s .........194 D1 Unadjusted Means, Standard D e v i a t i o n s , and Group S i z e s of Volume achievement P o s t t e s t Scores f o r Treatments by Conservation L e v e l s ................................... 200 D2 Unadjusted Means, Standard D e v i a t i o n s , and Group S i z e s of Volume achievement B e t e n t i o n Test Scores f o r Treatments by Conservation L e v e l s ..................... 200 ACKNOWLEDGEMENTS Sinc e r e a p p r e c i a t i o n i s expressed to Er..Douglas Owens, my t h e s i s s u p e r v i s o r , f o r h i s encouragement, guidance and a s s i s t a n c e i n the development of t h i s study. G r a t i t u d e i s a l s o extended to other members of my t h e s i s committee: Dr. P a t r i c i a A r l i n , P r o f e s s o r Thomas Bates, Dr. Edward Hobns and Dr. G a i l S p i t l e r . I n d i v i d u a l l y and c o l l e c t i v e l y , they gave generously of t h e i r time and e x p e r t i s e i n the f o r m u l a t i o n and completion of t h i s study. Thanks i s a l s o expressed t o Dr. Todd Rodgers and Dr. Thomas O'Shea f o r t h e i r s t a t i s t i c a l a s s i s t a n c e . I would l i k e a l s o t o thank the a d m i n i s t r a t o r s of D e l t a School D i s t r i c t and the p r i n c i p a l s , t e a c h e r s , s e c r e t a r i e s and students of J a r v i s Elementary School, Richardson Elementary School and C l i f f D r i v e Elementary School f o r t h e i r p a r t i c i p a t i o n i n t h i s study. A s p e c i a l note of thanks i s a l s o expressed t o my c o l l e a g u e s , John T a y l o r and Lynn Cannon, and to my f r i e n d s Beverley Johnson, Lee Herberts and Garry Roth f o r t h e i r h e l p i n t e s t i n g , c o d i n g , e d i t i n g and most of a l l ' f o r t h e i r encouragement and suggestions,. And f i n a l l y , I am g r a t e f u l t o the E d u c a t i o n a l Research I n s t i t u t e of B r i t i s h Columbia f o r p r o v i d i n g f i n a n c i a l support to c a r r y out the study. 1 CHAPTER I THE PROBLEM The process o f s e l e c t i n g * o r d e r i n g and t i m i n g t o p i c s i n the mathematics c u r r i c u l u m has caught the i n t e r e s t o f many mathematics edu c a t o r s . There has been, f o r example, major concern among c u r r i c u l u m a n a l y s t s about the necessary c o g n i t i v e a b i l i t i e s and a p p r o p r i a t e age l e v e l f o r p r e s e n t i n g volume concepts to elementary s c h o o l students. While the major i t y of textbooks c o n t a i n volume a c t i v i t i e s as e a r l y as grade t h r e e , many edu c a t o r s hold that most c h i l d r e n do not conserve volume u n t i l about age 12 ( O z g i r i s , 1964; Carpenter, 1975-b; E l k i n d , 1961-a). There has been a need, t h e r e f o r e , t o t h e o r e t i c a l l y and e x p e r i m e n t a l l y examine the p o s i t i o n s of these educators i n order t o j u s t i f y the present c u r r i c u l u m or suggest i t s m o d i f i c a t i o n . Such an examination can fo c u s on many asp e c t s of volume p r e s e n t a t i o n . The present study, however, d e a l s p a r t i c u l a r l y with the i n t r o d u c t i o n of the volume a l g o r i t h m f o r a r e c t a n g u l a r p a r a l l e l e p i p e d i . e . , "Volume = Length x Width x Height (V = L x W x H) ". Two widely used textbook s e r i e s i n B r i t i s h Columbia i n t r o d u c e the a l g o r i t h m "V = L x W x H" i n grade 5 (age 10) ( D i l l e y et a l . , 1974 and E i c h o l z e t a l . , 1974). Another s e r i e s i n t r o d u c e s the a l g o r i t h m f o r m a l l y i n grade 4 (age 9) ( E l l i o t et 2 a l , , 1974) and uses i t i n f o r m a l l y i n grade 3 (age 8) ( E l l i o t et a l . , 1975). T h i s l a s t s e r i e s , f o r example, i n v o l v e s t h i r d graders i n s i t u a t i o n s i n which they are t o compute the volume of a r e c t a n g u l a r p a r a l l e l e p i p e d given i t s l e n g t h , width and h e i g h t . Most proponents of P i a g e t ' s theory would disagree with such e a r l y i n t r o d u c t i o n of the a l g o r i t h m and c l a i m t h a t most c h i l d r e n do not develop the necessary c o g n i t i v e a b i l i t i e s f o r l e a r n i n g i t before grade 6 (age 11).,Piaget (Piaget e t a l . , 1960) h i m s e l f , f o r example, h o l d s t h a t " i t i s not u n t i l stage IV [ f o r m a l o p e r a t i o n a l ] t h a t c h i l d r e n understand how they can a r r i v e a t an area or volume simply by m u l t i p l y i n g boundary edges" (p. 408). Piaget (1960) adds i n h i s d i s c u s s i o n of volume c a l c u l a t i o n t h a t "knowledge l e a r n e d i n s c h o o l i n c r e a s i n g l y i n t e r f e r e s with the spontaneous development o f g e o m e t r i c a l n o t i o n s as c h i l d r e n grow o l d e r " (p. 381). Osborn (1976, pp. 27-28), l i k e w i s e , warned t h a t a premature s t r e s s of volume a l g o r i t h m s c r e a t e s a s e r i o u s l e a r n i n g problem;. There seems t o be some d i s c r e p a n c y between P i a g e t ' s p o s i t i o n and most s c h o o l programs regarding t h e : l e v e l at which the a l g o r i t h m "V = L x W x H" should be i n t r o d u c e d . T h i s d i s c r e p a n c y r a i s e s the i s s u e of whether or not present s c h o o l c u r r i c u l a are j u s t i f i e d i n p r e s e n t i n g the a l g o r i t h m at the grade 5 l e v e l s F a b r i c a n t (1975) suggested t h a t " t e a c h i n g of geometric formulas a t the elementary s c h o o l l e v e l has to be s e r i o u s l y s t u d i e d to see where such formulas would be most p r o f i t a b l y placed i n the c u r r i c u l u m " (pp. .6-7). I t was as a 3 r e s u l t of the concerns mentioned above that the present study o r i g i n a t e d . D e f i n i t i o n of..Terms I t i s necessary to c l a r i f y the usage of c e r t a i n terms which w i l l occur throughtout the study. The term volume r e f e r s t o the measure of the space d i s p l a c e d by a three dimensional o b j e c t . The o b j e c t may be any substance: s o l i d , l i q u i d or gas. Volume i s not to be confused with c a p a c i t y which r e f e r s t o the measure of the space e n c l o s e d by a three d i m e n s i o n a l o b j e c t . Even though " i n t e r n a l volume of a hollow c o n t a i n e r ... i s synonymous to c a p a c i t y " ( K e r s l a k e , 1976, p. 14), the term volume u s u a l l y r e f e r s to non-hollow o b j e c t s . The term c o n s e r v a t i o n r e f e r s to the concept that a c e r t a i n a t t r i b u t e of an o b j e c t (or o b j e c t s ) remains i n v a r i a n t under changes of other i r r e l e v a n t a t t r i b u t e s (Wohlwill and Lowe, 1962, p i 153). F o r example, the volume of a substance remains i n v a r i a n t r e g a r d l e s s of i t s shape and p o s i t i o n as l o n g as nothing i s added or taken away. L i k e w i s e , the numerousness of a s e t remains unchanged d u r i n g changes i n the s p a c i a l arrangement of the s e t as long as nothing i s added or taken away. The term r e c t a n g u l a r , p a r a l l e l e p i p e d which i s synonymous with cuboid (Webster's d i c t i o n a r y , e d i t e d by Gove, 1971, p. 550) i s i l l u s t r a t e d by the shape of f i l l e d boxesi These terms, however, have been confused with r e c t a n g u l a r prism and a 4 c l a r i f i c a t i o n i s needed. A r e c t a n g u l a r ^ p a r a l l e l e p i p e d i s a r i g h t r e c t a n g u l a r prism r a t h e r than j u s t a r e c t a n g u l a r prism (James and James, 1S76). Throughout t h i s paper cuboid w i l l be used i n t e r c h a n g a b l y with r e c t a n g u l a r p a r a l l e l e p i p e d . The term a l g o r i t h m r e f e r s to a procedure of ordered s t e p s t h a t guarantees a c o r r e c t r e s u l t i f the s t e p s are performed c o r r e c t l y and i n the proper order (Lewis and P a p a d i m i t r i o u , 1978).. Algorithms vary i n t h e i r l e v e l of d i f f i c u l t y . The f o u r b a s i c o p e r a t i o n s , a d d i t i o n , s u b t r a c t i o n * m u l t i p l i c a t i o n and d i v i s i o n have r a t h e r simple a l g o r i t h m s , while s o l v i n g systems of eguations by the use of i n v e r s e matrices i s a more complex one. The volume of a cuboid may be obtained by a p p l y i n g the a l g o r i t h m of f i n d i n g the measures of the three dimensions l e n g t h , width, and h e i g h t - of the cuboid and computing the product o f these t h r e e measures (V = L x W x H ) . _ The term l e a r n i n g the volume a l g o r i t h m r e f e r s to the mastery of the a l g o r i t h m "V = L x W x H" at the l e v e l s o f computation and comprehensioni The l e v e l of computation i n c l u d e s , f o r example, s i t u a t i o n s where students are asked to s t a t e the volume given diagrams of p a r t i t i o n e d cuboids, non-p a r t i t i o n e d cuboids with known dimensions, or a word d e s c r i p t i o n of the dimensions of cuboids* The l e v e l of comprehension i n c l u d e s , f o r example, s i t u a t i o n s where st u d e n t s are asked to s t a t e the t o t a l volume given a diagram o f attachments of cuboids with known dimensions or to s t a t e the volume of the cuboid r e s u l t i n g from proposed dimensional t r a n s f o r m a t i o n s on a given cuboid. 5 Statement of the Problem The purpose of t h i s study was t o determine the r e l a t i o n s h i p between the l e v e l of c o n s e r v a t i o n of volume and the degree t o which s i x t h grade students l e a r n the volume a l g o r i t h m f o r a cuboid "V = L x W x H". The l e v e l o f volume c o n s e r v a t i o n was determined using v a r i a t i o n s of t e s t s employed by P i a g e t ( P i a g e t , I n h e l d e r and Szeminska, 1960). The achievement on the usage of the a l g o r i t h m was based on o b j e c t i v e s and a c t i v i t i e s found i n widely used elementary textbook s e r i e s . The study a l s o provided i n f o r m a t i v e data with regard to the volume c o n s e r v a t i o n l e v e l of s i x t h grade c h i l d r e n . Furthermore, the study showed the r e l a t i o n s h i p s among mathematics achievement, l e v e l s of c o n s e r v a t i o n , and l e a r n i n g of the a l g o r i t h m "V = L x W x H M f o r the volume of a cuboid. In order to gain i n f o r m a t i o n about l e a r n i n g the a l g o r i t h m f o r the volume of a cuboid, t h r e e treatments were implemented. The f i r s t treatment c o n s i s t e d of t e a c h i n g the volume a l g o r i t h m of a cuboid "V = I x W x H" using a guided d i s c o v e r y method based on approaches of present s c h o o l programs. The second treatment c o n s i s t e d mainly of l e a r n i n g the t a s k of v a r y i n g f a c t o r s when the product i s c o n s t a n t . For example, given t h a t 36 = 2 X 3 X 6,- the . student would be able t o complete statements such as 36 = 4 X [ ] X [ J. T h i s task was supplemented by a b r i e f d i s c u s s i o n of the volume algorithm of a cuboid "V = L x W x H i " The t h i r d treatment served- as c o n t r o l treatment and c o n s i s t e d of t e a c h i n g v a r i o u s numeration systems. The g e n e r a l aims of t h i s study may b e ' l i s t e d as f o l l o w s : 6 1. To determine the v a r i o u s degrees to which c o n s e r v e r s , p a r t i a l c o n s e r v e r s , and non-conservers of volume l e a r n the volume a l g o r i t h m of a cuboid "V = L x W x H." 2. To determine the degree of e f f e c t i v e n e s s f o r each of the two t e a c h i n g methods on l e a r n i n g the volume a l g o r i t h m f o r a cuboid. 3. To determine the e f f e c t of l e a r n i n g the volume a l g o r i t h m of a cuboid on the t r a n s i t i o n from one volume c o n s e r v a t i o n l e v e l to another. 4. To determine the r e l a t i o n s h i p between sex and the l e v e l s of c o n s e r v a t i o n of volume. 5. To determine the r e l a t i o n s h i p between sex and the degree of l e a r n i n g the volume a l g o r i t h m f o r a cuboid. 6. To determine the r e l a t i o n s h i p between mathematics achievement and the l e v e l s of c o n s e r v a t i o n of volume. 7. To determine the r e l a t i o n s h i p between mathematics achievement and the degree of l e a r n i n g the volume a l g o r i t h m f o r a cuboid. F i n a l l y the r e s u l t s of t h i s study w i l l be u s e f u l i n v e r i f y i n g a spects of the developmental theory of Piaget and a c c o r d i n g to S i e g l e r and a t l a s (1976, p. 360) t r a i n i n g s t u d i e s (not u n l i k e t h i s one) have become a standard method f o r i n v e s t i g a t i n g c o g n i t i v e development. 7 J u s t i f i c a t i o n of the Problem The present study i s a consequence of a concern about a d i s c r e p a n c y between the p r e s e n t s c h o o l programs and the c o g n i t i v e theory of P i a g e t . S p e c i f i c a l l y , some s c h o o l textbooks i n t r o d u c e the volume algorithm of a cuboid as e a r l y as grade 3 (age 8) while P i a g e t and h i s f o l l o w e r s c l a i m t h a t most c h i l d r e n do not develop the c o n s e r v a t i o n of volume before age 11 (grade 6 ) . In such a predicament t h e r e seems t o be a need f o r r e s e a r c h i n order to j u s t i f y our present s c h o o l c u r r i c u l u m o r suggest i t s m o d i f i c a t i o n . DeVault expressed the need f o r such r e s e a r c h by n o t i n g the f o l l o w i n g : Needed now i s the r e s e a r c h t h a t w i l l make the l i n k i n the continuum between the r e s e a r c h of the b e h a v i o r a l s c i e n t i s t s and the work of the mathematicians who have designed new programs f o r s c h o o l s ... The s t u d i e s most l i k e l y t o produce u s e f u l r e s u l t s f o r c u r r i c u l u m work would be experimental s t u d i e s ... (DeVault> 1966, pp. 637-639) L i k e w i s e , S t e f f e and H i r s t e i n (1976) d i s c u s s e d c h i l d r e n ' s t h i n k i n g i n measurement s i t u a t i o n s and recommended the f o l l o w i n g : In planning the mathematics e x p e r i e n c e s .... the teacher should c o n s i d e r the stages of c o g n i t i v e development. The proposed content and the methods of p r e s e n t i n g t h a t content should a l s o be c o n s i d e r e d . ( S t e f f e and H i r s t e i n , 1976, p. 35) I m p l i c a t i o n s of P i a g e t ' s C o g n i t i v e Theory P i a g e t has f o r s e v e r a l decades t e s t e d , i n t e r v i e w e d and observed c h i l d r e n . His theory has become i n c r e a s i n g l y more i n f l u e n t i a l i n c u r r i c u l u m p l a n n i n g because "everybody i n e d u c a t i o n r e a l i z e s t h a t Piaget i s saying something t h a t i s 8 r e l e v a n t to the t e a c h i n g of c h i l d r e n " (Duckworth, 1964-b, p. 496) . A c e n t r a l theme i n the c o g n i t i v e theory of Piaget i s the attainment of c e r t a i n c o n s e r v a t i o n tasks t h a t are c o n s i d e r e d requirements f o r understanding of mathematical concepts (Piaget, 1941, p. 4 ) . For example "3 + 5" and "8" are two names f o r the same number. T h i s c o n s e r v a t i o n of number i s necessary f o r comprehension, g e n e r a l i z a t i o n and r e t e n t i o n of a d d i t i o n b a s i c f a c t s . I t has even been repo r t e d t h a t c o n s e r v a t i o n of number i s a b e t t e r p r e d i c t o r of success i n a d d i t i o n and s u b t r a c t i o n problem s o l v i n g than i s I n t e l l i g e n c e Quotient (I.Q.) (Van Engen, 1971). Van Engen, f u r t h e r , recommended t h a t " i t .would seem t h a t f o r these c h i l d r e n [nonconservers of number], the s c h o o l should c e n t e r t h e i r a t t e n t i o n on a c t i v i t i e s t h a t might enhance c o n s e r v a t i o n r a t h e r than on our t r a d i t i o n a l a r i t h m e t i c c u r r i c u l u m " (p. 48). The e f f e c t of number c o n s e r v a t i o n i n a d d i t i o n and s u b t r a c t i o n problem s o l v i n g i s p a r t i c u l a r l y r e l e v a n t to t h i s study. Even though care should be taken i n g e n e r a l i z i n g to volume concepts, the importance of number c o n s e r v a t i o n seems to suggest a p o s s i b l e analogy. T h i s study was designed t o f i n d the r e l a t i o n s h i p between c o n s e r v a t i o n of volume and a volume a l g o r i t h m . Volume c o n s e r v a t i o n seems to be an apparent n e c e s s i t y f o r volume measurement. To measure volume i s to compare a chosen u n i t of volume with the volume of an o b j e c t . I t i s evident t h a t volume c o n s e r v a t i o n of the u n i t and o f the o b j e c t t o be measured i s a requirement before, measurement can 9 be meaningful. P i a g e t (Piaget et a l . , 1960) seems t o h o l d the same p o s i t i o n r e g a r d i n g c a l c u l a t i o n s i n measurement i n c l u d i n g t h a t of volume. He c o n s i d e r s the concept of c o n s e r v a t i o n to be necessary f o r any meaningful computation i n both area and volume: ... C h i l d r e n a t t a i n a c e r t a i n kind of c o n s e r v a t i o n of area (and volume), based on the p r i m i t i v e c o n c e p t i o n of area (and volume) as t h a t which i s bounded by l i n e s (or f a c e s ) . That understanding comes long before the a b i l i t y to c a l c u l a t e areas and volumes by mathematical m u l t i p l i c a t i o n , i n v o l v i n g r e l a t i o n s between u n i t s of d i f f e r e n t powers i . . (p. 355) P i a g e t ' s volume experiments have a l s o r e v e a l e d t h a t most c h i l d r e n do not conserve volume before the f o r m a l o p e r a t i o n a l l e v e l and thus do not understand how they can a r r i v e at a volume o f a cuboid by simply m u l t i p l y i n g i t s dimensions. Piaget argues t h a t : The decomposition and redecomposition of a continuum are o p e r a t i o n s which belong to the l e v e l of f o r m a l o p e r a t i o n s . T h i s e x p l a i n s why i t i s not u n t i l stage IV that c h i l d r e n understand how they can a r r i v e at an area or a volume simply by m u l t i p l y i n g boundary edges.. (Piaget e t a l . , 1960, p. 408) On the other hand, P i a g e t i s not s a y i n g t h a t the i n t e l l e c t u a l development proceeds on i t s own r e g a r d l e s s of the s t i m u l i of the surroundings. In f a c t , and c o n t r a r y to what has been a t t r i b u t e d to him, P i a g e t c o n s i d e r s education to be a t o o l f o r c o g n i t i v e development; he only q u e s t i o n s the extent to which i t i s b e n e f i c i a l ( M o d g i l l , 1974, pp. 126-127). In other words, Piaget f a v o r s e d u c a t i o n t h a t l e a d s the c h i l d to d i s c o v e r y and r e j e c t s r o t e l e a r n i n g t h a t f o r c e s i n f o r m a t i o n on 10 t i e student who i s not ready f o r i t : T h i s i s a b i g danger of s c h o o l - f a l s e accommodation which s a t i s f i e s a c h i l d because i t agrees with a v e r b a l formula he has been given. T h i s i s a f a l s e e q u i l i b r a t i o n which s a t i s f i e s a c h i l d by accommodating to words, to a u t h o r i t y and not to o b j e c t s as they present themselves to him ... (Piaget, 1964, p.4) V a l i d a t i o n of some of P i a g e t ' s F i n d i n g s P i a g e t ' s f i n d i n g s i n c l u d i n g those of volume c o n s e r v a t i o n and computation have been examined by r e s e a r c h e r s throughout the world. E l k i n d (1961-a) , Carpenter (1975-b) and U z g i r i s (1964), f o r example, found t h a t at l e a s t 75% of American students do not develop c o n s e r v a t i o n of volume before age 11 (grade 6 ) . L i k e w i s e , L o v e l l and O g i l v i e (1961) found t h a t i t was not u n t i l age 14 t h a t 50% of B r i t i s h s t u d e n t s developed volume c o n s e r v a t i o n . More r e c e n t l y , A r l i n (1977) reported t h a t only about 30% of grade f i v e students i n B r i t i s h Columbia conserved volume. On the other hand, many educators b e l i e v e t h a t " a c q u i s i t i o n of formal s c i e n t i f i c r easoning may be f a r more dependent on s p e c i f i c i n s t r u c t i o n a l e x periences and f a r l e s s dependent on g e n e r a l maturation than hypothesized by I n h e l d e r and P i a g e t (1960)" ( S i e g l e r and A t l a s , 1976, p. 368). Graves (1972, p. 223), f o r example, c o n s i d e r e d education and experience t o be necessary f o r volume c o n s e r v a t i o n . L o v e l l (1971, p. 179) went f u r t h e r to suggest t h a t even seven- and e i g h t - y e a r o l d s (grade 2 and 3) can l e a r n how to use the a l g o r i t h m "V = L x W x H" i n order to c a l c u l a t e the volume. 11 Need f o r Research The task of a p p l y i n g p s y c h o l o g i c a l t h e o r i e s to s c h o o l c u r r i c u l u m depends, u n f o r t u n a t e l y , on p o l i c y makers r a t h e r than on e d u c a t i o n a l r e s e a r c h e r s . DeVault found t h a t o f t e n s c h o o l p o l i c i e s "are based on theory t h a t i s never t e s t e d i n i n s t r u c t i o n a l c o n t e x t s " (DeVault, 1966, p. 636). One reason f o r t h i s i s t h a t our awareness of the psychology of l e a r n i n g i s very l i m i t e d (Young, 1967, p. 40). I t has been r e p o r t e d t h a t the m a j o r i t y of students do not conserve volume before 13 or 14 years of age ( L o v e l l and O g i l v i e , 1961) . Meanwhile, i t has a l s o been noted t h a t the majority of a d u l t s do not conserve volume ( E l k i n d , 1962; Towler and Wheatly, 1971; Graves, 1972). The averages s t a t e d above have v a r i e d c o n s i d e r a b l y with c u l t u r e s and communities and may be m i s l e a d i n g i f used without v e r i f i c a t i o n i n c u r r i c u l u m planning (Fogelman, 1970). In any case, one should not n e c e s s a r i l y delay the i n t r o d u c t i o n of the volume a l g o r i t h m "V = L x W x H" u n t i l a l l students conserve volume. S t u d i e s , such as the ones mentioned above, i n d i c a t e t h a t one can not expect a l l s t u d e n t s to conserve.volume. There i s a l s o the danger of i n t r o d u c i n g the a l g o r i t h m too e a r l y and harming the process o f l e a r n i n g . I t seems, t h e r e f o r e , t h a t the matter of " p l a c i n g a c t i v i t i e s , which depend on volume c o n s e r v a t i o n , i n the proper grade i s p r e s e n t l y a matter of p r e f e r e n c e r a t h e r than exactness. I t was an i n t e n t i o n of the i n v e s t i g a t o r t o provide necessary data of the r e l a t i o n s h i p between volume c o n s e r v a t i o n and a volume a l g o r i t h m . . Such data was used f o r making 12 recommendations r e l a t e d to the j u s t i f i c a t i o n f o r t e a c h i n g the volume a l g o r i t h m p r i o r to grade 6. DeVault advocates t h a t " i t seems reasonable ... to a s s e r t t h a t the s t u d i e s most l i k e l y to produce u s e f u l r e s u l t s f o r c u r r i c u l u m work would be experimental s t u d i e s " (DeVault, 1966, p. 639) . 13 CHAPTER I I REVIEW OF RELATED LITERATURE Even though P i a g e t has f o r s e v e r a l decades t e s t e d , i n t e r v i e w e d and observed c h i l d r e n , h i s i n f l u e n c e i n North American e d u c a t i o n a l psychology and education i n g e n e r a l was, u n t i l the 1950's, l i m i t e d to a few i n d i v i d u a l s . In the l a s t two decades, however, i t has become i n c r e a s i n g l y c l e a r t h a t he i s "the foremost c o n t r i b u t o r to the f i e l d of i n t e l l e c t u a l development" (Ginsburg and opper, 1969, p. i x ) . Only aspects of P i a g e t ' s theory which are r e l e v a n t to t h i s study are reviewed here. I n t e r e s t e d readers may f i n d more comprehensive summaries of P i a g e t ' s views i n F l a v e l l (1963), Maier (1965), Ginsburg and Opper (1969) and Berlyne (1957).. Summary of P i a g e t ' s Theory P i a g e t ' s goals of e d u c a t i o n seem to be c o n s i s t e n t with h i s theory i n g e n e r a l and h i s philosophy of l e a r n i n g i n p a r t i c u l a r . His e d u c a t i o n a l g o a l s c o n s i s t of c r e a t i n g i n d i v i d u a l s who are a c t i v e , c r i t i c a l , c r e a t i v e , i n v e n t i v e and d i s c o v e r e r s ( P i a g e t , 1964, p. 5),. P i a g e t d i s t i n g u i s h e s between development of knowledge and l e a r n i n g . While development i s a spontaneous and 14 genuine process concerning the t o t a l i t y of knowledge, l e a r n i n g i s a process l i m i t e d to only a p a r t i c u l a r problem and caused by a teacher or an e x t e r n a l s t i m u l u s ( P i a g e t , 1964, p. 8). Piaget e x p l a i n s t h a t the essence of the : development of knowledge s t a r t s f i r s t , even i n young c h i l d r e n , by forming "schemes" which are organized p a t t e r n s of behavior and are based on the c h i l d ' s " a c t i o n s on" and "experience i n " h i s surroundings (Ginsburg and Opper, 1969, pp.. 21-22). These schemes are nourished and manifested by " o p e r a t i o n s " which c o n s i s t of i n t e r i o r i z e d a c t i o n s t h a t a c t , mentally or p h y s i c a l l y , on events or o b j e c t s by modifying them, i n t e r p r e t i n g them and understanding the way they are c o n s t r u c t e d ( P i a g e t , 1964, p. 8 ) . P i a g e t , f u r t h e r , p e r c e i v e s t h a t a given o p e r a t i o n does not e x i s t independently from other o p e r a t i o n s . Rather, " i t i s l i n k e d to other o p e r a t i o n s and as a r e s u l t i t i s always a p a r t of a t o t a l s t r u c t u r e " (Piaget, 1964, p. 9). ft. s t r u c t u r e i s , t h u s , an independent system of o p e r a t i o n s which i s governed and c l o s e d , ( i n the mathematical s e n s e ) , under c e r t a i n laws of t r a n s f o r m a t i o n . Mathematically, the interdependence of o p e r a t i o n s can be i l l u s t r a t e d i n many ways. For i n s t a n c e , some s t r u c t u r e s are s u b s t r u c t u r e s of l a r g e r s t r u c t u r e s as i n the case of the n a t u r a l number system being a s u b s t r u c t u r e of the r a t i o n a l , r e a l or complex s t r u c t u r e (Piaget, 1970, p. 23). P i a g e t d i d not attempt t o d e s c r i b e the most complicated and g e n e r a l s t r u c t u r e s but he t r i e d t o d i s c o v e r the s i m p l e s t ones t h a t i l l u s t r a t e the a c q u i s i t i o n of knowledge. For P i a g e t , 15 "... the c e n t r a l problem o f development i s t o understand the (processes o f ) f o r m a t i o n , e l a b o r a t i o n , o r g a n i z a t i o n and f u n c t i o n i n g o f these s t r u c t u r e s " ( P i a g e t , 1964, p. 9) . Piaget observed t h a t the processes of these s t r u c t u r e s f a l l i n t o f o u r main stages and substages which c h a r a c t e r i z e mental growth. The f i r s t stage, the sensorimotor, s t a r t s from b i r t h and conti n u e s u n t i l about two years of age* In t h i s stage the c h i l d grows t o d i s t i n g u i s h between h i m s e l f and h i s surroundings; he a l s o develops the permanence o f o b j e c t s even when they can no lon g e r be seen. The second stage, the p r e o p e r a t i o n a l , l a s t s u n t i l about the age of seven years. I t i s c h a r a c t e r i z e d by the c h i l d ' s use of symbols, by language development, and by growth i n i n t u i t i v e r e a s o n i n g . In the t h i r d s t age, the c o n c r e t e . o p e r a t i o n a l , which l a s t s u n t i l about age 11 or 12, t h e c h i l d l e a r n s t o group using c l a s s i f i c a t i o n as w e l l as s e r i a t i o n . The c h i l d i s capable a l s o o f c l a s s i f y i n g and s e r i a t i n g simultaneously,. However, he i s s t i l l l i m i t e d to t h i n k i n g only about o b j e c t s t h a t e x i s t and a c t i o n s t h a t are p o s s i b l e . The l a s t stage, the fo r m a l o p e r a t i o n a l , i s demonstrated by the c h i l d ' s c a p a b i l i t y t o d e f i n e concepts and to reason l o g i c a l l y , s y s t e m a t i c a l l y , and s y m b o l i c a l l y . The c h i l d can al s o perform o p e r a t i o n s on v e r b a l l y s t a t e d p r o p o s i t i o n s r a t h e r than only d i r e c t l y on r e a l i t y (Ginsburg and Opper, 1969). The ages mentioned i n the above paragraph are not i n any way f i x e d o r u n i v e r s a l ; they are the approximate average ages noted i n P i a g e t ' s s t u d i e s . What i s f i x e d , however, i s the order 16 of s u c c e s s i o n of the stages; a normal c h i l d develops through every one of the stages i n the order they are mentioned. The age at which c h i l d r e n reach a c e r t a i n stage may vary, amongst c h i l d r e n * from a few months to a few years and the p e r i o d one remains i n t h a t stage depends on h i s degree of i n t e l l i g e n c e and h i s s o c i a l m i l i e u (Piaget and I n h e l d e r , 1969, pp. „ 152-153) . The t r a n s i t i o n from one stage to the next i s not i n t e r c h a n g e a b l e but i n t e g r a t i v e . "Each (stage) r e s u l t s from the preceeding one, i n t e g r a t i n g i t as a subordinate s t r u c t u r e , and prepares f o r a subseguent one, i n t o which i t i s sooner or l a t e r i n t e g r a t e d " (Piaget and I n h e l d e r , 1969, p. 153) i The sensorimotor p e r c e p t i o n , f o r example, does not cease i n the f o l l o w i n g stages but i t c o n t i n u e s to f u n c t i o n i n developed thought and i t becomes i n t e g r a t e d i n i t s s t r u c t u r e s ( P i a g e t , 1973, p. 122). P i a g e t (Piaget and I n h e l d e r , 1969, pp. 154-155) r e c o g n i z e s f o u r f a c t o r s t h a t are r e s p o n s i b l e f o r the development from one stage t o the next: maturation, e x p e r i e n c e , s o c i a l t r a n s m i s s i o n and most i m p o r t a n t l y , " e q u i l i b r a t i o n " . Maturation c o n s i s t s of o r g a n i c growth, e s p e c i a l l y of the nervous system which p r o v i d e s p o s s i b i l i t i e s f o r development i f minimal experience i s a v a i l a b l e . The second f a c t o r , experience, can be p h y s i c a l or l o g i c a l - m a t h e m a t i c a l i The p h y s i c a l experience c o n s i s t s of knowledge ac q u i r e d by a b s t r a c t i n g the p h y s i c a l p r o p e r t i e s of o b j e c t s while the l o g i c a l - m a t h e m a t i c a l r e s u l t s by a b s t r a c t i n g a c t i o n s performed on the p h y s i c a l o b j e c t s r a t h e r than the p h y s i c a l o b j e c t s themselves. An example of a l o g i c a l -17 mathematical experience* o f t e n given by P i a g e t * concerns a c h i l d who was p l a y i n g with pebbles. He put h i s pebbles i n a row and counted te n , then he counted i n the other d i r e c t i o n and a l s o got ten. Next, he put the pebbles i n a c i r c l e and s t i l l counted ten. The c h i l d , thus, d i s c o v e r e d t h a t h i s pebbles add up to ten r e g a r d l e s s of t h e i r c o n f i g u r a t i o n ( P i a g e t , 1964, p. 12, P i a g e t , 1970, pp. 16-17). P i a g e t l a b e l s t h i s experience as l o g i c a l - m a t h e m a t i c a l , which i s o b v i o u s l y independent o f and s u p e r i o r t o the p h y s i c a l experience of the pebbles themselves. The t h i r d f a c t o r , s o c i a l t r a n s m i s s i o n and i n t e r a c t i o n , i s i l l u s t r a t e d by the v e r b a l i n s t r u c t i o n s a c q u i r e d by the c h i l d i n the process of development and formal e d u c a t i o n ( P u l a s k i , '1971, pp. 9-11). The t h r e e f a c t o r s d e s c r i b e d above are c o n s i d e r e d by P i a g e t t o be necessary but not s u f f i c i e n t f o r mental development. The most important f a c t o r f o r development i s " e q u i l i b r a t i o n " which c o n s i s t s of "a s e r i e s of a c t i v e compensations on the p a r t of the s u b j e c t i n response t o e x t e r n a l d i s t u r b a n c e s and adjustment (Piaget and I n h e l d e r , 1969, p. 157)." The term " e q u i l i b r i u m " , i n i n t e l l e c t u a l c o n n o t a t i o n , r e f e r s to an a c t i v e s t a t e of an open system of s t r u c t u r e s which i n t e r a c t s with the environment and m o d i f i e s i t s e l f a c c o r d i n g l y (Ginsburg and Opper, 1969, p. 172),. Piaget (1975, p. 170) and s i m i l a r l y F e s t i n g e r (1957), e x p l a i n t h a t the human being s t r i v e s toward e q u i l i b r i u m and harmony amongst h i s knowledge, o p i n i o n s , a t t i t u d e s and behaviour. P i a g e t c l a i m s , however, t h a t t h i s e q u i l i b r i u m i s never a t t a i n e d but i s c o n t i n u o u s l y improved (1975, p. 23) 18 through the c u r i o s i t y of e x p l o r i n g and the c o n s t r u c t i o n of new i n f o r m a t i o n (1975, p. 170). Thus, "a c o g n i t i v e system which has a t t a i n e d a high degree of e q u i l i b r i u m i s not at r e s t . I t i n t e r a c t s with the environment i n terms of i t s s t r u c t u r e s i_ a s s i m i l a t i o n ] and i t can modify i t s e l f i n l i n e with environmental demands |accomodation J" (Ginsburg and Opper, 1969, p. 172). A s s i m i l a t i o n and accomodation are two i n s e p a r a b l e aspects of every a c t ; on one hand the l e a r n e r f i l t e r s the i n p u t of the environment i n t o h i s own s t r u c t u r e s and on the other hand he m o d i f i e s h i s s t r u c t u r e s i n order to f i t the pressure of r e a l i t y (Piaget and I n h e l d e r , 1969, pp. 4-6) . The above d e s c r i p t i o n of the stages and f a c t o r s i n f l u e n c i n g development seem t o warn educators t h a t not o n l y does the c h i l d " t h i n k l e s s e f f i c i e n t l y than the a d u l t , but he t h i n k s d i f f e r e n t l y " (Ginsburg and Opper, 1969, p. 8) depending upon h i s mental s t r u c t u r e s and c a p a b i l i t i e s . In f a c t , P i a g e t p o s t u l a t e d t h a t the c a p a b i l i t i e s of understanding mathematical concepts depends upon the attainment of s p e c i f i c " c o n s e r v a t i o n " a t t r i b u t e s (Piaget, 1941, p. 4). C o n s e r v a t i o n of an a t t r i b u t e r e f e r s t o the r e a l i z a t i o n t h a t t h i s a t t r i b u t e remains i n v a r i a n t under changes of other i r r e l e v a n t a t t r i b u t e s . P i a get has made a c o n s i d e r a b l e e f f o r t to determine the approximate ages at which v a r i o u s c o n s e r v a t i o n t a s k s a r e achieved and the means by which the mind c o n s t r u c t s the n o t i o n s of these t a s k s . He has not been n e a r l y as s u c c e s s f u l i n the l a t t e r as i n the former., H i s f i n d i n g s have r e v e a l e d t h a t c h i l d r e n i n the p r e o p e r a t i o n a l 19 stage r e l y h e a v i l y on t h e i r immediate p e r c e p t i o n i n a c q u i r i n g knowledge about t h e i r surroundings..He f u r t h e r r e p o r t e d t h a t , on the average, h i s s u b j e c t s conserved number at about age s i x , l e n g t h a t seven* substance at e i g h t , weight at nine and volume at eleven (Piaget and I n h e l d e r , 1969, p. 99). P i a g e t i s not s a y i n g , however, t h a t the i n t e l l e c t u a l c a p a b i l i t i e s of the c h i l d develop on t h e i r own r e g a r d l e s s of the s t i m u l i of the surroundings. He i s only s a y i n g t h a t one can not i n c r e a s e the understanding of the c h i l d by j u s t t e l l i n g him; understanding n e c e s s i t a t e s c o n d i t i o n s i n which the c h i l d experiments, manipulates, and i n t e r p r e t s (Duckworth, 1964-a, p. 2),. P i a g e t h o l d s , on the other hand, t h a t the c h i l d ' s i n t e r p r e t a t i o n and response depend on h i s r e a d i n e s s l e v e l and h i s mental s t r u c t u r e s . P i a g e t , consequently, does not encourage a c c e l e r a t i o n of l e a r n i n g and he charges Americans of u n p r o f i t a b l y doing so: T e l l an American t h a t a c h i l d develops a c e r t a i n way of t h i n k i n g at seven, and he immediately s e t s about to t r y to develop those same ways of t h i n k i n g a t s i x or even f i v e y e a rs of age ... . Most of the r e s e a r c h ... hasn't worked because experimenters have not paid a t t e n t i o n to the e q u i l i b r i u m theory ... . Learning a f a c t by r e i n f o r cement does not i n and of i t s e l f r e s u l t i n mental a d a p t a t i o n . (Piaget, 1967, p. 343) P i a g e t ' s S t u d i e s of Volume P i a g e t , I n h e l d e r , and Szeminska (1960) s t u d i e d r a t h e r thoroughly the s u b j e c t of c o n s e r v a t i o n and measurement of volumei They showed c h i l d r e n a s o l i d n o n p a r t i t i o n e d cuboid of base 3 cm x 3 cm and height 4 cm and they t o l d them t h a t the 20 block was a condemned house b u i l t on an i s l a n d . They f u r t h e r asked the c h i l d r e n t o use centimetre cubes i n order to b u i l d p o s s i b l e houses t h a t have the same space; the new houses were to be b u i l t on i s l a n d s of 2 cm x 2 cm, 2 cm x 3 cm, 1 cm x 1 cm, or 3 cm x 4 cm (pp^ 355-357). Pi a g e t e t a l . (i960) a l s o used an a u x i l i a r y method. They showed c h i l d r e n s e v e r a l r e c t a n g u l a r p a r a l l e l e p i p e d s and asked them t o compare p a i r s of them: "are these two as b i g as one another? I s t h e r e as much wood i n each of them?" (p. 357). An a l t e r n a t i v e f o r t h i s method was t o give c h i l d r e n a c e r t a i n cuboid and ask them to f i n d * out of a dozen cuboids, another t h a t had the same s i z e or room (p. 357). The p r e l i m i n a r y two methods d e s c r i b e d above were f o l l o w e d by f u r t h e r q u e s t i o n i n g . The experimenters used the cubes o f a p a r t i t i o n e d r e c t a n g u l a r p a r a l l e l e p i p e d t o b u i l d another of d i f f e r e n t base while the s u b j e c t watched. Then t h e s u b j e c t was asked: (a) whether the new and o l d houses had the same room or which had more, (b) whether one can use the same cubes i n the present house i n order to b u i l d a new one t h a t had as much space as the o l d one and looked e x a c t l y l i k e i t . The experimenters t r i e d c o n t i n u o u s l y t o d i s c o v e r i f the c h i l d r e l i e d t o t a l l y on the c o n s e r v a t i o n of the number of cubes or i f he c o n s i d e r e d the t o t a l volume and conserved i t (pp..357-358). The experimenters always checked t h e i r r e s u l t s by using a water displacement technique. They b u i l t a metal cuboid of 3 cm x 3 cm x 4 cm i n the bottom of a c o n t a i n e r while the c h i l d looked and observed the r i s e i n the water l e v e l . They then 21 asked the c h i l d i f he thought the l e v e l of the water would change i f the arrangement of the cubes was modified to 2 cm x 1 cm x 18 cm or t o 2 cm x 2 cm x 9 cm. P i a g e t et a l . noted i n t h e i r experiments t h r e e l e v e l s of understanding the volume concept (1960, pp. 358-385). These l e v e l s are b r i e f l y d e s c r i b e d below: Level, 1 (age 8 or 9) : C h i l d r e n conserved i n t e r i o r volume i . e . , the q u a n t i t y of matter c o n t a i n e d i n s i d e the boundary. They a l s o showed understanding of the l o g i c a l (not mathematical) r e l a t i o n s h i p s between dimensions, that i s , when asked to r e c o n s t r u c t a house on a s m a l l e r base they c o n s t r u c t e d i t t a l l e r than the o r i g i n a l , though not t a l l enough. The c h i l d r e n d i d not show c o n s e r v a t i o n of volume i n the sense of space occupied or water d i s p l a c e d . L e v e l 2 (age 9 or 10): C h i l d r e n of t h i s l e v e l showed progress over those of the p r e v i o u s one* They s t a r t e d to measure c o r r e c t l y by using the unit-cubes and e x p r e s s i n g measures i n m e t r i c a l r e l a t i o n s h i p s such as "twice as much" and " n e a r l y three times as h i g h . " They c o u l d a l s o copy volume c o r r e c t l y but they c o u l d not meaningfully c a l c u l a t e i t by m u l t i p l y i n g the l e n g t h , width and h e i g h t . In f a c t , when asked about the volume or the space occupied by an o b j e c t they equated the volume with the number of cubes necessary to surround the o b j e c t . They s t i l l d i d not conserve volume i n terms of space occupied or water d i s p l a c e d even though they r e c o g n i z e d t h a t the rearrangement of the u n i t - c u b e s d i d not a l t e r the i n t e r i o r volume. 22 L e v e l 3 (age 12 and above): C h i l d r e n of t h i s l e v e l e s t a b l i s h e d a r e l a t i o n s h i p between the area of the base and the volume. They d i s c o v e r e d volume i n terms of the product of l e n g t h , width, and h e i g h t ; and t h a t two volumes were equal i f the product of t h e i r r e s p e c t i v e l i n e a r dimensions were equal. C h i l d r e n of t h i s stage a l s o conserved volume with r e s p e c t to the surrounding space as i n the water displacement. The experiments and i n t e r p r e t a t i o n s of Piaget et a l . (1960) seem to say t h a t the c o n s e r v a t i o n of volume i s w e l l developed when the c h i l d a c q u i r e s i t s three meanings i . e . , (1) c o n s e r v a t i o n of i n t e r i o r volume, (2) c o n s e r v a t i o n of occupied space and (3) c o n s e r v a t i o n of complementary volume or water d i s p l a c e d * P i aget et a l . (1960) concluded t h a t the complete n o t i o n of volume c o n s e r v a t i o n and measurement i n v o l v e s m u l t i p l i c a t i o n of three l e n g t h s and n e c e s s i t a t e s the formal o p e r a t i o n a l grasp of the c o n t i n u i t y of space. "The decomposition and redecomposition of a continuum are o p e r a t i o n s which belong t o the l e v e l of f o r m a l o p e r a t i o n s * T h i s e x p l a i n s why i t i s not u n t i l stage IV t h a t c h i l d r e n understand how they can a r r i v e at an area or a volume simply by m u l t i p l y i n g boundary edges" (p. 408),.. 23 Reactions t o P i a g e t ' s Theory and.Experiments The r e a c t i o n s of p s y c h o l o g i s t s and educators, r e g a r d i n g P i a g e t ' s methods o f t e s t i n g and h i s c o n c l u s i o n s , have v a r i e d from c r i t i c i s m t o p r a i s e . L i k e w i s e the v a l i d a t i o n experiments have shown i n c o n s i s t e n t r e s u l t s . P i n a r d and Laurendeau (1964) looked f o r the e x i s t e n c e of P i a g e t ' s g e n e r a l stages of mental development i n a p o p u l a t i o n of French-Canadians. They g e n e r a l l y confirmed the e x i s t e n c e of P i a g e t ' s s t a g e s of mental development but they e l i m i n a t e d some substages and added o t h e r s . The data they c o l l e c t e d from 700 c h i l d r e n i n Quebec r e v e a l e d t h a t French-Canadian students a t t a i n e d the P i a g e t i a n stages a year or two l a t e r than s t u d e n t s used by Piaget i n Geneva. Carpenter (1975-a) r e p o r t e d t h a t h i s own study r e v e a l e d t h a t i n the case of f i r s t and second graders "measurement concepts begin to appear i n young c h i l d r e n e a r l i e r than P i a g e t et a l . (1960) concluded" (pp. 3-13) i In the same year however, he (Carpenter, 1975-b) pu b l i s h e d c o n t r a d i c t o r y r e s u l t s . He d e c l a r e d t h a t the r e s u l t s of the N a t i o n a l Assessment and Michigan S t a t e Assessment showed t h a t "on the whole measurement concepts develop somewhat l a t e r i n average American s t u d e n t s than i n d i c a t e d by the measurement s t u d i e s of P i a g e t , I n h e l d e r and Szeminska" (pp. 501-507) . F l a v e l l (1963) reviewed P i a g e t ' s w r i t i n g and questioned h i s work. He c r i t i c i s e d Piaget mainly on (1) the gap between the f a c t s d e s c r i b e d i n the experiments and the theory he concluded and (2) the use of the q u a l i t a t i v e method and the 24 r o l e of the language i n the i n t e r p r e t a t i o n of data,. P i a g e t responded t o F l a v e l l ' s c r i t i c i s m s . He e x p l a i n e d t h a t h i s method of study i s e p i s t e m o l o g i c a l r a t h e r than p s y c h o l o g i c a l and he admitted t h a t h i s r e s e a r c h i s f a r from complete. He even encouraged p s y c h o l o g i s t s and educators to c a r r y on f u r t h e r s t a t i s t i c a l l y sound s t u d i e s under c o n t r o l l e d c o n d i t i o n s ( P i a g e t , 1963, p. i x ) . F i n a l l y , most educators seem t o approve the P i a g e t i a n order of the stages of development, but they appear to be d i v i d e d r e g a r d i n g the degree of a c c e l e r a t i n g t h e development and p a r t i c u l a r l y the c o n s e r v a t i o n t a s k s . Many s i d e with P i a g e t i n r e s p e c t i n g the l e v e l g u i d e l i n e s and c o n s i d e r l e a r n i n g e x p e r i e n c e s t o be necessary but not s u f f i c i e n t f o r c o n s e r v a t i o n t a s k s . Other educators c h a l l e n g e P i a g e t ' s l e v e l g u i d e l i n e s and c o n s i d e r the c o n s e r v a t i o n t a s k s to be a b i l i t i e s which are a c g u i r e d through a process of cumulative l e a r n i n g . Shulman (1971) claimed t h a t those who s i d e with Piaget 1 seem to outnumber those who do not. However, the q u e s t i o n of a c c e l e r a t i n g the a q u i s i t i o n of c o n s e r v a t i o n t a s k s and thus the s u i t a b l e ages f o r p r e s e n t i n g v a r i o u s a c t i v i t i e s i s by no means r e s o l v e d . 25 Training,, Experiments Educators have been c h a l l e n g e d , f o r decades, by the developmental theory of P i a g e t t o take a stand r e g a r d i n g i s s u e s r e l a t e d to education; f o r example, the r o l e of t r a i n i n g i n c o n s e r v a t i o n tasks,. A number of r e s e a r c h e r s , consequently, have s e r i o u s l y s t u d i e d t h i s s u b j e c t and have drawn t h e i r own c o n c l u s i o n s . Many have claimed success i n t h e i r c o n s e r v a t i o n t r a i n i n g ; o t hers have repo r t e d f a i l u r e . The f o l l o w i n g two s e c t i o n s of t h i s paper w i l l summarize some.of these t r a i n i n g s t u d i e s i n c o n s e r v a t i o n tasks other than volume, and i n volume. D i s c u s s i o n of T r a i n i n g Experiments other than Volume P i a g e t (1970, p. 13) set out "to e x p l a i n how the t r a n s i t i o n i s made from a lower l e v e l of knowledge to a l e v e l t h a t i s judged to be h i g h e r " using h i s e q u i l i b r a t i o n t h e o r y . U n f o r t u n a t e l y , t h e r e has been l i t t l e evidence v a l i d a t i n g t h i s theory;. He h i m s e l f admitted (seminar, the C a t h o l i c U n i v e r s i t y of America) t h a t the e q u i l i b r a t i o n t h e o r y does not s u f f i c i e n t l y e x p l a i n the t r a n s i t i o n between c o g n i t i v e development stages ( B e i l i n * 1971, p. 84)* S p e c i f i c a l l y , l i t t l e seems t o be known about the laws of t r a n s i t i o n from non-conservation to c o n s e r v a t i o n (Wohlwill and Lowe, 1962, p. 153). T r a i n i n g experiments have been able to suggest and t e s t v a r i o u s ways thought t o a i d the c o g n i t i v e development from non-conservation to c o n s e r v a t i o n * Even though the u l t i m a t e purpose of t h i s study i s to i n v e s t i g a t e the l e a r n i n g of a volume a l g o r i t h m i t seems 26 necessary to review t r a i n i n g experiments i n areas other than volume whenever they are r e l e v a n t . The s t u d i e s reviewed, t h e r e f o r e , i n c l u d e t r a i n i n g f o r c o n s e r v a t i o n of substance, weight, volume, and number. I t i s i n number c o n s e r v a t i o n t r a i n i n g t h a t most r e s e a r c h e r s have occupied themselves and many have claimed success. U n f o r t u n a t e l y , r e s e a r c h s t u d i e s have not been c o n s i s t e n t with regard to the e f f e c t i v e n e s s of c o n s e r v a t i o n t r a i n i n g . In f a c t , Rotbenberg and Orost (1969) repo r t e d t h a t " f o r every s u c c e s s f u l study there are o t h e r s t e s t i n g the same types of t r a i n i n g which r e p o r t no s i g n i f i c a n t t r a n s f e r - t o - t r a i n i n g " (p. 70). T h i s i s not unusual i n e d u c a t i o n a l r e s e a r c h because o f d i f f e r e n c e s i n procedures and t e s t c o n d i t i o n s and because of v a r i e d emphasis on s u b j e c t responses.. Mermelstein e t a l . (1967), f o r example, employed what appeared t o be s u c c e s s f u l procedures o f substance c o n s e r v a t i o n i n f o u r major s t u d i e s (Smedslund, 1961; Bruner, 1964, B e i l i n , 1965; S i g e l , 1966). T h e i r r e s u l t s showed t h a t none o f the t r a i n i n g procedures c o u l d induce substance c o n s e r v a t i o n . Nevertheless, s t u d i e s have r a i s e d important aspects of the process of a c q u i s i t i o n of c o n s e r v a t i o n . The most important of these a s p e c t s are adding-s u b t r a c t i n g , language and v e r b a l e x p l a n a t i o n , s c r e e n i n g , e x t i n c t i o n and r e v e r s i b i l i t y . A summary of what r e s e a r c h has shown r e g a r d i n g each of these a s p e c t s w i l l be d i s c u s s e d h e r e a f t e r . H o h l w i l l (1959) was the f i r s t t o study the e f f e c t o f " a d d i t i o n - s u b t r a c t i o n " treatment on the a c q u i s i t i o n of number 27 c o n s e r v a t i o n . The treatment c o n s i s t e d of p l a y i n g a game of matching a given s e t of o b j e c t s with one of t h r e e p i c t u r e s r e p r e s e n t i n g s i x , seven or e i g h t elements. The experimenter changed the c o n f i g u r a t i o n of o b j e c t s while adding or t a k i n g away one element. Wohlwill reported t h a t t h i s treatment was s u c c e s s f u l i n i n d u c i n g number c o n s e r v a t i o n but he a s s e r t e d t h a t the s u b j e c t s who b e n e f i t e d the most were those who were "on the doorstep" of c o n s e r v a t i o n * L a t e r , Wohlwill and Lowe (1962) and Bothenberg and Orost (1969) used s i m i l a r " a d d i t i o n - s u b t r a c t i o n " t r a i n i n g to t h a t of W o h l w i l l and a l s o r e p o r t e d success i n i n d u c i n g number conservation* As f a r as the e f f e c t of language i s concerned, i t does not seem to be c l e a r how t h i s f a c t o r c o n t r i b u t e s t o i n t e l l e c t u a l development. Language has been used mainly i n v e r b a l l y i n s t r u c t i n g the s u b j e c t s , t h e i r v e r b a l d e s c r i p t i o n of the changes i n an a t t r i b u t e , o r i n demanding v e r b a l e x p l a n a t i o n s f o r t h e i r responses. Sonstroem (1966) advocated t h a t " i t i s when the c h i l d i s both saying and doing t h a t he l e a r n s not to b e l i e v e f u l l y what he i s s e e i n g " (p. 224). B e i l i n (1965) a l s o reported success i n c o n s e r v a t i o n t r a i n i n g when he v e r b a l l y e x p l a i n e d t o the c h i l d r e n the law of c o n s e r v a t i o n a f t e r each u n s u c c e s s f u l response. Mermelstein e t a l . . (1967), however, r e p l i c a t e d the above experiments and found c o n t r a d i c t o r y r e s u l t s . They observed t h a t language t r a i n i n g i n t e r f e r e d with r a t h e r than f a c i l i t a t e d the substance c o n s e r v a t i o n p r o c e s s e s . Moreover, the e f f e c t of most t r a i n i n g procedures became i n s i g n i f i c a n t when v e r b a l e x p l a n a t i o n was demanded from the 28 s u b j e c t s who newly a c q u i r e d c o n s e r v a t i o n (Sonstroem, 1966; Wohlwill and Lowe, 1962; R o l l , 1970). The e f f e c t of s c r e e n i n g m i s l e a d i n g cues i s c l o s e l y r e l a t e d to the e f f e c t of language* S u b j e c t s who are l e d by m i s l e a d i n g aspects are put i n a s i t u a t i o n where they respond v e r b a l l y to a non-perceived a c t i o n * Bruner (1966, pp. 183-207) used the s c r e e n i n g technique i n a s k i n g f o u r - to seven-year-olds to p r e d i c t the l e v e l of the water to be poured from a c o n t a i n e r to another and then watch the a t t a i n e d l e v e l . He claimed (p. 235) t h a t i t i s when both e n a c t i v e and symbolic r e p r e s e n t a t i o n s agree t h a t the i k o n i c y i e l d s and c o n s e r v a t i o n i s achieved. He f u r t h e r claimed success i n i n d u c i n g substance c o n s e r v a t i o n with h i s f i v e - to seven-year-olds but not with the f o u r - y e a r - o l d s . Bruner's approach has been under a t t a c k by other r e s e a r c h e r s i n c l u d i n g P i a g e t h i m s e l f . P i a g e t (1968) accused Bruner of i n d u c i n g a pseudoconservation and f o r c i n g the v e r b a l i z a t i o n o f i d e n t i t y * He argued t h a t c o n s e r v a t i o n shows understanding of i d e n t i t y and not v i c e v e r s a . Moreover, Piaget advocated t h a t the mental s t r u c t u r e precedes language development r a t h e r than f o l l o w s i t . . E x p e r i m e n t a l l y , S t r a u s s and Langer (1970) r e p o r t e d r e j e c t i o n of Bruner's h y p o t h e s i s t h a t s c r e e n i n g m i s l e a d i n g cues induces substance c o n s e r v a t i o n * The e f f e c t of e x t i n c t i o n has a l s o been a source of c o n f l i c t i n the r e s e a r c h l i t e r a t u r e . E x t i n c t i o n i s measured by the degree to which conservers r e s i s t d e l i b e r a t e c o n f u s i o n by the r e s e a r c h e r * Smedslund (1961) reported t h a t t r a i n e d c o n s e r v e r s d i d not r e s i s t e x t i n c t i o n as d i d n a t u r a l c o n s e r v e r s . 29 H a l l and K i n g s l e y (1968) r e j e c t e d Smedslund's c l a i m t h a t n a t u r a l c onservers r e s i s t e x t i n c t i o n more than t r a i n e d ones do. H a l l and K i n g s l e y r e p l i c a t e d Smedslund's experiment and r e p o r t e d t h a t none of t h e i r 17 n a t u r a l c o n s e r v e r s r e s i s t e d e x t i n c t i o n . There seems to be a promising t r e n d i n i n d u c i n g c o n s e r v a t i o n by r e v e r s i b i l i t y t r a i n i n g . P i a g e t (1968) e x p l a i n s t h a t r e v e r s i b i l i t y has two forms, r e v e r s i b i l i t y by i n v e r s i o n -negation and r e v e r s i b i l i t y by r e c i p r o c i t y . I n v e r s i o n - n e g a t i o n i n c l u d e s the mental o p e r a t i o n of r e t u r n i n g t o the o r i g i n a l s t a t e while r e c i p r o c i t y c o n s i s t s of compensating v a r i a t i o n s i n r e l a t e d a t t r i b u t e s * Wallach and S p r o t t (1964) employed what seemed to be s u c c e s s f u l a p p l i c a t i o n of r e v e r s i b i l i t y , v i a i n v e r s i o n - n e g a t i o n * f o r number c o n s e r v a t i o n . They used the technique of f i t t i n g d o l l s back i n t o t h e i r beds a f t e r they have been c l u s t e r e d or s c a t t e r e d . The experimenters d i d not p r o v i d e t r a i n i n g f o r r e v e r s i b i l i t y but gave those s u b j e c t s who a l r e a d y knew i t i n r e a l l i f e a chance t o apply i t i n the number c o n s e r v a t i o n process.. In a l a t e r study, Wallach, Wall and Anderson (1967) checked Wallach and S p r o t t * s (1964) r e s u l t s and a p p l i e d them t o c o n s e r v a t i o n of continuous l i q u i d q u a n t i t y . They concluded t h a t " r e v e r s i b i l i t y as w e l l as not u s i n g m i s l e a d i n g p e r c e p t u a l cues would seem to be necessary f o r c o n s e r v a t i o n " (p. 441). R o l l ' s (1970) f i n d i n g s , however, confirmed those of Wallach and S p r o t t (1964) and Wallach et a l * (1967). R o l l (1970) a s s e r t e d t h a t i n v e r s i o n - n e g a t i o n r e v e r s i b i l i t y r e s u l t s i n an i n c r e a s e of number c o n s e r v a t i o n i f 30 no v e r b a l e x p l a n a t i o n i s r e q u i r e d . D i s c u s s i o n of Experiments I n v o l v i n g Volume, Co n s e r v a t i o n Tasks E l k i n d (1961-a) r e p l i c a t e d some of P i a g e t ' s c o n s e r v a t i o n experiments. H i s r e s u l t s agreed with P i a g e t ' s f o r mass and weight but not f o r volume. The data he c o l l e c t e d from a sample of 175 elementary students i n Newton, Massachusetts, r e v e a l e d t h a t "the c o n s e r v a t i o n of volume d i d not i n most cases (75%) appear before the age of 11" (p. 225). Carpenter (1975-b) re p o r t e d s i m i l a r r e s u l t s from the data c o l l e c t e d i n the N a t i o n a l Assessment of E d u c a t i o n a l Progress (NAEP). Only 6% o f the 9-year-olds, 21% of the 13-year o l d s and 43% of the 17-y e a r - o l d s gave c o r r e c t answers to the volume of a p i c t u r e d s o l i d which i s p a r t i t i o n e d i n t o u n i t cubes. E l k i n d (1961-rb) extended h i s r e p l i c a t i o n to j u n i o r and s e n i o r h i g h s c h o o l students. The data of 469 s u b j e c t s from Newton, Massachusetts, showed t h a t 95% to 100% of high s c h o o l graduates a t t a i n c o n s e r v a t i o n of mass and weight but only 68% of them reach the c o n s e r v a t i o n of volume* On the average, however, only 47% of the secondary students conserve volume. Moreover, E l k i n d found t h a t , i n a l l secondary grades, I.Q. s c o r e s and age were p o s i t i v e l y c o r r e l a t e d with volume c o n s e r v a t i o n and a s i g n i f i c a n t l y higher number of boys than g i r l s conserved volume. Nadel and Schoeppe (1973) r e p l i c a t e d E l k i n d ' s (1961-b) study on 28 eighth-grade females i n L e v i t t o w n , New York. T h e i r " r e s u l t s are s t r i k i n g l y p a r a l l e l to those obtained by E l k i n d f o r the same mean age group" (p. 309). 31 In volume, p a r t i c u l a r l y , Nadel and Shoeppe found t h a t o n l y 29% of t h e i r s u b j e c t s , whose mean age was 13 years 5 months, had reached the c o n c e p t i o n of volume c o n s e r v a t i o n (pi 309) E l k i n d (1962), f i n a l l y , extended h i s t e s t i n g to young a d u l t s . He chose a sample of 240 c o l l e g e students from Massachusetts whose ages v a r i e d between 17 and 37 years. His r e s u l t s were c o n s i s t e n t with h i s p r e v i o u s f i n d i n g s i n t h a t 92% of those c o l l e g e students conserved mass and weight but only 58% conserved volume. Age as w e l l as I. Q. scor e . were p o s i t i v e l y c o r r e l a t e d t o volume c o n s e r v a t i o n and s i g n i f i c a n t l y more boys conserved volume than d i d g i r l s i n a l l age groups. Towler and Wheatley (1971) r e p l i c a t e d E l k i n d ' s (1962) l a s t study on 71 c o l l e g e students at Purdue U n i v e r s i t y and they confirmed E l k i n d ' s r e s u l t s . L o v e l l and O g i l v i e (1961) used P i a g e t i a n methods to que s t i o n 191 B r i t i s h students i n grades 6 t o 9 about v a r i o u s volume concepts. They found t h a t P i a g e t (Piaget e t a l . , 1960) was c e r t a i n l y c o r r e c t when he a s s e r t e d t h a t a w e l l developed concept of volume can not be a t t a i n e d , on the average, be f o r e age 11 or 12. In f a c t , " i t appears t h a t not u n t i l the f o u r t h year (age 14) of the j u n i o r s c h o o l do 50 per cent of p u p i l s r e a l i z e t h a t the amount of water d i s p l a c e d by a s i n g l e cube i s independent of the s i z e of the f u l l c o n t a i n e r " (p. 124). On the other hand, the r e s e a r c h e r s b e l i e v e d t h a t proper s c h o o l t r a i n i n g can speed up the a c q u i s i t i o n of the volume concepts. They even assumed t h a t i t i s p o s s i b l e t o " l e a r n " how t o c a l c u l a t e i n t e r i o r volume as w e l l as occupied volume, before 32 the concept of volume (in terms of water displacement) i s developed; and t r u s t e d t h a t such a c t i v i t i e s may fo c u s a t t e n t i o n on c o n s e r v a t i o n of both i n t e r i o r and occupied volume. L o v e l l (1971) suggested t h a t even some seven and e i g h t year o l d s can l e a r n how to use the a l g o r i t h m "V = L x W x H" i n order to c a l c u l a t e the i n t e r i o r o r occupied volume of a cuboid (p. 179). O z g i r i s (1964) t e s t e d P i a g e t ' s f i n d i n g s of the ordered sequence of c o n s e r v a t i o n of substance, weight* and volume with r e s p e c t t o the v a r i a t i o n of m a t e r i a l s used i n the experiments. The data c o l l e c t e d from 120 elementary students i n I l l i n o i s supported P i a g e t ' s sequence and the average ages he found i n c o n s e r v a t i o n of substance and weight but not of volume. Only 20% of O z g i r i s ' s i x t h graders, who had a mean age of 12 years 2 months, conserved volume* Bat-Haee (1971) a l s o r e p o r t e d support of the above sequence of P i a g e t i a n c o n s e r v a t i o n tasks i n the data he c o l l e c t e d from 181 p u p i l s i n M i s s o u r i . Bat-haee, f u r t h e r s t a t e d t h a t c o n s e r v a t i o n performance was p o s i t i v e l y r e l a t e d t o I*Q. score and age; i t was independent of sex. Graves (1972) i n v e s t i g a t e d the e f f e c t of race and sex on the degree of c o n s e r v a t i o n of mass, weight* and volume. She t e s t e d 120 a d u l t s ; 30 of whom were white males, 30 white females, 30 black males, and 30 black females. The s u b j e c t s were a l l e n r o l l e d i n adult b a s i c e d u c a t i o n c l a s s e s and t h e i r grade l e v e l s v a r i e d from one to e i g h t . Graves found t h a t i n volume c o n s e r v a t i o n t a s k s , whites scored s i g n i f i c a n t l y higher than b l a c k s and males scored s i g n i f i c a n t l y higher than females. Moreover, 78% of a l l s u b j e c t a d u l t s conserved mass, 67% weight 33 but only 24% volume. Graves concluded t h a t i n the case of volume, maturation alone can not e x p l a i n development* " I t may be t h a t e d u c a t i o n a l or p r a c t i c a l e x p e r i e n c e s i n the s c i e n c e s and mathematics are necessary before an i n d i v i d u a l can a t t a i n c o n s e r v a t i o n ... of volume" (Graves, 1972, p. 223). P r i c e - W i l l i a m s , Gordon and Eaminez (1969) s t u d i e d t h e r o l e of experience and p a r t i c u l a r l y manipulation i n v a r i o u s c o n s e r v a t i o n t a s k s . They a d m i n i s t e r e d Spanish v e r s i o n s of P i a g e t i a n c o n s e r v a t i o n experiments i n number, l i q u i d , substance, weight, and volume. T h e i r sample c o n s i s t e d of a t o t a l of 56 Mexican students whose ages v a r i e d between 6 and 9 years; h a l f of the students l i v e d i n pottery-making f a m i l i e s . Those c h i l d r e n who had experi e n c e i n pottery-making scored s i g n i f i c a n t l y higher i n substance c o n s e r v a t i o n t a s k s than the othe r s u b j e c t s . The same c h i l d r e n scored higher but not s i g n i f i c a n t l y h i g h e r than the r e s t of the s u b j e c t s i n a l l four remaining c o n s e r v a t i o n t a s k s . . P r i c e - W i l l i a m s e t a l . concluded t h a t t h e i r study suggested " t h a t the r o l e of s k i l l s i n c o g n i t i v e growth may be a very important f a c t o r " (p. 769). 34 Summary and I m p l i c a t i o n s of L i t e r a t u r e Reviewed ftqe of S u b j e c t s Within c e r t a i n l i m i t s , age does not seem t o a f f e c t the process of c o n s e r v a t i o n t r a i n i n g . In number c o n s e r v a t i o n , f o r example, Rothenberg and Orost (1969) succeeded with younger c h i l d r e n while Wohlwill and Low (1962) f a i l e d with o l d e r ones. What appears to be important i s the c o g n i t i v e l e v e l of the s u b j e c t i n v o l v e d . Inhelder ( c i t e d i n Modgil, 1974, pp. 125-126) reported t h a t i n one of her experiments 12.5% of the p r e o p e r a t i o n a l s u b j e c t s progressed to an i n t e r m e d i a t e l e v e l while 75% o f the i n t e r m e d i a t e l e v e l s u b j e c t s advanced t o the o p e r a t i o n a l l e v e l . T h i s p a r t i c u l a r f i n d i n g can be seen a c r o s s many s t u d i e s ; Wohlwill (1959), B e i l i n (1965), Strauss and Langer (1970) and others have noted t h a t t r a i n i n g i s most s u c c e s s f u l with c h i l d r e n who are " i n possession of the proper c o g n i t i v e s t r u c t u r e but have not yet reached e q u i l i b r a t i o n " ( K i n g s l e y and H a l l , 1967). These s u b j e c t s are s a i d t o be at a t r a n s i t i o n a l stage and standing "on the doorstep" of c o n s e r v a t i o n . For volume, however, the reviewed s t u d i e s i n d i c a t e t h a t the expected percentage of c o n s e r v e r s among 11 and 12 year o l d s (grade 6 and 7) v a r i e s from 20% to 25% ( U z g i r i s , 1964; E l k i n d , 1961-a; Carpenter, 1975-b)i T h i s expected percentage of volume c o n s e r v e r s i s p a r t i c u l a r l y important to t h i s study. The nature of t h i s study n e c e s s i t a t e s the i n c l u s i o n of volume c o n s e r v e r s 35 i n the sample. Grade 6 seems t o be a reasonable c h o i c e f o r t h i s study because (1) i t i s j u s t one grade higher than grade 5 wherein most textbook s e r i e s i n t r o d u c e the volume algorithm f o r a cuboid and (2) one can expect t o f i n d a s u f f i c i e n t number of conservers f o r the study. Nature of Volume Conservation T e s t s The value given to the s u b j e c t s ' j u s t i f i c a t i o n f o r t h e i r responses while i n f e r r i n g t h e i r developmental stages ( i . e . , c o n s e r v a t i o n l e v e l s ) seems to have caused a c o n s i d e r a b l e d i s c u s s i o n i n the l i t e r a t u r e * Some of P i a g e t ' s main c o l l a b o r a t o r s , f o r example, have e x p l a i n e d t h a t i n P i a g e t i a n experiments " s p e c i a l a t t e n t i o n should be p a i d to the c h i l d ' s j u s t i f i c a t i o n of h i s answers (Inhelder and S i n c l a i r , 1969, p.5)." Others have c r i t i c i z e d the eminance of language i n P i a g e t i a n type experiments and, s p e c i f i c a l l y have d i s a g r e e d with P i a g e t ' s emphasis on the c h i l d ' s v e r b a l e x p l a n a t i o n of h i s a c t i o n s and d e c i s i o n s ( F l a v e l l , 1963). In order to e l i m i n a t e t h i s p r o c e d u r a l problem, some r e s e a r c h e r s have conducted r e s e a r c h i n which they d i d not r e q u i r e j u s t i f i c a t i o n f o r s t u d e n t s ' responses, and i n f a c t have attempted t o avoid v e r b a l i n s t r u c t i o n s through p r e t r a i n n i n g procedures (Braine, 1959, Bever, Mehler and E p s t e i n , 1968, f o r example)* Calhoun (1971) re p o r t e d d i f f i c u l t y i n a s s e s s i n g c h i l d r e n ' s number c o n s e r v a t i o n using t h e i r v e r b a l responses and recommended the use of t o t a l l y n onverbal procedures f o r such assessment. Researchers who favour j u s t i f i c a t i o n seem to b e l i e v e t h a t 36 without such j u s t i f i c a t i o n type I e r r o r may be caused by mistakenly i n f e r r i n g a higher developmental stage to s u b j e c t s . Subjects may respond c o r r e c t l y by c o n c e n t r a t i n g on i r r e l e v a n t a t t r i b u t e s (Smedslund, 1963 and 1969). For example, a c h i l d might respond c o r r e c t l y by simply chosing the f i r s t (or l a s t ) a l t e r n a t i v e o r , i n case of volume c o n s e r v a t i o n t e s t i n g , by f o c u s i n g on the substance or weight a t t r i b u t e . Research i s reported which has shown t h a t more s u b j e c t s are c l a s s i f i e d as c o n s e r v e r s when the j u s t i f i c a t i o n c r i t e r i o n was not used than when i t was ( B r a i n e r d , 1973, p. 174). For example. R o l l (1970) r e p o r t s t h a t few of h i s t r a i n e d number co n s e r v e r s showed v e r b a l awareness of c o n s e r v a t i o n . W o h l w i l l and Lowe (1962) i n d i c a t e t h a t u s i n g t h e i r nonverbal t r a i n i n g procedures f o r number c o n s e r v a t i o n the i n c r e a s e i n c o n s e r v a t i o n responses became n e g l i g i b l e when v e r b a l j u s t i f i c a t i o n s were demanded. Thus the argument i s t h a t j u s t i f i c a t i o n s of st u d e n t s ' responses may f u r t h e r r e v e a l t h e i r developmental l e v e l and reduce type I e r r o r . Researchers who oppose the requirement o f a v e r b a l e x p l a n a t i o n f o r i n f e r r i n g the s u b j e c t ' s developmental l e v e l , argue t h a t type I I e r r o r s may be made by using c r i t e r i a which are too s t r i n g e n t . The argument of those r e s e a r c h e r s seems to be based on the theory of P i a g e t h i m s e l f . F l a v e l l (1963), f o r example, observed t h a t i n the theory of P i a g e t "language behavior i s here t r e a t e d as a dependent v a r i a b l e with c o g n i t i o n as the independent v a r i a b l e (p. 271)." B r a i n e r d (1973) a l s o e x p l a i n s t h a t " P i a g e t has long maintained t h a t ... c o g n i t i v e 37 s t r u c t u r e s o r i g i n a t e i n l o g i c r a t h e r than language" and f u r t h e r , c o n s i d e r s adequate e x p l a n a t i o n s s u f f i c i e n t but not necessary c o n d i t i o n s f o r i n f e r e n c e of c o g n i t i v e s t r u c t u r e s (p. 177). B r a i n e r d holds t h a t i f one choses t o employ the e x p l a n a t i o n c r i t e r i o n , then one unduly r e s t r i c t s the b e h a v i o r a l domain to which the t h e o r e t i c a l c o n s t r u c t ( s t r u c t u r e ) a p p l i e s (p. 177). B r a i n e r d concludes t h a t "from the standpoint of P i a g e t ' s theory the judgement c r i t e r i o n r i s k s only the usual "extraneous" type I and type I I e r r o r s but does not r i s k any b u i l t i n source of e r r o r as does the e x p l a n a t i o n c r i t e r i o n (p.178). Hobbs (1975) s i d e d with B r a i n e r d while d i s c u s s i n g the necessary and s u f f i c i e n t s u b j e c t ' s behaviour f o r determining developmental l e v e l and a p p l i e d h i s c o n c l u s i o n s t o volume c o n s e r v a t i o n t e s t i n g . He e x p l a i n e d t h a t even though i n P i a g e t i a n type t e s t i n g a mistake i s not taken at f a c e value and s u b j e c t s are given repeated chances to g i v e the c o r r e c t answers, the p e r s i s t e n t s u b j e c t i v i t y of the experimenter i s l i a b l e t o cause i n c o r r e c t i n f e r e n c e of c o n s e r v a t i o n l e v e l s (Hobbs, 1975, p. 272). P i a g e t ' s p o s i t i o n , r e g a r d i n g the v e r b a l j u s t i f i c a t i o n of the s u b j e c t ' s a c t i o n * does not seem to oppose Brainerd's (1973) or Hobbs' (1975) p o s i t i o n . Piaget (1963) e x p l a i n e d t h a t the nature of h i s e p i s t e m o l o g i c a l s t u d i e s n e c e s s i t a t e s i n t e r a c t i o n with the s u b j e c t s i n order to "... unearth what i s o r i g i n a l and e a s i l y overlooked ... and to J. use] methods, i n c l u d i n g v e r b a l ones, which are as f r e e and f l e x i b l e as p o s s i b l e * " Furthermore, 38 he encouraged educators to conduct s t u d i e s under c o n t r o l l e d c o n d i t i o n s ( P i a g e t , 1963, p. i x ) . More r e c e n t l y , Piaget (1973) exp l a i n e d t h a t In f a c t i t i s a very general p s y c h o l o g i c a l law t h a t the c h i l d can do something i n a c t i o n long before he r e a l l y becomes 'aware 1 of what i s i n v o l v e d - 'awareness' occurs l o n g a f t e r the a c t i o n * In other words, the s u b j e c t possesses f a r g r e a t e r i n t e l l e c t u a l powers than he a c t u a l l y c o n s c i o u s l y uses. (Pia g e t , 1973, p. 86) Hobbs (1975) developed and used a d i s p l a c e d volume c o n s e r v a t i o n t e s t based on judgement alone. In the f i r s t p art of the t e s t s u b j e c t s were shown e x p e r i m e n t a l l y t h a t o b j e c t s occupy space i n water and cause the water l e v e l t o r i s e and t h a t the space occupied v a r i e s d i r e c t l y with the s i z e of o b j e c t s . The second part of the t e s t was designed to d e t e c t those who p e r s i s t e n t l y judged that weight, r a t h e r than s i z e , i s d i r e c t l y r e l a t e d t o the space occupied. The l a s t p a r t o f the t e s t c o n s i s t e d of p r e s e n t i n g b a l l s of the same s i z e and g l a s s e s of water to the same l e v e l , immersing one of the b a l l s i n water, t r a n s f o r m i n g each of the other b a l l s i n t u r n and q u e s t i o n i n g the student about the a n t i c i p a t e d water l e v e l i n the other g l a s s i f the transformed b a l l was immersed i n i t . The procedures used by Hobbs (1975) were p a r t i c u l a r l y u s e f u l i n the development of the Volume Cons e r v a t i o n T e s t used i n t h i s study. The p o s i t i o n of Piaget (1973) h i m s e l f and the i n t e r p r e t a t i o n s of F l a v e l l (1963) and B r a i n e r d (1973) to P i a g e t ' s theory are the b a s i s f o r the j u s t i f i c a t i o n o f the judgement-based Volume C o n s e r v a t i o n Test used i n t h i s study. In the development of the Volume Cons e r v a t i o n T e s t , f u r t h e r care was taken to reduce type I e r r o r caused by s t u d e n t s ' 39 c o n c e n t r a t i o n on i r r e l e v a n t a t t r i b u t e s . For example, no qu e s t i o n s t h a t allowed s u b j e c t s ' guessing were asked. In f a c t the response format c o n s i s t e d of darkening one of f i v e broken l i n e s to whichever the student thought the water rose* F u r t h e r , an e f f o r t was made to d e t e c t the students who c o n c e n t r a t e d on weight r a t h e r than volume and c l a s s i f y those students as nonconservers of volume. The c r i t e r i a used i n developing the items of the Volume C o n s e r v a t i o n Test used i n t h i s study were based on P i a g e t ' s t e s t i n g procedures of volume c o n s e r v a t i o n (Piaget e t a l . , 1960, Piaget and I n h e l d e r , 1968). The v a r i e t y of t r a n s f o r m a t i o n s t o p l a s t i c e n e b a l l s , the p r e p a r a t i o n of s u b j e c t s f o r the t e s t and d e t e c t i n g the c o n s e r v e r s of weight but not volume were ad a p t a t i o n s of Hobbs' (1975) procedures i n h i s volume c o n s e r v a t i o n t e s t i n g . The p r o t o c o l o f the Volume Cons e r v a t i o n T e s t used i n t h i s study i s presented i n d e t a i l i n Chapter I I I . Choice, of Treatments The concept underlying treatment procedures i s a very important i s s u e because t h i s study i s designed t o show the r e l a t i o n s h i p between t r a i n i n g on the usage of the volume a l g o r i t h m "V = L x W x H" and volume c o n s e r v a t i o n . A concern has been expressed about the necessary m u l t i p l i c a t i o n a b i l i t i e s i n v o l v e d i n the c a l c u l a t i o n o f the volume of a cub o i d ( S p i t l e r , 1977). I t i s c o n j e c t u r e d t h a t when students are p r o f i c i e n t i n v a r y i n g f a c t o r s of a f i x e d product they can r a p i d l y p r e d i c t , determine and compare volumes or dimensions o f cuboids* For 40 example, i t i s probable t h a t students with such p r o f i c i e n c y would s u c c e s s f u l l y solve P i a g e t ' s i s l a n d problem i n which s u b j e c t s are t o p r e d i c t the h e i g h t of a replacement f o r a condemned b u i l d i n g to be b u i l t on a base d i f f e r e n t from the o r i g i n a l . In other words i n t h i s treatment f a c t o r manipulation i s c o n s i d e r e d to be the key f o r volume c a l c u l a t i o n and c o n s e r v a t i o n . T h i s c o n j e c t u r e p r o v i d e s the b a s i s f o r the m u l t i p l i c a t i o n treatment of the study. An o u t l i n e i s given i n Chapter I I I and d e t a i l s are provided i n Appendix A. The c o n j e c t u r e mentioned above may be j u s t i f i e d w i t h i n the cojitext of the c o g n i t i v e theory of P i a g e t . I n h e l d e r and P i a g e t (1958) a s s e r t e d t h a t c o n s e r v a t i o n presumes r e v e r s i b i l i t y by i n v e r s i o n - n e g a t i o n or by r e c i p r o c i t y . R e c i p r o c i t y "... i s analogous t o compensating changes i n one a f f i r m a t i o n by equal and o p p o s i t e changes i n a r e l a t e d a f f i r m a t i o n " (Brainerd, 1970, p. 227). The procedure proposed i n the previous paragraph i s intended to t r a i n s t udents i n compensating one o r more f a c t o r s i n m u l t i p l i c a t i o n with r e s p e c t to v a r i a t i o n s i n other f a c t o r s . In f a c t the e f f e c t i v e n e s s of r e v e r s i b i l i t y v i a i n v e r s i o n -negation seems the most promising of t r a i n i n g procedures. Piaget c a u t i o n e d , however, a g a i n s t i n a p p r o p r i a t e g e n e r a l i z a t i o n a c r o s s v a r i o u s c o n s e r v a t i o n s . He e x p l a i n e d t h a t the f i r s t - o r d e r c o n s e r v a t i o n (number, l e n g t h , substance, weight, and area) are an index of c o n c r e t e o p e r a t i o n a l l e v e l and t h a t second-order c o n s e r v a t i o n s (volume, d e n s i t y , momentum, and r e c t i l i n e a r motion) are an index of formal o p e r a t i o n a l l e v e l * He added t h a t the f i r s t - o r d e r c o n s e r v a t i o n s r e q u i r e only s u c c e s s i v e 41 a p p l i c a t i o n o f the two a s p e c t s o f r e v e r s i b i l i t y ( i n v e r s i o n -negation and compensation), while the second-order ones n e c e s s i t a t e simultaneous a p p l i c a t i o n of both aspects (Inhelder and P i a g e t , 1958, p. 320). I n h e l d e r ( c i t e d i n Green, Ford and Flamer, 1971) a l s o objected to the s e p a r a t i o n of v a r i o u s a s p e c t s of r e v e r s i b i l i t y . She held t h a t emphasizing one aspect o f r e v e r s i b i l i t y , at the expense of the other, c o u l d harm the s u b j e c t s 1 l e a r n i n g . P i a g e t f u r t h e r maintains t h a t p r o f i c i e n c y i n number manipulation does not l e a d to understanding of volume i f c o n s e r v a t i o n i s not achieved. He h o l d s t h a t " i t i s one t h i n g to m u l t i p l y two numbers together and q u i t e another t o m u l t i p l y two l e n g t h s or three lengths and understand t h a t t h e i r product i s an area or a volume. The l a t t e r i n v o l v e s the c o n t i n u i t y o f space ..." (Piaget, et a l . , 1960, p. 408). There seems t o be a p o s s i b i l i t y t h a t the m u l t i p l i c a t i o n treatment would l e a d to a l i m i t e d and temporary l e a r n i n g of the volume a l g o r i t h m and of c o n s e r v a t i o n . Piaget would examine the e f f e c t i v e n e s s of such l e a r n i n g with r e s p e c t t o t h r e e c r i t e r i a : r e t e n t i o n , g e n e r a l i z a t i o n , and c o g n i t i v e l e v e l of s u b j e c t s before such t r a i n i n g (Piaget, 1964, pp. 17-18). The r e s u l t s of t h i s study are expected to r e v e a l the e f f e c t of m u l t i p l i c a t i o n s k i l l s i n the l e a r n i n g of the volume a l g o r i t h m . The other experimental treatment, l a b e l e d volume treatment, was designed to teach the volume al g o r i t h m f o r a cuboid "V = L x W x H" using an approach t h a t resembles those of s c h o o l programs used i n North America and p a r t i c u l a r l y i n 42 B r i t i s h Columbia. Such resemblance was necessary i n order to apply the r e s u l t s of the study to s c h o o l programs. On the other hand, t h i s treatment' was c o n s i d e r e d an improvement over s c h o o l approaches because i t was more comprehensive and r e q u i r e d more st u d e n t s ' a c t i v e involvement, f o r example i n b u i l d i n g with cubes, than i s normal i n the elementary s c h o o l classroom. Furthermore, the s e q u e n t i a l p r o g r e s s of a c t i v i t i e s used i n t h i s treatment, comparison, o r d e r i n g , c o u n t i n g of cubes, a l g o r i t h m , was c o n s i s t e n t with ether models f o r the t e a c h i n g of volume i n p a r t i c u l a r ( E l l i o t e t a l . , Teacher's guidebook, 1974, v. 4, p. 4) and of measurement i n g e n e r a l (Thyer and Maggs, 1971). P r e l i m i n a r y a c t i v i t i e s of t h i s treatment i n v o l v e d m a c r o s c o p i c a l d i r e c t comparison and d i r e c t o r d e r i n g o f c l o s e d boxes (cuboids) with r e s p e c t to t h e i r volume as w e l l as b u i l d i n g with u n i t cubes models of p o l y h e d r a l s some of whose u n i t s may not be v i s i b l e , c o u n t i n g the number of cubes and s t a t i n g the volume (s) ( E i c h o l z et a l . , 1974, v. 5, p. .276; D i l l e y e t a l . , 1974, v. 6, p. 110; E l l i o t et a l . , 1974, v. 4, p. 145)..Later, students used nonstandard then standard u n i t b l o c k s to b u i l d s i m i l a r cuboids t o the ones g i v e n , counted the number o f blocks and i n d i r e c t l y compared then ordered those cuboids. . The volume al g o r i t h m "7 = L x W x H" was i n t r o d u c e d as a s i m p l i f i c a t i o n ( E i c h o l z et a l . , 1974, v. 5, p. 276) of c o u n t i n g cubes i . e . , l e n g t h X width y i e l d e d the number of cubes i n one l a y e r and l e n g t h X width X height gave the t o t a l number of cubes ( D i l l e y et a l . , 1974, v. 5, p. 134; E l l i o t e t a l . , 1974, 43 v. 4, p. 143). T h i s treatment ended by a p p l y i n g the volume a l g o r i t h m to v a r i o u s cases. F o r example, the a l g o r i t h m was a p p l i e d to attachments of cuboids ( D i l l e y et a l * , 1974, v. 6, p. 111), t o p a r t i a l l y covered cuboids ( E i c h o l z e t a l . , 1974, v. 6, p. 272) and to proposed dimensional t r a n s f o r m a t i o n s ( E i c h o l z et a l . , 1974;, v. 6, p. 273)..An o u t l i n e of the main a c t i v i t i e s o f t h i s treatment i s given i n Chapter I I I and d e t a i l e d l e s s o n plans are provided i n Appendix A. C o n c l u s i o n There seems t o be a growing b e l i e f - among educators t h a t i. i n t e l l e c t u a l development* a t l e a s t f o r t r a n s i t i o n a l s u b j e c t s , can be a c c e l e r a t e d with proper t r a i n i n g . F l a v e l l and H i l l (1969) summarized: The e a r l y P i a g e t i a n t r a i n i n g s t u d i e s had n e g a t i v e outcomes, but the p i c t u r e i s now changing. I f our r e a d i n g of r e c e n t t r e n d s i s c o r r e c t , few on e i t h e r side of the A t l a n t i c would now maintain t h a t one cannot by any pedagogic means measurably %>ur, s o l i d i f y , or otherwise f u r t h e r the c h i l d * s concrete-c"jferational progress, (p. 19) Even those who are i n accord with the P i a g e t i a n theory ho l d t h a t l e a r n i n g can a c c e l e r a t e development but they maintain t h a t i t does not i n i t i a t e i t (Saedslund, 1961; H a l f o r d and F u l l e r s t o n , 1970). p i a g e t h i m s e l f c o n s i d e r s education to be a t o o l f o r stage a c c e l e r a t i o n : But i t remains t o be decided to what extent i t [ e d u c a t i o n ] i s b e n e f i c i a l . . . Consequently, i t i s h i g h l y probable t h a t t h e r e i s an optimum r a t e of development, to exceed or f a l l 44 behind which would be e q u a l l y harmful. But we do not know i t s laws, and on t h i s p o i n t a^ s w ell i t w i l l be up to f u t u r e r e s e a r c h to e n l i g h t e n us. (Piaget, 1972, quoted by f l o d g i l , 1974, pp. 126-127) 45 CHAPTER I I I PROCEDURES T h i s chapter i n c l u d e s d i s c u s s i o n and d e s c r i p t i o n of f o u r major c o n s i d e r a t i o n s : the choice of s u b j e c t s , d e s c r i p t i o n of the treatments, the p r e p a r a t i o n o f t e s t s and the way the t e s t s and treatments were conducted. Subje c t s The study was conducted on s i x t h grade students i n a suburban s c h o o l d i s t r i c t o f the lcwer mainland of B r i t i s h Columbia. These s u b j e c t s f o l l o w e d a mathematics program t y p i c a l of those used i n North America. In 1971, the annual average f a m i l y income i n that d i s t r i c t was $11 033 while the average f a m i l y income i n the Vancouver m e t r o p o l i t a n area was $10 664 ( S t a t i s t i c s Canada, 1974). These s u b j e c t s were of ages and socioeconomic s t a t u s s i m i l a r t o those of most suburban grade 6 students i n North America.. The sample chosen, t h e r e f o r e , appeared t o be r e p r e s e n t a t i v e of the p o p u l a t i o n of suburban grade 6 students i n North America. ., The s u b j e c t s c o n s i s t e d of 171 students of seven grade 6 c l a s s e s i n th r e e s c h o o l s . Each of two s c h o o l s had two f u l l 46 c l a s s e s of grade 6, while the t h i r d s c h o o l had one f u l l c l a s s of grade 6, one c l a s s of grade 5 and grade 6 combined, and one c l a s s of grade 7 and grade 6 combined. Subjects who missed any t e s t or treatment day were e l i m i n a t e d from the study..The f i n a l sample was 105 s t u d e n t s . D e s c r i p t i o n of the, Treatments The study i n c l u d e d two e x p e r i m e n t a l groups and one c o n t r o l group. The two experimental groups underwent two d i f f e r e n t treatments which were both aimed at a r r i v i n g at the volume a l g o r i t h m of a cuboid i . e . , V = L x W x H . The treatment of the c o n t r o l group c o n s i s t e d of l e a r n i n g v a r i o u s numeration systems. The t h r e e treatments were b e l i e v e d t o be of about the same l e v e l o f d i f f i c u l t y and r e g u i r e d about the same amount o f time. Each of the t h r e e treatments i s d e s c r i b e d i n d e t a i l below. Volume Treatment The aim of t h i s treatment was to teach the volume a l g o r i t h m f o r a cuboid "V = L x W x H" using a guided d i s c o v e r y method based on approaches of present s c h o o l programs. In such programs volume l e s s o n s i n c l u d e a c t i v i t i e s f o r f i n d i n g or computing the volume of cuboids by c o u n t i n g cubes or by using the a l g o r i t h m , " V = L x W x H . " The main concepts and a c t i v i t i e s of t h i s treatment are o u t l i n e d below. Complete and d e t a i l e d l e s s o n plans are provided i n Appendix A. 47 1. D i r e c t comparison of o b j e c t s . 2. D i r e c t o r d e r i n g of o b j e c t s . 3. I n d i r e c t comparison of c l o s e d boxes. 4. Standard u n i t s : m3, dm3 and cm 3. 5. I n d i r e c t o r d e r i n g of c l o s e d boxes. 6. Volume of p o l y h e d r a l models b u i l t from u n i t cubes. 7. Volume of p a r t i t i o n e d and n o n - p a r t i t i o n e d cuboids. 8. Algorithm f o r the volume of a cuboid "V = L X W X.H." 9. A p p l i c a t i o n of the volume a l g o r i t h m to cuboids and diagrams of cuboids. 10. A p p l i c a t i o n of the volume a l g o r i t h m t o the f o l l o w i n g cases: a. Word d e s c r i p t i o n of c u b o i d s . b. Cuboids touching s i d e by s i d e . c. Diagrams of cuboids with some u n i t cubes attached or removed. d. Attachments of h a l f cubes to cuboids. e. Diagrams of p a r t i a l l y covered cuboids. 11. A p p l i c a t i o n of the volume a l g o r i t h m to cuboids with proposed dimensional t r a n s f o r m a t i o n s . M u l t i p l i c a t i o n Treatment T h i s treatment, l i k e the volume treatment, was aimed at t e a c h i n g the volume algorithm of a cuboid "V = L x W x H". The emphasis here was on developing the s k i l l s of v a r y i n g two and t h r e e f a c t o r s provided t h a t t h e i r product remained f i x e d . For example, given t h a t 24 = 2 X 3 X 4, the students were t r a i n e d i n completing statements such as 24 = 6 X £ ].I { ]. T h i s t a s k was f o l l o w e d by a b r i e f d i s c u s s i o n of the volume of a cuboid "V = L x W x H". The d e t a i l s o f t h i s treatment are provided i n Appendix A; a b r i e f o u t l i n e i s . g i v e n below. 48 il . Review of the commutative and a s s o c i a t i v e p r i n c i p l e s . 2. P r e s c r i b i n g the range f o r the missing f a c t o r i n an i n e q u a l i t y i n v o l v i n g two f a c t o r s at each of i t s s i d e s . 3. P r e s c r i b i n g the range f o r the missing f a c t o r i n an i n e q u a l i t y i n v o l v i n g t h r e e f a c t o r s at each o f i t s s i d e s . 4. E f f e c t on the product of two f a c t o r s when these f a c t o r s are changed a d d i t i v e l y or m u l t i p l i c a t i v e l y . (Note: " a d d i t i v e l y " and " m u l t i p l i c a t i v e l y " w i l l subsume decrease as w e l l as inc r e a s e . ) 5. E f f e c t on the product of three f a c t o r s when these f a c t o r s are changed a d d i t i v e l y or m u l t i p l i c a t i v e l y . (Note: " a d d i t i v e l y " and " m u l t i p l i c a t i v e l y " w i l l subsume decrease as w e l l as i n c r e a s e . ) 6. E f f e c t on one of two f a c t o r s when the other f a c t o r i s changed and the product i s f i x e d 7. E f f e c t on twc (or one) of the three f a c t o r s when one (or two) of these f a c t o r s i s (are) changed and the product i s f i x e d . 8 . , C l a r i f i c a t i o n of the concept of volume. 9. .Algorithm f o r the volume of a cuboid "V = L X W X H." (IQ. . A p p l i c a t i o n o f the volume a l g o r i t h m t o p a r t i t i o n e d , p a r t i a l l y p a r t i t i o n e d and p a r t i a l l y covered cuboids. 1 1 . . A p p l i c a t i o n of the volume algorithm to cuboids with proposed dimensional t r a n s f o r m a t i o n s . C o n t r o l Treatment T h i s treatment was given to the c o n t r o l group f o r the purpose of c o n t r o l l i n g f o r any "Hawthorne", maturation, h i s t o r y and s e n s i t i z a t i o n e f f e c t s . The treatment c o n s i s t e d of an i n s t r u c t i o n a l u n i t on numeration systems and was b e l i e v e d t o be of about the same l e v e l of d i f f i c u l t y as the treatment o f f e r e d to the two experimental groups. The f o l l o w i n g i s a g e n e r a l o u t l i n e f o r t h i s treatment ( d e t a i l e d l e s s o n p l a n s may be found 49 i n Appendix A) . 1. .Review of Base 10 p l a c e value concepts 2. Bundling i n f i v e s and e x p r e s s i n g numbers i n Base 5 3. Counting i n Base 5 4. Converting numerals from Base 10 to Base 5 5. Bundling i n s i x e s and e x p r e s s i n g numbers i n Base 6 6. Counting i n Ease 6 7. Converting numerals from Base 10 to Base 6 8. C o n v e r t i n g numerals from Base 5 to Base 10 9. A d d i t i o n i n Base 5 with and without renaming 1Q. S u b t r a c t i o n i n Base 5 with and without renaming D e s c r i p t i o n . o f T e s t s Three d i f f e r e n t t e s t s were administered as p r e t e s t s , p o s t t e s t s and r e t e n t i o n t e s t s . The p r e t e s t s c o n s i s t e d of the Volume Achievement T e s t , the Volume Co n s e r v a t i o n Test and a p o r t i o n of the S t a n f o r d Achievement T e s t (SAT). The p o s t t e s t s and the r e t e n t i o n t e s t s c o n s i s t e d of the Volume Achievement T e s t , the Volume Con s e r v a t i o n Test and the M u l t i p l i c a t i o n Achievement T e s t . The Volume C o n s e r v a t i o n T e s t , the Volume Achievement Test and the M u l t i p l i c a t i o n Achievement Test were p i l o t e d using a f i f t h grade c l a s s , a s i x t h grade c l a s s and a seventh grade c l a s s . The p i l o t r e s u l t s were used to r e v i s e the c l a s s i f i c a t i o n scheme of c o n s e r v a t i o n l e v e l s , t o confirm the s u i t a b i l i t y of the grade l e v e l ( s ixth) chosen f o r the major study and to improve the t e s t i n g instruments. Each of the 50 r e v i s e d t e s t s w i l l be d e s c r i b e d i n t u r n and c o p i e s of the t e s t s except the Volume Conservation Test are i n c l u d e d i n Appendix E. The Volume Conservation Test i s completely d e s c r i b e d below and the answer sheets are given i n Appendix E. The Volume Conservation Test was based on procedures used by Piaget (1960), Piaget and I n h e l d e r (1968) and Hobbs (1975) i n t h e i r d e t e c t i o n t e s t s f o r volume c o n s e r v a t i o n l e v e l s . In t h i s t e s t the experimenter e x p l a i n e d the procedures and demonstrated the t a s k s t o the c l a s s as a group. During t h i s t e s t an e f f o r t was made to keep minimal the i n t e r a c t i o n among student s , and between students and the experimenter. Each student responded on a separate answer sheet by darkening a l i n e t o show the judgement. Terms such as "more" or " l e s s " were avoided as much as p o s s i b l e . The f i r s t p a r t (pages 2 and 3, see Appendix E) of the t e s t was intended t o g i v e the s t u d e n t s f a m i l i a r i t y with the t e s t procedures. The second part (pages 4-6) was used to i d e n t i f y those s u b j e c t s that a s s o c i a t e volume with weight. The t h i r d part (pages 7-11) was designed to c l a s s i f y the s u b j e c t s i n one of the t h r e e c a t e g o r i e s , nonconservation, p a r t i a l c o n s e r v a t i o n and c o n s e r v a t i o n . In the l a s t p a r t (pages 11 and 12) the students were asked t o give reasons f o r t h e i r judgement; t h i s provided v a l i d i t y i n f o r m a t i o n f o r the c l a s s i f i c a t i o n i n the t h i r d part of the t e s t . The f o l l o w i n g i s a d e s c r i p t i o n of the c o n s e r v a t i o n t e s t . 1, The experimenter d i s p l a y e d , side by s i d e , t h r e e i d e n t i c a l t e s t tubes p a r t i a l l y f i l l e d to the same l e v e l with c o l o u r e d water. The l e v e l s were marked around the tubes. The experimenter t o l d the group t h a t the l e v e l s were the same. The experimenter then d i s p l a y e d two i d e n t i c a l b a l l s and a l a r g e r b a l l of p l a s t i c i n e c l o s e to 51 the tubes. He t o l d the group that two of the b a l l s were the same and the t h i r d was l a r g e r . He asked a student to come forward and c o n f i r m t h a t the water l e v e l s i n the tubes were the same, t h a t two of the b a l l s were the same and t h a t the t h i r d was l a r g e r . I f the student d i s a g r e e d , the experimenter asked him to a d j u s t the amount of water or p l a s t i c i n e by adding or d e l e t i n g . Each student was given a p e n c i l and an answer booklet which c o n s i s t e d of 12 d i f f e r e n t c o l o u r e d answer sheets. 2. The experimenter asked, "What w i l l happen i f I put t h i s b a l l ( r i g h t ) i n t o t h i s tube ( r i g h t ) ? Where w i l l the water l e v e l be?" The experimenter put the b a l l i n one o f the tubes, the water rose and students were i n s t r u c t e d t o t u r n to page 2 and observe the drawn r e s u l t . T h i s question was s u g g e s t i v e by nature and was intended to f a m i l i a r i z e s t u d e n t s with the q u e s t i o n i n g and answering process. 3. The experimenter asked, "What w i l l happen i f I put t h i s other b a l l (middle) i n t o t h i s tube (middle)? Where w i l l the water l e v e l be a f t e r I put the b a l l i n ? Darken the l i n e nearest to where the water l e v e l w i l l be a f t e r I put the b a l l i n . " The s u b j e c t s darkened a l i n e showing t h e i r judgement and turned to page 3. Then the experimenter put the b a l l i n the tube and pointed out the l e v e l to the s t u d e n t s . 4. Step 3 was repeated u s i n g the t h i r d tube and the l a r g e r p l a s t i c i n e b a l l than the ones i n the f i r s t two tubes; students used page 3 f o r t h e i r responses. 5. A l l tubes and b a l l s were removed. The experimenter d i s p l a y e d two new tubes p a r t i a l l y f i l l e d with the same amount of water, a p l a s t i c i n e b a l l , and a s t e e l b a l l of the same s i z e as the p l a s t i c i n e b a l l . The experimenter pointed out t h a t the b a l l s were of the same s i z e . A d i f f e r e n t student was c a l l e d on f o r v e r i f i c a t i o n . The student was a l s o asked t o compare the weight of the s t e e l b a l l and the p l a s t i c i n e b a l l using a double-pan balance s c a l e . The experimenter put the p l a s t i c i n e b a l l i n one of the tubes, the water rose. S i m i l a r g uestions t o the ones i n s e c t i o n 3 above were asked using the s t e e l b a l l . Students responded on page 4, then the experimenter put the s t e e l b a l l i n the tube. 6. Step 5 was repeated using a s t e e l b a l l which was s m a l l e r but h e a v i e r than a b a l l of p l a s t i c i n e . Students responded on page 5, turned to page 6 and the experimenter put the s t e e l b a l l i n the tube. 7. A l l tubes and b a l l s were removed. Two new t e s t tubes p a r t i a l l y f i l l e d with water were d i s p l a y e d . The experimenter presented two cubes; one made of g l a s s , the other of aluminum..Both cubes had the same s i z e but one 52 was h e a v i e r than the other. A student came forward and confirmed these f a c t s . The experimenter put the g l a s s cube i n one of the tubes and the water rose to a c e r t a i n l e v e l . The experimenter s a i d , "From now on I w i l l not show you the answers. I f I put the aluminum cube i n t o t h i s other tube, where w i l l the water l e v e l be? Darken the l i n e n e a r e s t to where the water l e v e l w i l l be." Students were i n s t r u c t e d to respond and then t u r n to page 7. The experimenter d i d not put the aluminum cube i n the tube. Subjects who f a i l e d two of the g u e s t i o n s on pages 4, 5 and 6 demonstrated evidence of a s s o c i a t i n g volume with weight. These s u b j e c t s were c l a s s i f i e d as nonconservers of volume. The second p a r t of the t e s t immediately f o l l o w e d . 8. The experimenter presented two t e s t tubes, a n d , f i v e p l a s t i c i n e b a l l s of the same s i z e . He put one of the b a l l s i n one of the tubes and the water rose. He then r o l l e d one of the b a l l s i n t o a sausage shape i n f r o n t of the group. The experimenter then s a i d , " I f I put the sausage i n t o t h i s o t h e r tube, where w i l l the water l e v e l be? Darken the l i n e n e a r e s t to where the water l e v e l w i l l be." The c h i l d r e n responded on page 7, turned to page 8 but the experimenter d i d not put the sausage i n the tube. 9. Step 8 was repeated by t r a n s f o r m i n g one of the b a l l s i n t o nine or ten s m a l l p i e c e s , one of the b a l l s i n t o a s m a l l p i e c e and a l a r g e p i e c e , and f i n a l l y the l a s t b a l l i n t o t hree s i m i l a r but f l a t t e n e d pieces. In each of the above mentioned t r a n s f o r m a t i o n s the students used a separate answer sheet (pages 8,9,10) t o darken a l i n e i n d i c a t i n g t h e i r judgement. 1 0 . . A l l tubes and b a l l s were removed. Two beakers with the same amount of water and two p l a s t i c i n e b a l l s of the same s i z e were presented. A student came forward and confirmed these f a c t s . The experimenter put one of the b a l l s i n t o one of the beakers. He transformed the other b a l l i n t o a "pancake" shape i n f r o n t of the group. He then s a i d , " I f I put the 'pancake' i n t o t h i s other beaker, where w i l l the water l e v e l be? Darken the l i n e n e a r e s t t o where the water l e v e l w i l l be." The c h i l d r e n responded (page 11). The experimenter d i d not put the 'pancake' i n t h e beaker but he s a i d , " I f you i n d i c a t e d t h a t the l e v e l of the water w i l l be h i g h e r than the l e v e l i n the other beaker, e x p l a i n why i t w i l l be h i g h e r . I f you i n d i c a t e d t h a t the l e v e l w i l l be lower, e x p l a i n why i t w i l l be so. And i f you i n d i c a t e d t h a t the water l e v e l w i l l be the same as the other beaker, e x p l a i n why i t w i l l be the same." The students responded and turned to page 12. 53 11. The beaker c o n t a i n i n g the b a l l was r e p l a c e d by an i d e n t i c a l beaker with water at the same l e v e l . Step 10 was repeated u s i n g two i d e n t i c a l boxes made o f marble. The experimenter put the f i r s t box and i t s detached top s i m u l t a n e o u s l y i n t o one of the beakers. They sank. He then p l a c e d the top of the other box on the box and s a i d , " I f I s e a l the top to the box, put the box i n the other beaker and i t s i n k s , where w i l l the water l e v e l be? Darken the l i n e n e arest to where the water l e v e l w i l l be." The experimenter asked f o r reasons as i n step 10. Success i n the above t e s t was measured by s c o r i n g the f i v e responses on sheets 7 through 11. Each item was c o n s i d e r e d c o r r e c t i f the c o r r e c t l i n e was darkened. Students who succeeded i n a l l 5 responses were c l a s s i f i e d as c o n s e r v e r s . Those who succeeded i n one or none were c l a s s i f i e d as nonconservers. The r e s t of the students were c l a s s i f i e d as p a r t i a l c o n s e r v e r s . The comments w r i t t e n by the students on pages 11 and 12 were only used to r e v e a l the degree of c o n s i s t e n c y between the students* judgement and r e a s o n i n g . However, due t o d i f f i c u l t i e s i n i n t e r p r e t i n g v e r b a l communications these comments were not c o n s i d e r e d i n the c o n s e r v a t i o n c l a s s i f i c a t i o n . Furthermore, the guestion on page 12 concerning the two marble boxes was not a p a r t of the c o n s e r v a t i o n c l a s s i f i c a t i o n scheme. T h i s g u e s t i o n i n v o l v e d two unequal g u a n t i t i e s and was not c o n s i s t e n t with the P i a g e t i a n g u e s t i o n i n g p r o t o c o l . I t was i n c l u d e d because i t was b e l i e v e d t h a t students are a b l e more e a s i l y t o g i v e reasons f o r i n e q u a l i t y than f o r e q u a l i t y . The purpose of the Mathematics Achievement Te s t was to r e v e a l any p o s s i b l e c o r r e l a t i o n between volume achievement s c o r e s and g e n e r a l mathematics achievement. The a r i t h m e t i c computation s e c t i o n of the S t a n f o r d Achievement Te s t was used 54 (Madden et a l . , 1973). Two r e l i a b i l i t y c o e f f i c i e n t s , the s p l i t h a l f e s t imate and the c o e f f i c i e n t based on Kuder-Eichardson formula, are s t a t e d i n the manual to be 0.90 with a standard P e r r o r of 2.9. T h i s Mathematics Achievement T e s t c o n t a i n s 45 m u l t i p l e c h o i c e items and was administered, as recommended, i n a maximum of 35 minutes. The score of each s u b j e c t on t h i s t e s t was determined by the number of c o r r e c t responses. The items of the Volume Achievement Test are v a r i a t i o n s of qu e s t i o n s found i n the three textbook s e r i e s . Heath Elementary Mathematics, I n v e s t i g a t i n g School, Mathematics, and P r o j e c t  Mathematics. The t e s t was composed of 27 qu e s t i o n s f o r each of which the student was given a drawing of a cuboid, or an attachment of cuboi d s , and asked t o f i n d the volume. Measurements of dimensions were given i n the form of number without u n i t s . Scores on t h i s t e s t were a l s o determined by the number of c o r r e c t responses. The M u l t i p l i c a t i o n Achievement T e s t c o n s i s t e d of 11 m u l t i p l e c h o i c e items and 9 s h o r t answer items. In some o f the items students were given a s s e r t i o n s of the form a X b = c X d, a X b > c X d , a X b X c = d X e X f , a X b X c > d X e X f , where one of the f a c t o r s was not given and were asked to p r e d i c t t h i s missing f a c t o r . In other items students were asked t o p r e d i c t changes, such as i n c r e a s e , decrease or d o u b l i n g , i n products when f a c t o r s changed a d d i t i v e l y or m u l t i p l i c a t i v e l y . Students' scores were determined by the number of t h e i r c o r r e c t responses. 55 I n s t r u c t i o n and Testing^Procedures I n s t r u c t o r s Three male i n s t r u c t o r s who were a l l c e r t i f i e d t e a c h e r s were used t o c a r r y out the t h r e e treatments. The choice of the same sex i n s t r u c t o r s was made i n order to exclude the t e a c h e r sex v a r i a b l e . The i n s t r u c t o r s were randomized i n such a way t h a t each of them taught a l l treatments. The i n v e s t i g a t o r d i d not teach any c f the treatments. He t r a i n e d the i n s t r u c t o r s b e f o r e the treatments and provided p r i n t e d guides and m a t e r i a l s f o r d a i l y i n s t r u c t i o n . He a l s o met with the i n s t r u c t o r s as a group every morning and afternoon d u r i n g the treatment p e r i o d i n order to review l e s s o n s , handle problems and i n s u r e u n i f o r m i t y o f i n s t r u c t i o n . The i n s t r u c t o r s only c a r r i e d out the 4-day treatments; they d i d not a d m i n i s t e r any of the t e s t s . The i n v e s t i g a t o r administered the p r e t e s t , p o s t t e s t and r e t e n t i o n t e s t o f volume c o n s e r v a t i o n . Two female t e a c h e r s a s s i s t e d i n a d m i n i s t e r i n g the Mathematics Achievement P r e t e s t and the Volume Achievement P r e t e s t , p o s t t e s t and r e t e n t i o n t e s t . Schedule of I n s t r u c t i o n and T e s t i n g In the beginning of the experiment the students were t o l d t h a t the reason f o r i n c l u d i n g them i n the study was to l e a r n more about the way grade 6 students l e a r n mathematics. They were a l s o informed t h a t the outcome of the study would not 56 a f f e c t t h e i r grades a t school nor would i t serve f o r i n d i v i d u a l d i a g n o s i s or e v a l u a t i o n . The classroom teachers were requested not to teach any mathematics d u r i n g the p e r i o d of treatment nor to d i s c u s s the treatment t o p i c s with t h e i r s tudents. The experiment began by g i v i n g the p r e t e s t s : mathematics achievement, volume achievement and volume c o n s e r v a t i o n . The Volume Conservation T e s t was used to c l a s s i f y students as c o n s e r v e r s , p a r t i a l c o n s e r v e r s and nonconservers. The names of the s u b j e c t s of a l l c l a s s e s a t each s c h o o l were l i s t e d i n i t i a l l y a c c o r d i n g t o c o n s e r v a t i o n l e v e l . Then the names of p u p i l s w i t h i n each c o n s e r v a t i o n group were:randomized ac r o s s the t h r e e treatments. Boys and g i r l s were randomized s e p a r a t e l y i n order to balance f o r sex., T h i s procedure determined the s u b j e c t s of each treatment a t each of the s c h o o l s . Three s c h o o l days at the beginning of the experiment were r e s e r v e d f o r the p r e t e s t s and randomization. The treatments began a f t e r the three p r e t e s t i n g days and l a s t e d f o r f o u r c o n s e c u t i v e s c h o o l days. At each s c h o o l the students of both c l a s s e s were taught i n the t h r e e predetermined groups f o r the same c l a s s p e r i o d . On each of the four days i n s t r u c t o r s moved t c a l l t h r e e s c h o o l s i n such a way t h a t they gave at each s c h o o l one c l a s s p e r i o d of i n s t r u c t i o n before r e c e s s , one a f t e r r e c e s s and one i n the a f t e r n o o n . Two days a f t e r the treatments the Volume Co n s e r v a t i o n Test, the Volume Achievement T e s t and the M u l t i p l i c a t i o n Achievement Te s t were administered. Three c o n s e c u t i v e s c h o o l days were reserved f o r these p o s t t e s t s . The r e t e n t i o n t e s t s of 5 7 volume c o n s e r v a t i o n , volume achievement and m u l t i p l i c a t i o n achievement were given seven weeks a f t e r the p o s t t e s t s . During these seven weeks the classroom t e a c h e r s resumed r e g u l a r classroom i n s t r u c t i o n i n mathematics. Design of the Study The b a s i c concern of t h i s study n e c e s s i t a t e d c o n s i d e r i n g two main f a c t o r s . One f a c t o r was made up of the three l e v e l s of volume c o n s e r v a t i o n , while the other was composed of the t h r e e treatments. There were a l s o f o u r dependent v a r i a b l e s , the Volume achievement P o s t t e s t , the Volume achievement Retention Test, the M u l t i p l i c a t i o n Achievement P o s t t e s t and the M u l t i p l i c a t i o n ! Achievement R e t e n t i o n T e s t . Scores on the Volume Achievement P r e t e s t and Mathematics Achievement P r e t e s t (SAT) were used as c o v a r i a t e s . A schematic r e p r e s e n t a t i o n o f the design i s g i v e n i n Table 3-1. Table 3.1 Experimental Design Conservation l e v e l s M u l t i p l i c a t i o n Volume C o n t r o l and treatments Nonconservers P a r t i a l c onservers Conservers A P r e t e s t - P o s t t e s t C o n t r o l Group design which appears i n Campbell and S t a n l e y (1963, p. 13) was used with b l o c k i n g on 58 the c o n s e r v a t i o n f a c t o r . . S c h e m a t i c a l l y the design i s as f o l l o w s : Randomized G1 P r e t e s t Treatment 1 assignment G2 P r e t e s t Treatment 2 to groups G3 P r e t e s t Treatment 3 Hypotheses While s e a r c h i n g f o r answers to the aims of the study c e r t a i n s t a t i s t i c a l hypotheses i n n u l l form were t e s t e d . Some of the hypotheses l i s t e d below were deduced from the design d e s c r i b e d above while the others were based on the aims of the study. H 1. There a r e no s i g n i f i c a n t d i f f e r e n c e s i n volume achievement s c o r e s , on the p o s t t e s t and r e t e n t i o n t e s t , among c o n s e r v a t i o n groups. H 2. There are no s i g n i f i c a n t d i f f e r e n c e s i n volume achievement s c o r e s , on the p o s t t e s t and r e t e n t i o n t e s t , among treatment groups. H 3. There are no s i g n i f i c a n t i n t e r a c t i o n s i n volume achievement s c o r e s , on the p o s t t e s t and r e t e n t i o n t e s t , between c o n s e r v a t i o n and treatment. P o s t t e s t Retention Test P o s t t e s t Retention Test P o s t t e s t Retention Test 59 H 4. There are no s i g n i f i c a n t d i f f e r e n c e s i n m u l t i p l i c a t i o n achievement s c o r e s , on the p o s t t e s t and r e t e n t i o n t e s t , among treatment groups. H 5. The t r a n s i t i o n from a lower l e v e l of c o n s e r v a t i o n on the p r e t e s t to a higher l e v e l of c o n s e r v a t i o n , on the p o s t t e s t and r e t e n t i o n t e s t , i s independent of volume achievement scores on the p o s t t e s t and r e t e n t i o n t e s t r e s p e c t i v e l y . H 6. The t r a n s i t i o n from a lower l e v e l of c o n s e r v a t i o n on the p r e t e s t t o a higher l e v e l of c o n s e r v a t i o n , on the p o s t t e s t and r e t e n t i o n t e s t , i s independent of treatments. H 7. The t r a n s i t i o n from a hi g h e r l e v e l of c o n s e r v a t i o n on the p r e t e s t t o a lower l e v e l of c o n s e r v a t i o n , on the p o s t t e s t and r e t e n t i o n t e s t , i s independent of volume achievement s c o r e s on the p o s t t e s t and r e t e n t i o n t e s t r e s p e c t i v e l y . H 8. The t r a n s i t i o n from a h i g h e r l e v e l of conservation, on the p r e t e s t t o a lower l e v e l of c o n s e r v a t i o n , on the p o s t t e s t and r e t e n t i o n t e s t , i s independent of treatments. H 9. The volume achievement s c o r e s , on the p o s t t e s t and r e t e n t i o n t e s t , are not r e l a t e d t o mathematics achievement s c o r e s . 60 H10. The i n i t i a l l e v e l of c o n s e r v a t i o n i s independent of mathematics achievement s c o r e s . . H11. The i n i t i a l l e v e l o f c o n s e r v a t i o n i s not r e l a t e d to sex. H12. The volume achievement s c o r e s on the p o s t t e s t are not r e l a t e d to sex. S t a t i s t i c a l Analyses In order t o t e s t hypotheses 1-4, each dependent v a r i a b l e p o s t t e s t score and r e t e n t i o n t e s t s core, was analyzed s e p a r a t e l y by using a 3 X 3 f u l l y c r o s s e d two-way a n a l y s i s o f c o v a r i a n c e . The a n a l y s i s was c a r r i e d out using the computer program BMDP2V. In the cases where s i g n i f i c a n t d i f f e r e n c e s were found a c r o s s treatments and c o n s e r v a t i o n l e v e l s , S c h e f f e post hoc comparisons were made to determine which groups d i f f e r e d s i g n i f i c a n t l y . Hypotheses 5 and 7 d e a l t with two v a r i a b l e s , one o f which i n v o l v e d a f o r c e d dichotomy while the other was measured on an i n t e r v a l s c a l e . A b i s e r i a l c o r r e l a t i o n c o e f f i c i e n t as recommended by Gl a s s and Stanley (1970, p. 168) was used f o r t e s t i n g these hypotheses. Hypotheses 6 and 8 were t e s t e d u s i n g frequency t a b l e s and the C h i Square s t a t i s t i c . . However, s i n c e the number of t r a n s i t i o n a l cases amonq the t h r e e c o n s e r v a t i o n l e v e l s was expected t o be sm a l l i t was not a p p r o p r i a t e to use 61 the number of t r a n s i t i o n a l cases between each p a i r of l e v e l s . Instead t r a n s i t i o n s were c l a s s i f i e d as "change" vs "no change" f o r each treatment i n order t o allow f o r Yates' c o r r e c t i o n f o r c o n t i n u i t y to be a p p l i e d (Glass and Stanley, 1970, p. 332). Hypothesis 11 d e a l t with a nominal v a r i a b l e and an o r d i n a l v a r i a b l e . The Wilcoxon two-sample t e s t using t i e d s c o r e s as recommended by M a r a s c u i l o and McSweeny was used f o r t e s t i n g o f t h i s h y p o t h e s i s (1977, p. 267). Hypothesis 9 d e a l t with two v a r i a b l e s on i n t e r v a l s c a l e s and the recommended a n a l y s i s i s the usage of Pearson's product-moment c o r r e l a t i o n c o e f f i c i e n t (Glass and S t a n l e y , 1970, pp. 109-113). Hypothesis 10 i n c l u d e d v a r i a b l e s cn o r d i n a l and i n t e r v a l s c a l e s . T h i s h y p o t h e s i s was t e s t e d using K e n d a l l ' s Tau c o r r e l a t i o n c o e f f i c i e n t (Glass and S t a n l e y , 1970, pp..176-178). F i n a l l y , hypothesis 12 i n v o l v e d a t r u l y dichotomous v a r i a b l e and a v a r i a b l e on an i n t e r v a l s c a l e . I t was t e s t e d using a p o i n t - b i s e r i a l c o r r e l a t i o n c o e f f i c i e n t (Glass and S t a n l e y , 1970, pp. 163-164). 62 CHAPTER IV RESULT S T h i s chapter c o n t a i n s the r e s u l t s of the study. Included are s e c t i o n s on t e s t a nalyses, p r e l i m i n a r y study, a n a l y s i s of c o v a r i a n c e , c o r r e l a t i o n study, t e s t s f o r independence and post hoc q u a l i t a t i v e a n a l y s e s . The s e c t i o n of p r e l i m i n a r y study c o n t a i n s two p a r t s , the v a r i a b l e c o v a r i a t e s and the i n s t r u c t o r e f f e c t . . Tests R e l i a b i l i t i e s and Item A n a l y s i s Three d i f f e r e n t t e s t s were administered as p r e t e s t s , p o s t t e s t s and r e t e n t i o n t e s t s . The p r e t e s t s c o n s i s t e d of the Volume Achievement T e s t , the Volume Conservation Test and the a r i t h m e t i c computation s e c t i o n of the Stanfo r d Achievement T e s t (SAT). The p o s t t e s t s and the r e t e n t i o n t e s t s c o n s i s t e d of the Volume Achievement Test, the Volume Co n s e r v a t i o n Test and the M u l t i p l i c a t i o n Achievement Test. The t e s t r e l i a b i l i t i e s and a summary of item a n a l y s i s w i l l be reported f o r the Volume Achievement T e s t , the Volume Co n s e r v a t i o n Test and the m u l t i p l i c a t i o n achievement t e s t . A complete l i s t o f item s t a t i s t i c s can be found i n Appendix B. 63 Volume Achievement Test The Hoyt estimates of r e l i a b i l i t i e s ( i n t e r n a l c o n s i s t e n c y ) i n the p r e t e s t , p o s t t e s t and r e t e n t i o n t e s t were 0.94, 0.95 and 0.94 r e s p e c t i v e l y . The d i f f i c u l t y l e v e l , as d e f i n e d by the percent of s u b j e c t s responding c o r r e c t l y , ranged from 8.2 to 80.7 i n the p r e t e s t , from 23.4 to 78.4 i n the p o s t t e s t and from 24.6 to 87.1 i n the r e t e n t i o n t e s t . The item p o i n t - b i s e r i a l c o r r e l a t i o n c o e f f i c i e n t s ranged from 0.35 t o 0.74 i n the p r e t e s t s , from 0.35 to 0.79 i n the p o s t t e s t and from 0.39 to 0.78 i n the r e t e n t i o n t e s t . Volume C o n s e r v a t i o n Test The Hoyt estimate of " r e l i a b i l i t i e s i n the p r e t e s t , p o s t t e s t and r e t e n t i o n t e s t were 0.78, 0.85 and 0.82 r e s p e c t i v e l y . . The d i f f i c u l t y l e v e l ranged from 35.7% to 76.6% i n the p r e t e s t , from 59.6% to 86.0% i n the p o s t t e s t and from 68.4% t o 91.2% i n the r e t e n t i o n t e s t . The item p o i n t - b i s e r i a l c o r r e l a t i o n c o e f f i c i e n t s ranged from 0.46 to 0.78 i n the p r e t e s t , from 0.46 to 0.75 i n the p o s t t e s t and from 0.33 to 0.77 i n the r e t e n t i o n t e s t . M u l t i p l i c a t i o n Achievement T e s t The Hoyt estimate of r e l i a b i l i t i e s i n the p o s t t e s t and r e t e n t i o n t e s t were 0.86 and 0.79 r e s p e c t i v e l y . . T h e d i f f i c u l t y l e v e l ranged from 8.2% to 85.4% i n the p o s t t e s t and from 7.6% t o 93.0% i n the r e t e n t i o n t e s t . The item p o i n t - b i s e r i a l c o r r e l a t i o n c o e f f i c i e n t s ranged from 0.34 to 0.67 i n the 64 p o s t t e s t and from 0.34 t o 0.55 i n the r e t e n t i o n t e s t . P r e l i m i n a r y A n a l y s i s C o v a r i a t e s The S t a n f o r d Achievement Test (SAT) and the Volume Achievement P r e t e s t were co n s i d e r e d as p o s s i b l e c o v a r i a t e s f o r the f o u r dependent v a r i a b l e s , Volume Achievement P o s t t e s t , Volume Achievement Bet e n t i o n T e s t , M u l t i p l i c a t i o n Achievement P o s t t e s t and M u l t i p l i c a t i o n Achievement B e t e n t i o n T e s t . To determine the c o v a r i a t e s a sequence of m u l t i p l e step-wise r e g r e s s i o n a n a l y s i s was conducted using a 5% i n c l u s i o n and a 558 d e l e t i o n l e v e l s . The a n a l y s i s was c a r r i e d out using the computer program BMD02B which a u t o m a t i c a l l y removes any v a r i a b l e when i t s s i g n i f i c a n c e l e v e l becomes too low. The r e s u l t s of the a n a l y s i s r e v e a l e d t h a t o n l y the Volume Achievement P r e t e s t entered as a c o v a r i a t e f o r both the Volume Achievement P o s t t e s t (F=78.63, df= 1,103) and the Volume Achievement B e t e n t i o n Test (F=58.39,' df=1,103).. The M u l t i p l i c a t i o n P o s t t e s t , however, had two c o v a r i a t e s , the SAT (F=35. 31, df=1,103) and the Volume Achievement P r e t e s t (F=11.33, df=2,102). S i m i l a r l y , the M u l t i p l i c a t i o n B e t e n t i o n Test had two c o v a r i a t e s the SAT (F=43.35, df=1,103) and the Volume Achievement P r e t e s t (F=21.81, df=2,102). Table 4.1 summarizes these r e s u l t s . 65 Table 4.1 F Values f o r E n t e r i n g of C o v a r i a t e s Dependent V a r i a b l e C o v a r i a t e df F Value Vol Ach Post Mul Ach Post Vol Ach Bet Mul Ach Ret Vol Ach Pre Vol Ach Pre SAT Vol Ach Pre SAT Vol Ach Pre 1,103 1,103 2,102 1,103 1,103 2, 102 78.63 35.31 11.33 85. 39 43.35 21.81 Vol Ach Pre: Volume Achievement P r e t e s t Vol Ach Post: Volume Achievement P o s t t e s t V o l Ach Ret: Volume Achievement Retention t e s t Mul Ach Post: M u l t i p l i c a t i o n Achievement P o s t t e s t Mul Ach Ret: M u l t i p l i c a t i o n Achievement Rete n t i o n t e s t SAT: Stanfo r d Achievemt Test Assumptions of A n a l y s i s of Covariance A n a l y s i s of c o v a r i a n c e as recommended by Winer (1971, pp. 752-753) was p r e f e r e d t o a n a l y s i s of varia n c e s i n c e b l o c k i n g was f e a s i b l e on c o n s e r v a t i o n l e v e l o n l y while past achievement i n computation and volume c a l c u l a t i o n were b e l i e v e d to be r e l a t e d t o f u t u r e volume achievement. In the a n a l y s i s of co v a r i a n c e adjustment of dependent v a r i a b l e s ( i . e . , volume achievement and m u l t i p l i c a t i o n achievement) f o r the r e g r e s s i o n on c o v a r i a t e s ( i . e . , SAT and p r e t e s t volume achievement) was intended to reduce b i a s and i n c r e a s e the accuracy o f the treatment e f f e c t (Cochran, 1957, p. 262). Fu r t h e r j u s t i f i c a t i o n s f o r using a n a l y s i s of c o v a r i a n c e are r e l a t e d to the f u l f i l l m e n t of c e r t a i n assumptions ( E l a s h o f f , 1969, p. 385): 1. Randomization was f u l f i l l e d s i n c e i n d i v i d u a l s were assigned randomly to groups and groups were assigned randomly to treatments. 2. C o v a r i a t e s , having been measured at the beginning of the experiment, were independent 66 of treatments. 3. C o v a r i a t e s are measured a c c u r a t e l y u sing the s t a n d a r d i z e d SAT (Kuder-Eichardson r e l i a b i l i t y c o e f f i c i e n t = 0.90) and the Volume Achievement T e s t (Hoyt estimate of r e l i a b i l i t y = 0.94). 4. The r e g r e s s i o n of the dependent v a r i a b l e s on the c o v a r i a t e s was b e l i e v e d t o be l i n e a r . In f a c t i n the p r e l i m i n a r y a n a l y s i s , a l i n e a r r e g r e s s i o n model was used to determine the r e l a t i o n s h i p between dependent v a r i a b l e s and c o v a r i a t e s . The s i g n i f i c a n t r e l a t i o n s h i p found confirmed the assumption of l i n e a r i t y . 5. F u l f i l l m e n t of the assumption of homogeneity of c o v a r i a n c e and no treatment-slope i n t e r a c t i o n was done through comparing s c a t t e r p l o t s of the dependent v a r i a b l e s versus c o v a r i a t e s f o r each treatment group ( E l a s h o f f , 1969, p. 392). Winer (1971) f u r t h e r c l a i m s t h a t "there evidence to i n d i c a t e t h a t the a n a l y s i s of c o v a r i a n c e i s robust with r e s p e c t t o homogeneity assumptions on ... r e g r e s s i o n c o e f f i c i e n t s (p.772)." 6. The assumption of normal d i s t r i b u t i o n of dependent v a r i a b l e s w i t h i n each treatment group at each c o n s e r v a t i o n l e v e l was not t e s t e d because of the s m a l l sumber of s u b j e c t s i n some c e l l s . E l a s h o f f (1969) e x p l a i n s t h a t the f u l f i l l m e n t of t h i s assumption i . e . , n o r m a l i t y i s r e q u i r e d f o r " s t a t i s t i c a l convenience" only (p. 386). I n s t r u c t o r . E f f e c t The proposed experimental (Treatments X Conservation problem i n a treatment group necessary the i n c l u s i o n of design i n chapter 3 was 3X3 L e v e l s ) . However, a behaviour i n one of the s c h o o l s made i n s t r u c t o r s as a f a c t o r . The 67 e x perimental design became 3X3X3 (Treatments X Co n s e r v a t i o n L e v e l s X I n s t r u c t o r s ) . A n a l y s i s of co v a r i a n c e were conducted to determine the e f f e c t of i n s t r u c t o r s i n both the Volume Achievement P o s t t e s t and the M u l t i p l i c a t i o n Achievement P o s t t e s t , . Since the i n t e n t of the 3-way a n a l y s i s was to i n s u r e t h a t any i n s t r u c t o r e f f e c t w i l l be i d e n t i f i e d , the s i g n i f i c a n c e l e v e l f o r the r e j e c t i o n of the n u l l hypothesis was set at 0.10. The main e f f e c t of i n s t r u c t o r s and the 3-way i n t e r a c t i o n s were found to be n o n s i g n i f i c a n t (p < 0.10).. The only 2-way i n t e r a c t i o n , found t o be s i g n i f i c a n t (p < 0. (10) was I n s t r u c t o r s X Treatments on the M u l t i p l i c a t i o n Achievement P o s t t e s t . A summary of the a n a l y s i s o f c o v a r i a n c e f o r the i n s t r u c t o r s e f f e c t i s presented i n Table 4.2. Since the main e f f e c t of i n s t r u c t o r s was not s i g n i f i c a n t and j u s t a s i n g l e 2-way i n t e r a c t i o n i n v o l v i n g i n s t r u c t o r s was s i g n i f i c a n t , the e f f e c t of i n s t r u c t o r s was not thought to be stro n g enough to n e c e s s i t a t e r e s t r u c t u r i n g the o r i g i n a l 2-way design o f the study. I t was t h e r e f o r e p o s s i b l e t o pool a c r o s s i n s t r u c t o r s and reduce the design t o 3X3 (Treatments X Conse r v a t i o n Levels) as was suggested i n chapter 3. 68 Table 4.2 A n a l y s i s of Covariance - I n s t r u c t o r s E f f e c t Source o f V a r i a t i o n df V o l Ach Post Main E f f e c t I n s t r 2 Tr 2 Cons 2 2- way I n t e r a c t i o n s I n s t r X Tr 4 I n s t r X Cons 4 Cons X Tr 4 3- way I n t e r a c t i o n s 8 E r r o r 76 MS 39.54 517.02 93. 24 12.77 52. 10 25.58 21.02 25.65 s i g n i f i c a n c e il.54 20.15 3.63 0.50 2.03 1.00 0.82 0. 22 0.00 0.03 0.74 0. 10 0.41 0. 59 Mul Ach Post Main E f f e c t I n s t r Tr Cons 2- way I n t e r a c t i o n s I n s t r X Tr I n s t r X Cons Cons X Tr 3- way I n t e r a c t i o n s E r r o r 2 2 2 4 4 4 8 76 11.40 71.80 25.10 20.55 4.58 16.09 13.00 8.86 1.29 8.10 2.83 2.32 0.52 1.82 1.47 0.28 0.01 0.07 0.07* 0.72 0. 14 0. 18 * p < 0.10 ( I n s t r u c t o r e f f e c t only) I n s t r : I n s t r u c t o r ; T r : Treatment; Cons: Co n s e r v a t i o n L e v e l A n a l y s i s of Covariance 1-3 The t e s t of hypothesis 1 (p. 58) r e v e a l e d t h a t a s i g n i f i c a n t (p < 0.05) d i f f e r e n c e was found i n Volume Achievement P o s t t e s t s c o r e s among c o n s e r v a t i o n groups. The ANCOVA f o r the Volume Achievement P o s t t e s t s c o r e s can be found i n Table 4.3. Post hoc a n a l y s i s using S h e f f e ' s method of 69 m u l t i p l e comparisons showed a s i g n i f i c a n t (p < 0.05) s u p e r i o r i t y o f the conservers group over the p a r t i a l - c o n s e r v e r s group. No s i g n i f i c a n t d i f f e r e n c e was found among any other c o n s e r v a t i o n groups at the 0.05 l e v e l . There was no s i g n i f i c a n t d i f f e r e n c e found i n Volume Achievement E e t e n t i o n T e s t s c o r e s among c o n s e r v a t i o n groups at the 0.05 l e v e l . The ANCOVA f o r the Volume Achievement E e t e n t i o n T e s t s c o r e s can be found i n Table 4.4. The means, standard d e v i a t i o n s and group s i z e s of treatments by c o n s e r v a t i o n l e v e l s f o r the Volume Achievement P r e t e s t can be found i n Table 4.5. The unadjusted means, standard d e v i a t i o n s and group s i z e s of treatments by c o n s e r v a t i o n l e v e l s f o r the Volume Achievement P o s t t e s t and Volume Achievement E e t e n t i o n Test can be found i n Appendix D. The a d j u s t e d means, standard d e v i a t i o n s and group s i z e s of treatments by c o n s e r v a t i o n l e v e l s f o r the Volume Achievement P o s t t e s t and Volume Achievement E e t e n t i o n T e s t can be found i n Tables 4.6 and 4.7 r e s p e c t i v e l y . I n a l l of these t a b l e s the marginal values f o r the means and standard d e v i a t i o n s are determined by c a l c u l a t i n g the weighted average of the c e l l v a l u e s . The t e s t of hy p o t h e s i s 2 (p. 58) rev e a l e d t h a t a s i g n i f i c a n t (p < 0.001) d i f f e r e n c e was found i n Volume Achievement P o s t t e s t scores among treatment groups (see Table 4.3). Post hoc a n a l y s i s using S h e f f e ' s method of m u l t i p l e comparisons showed a s i g n i f i c a n t s u p e r i o r i t y of volume treatment over m u l t i p l i c a t i o n treatment.(p < 0.01) and over 70 c o n t r o l treatment (p < 0.01). No s i g n i f i c a n t d i f f e r e n c e was found between m u l t i p l i c a t i o n treatment and c o n t r o l treatment at the 0.05 l e v e l . S i m i l a r l y , a s i g n i f i c a n t (p < 0.001) d i f f e r e n c e was found i n Volume Achievement R e t e n t i o n Test scores among treatment groups (see Table 4.4). Post hoc a n a l y s i s using She f f e ' s method of m u l t i p l e comparisons showed a s i g n i f i c a n t s u p e r i o r i t y of volume treatment over m u l t i p l i c a t i o n treatment (p < 0.01) and over c o n t r o l treatment (p < 0.01) . No s i g n i f i c a n t d i f f e r e n c e was found between m u l t i p l i c a t i o n treatment and c o n t r o l treatment at the 0.05 l e v e l . . The t e s t of hypothesis" 3 (p. 58) r e v e a l e d t h a t no s i g n i f i c a n t i n t e r a c t i o n was found i n Volume Achievement P o s t t e s t s c o r e s between c o n s e r v a t i o n l e v e l s and treatments at the 0.05 l e v e l . S i m i l a r l y , no s i g n i f i c a n t i n t e r a c t i o n was found i n Volume Achievement R e t e n t i o n Test s c o r e s between c o n s e r v a t i o n l e v e l s and treatments at the 0.05 l e v e l . The a d j u s t e d means on the Volume Achievement P o s t t e s t were 13.06 (48%) f o r the nonconservers, 10.82 (40%) f o r the p a r t i a l c o n s e r v e r s , 15.18 (56%) f o r the c o n s e r v e r s , 17.42 (65%) f o r the volume treatment group, 12.30 (46%) f o r the m u l t i p l i c a t i o n treatment group and 9.87 (37%) f o r the c o n t r o l group. L i k e w i s e the a d j u s t e d means on the Volume Achievement Retention Test were 14. 52 (54%) f o r the nonconservers, 11. 23 (42%) f o r the p a r t i a l c o n s e r v e r s , 14.87 (55%) f o r the c o n s e r v e r s , 17.49 (65%) f o r the volume treatment group, 13.51 (50%) f o r the m u l t i p l i c a t i o n treatment group and 10.89 (40%) f o r the c o n t r o l group. 71 Table 4 .3 A n a l y s i s of Covariance of Volume Achievement P o s t t e s t Scores — =—— —— —— — — = — 3 ==—————= ——=== =--------== -----:==== = = Source df MS F Conservation L e v e l 2 88. 88 3. 20* Treatment 2 339.56 12. 24** Conservation L e v e l X Treatment 4 31. 50 1. 14 E r r o r 95 27.74 * p < 0.05 ** p < 0.001 Table 4.4 A n a l y s i s of Covariance of Volume Achievement E e t e n t i o n Scores Source df MS F Con s e r v a t i o n L e v e l 2 74. 16 2.67 Treatment 2 217.01 7.81* Cons e r v a t i o n L e v e l X Treatment 4 23.35 0.84 E r r o r 95 27. 79 * p < 0.001 Table 4.5 Means, Standard D e v i a t i o n s , and Group S i z e s of Volume Achievement P r e t e s t Scores f o r Treatments by Conservation L e v e l s (Maximum Score = 27) Conser v a t i o n •••Treatments • „, Lev e l Volume M u l t i p l i c a t i o n C o n t r o l T o t a l Non-conservers 7.21 7 .61 4.45 6. 37 (5. 23) (4 .88) (3.98) (4.68) 19 18 20 57 P a r t i a l - c o n s e r v e r s 10.17 8 .67 5.75 8. 50 (8. 23) (7 -71) (4.50) (7. 10) 6 6 4 16 Conservers 9.18 10 .60 14.00 10. 75 • (6.85) (6 .17) (8.85) (6.9D 11 15 6 32 T o t a l 8.31 8 .92 6.53 8. 03 (6. 23) (5 .81) (5.02) (5.73) 36 39 30 105 72 Table 4.6 Adjusted Means, Standard D e v i a t i o n s , and Group S i z e s of Volume Achievement P o s t t e s t Scores f o r Treatments by Conservation L e v e l s (Maximum Score = 27) Conservation • Treatments L e v e l Volume M u l t i p l i c a t i o n C o n t r o l T o t a l Non-conservers 18.33 11 .60 9.35 13.06 (5.63) (7 .27) (6.87) (6.58) 19 18 20 57 P a r t i a l - c o n s e r v e r s 15.49 7 .83 8.29 10.82 (7.41) (10.17) (7.05) (8.36) 6 6 4 16 Conservers ~ 16.91 14 .92 12*65 15. 18 (4.29) (7 .54) (8.59) (6. 62) 11 15 6 32 T o t a l 17.42 12 .30 9.87 13.36 (5.52) (7 .82) (7.24) (6.86) 36 39 30 105 Table 4.7 Adjusted Means, Standard D e v i a t i o n s , and Group S i z e s of Volume Achievement Retention Test Scores f o r Treatments by Conservation L e v e l s (Maximum Score = 27) Cons e r v a t i o n ..Treatments L e v e l Volume M u l t i p l i c a t i o n C o n t r o l T o t a l Non-conservers 19.00 13 .91 10 .81 14.52 (6.51) (6 .34) (6 .90) (6.59) 19 •18 20 57 P a r t i a l - c o n s e r v e r s 14.04 8 .96 10 .41 11. 23 (8. 13) (10.00) (8 .66) (8.96) 6 6 4 16 Conservers 16.76 14 .84 11 .48 14. 87 (5.08) (7 .96) (9 .05) (7.17) 11 15 6 32 T o t a l 17.49 13 .51 10 .89 14. 12 (6.34) (7 .53) (7 .56) (7.13) 36 39 30 105 73 Hypothesis 4 The t e s t of h y p o t h e s i s 4 (p* 59) r e v e a l e d t h a t a s i g n i f i c a n t d i f f e r e n c e (p < 0.001) was found i n M u l t i p l i c a t i o n Achievement P o s t t e s t s c o r e s among treatment groups. No s i g n i f i c a n t d i f f e r e n c e was found i n M u l t i p l i c a t i o n Achievement Retention s c o r e s among treatment groups at the 0.05 l e v e l . The ANCOVA f o r the M u l t i p l i c a t i o n P o s t t e s t s c o r e s can be found i n Table 4.8 and f o r the r e t e n t i o n t e s t s c o r e s i n T a b l e 4.9. Post hoc a n a l y s i s using S h e f f e ' s m u l t i p l e comparisons, a t the p o s t t e s t l e v e l , showed a s i g n i f i c a n t s u p e r i o r i t y of m u l t i p l i c a t i o n treatment over volume treatment (p < 0.01) and over c o n t r o l treatment (p < 0.01). The means, standard d e v i a t i o n s and group s i z e s of M u l t i p l i c a t i o n Achievement P o s t t e s t and Retention t e s t s c o r e s f o r treatments can be found i n Table 4.10. Table 4.8 A n a l y s i s of Covariance of M u l t i p l i c a t i o n P o s t t e s t Scores Source df MS F C o n s e r v a t i o n L e v e l 2 23.46 2.46 Treatment 2 98.54 10.33* Cons e r v a t i o n L e v e l X Treatment 4 19.16 2.01 E r r o r 94 9.54 * p < Q. 001 74 Table 4.9 A n a l y s i s of Covariance of M u l t i p l i c a t i o n R e t e n t i o n Scores Source df MS F 1 w~— —• , , . , , C o n s e r v a t i o n L e v e l 2 4.62 0.68 Treatment 2 7.81 1.15 Cons e r v a t i o n L e v e l X Treatment 4 3.42 0.51 E r r o r 94 6.76 Table 4. 10 Adjusted Means, Standard D e v i a t i o n s , and Group S i z e s of M u l t i p l i c a t i o n Achievement P o s t t e s t and Retention T e s t Scores f o r Treatments (Maximum Score = 20) Treatments Test Volume M u l t i p l i c a t i o n C o n t r o l T o t a l P o s t t e s t 9.67 12.59 10.29 10.9 3 (3.69) (3.01) (4.08) (3. 55) 36 39 30 105 Reten t i o n Test 10.65 11.34 10.76 10.95 (2.99) (3.53) (3.14) (3.23) 36 39 30 105 C o r r e l a t i o n Study Hypotheses 5 and, 7 B i s e r i a l c o r r e l a t i o n c o e f f i c i e n t s were c a l c u l a t e d f o r t e s t i n g of hypotheses 5 and 7 (p. 59). The t e s t of hy p o t h e s i s 5 r e v e a l e d t h a t the t r a n s i t i o n from a lower t o a h i g h e r l e v e l of 75 c o n s e r v a t i o n between the p r e t e s t and the p o s t e s t was found independent of volume achievement scores at the 0.05 lev e l . . S i m i l a r l y , the t r a n s i t i o n from a lower to a higher l e v e l of c o n s e r v a t i o n between the p r e t e s t and the r e t e n t i o n t e s t was found independent of volume achievement s c o r e s at the 0.05 l e v e l . Table 4.11 shows the b i s e r i a l c o r r e l a t i o n c o e f f i c i e n t s between volume achievement s c o r e s and t r a n s i t i o n to a higher or lower l e v e l of c o n s e r v a t i o n at the p o s t t e s t and the r e t e n t i o n t e s t l e v e l s . Table 4.12 shows the c e l l s i z e s of c o n s e r v a t i o n l e v e l s X treatments on the p r e t e s t , p o s t t e s t and r e t e n t i o n t e s t . Table 4. 13 shows the number of s t u d e n t s whose c o n s e r v a t i o n l e v e l s went up, down or remained the same between the p r e t e s t and each of the p o s t t e s t and the r e t e n t i o n t e s t . L i k e w i s e , the t e s t of h y p o t h e s i s 7 r e v e a l e d t h a t the t r a n s i t i o n from a higher t o a lower l e v e l of c o n s e r v a t i o n i s found independent of volume achievement s c o r e s between "the p r e t e s t and the p o s t t e s t and between the p r e t e s t and the r e t e n t i o n t e s t at the 0.05 l e v e l (see Table 4.11). Even though the t r a n s i t i o n from one c o n s e r v a t i o n l e v e l to another was independent of volume achievement s c o r e s t h e r e was a g e n e r a l improvement i n the s u b j e c t s ' c o n s e r v a t i o n l e v e l s ; In the p r e t e s t there were 57 nonconservers, 16 p a r t i a l c o n s e r v e r s and 32 c o n s e r v e r s while i n the r e t e n t i o n t e s t t h e r e were 36 nonconservers, 13 p a r t i a l conservers and 56 c o n s e r v e r s (see Table 4. 12). 76 Table 4.11 B i s e r i a l C o r r e l a t i o n C o e f f i c i e n t s Between Volume Achievement Scores on the P o s t t e s t , R e t e n t i o n Test and T r a n s i t i o n to a Higher o r Lower L e v e l of Conservation T r a n s i t i o n Higher T r a n s i t i o n Lower P o s t t e s t - p r e t e s t 0. 13 (0.18)1 0.03 (0.79) Retention T e s t - P r e t e s t 0. 09 (0.34) 0.03 (0.77) i Number i n ( ) i n d i c a t e s the s i g n i f i c a n c e l e v e l Table 4.12 C e l l S i z e s of Con s e r v a t i o n L e v e l s X Treatments on the P r e t e s t , P o s t t e s t and Retention T e s t C o n s e r v a t i o n _ Treatments L e v e l Volume M u l t i p l i c a t i o n C o n t r o l T o t a l Non-conservers 19* 18 20 57 19 10 17 46 13 11 12 36 P a r t i a l - c o n s e r v e r s 6 6 4 16 3 8 4 15 6 5 2 13 Conservers 11 15 6 32 14 21 9 44 17 23 16 56 T o t a l 36 39 30 105 36 39 30 105 36 39 30 105 * The f i r s t number r e f e r s to the p r e s t e s t The second number r e f e r s to the p o s t t e s t The t h i r d number r e f e r s to the r e t e n t i o n t e s t 77 Table 4.13 Number o f Subjects Whose C o n s e r v a t i o n L e v e l s Changed/ Did Not Change Between the P r e t e s t and each o f the Post and E e t e n t i o n T e s t L e v e l Op L e v e l Down Same L e v e l T o t a l P o s t t e s t 23 11 73 105 Be t e n t i o n Test 32 8 65 105 Hypotheses 9 and 12 Pearson's product-moment c o r r e l a t i o n c o e f f i c i e n t was c a l c u l a t e d f o r t e s t i n g of h y p o t h e s i s 9 (p. 59). The t e s t r e v e a l e d t h a t the Volume Achievement P o s t t e s t scores were found to be s i g n i f i c a n t l y c o r r e l a t e d (r = 0.35, p < 0.001) to the p r e t e s t mathematics achievement s c o r e s measured by SAT. S i m i l a r l y , Volume Achievement Bet e n t i o n Test s c o r e s were found to be s i g n i f i c a n t l y c o r r e l a t e d (r = 0.37, p < 0.001) to the p r e t e s t SAT s c o r e s . Table 4.14 summarizes the c o r r e l a t i o n c o e f f i c i e n t s and the s i g n i f i c a n t l e v e l s between the volume achievement scores and the SAT sc o r e s . P o i n t b i s e r i a l c o r r e l a t i o n c o e f f i c i e n t s were c a l c u l a t e d f o r t e s t i n g of hy p o t h e s i s 12 (p. .60). The t e s t r e v e a l e d t h a t the volume achievement scores on the p o s t t e s t and the r e t e n t i o n t e s t were not found t o be c o r r e l a t e d t o sex. Table 4.13 summarizes the p o i n t - t i s e r i a l c o r r e l a t i o n c o e f f i c i e n t s and the s i g n i f i c a n t l e v e l s between the volume achievement scores and sex. 78 Table '4. 14 Pearson's Product Moment C o r r e l a t i o n C o e f f i e n t s and S i g n i f i c a n c e L e v e l s Between Volume Achievement Scores and SAT Scores on the P o s t t e s t and the R e t e n t i o n T e s t SAT S i g n i f i c a n c e V o1u me~Ach i eve m e n t ~ P o s t t e s t 0. 35 5700026 Volume Achievement Retention Test 0.37 0.00011 Table 4.15 P o i n t - B i s e r i a l C o r r e l a t i o n C o e f f i e n t s and S i g n i f i c a n c e L e v e l s Between Volume Achievement Scores and Sex on the P o s t t e s t and the R e t e n t i o n Test Sex S i g n i f i c a n c e Volume Achievement P o s t t e s t 0.14 0.(|3 Volume Achievement Retention Test 0.12 0.23 Hypothesis 10 K e n d a l l ' s Tau c o r r e l a t i o n c o e f f i c i e n t was c a l c u l a t e d f o r t e s t i n g of h y p o t h e s i s 10 (p. 60). The t e s t showed t h a t the i n i t i a l l e v e l of c o n s e r v a t i o n was found to be independent (Ken d a l l ' s Tau = 0.09, p = 0.08) of mathematics achievement sc o r e s measured by the computation s e c t i o n of SAT. The sample used f o r t e s t i n g of t h i s h y p o t h e s i s c o n s i s t e d of a l l 146 s u b j e c t s who took the Volume Co n s e r v a t i o n P r e t e s t r e g a r d l e s s of t h e i r p a r t i c i p a t i o n i n ot h e r p a r t s of the experiment. A l l other hypotheses were t e s t e d using the data of the 105 s u b j e c t s who d i d not miss any t e s t or treatment day. 79 Tests of Independence Hypotheses 6 and 8 Frequency t a b l e s were made and C h i Square s t a t i s t i c s were c a l c u l a t e d f o r t e s t i n g of hypotheses 6 and 8 (p. 59). The t e s t of h y p o t h e s i s 6 revealed that the t r a n s i t i o n from a lower to a hi g h e r l e v e l o f c o n s e r v a t i o n between the p r e t e s t and the p o s t t e s t was found independent of treatments at the 0.05 l e v e l ; the Chi Square f o r t r a n s i t i o n up or no t r a n s i t i o n up versus treatments was 0.93 with d f = 2. S i m i l a r l y , the t r a n s i t i o n from a lower to a high e r l e v e l of c o n s e r v a t i o n between the p r e t e s t and the r e t e n t i o n t e s t was found independent of treatments at the 0.05 l e v e l with C h i square of 0.97 and df = 2. The f r e q u e n c i e s of t r a n s i t i o n up versus treatments between the p r e t e s t and each of the p o s t t e s t and the r e t e n t i o n t e s t are repr e s e n t e d i n Table 4.16. The f r e q u e n c i e s of t r a n s i t i o n up or no t r a n s i t i o n up versus treatments between the p r e t e s t and each of the p o s t t e s t and the r e t e n t i o n t e s t are represented i n Table 4.17. L i k e w i s e , the t e s t of hypothesis 8 r e v e a l e d t h a t the t r a n s i t i o n from a higher to a lower l e v e l of c o n s e r v a t i o n between the p r e t e s t and p o s t t e s t was found independent of treatments at the 0.05 l e v e l ; the Chi square f o r t r a n s i t i o n down or no t r a n s i t i o n down versus treatments was 0.91 with df = 2. L i k e w i s e , the t r a n s i t i o n from a higher t o a lower l e v e l of c o n s e r v a t i o n between the p r e t e s t and the r e t e n t i o n t e s t was 80 found independent of treatments at the 0.05 l e v e l with C h i square o f 3.48 and df = 2. The f r e q u e n c i e s of t r a n s i t i o n down or no t r a n s i t i o n down versus treatments between the p r e t e s t and each of the p o s t t e s t and the r e t e n t i o n t e s t a r e . r e p r e s e n t e d i n Table 4.18. Table 4. 16 Contingency Table: T r a n s i t i o n Up, Down, or Staying the Same Versus Treatments Between P r e t e s t - P o s t t e s t and P r e t e s t - R e t e n t i o n T e s t Treatment T r a n s i t i o n Up T r a n s i t i o n Down No T r a n s i t i o n Volume Post 6 5 25 Retention 11 4 21 M u l t i p l i c a t i o n Post 10 4 25 R e t e n t i o n 10 4 25 C o n t r o l Post 7 2 21 R e t e n t i o n 11 0 19 Table 4. 17 Contingency Table: T r a n s i t i o n Up Versus Treatments on the P o s t t e s t and the Retention T e s t Treatment T r a n s i t i o n Up No T r a n s i t i o n Up Volume Post 6 30 R e t e n t i o n 11 25 M u l t i p l i c a t i o n Post 10 29 R e t e n t i o n 10 29 C o n t r o l Post 7 23 R e t e n t i o n 11 19 P o s t t e s t : "chi sguare = 0.93, df = 2, p = 0.63 E e t e n t i o n t e s t : Chi sguare = 0.97, df = 2, p = 0.61 81 Table 4.18 Contingency Table: T r a n s i t i o n Down Versus Treatments on the P o s t t e s t and the Re t e n t i o n Test Treatment T r a n s i t i o n Down No T r a n s i t i o n Volume Post 5 31 Retent i o n 4 32 M u l t i p l i c a t i o n Post 4 35 Retent i o n 4 35 C o n t r o l Post 2 28 Ret e n t i o n 0 30 P o s t t e s t : Chi square = 0.91, df = 2, p = 0.63 Retention t e s t : C h i square = 3.48, df = 2, p = 0. 18 The r e s u l t s r e p o r t e d i n t h i s s e c t i o n are r e l a t e d d i r e c t l y to the hypotheses of the study. However, a d d i t i o n a l f i n d i n q s about the t r a n s i t i o n between c o n s e r v a t i o n l e v e l s are i n c l u d e d i n the s e c t i o n of Post Hoc Q u a l i t a t i v e Analyses. Hypotheses 11 The Wilcoxon two-sample t e s t using t i e d s c o r e s was a p p l i e d f o r t e s t i n g of hypothesis 11 (p. 60). The t e s t r e v e a l e d t h a t the i n i t i a l l e v e l of c o n s e r v a t i o n of the males' was found s i g n i f i c a n t l y (p < 0.05) b e t t e r than t h a t of the females. The f r e q u e n c i e s of the p r e t e s t c o n s e r v a t i o n l e v e l s versus sex i s represented i n Table 4.19. The sample used f o r t e s t i n g o f t h i s h y p o t h e s i s c o n s i s t e d of the 150 s u b j e c t s t o whom the p r e t e s t 82 Table 4.19 Contingency T a b l e : P r e t e s t C o n s e r v a t i o n L e v e l Versus Sex Conservation L e v e l Males Females T o t a l Nonconservers 33 45 78 P a r t i a l Conservers 13 9 22 Conservers 30 20 50 T o t a l 76 74 150 was administered. There were 76 males and 74 females. The nonconservers were 33 males and 45 females, the p a r t i a l c o nservers were 13 males and 9 females and the conservers were 30 males and 20 females. Post Hoc p u a l i t a t i v e Analyses The r e s u l t s r e p o r t e d i n the previous s e c t i o n s o f t h i s chapter r e l a t e d i r e c t l y to the hypotheses of t h i s study. However, some a d d i t i o n a l f i n d i n g s which seem to be of s i g n i f i c a n c e are r e p o r t e d i n t h i s s e c t i o n . T r a n s i t i o n between C o n s e r v a t i p n ^ L e y e l s The t e s t s o f hypotheses 6 and 8 r e v e a l e d t h a t the t r a n s i t i o n between c o n s e r v a t i o n l e v e l s from p r e t e s t to p o s t t e s t , p r e t e s t t o r e t e n t i o n t e s t and p o s t t e s t t o r e t e n t i o n t e s t was independent of treatments a t the 0.05 l e v e l . The f o l l o w i n g , however, are o b s e r v a t i o n s based on the d e t a i l e d contingency t a b l e s of t r a n s i t i o n among c o n s e r v a t i o n l e v e l s . These t a b l e s are i n c l u d e d as Ta b l e s 4.20, 4.21 and 4.22. 83 1. The r e g r e s s i o n of conservers to nonconservers o c c u r r e d very r a r e l y . Only two of 32 (6.25%) c o n s e r v e r s r e g r e s s e d to nonconservers between the p r e t e s t and the p o s t t e s t . None of the 32 c o n s e r v e r s regressed to nonconservers between the p r e t e s t and r e t e n t i o n t e s t . . S i m i l a r l y , none of the 44 c o n s e r v e r s r e g r e s s e d t o nonconservers between the p o s t t e s t and the r e t e n t i o n t e s t . 2. There does not seem t o be any observable d i f f e r e n c e i n the progress of nonconservers and p a r t i a l c o n s e r v e r s to higher l e v e l s of c o n s e r v a t i o n . For example between the p r e t e s t and the p o s t t e s t 19 of 57 (33%) nonconservers and 4 of 14 (29%) p a r t i a l c o n s e rvers progressed to a higher l e v e l of c o n s e r v a t i o n . .3. Even though there was a g e n e r a l improvement of c o n s e r v a t i o n l e v e l s among a l l treatment groups, the c o n t r o l group seemed to have undergone a steady progress with r e s p e c t to c o n s e r v a t i o n l e v e l s . That i s , s u b j e c t s i n a l l groups progressed and r e g r e s s e d but those i n the c o n t r o l group d i d not seem to r e g r e s s as much as the s u b j e c t s i n the o t h e r two treatment groups. The two of 30 s u b j e c t s (6.66%) i n the c o n t r o l group who r e g r e s s e d between the p r e t e s t and p o s t t e s t , progressed back to t h e i r o r i g i n a l l e v e l i n the r e t e n t i o n t e s t . Two other s u b j e c t s i n the c o n t r o l group regressed between the p o s t t e s t and the r e t e n t i o n t e s t . In the volume' and m u l t i p l i c a t i o n treatments there were f o u r i n each who r e g r e s s e d from p r e t e s t to r e t e n t i o n t e s t . The h i g h s t a b i l i t y of the c o n s e r v a t i o n l e v e l of c o n s e r v e r s throughout the experiment i s not s u r p r i s i n g . . Some r e s e a r c h 84 r e p o r t e d i n chapter 2 i n d i c a t e d t h a t n a t u r a l c o n s e r v e r s showed s t a b i l i t y of t h e i r c o n s e r v a t i o n l e v e l and even r e s i s t e d m i s l e a d i n g cues. However, i t was c u r i o u s t o note t h a t the c o n s e r v a t i o n l e v e l of s u b j e c t s i n the c o n t r o l group d i d not seem to r e g r e s s as much as the l e v e l of s u b j e c t s i n the other treatment groups. The i n s t a b i l i t y of the other two groups c o u l d be e x p l a i n e d by the. i n f l u e n c e of experience i n volume a c t i v i t i e s on the p a r t i a l c o n s e r v e r s . Tables 4.20, 4.21, and 4.22 show t h a t most of those who regressed were p a r t i a l c o n s e r v e r s i n the volume and m u l t i p l i c a t i o n groups. The experience i n volume a c t i v i t i e s c ould have d i s t u r b e d the p a r t i a l c o n s e r v e r s so t h a t they i n c o r r e c t l y a p p l i e d knowledge acq u i r e d i n the treatments to volume c o n s e r v a t i o n t a s k s . Those who were i n the c o n t r o l group c o u l d have used t h e i r i n t u i t i v e understanding of volume c o n s e r v a t i o n . 85 Table 4.20 Contingency Table: P r e t e s t - P o s t t e s t T r a n s i t i o n Among Conservation L e v e l s by Treatments Post Test P r e t e s t Nonconservers P a r t i a l Conservers Conservers Volume Treatment Nonconservers 15 1 3 P a r t i a l c o n s e r v e r s 3 1 2 Conservers 1 1 9 M u l t i p l i c a t i o n Treatment Nonconservers 8 2 8 P a r t i a l c o n s ervers 2 4 0 Conservers 0 2 13 C o n t r o l Treatment Nonconservers 15 3 2 P a r t i a l c o n s e r v e r s 1 1 2 Conservers 1 0 5 Table 4.21 Contingency Table: P r e t e s t - R e t e n t i o n Test T r a n s i t i o n Among Conservation L e v e l s by Treatments Retention Test P r e t e s t Nonconservers P a r t i a l Conservers Conservers Volume Treatment Nonconservers P a r t i a l c o n s e r v e r s Conservers 11 2 0 3 1 2 5 3 9 l u l t i p l i c a t i o n Treatment Nonconservers 8 P a r t i a l c o n s ervers 3 Conservers 0 1 3 1 9 0 14 C o n t r o l Treatment Nonconservers P a r t i a l c o n s e r v e r s Conservers 12 0 0 1 1 0 7 3 6 86 Table 4.22 Contingency Table: P o s t t e s t - R e t e n t i o n Test T r a n s i t i o n ftmong Conservation L e v e l s by Treatments t B e t e n t i o n T e s t . P o s t t e s t Nonconservers P a r t i a l Conservers Conservers Volume Treatment Non-cqnservers 12 4 3 P a r t i a l - c o n s e r v e r s 0 2 1 Conservers 0 1 13 M u l t i p l i c a t i o n Treatment Non-conservers 8 1 1 P a r t i a l - c o n s e r v e r s 2 4 2 Conservers 0 1 20 C o n t r o l Treatment Non-conservers 10 2 5 P a r t i a l - c o n s e r v e r s 2 0 2 Conservers 0 0 9 Students' Seasons f o r t h e i r Responses on Question 11 Question 11 of the Volume Conservation T e s t was a p a r t of the c o n s e r v a t i o n c l a s s i f i c a t i o n scheme. In t h i s guestion the students were asked to g i v e reasons f o r t h e i r responses. The reasons were intended t o provide v a l i d i t y i n f o r m a t i o n f o r the c l a s s i f i c a t i o n scheme. The reason given by each student f o r the response on q u e s t i o n 11 was f i r s t c l a s s i f i e d as c o n s i s t e n t , i n c o n s i s t e n t , or u n c l a s s i f i a b l e . A reason was c l a s s i f i e d as c o n s i s t e n t i f i t d i d not c o n t r a d i c t the response. A reason was coded as u n c l a s s i f i a b l e i f i t was not p o s s i b l e t o understand the reason given by the student. The c o n s i s t e n t and i n c o n s i s t e n t responses were f u r t h e r c l a s s i f i e d a c c o r d i n g to the f o l l o w i n g nine a t t r i b u t e s : 1. Same: general use of the term "same" with r e f e r e n c e t o the 87 two b a l l s without s p e c i f y i n g the a t t r i b u t e . Example: "The two b a l l s are s t i l l the same." 2. S i z e : g e n e r a l use of the term " s i z e " . 3. Volume: s p e c i f i c use of the term "volume". 4. Amount: s p e c i f i c use of the term "amount". 5. Boom: s p e c i f i c use of the term "room". 6. Mass: s p e c i f i c use of the term "mass". 7. Weigbt: s p e c i f i c use of the term "weight". Example: " I t would be the same because i t ' s got the same weight. 1 1 8. Shape: s p e c i f i c use of the term "shape". 9. Other: r e f e r e n c e t o a reason other than the above. Examples: "Eecause they look the same l e n g t h " . "Because the b a l l was heated." There was only one u n c l a s s i f i a b l e response i n the p r e t e s t , none i n the p o s t t e s t and none i n the r e t e n t i o n t e s t . T h i s u n c l a s s i f i a b l e response was e l i m i n a t e d from the data of p r e t e s t . Observations based on Tables 4.23, 4. 24 and 4.25 can be made about 104 c l a s s i f i a b l e responses i n the p r e t e s t , 105 i n the p o s t t e s t and 105 i n the r e t e n t i o n t e s t , . There were ten i n c o n s i s t e n t responses i n the p r e t e s t , ten i n the p o s t t e s t and two i n the r e t e n t i o n t e s t . A l l of these responses except two c o n s i s t e d of an i n c o r r e c t response and a reason which might support the c o r r e c t response of " e q u i v a l e n c e " or "same water l e v e l . " 1. In the p r e t e s t , p o s t t e s t and r e t e n t i o n t e s t the most frequent reason given f o r a c o r r e c t response was r e l a t e d to s i z e . The second and t h i r d most fr e q u e n t reasons given f o r a 88 c o r r e c t response were weight and amount r e s p e c t i v e l y . The reasons f o r a c o r r e c t response d i d not seem to be a f f e c t e d by the c o n s e r v a t i o n l e v e l of students. 2. The reasons given by c o n s e r v e r s f o r t h e i r c o r r e c t responses seem to be more evenly d i s t r i b u t e d among the v a r i o u s a t t r i b u t e s than the reasons given by nonconservers and p a r t i a l c o n s e r v e r s . 3. The reason given most f r e q u e n t l y f o r i n c o r r e c t responses i n the p r e t e s t , p o s t t e s t and r e t e n t i o n t e s t r e l a t e d to shape, room and weight. Table 4.23 Number of Students with Respect to t h e i r Reason f o r t h e i r Response on Question 11 of the Volume Con s e r v a t i o n P r e t e s t Reasons Response 1 2 3 4 5 6 7 8 9 Question 11 C o r r e c t Nonconservers 2 10 0 7 0 0 7 1 0 P a r t i a l c o n s e r v e r s 1 2 0 1 0 0 2 0 0 Conservers 0 12 4 5 0 1 9 0 1 Question 11 I n c o r r e c t Nonconservers 0 42 0 1i 4 0 5 i 8 i 72 P a r t i a l c o n s e r v e r s 22 11 2 0 4 0 0 1 0 Conservers 0 0 0 0 0 0 0 0 0 Code f o r t a b l e s 4.21, 4.22 , 4.23 :: 1. Same 2» S i z e 3. Volume 4. amount 5. Room 6. , Mass 7. Weight 8. . Shape 9. Other i means 1 i n c o n s i s t e n t response included 2 means 2 i n c o n s i s t e n t responses ; i n c l u d e d . 89 Table 4.24 Number of Students with Respect to t h e i r Reason f o r t h e i r Response on Question 11 of the Volume Conservation P o s t t e s t Reasons  Response 1~T~2 3 4 5 6 7 8 ~~9 Question 11 C o r r e c t Nonconservers 2 15 1 1 0 0 5 1 2* P a r t i a l c o n s e r v e r s 1 2 0 0 0 0 3 2* 0 Conservers 4 17 4 10 2 0 5 0 2 Question 11 I n c o r r e c t Nonconservers 0 1 * 1* 0 1 0 7* 3 6 P a r t i a l c onservers 1* 11 0 0 1 0 1 1 2 Conservers 0 0 0 0 0 0 0 0 0 * means 1 i n c o n s i s t e n t response i n c l u d e d . * means 4 i n c o n s i s t e n t responses i n c l u d e d . Table 4.25 Number of Students with Respect to t h e i r Reason f o r t h e i r Response on Question 11 of the Volume Conservation Retention Test Reasons Response 1 2 3 4 5 6 7 8 9 Question 11 C o r r e c t ,_..... — f -Nonconservers 1 8 0 1 0 0 6 0 1 P a r t i a l c o n s e r v e r s 0 1 0 0 0 0 3 0 1 Conservers 5 24 5 8 1 2 9 0 2 Question 11 I n c o r r e c t Nonconservers 0 0 0 0 4 0 5i 6 4 i P a r t i a l c o n s e r v e r s 0 1 0 0 0 0 0 3 4 Conservers 0 0 0 0 0 0 0 0 0 i means 1 i n c o n s i s t e n t response i n c l u d e d . Most of those who answered q u e s t i o n 11 c o r r e c t l y gave a reason r e l a t e d t o s i z e . However, a c o n s i d e r a b l e number of s u b j e c t s , i n c l u d i n g nine conservers i n each of the p r e t e s t and r e t e n t i o n t e s t , who answered q u e s t i o n 11 c o r r e c t l y gave a reason r e l a t e d t o weight. I f those nine c l a s s i f i e d as c o n s e r v e r s are only weight conservers and not volume conservers 90 there i s doubt about the v a l i d i t y of the c o n s e r v a t i o n t e s t used i n t h i s study. However, an e a r l i e r part of the c o n s e r v a t i o n t e s t (items 4, 5 and 6) was designed to d e t e c t those who were only weight c o n s e r v e r s . These weight c o n s e r v e r s were c l a s s i f i e d as nonconservers of volume. Language f a c t o r s could have prevented some of those s u b j e c t s from e x p r e s s i n g t h e i r reason more a p p r o p i a t e l y . The reasons given by students f o r t h e i r responses to g u e s t i o n 11 do not seem to provide s u f f i c i e n t i n f o r m a t i o n f o r c o n c l u s i v e evidence about the v a l i d i t y of the c l a s s i f i c a t i o n scheme used i n the Volume Con s e r v a t i o n Test. C o n s i s t e n c y between Leve l s of Conservation^and Responses t o  Question 12 of, the Volume Conservation Test Question 12 of the Volume C o n s e r v a t i o n Test concerning the two marble boxes was not a p a r t of the c o n s e r v a t i o n c l a s s i f i c a t i o n scheme. This g u e s t i o n i n v o l v e d two unequal q u a n t i t i e s and was not, t h e r e f o r e , t y p i c a l of the u s u a l P i a g e t i a n q u e s t i o n i n g p r o t o c o l . I t was i n c l u d e d because i t was b e l i e v e d t h a t students might be able to g i v e reasons f o r i n e q u a l i t y more e a s i l y than f o r e q u a l i t y . The f o l l o w i n g paragraph r e p o r t s o b s e r v a t i o n s about the c o n s i s t e n c y found between responses t c g u e s t i o n 12 and the l e v e l s of c o n s e r v a t i o n . The reasons given by the students f o r t h e i r responses are r e p o r t e d i n the f o l l o w i n g s e c t i o n of t h i s chapter. The numbers of students who answered g u e s t i o n 12 c o r r e c t l y or i n c o r r e c t l y are represented i n Table 4.26 f o r a l l students 91 who took that p a r t i c u l a r t e s t . On the p r e t e s t and p o s t t e s t t h e r e was a p o s i t i v e r e l a t i o n s h i p between the l e v e l of c o n s e r v a t i o n of s u b j e c t s and t h e i r responses on quest i o n 12. For the p r e t e s t Chi Square was 7.82 (p < 0.05) and f o r the p o s t t e s t Chi Square was 15.59 (p < 0.001). . I t was s u r p r i s i n g t h a t there was not a s i g n i f i c a n t l y p o s i t i v e r e l a t i o n s h i p between the l e v e l of c o n s e r v a t i o n of s u b j e c t s and t h e i r responses on qu e s t i o n 12 i n the r e t e n t i o n t e s t at the 0.05 l e v e l . There was a g e n e r a l improvement of c o r r e c t answers on questi o n 12 among a l l c o n s e r v a t i o n groups between the p r e t e s t and the p o s t t e s t . . In the p r e t e s t 29 of 78 ( 3 7 % ) nonconservers, 14 of 22 (64%) p a r t i a l c o n s ervers and 29 of 50 (5 8%) c o n s e r v e r s answered g u e s t i o n 12 c o r r e c t l y . In the p o s t t e s t 32 o f 77 (42%) nonconservers, 16 of 21 (76%) p a r t i a l c o n s ervers and 35 of 48 (73%) c o n s e r v e r s answered i t c o r r e c t l y . In the r e t e n t i o n t e s t , 27 o f 53 (51%) nonconservers, 14 of 24 (58%) p a r t i a l c o n s e r v e r s and 60 of 93 (65%) conservers answered i t c o r r e c t l y . I t appears t h a t between the p r e t e s t and p o s t t e s t there was a g e n e r a l improvement among a l l c o n s e r v a t i o n groups i n answering q u e s t i o n 12. The con s e r v e r s seem t o have improved the most (15% improvement) f o l l o w e d by p a r t i a l c o n s e r v e r s (12%) fol l o w e d by the nonconservers ( 5 % ) . On the o t h e r hand the number of nonconservers who answered q u e s t i o n 12 c o r r e c t l y seems to have i n c r e a s e d s t e a d i l y i n the p r e t e s t , p o s t t e s t and r e t e n t i o n t e s t . The per c e n t s of c o n s e r v e r s and p a r t i a l c o n s e r v e r s who answered i t c o r r e c t l y seem to have i n c r e a s e d i n 92 the p o s t t e s t and then decreased i n the r e t e n t i o n t e s t . The i n c r e a s e d percents of c o r r e c t responses of these two groups f o l l o w e d by decreased percents c o u l d p o s s i b l y be a t t r i b u t e d t o the tendency of r e g r e s s i o n toward the mean i n s u c c e s s i v e o b s e r v a t i o n s . Table 4.26 Contingency Table: C o r r e c t or I n c o r r e c t Response on Item 12 of the Volume C o n s e r v a t i o n P r e t e s t , P o s t t e s t and Retention T e s t Versus Conservation L e v e l s Test C o r r e c t I n c o r r e c t P r e t e s t Nonconservers 29 49 P a r t i a l c o n s e r v e r s 14 8 Conservers 29 21 P o s t t e s t Nonconservers 32 45 P a r t i a l c o n s e r v e r s 16 5 Conservers 35 13 Re t e n t i o n Test Nonconservers 27 26 P a r t i a l c o n s e r v e r s 14 10 Conservers 60 33 P r e t e s t : N = 150, C h i square = 7.82, df = 2, p = 0.0201 P o s t t e s t : N = 146, C h i square = 15.59, df = 2, p = 0.0004 Ret e n t i o n t e s t : N = 170, C h i square = 2.59, df = 2, p = 0.2735 Students' Reasons f o r t h e i r Responses on Question 12 A l l the reasons given by the students f o r t h e i r judgement on q u e s t i o n 12 were f i r s t c l a s s i f i e d as c o n s i s t e n t , i n c o n s i s t e n t , u n c l a s s i f i a b l e , or no response. There were very few cases of no response. There were none i n the p r e t e s t , one i n the p o s t t e s t and one i n the r e t e n t i o n t e s t . There were only two u n c i a s s i f i a b l e responses i n the p r e t e s t , none i n the 93 p o s t t e s t and t h r e e i n the r e t e n t i o n t e s t . The data of Tables 4.27, 4.28 and 4.29 summarize a l l reasons g i v e n by students e x c l u d i n g the no response and u n c l a s s i f i a b l e cases,. Tables 4.27, 4.28 and 4.29 c o n t a i n 103 c l a s s i f i a b l e responses of the p r e t e s t , 104 responses of the p o s t t e s t and 101 responses of the r e t e n t i o n t e s t . The c o n s i s t e n t and i n c o n s i s t e n t reasons were c l a s s i f i e d a c c o r d i n g to the f o l l o w i n g ten a t t r i b u t e s : 1. Weight: s p e c i f i c use of the term "weight". Example: "Because i t i s the same weight." 2. Amount: s p e c i f i c use of the term "amount". 3. S i z e : s p e c i f i c use of the term " s i z e " . 4. Boom: s p e c i f i c use of the term "room". 5. Shape: s p e c i f i c use of the term "shape". 6. Space: s p e c i f i c use of the term "space". 7.. Same: general r e f e r e n c e to "the term "same" without s p e c i f y i n g any ether a t t r i b u t e . 8. C l o s e d c o n t a i n e r : r e f e r e n c e t o the f a c t t h a t when the box i s c l o s e d water w i l l not go i n i t . Example: " I t w i l l be lower because i t has a l i d and water won't go i n . " 9. Open c o n t a i n e r : r e f e r e n c e t o the f a c t t h a t the water went i n one of the boxes because i t was open. 10. A i r i n s i d e : r e f e r e n c e to the f a c t t h a t the c l o s e d box keeps the a i r i n s i d e i t . 0. Other: r e f e r e n c e to a reason other than the above. Examples: "Because i t i s b i g g e r " . "Because more p r e s s u r e w i l l be on the water than the open box." 94 There were f o u r i n c o n s i s t e n t responses i n the p r e t e s t , f i v e i n the p o s t t e s t and s i x i n the r e t e n t i o n t e s t , i l l of these responses except two c o n s i s t e d of an i n c o r r e c t response and a reason which might support the c o r r e c t response of i n e q u a l i t y or d i f f e r e n t water l e v e l s . 1. In the p r e t e s t , p o s t t e s t and r e t e n t i o n t e s t most of those who answered c o r r e c t l y gave a reason r e l a t e d to the f a c t s t h a t water can not go i n t o the c l o s e d box, t h a t the water went i n the open box, or t h a t there i s a i r i n s i d e the c l o s e d box. The most f r e q u e n t l y given reason was t h a t water can not go i n t o the c l o s e d box. 2. In the p r e t e s t , p o s t t e s t and r e t e n t i o n t e s t most of those who gave an i n c o r r e c t response gave a reason r e l a t e d to weight, amount or s i z e . The reason r e l a t e d to weight was the most f r e g u e n t l y g i v e n . 3. There does not seem t o be any observable d i f f e r e n c e i n the frequency of reasons given by the three c o n s e r v a t i o n groups f o r t h e i r c o r r e c t or i n c o r r e c t responses. 95 Table 4.27 Number of Students with Bespect to t h e i r Season f o r t h e i r Response on Question 12 of the Volume Conservation P r e t e s t : Reasons ,„,,,• Response 1 2 3 4 5 6 7 8 9 10 0 Question 12 C o r r e c t Nonconservers 5 0 0 0 0 1 0 11 1 3 1 P a r t i a l c o n s ervers 1 0 0 0 0 0 0 6 3 1 0 Conservers 1 0 0 1 0 1 0 9 3 3 1 Question 12 I n c o r r e c t ' Nonconservers 142 0 2 1 0 0 4 91 1 0 2 P a r t i a l c o n s ervers 1 0 1 0 0 0 1 2 1 0 0 Conservers 7 0 2 0 0 0 0 2i 0 1 0 Code f o r t a b l e s 4.24, 4.25, 4.26: 1. Weight 2. Amount 3. S i z e 4._Room 5. Shape 6. Space 7. Same 8. Water does not go i n , box c l o s e d , e t c . .9. Water went i n , box open, e t c . .10. A i r i n s i d e 0. Other * means 1 i n c o n s i s t e n t response i n c l u d e d . 2 means 2 i n c o n s i s t e n t responses i n c l u d e d . Table 4.28 Number of Students with Respect to t h e i r Reason f o r t h e i r Response on Question 12 of the Volume Con s e r v a t i o n P o s t t e s t Reasons Response 1 2 3 4 5 6 7 8 9 10 0 Question 12 C o r r e c t Nonconservers 0 0 0 0 0 0 0 17 5 2 2 P a r t i a l c o n s e r v e r s 1 0 0 0 0 1 0 5 1 1 0 Conservers 1 0 0 5 0 3 0 12 4 3 0 Question 12 i n c o r r e c t Nonconservers 62 1 2 0 0 0 2 1 2i 0 5i P a r t i a l c o n s e r v e r s 2 0 2 0 0 0 1 1i 0 0 0 Conservers 1 2 6 0 1 0 2 2 0 0 2 1 means 1 i n c o n s i s t e n t response i n c l u d e d . 2 means 2 i n c o n s i s t e n t responses i n c l u d e d . 96 Table 4.29 Number of Students with Respect to t h e i r Reason f o r t h e i r Response on Question 12 of the Volume C o n s e r v a t i o n Retention Test . Reasons  Eesponse 1 2 3 4 ~5 6 7 8 9 10 0 Question 12 C o r r e c t Nonconservers 2 0 0 0 0 1 0 7 6 2 0 P a r t i a l c onservers 0 0 0 0 0 1 0 5 0 2 1 Conservers 0 0 0 1 0 2 0 10 10* 4 6* Question 12 I n c o r r e c t Nonconservers 7 P a r t i a l c o n s e r v e r s 0 Conservers 3 1 means 1 i n c o n s i s t e n t response i n c l u d e d . 2 means 2 i n c o n s i s t e n t responses i n c l u d e d . Most students who responded c o r r e c t l y to question 12 could give an e x p l i c i t reason which was r e l a t e d to the f a c t t h a t one of the c o n t a i n e r s was open (closed) and water would (would not) go i n s i d e i t . In the p r e t e s t 40 of 52 (77%) s u b j e c t s who answered c o r r e c t l y gave reason 8, 9 or 10; i n the p o s t t e s t 50 of 63 (79%) s u b j e c t s who answered c o r r e c t l y gave reason 8, 9 or 10; i n the r e t e n t i o n t e s t 46 of 66 (77%) s u b j e c t s who answered c o r r e c t l y gave reason 8, 9 or 10. On the o t h e r hand, i t was p r e v i o u s l y noted t h a t most of those who responded c o r r e c t l y to q u e s t i o n 11 gave an imprecise reason t h a t was r e l a t e d t o s i z e . I t appears t h a t i t was e a s i e r f o r students who responded c o r r e c t l y to g i v e more e x p l i c i t reasons about i n e q u a l i t y of volumes i n question 12 than about e q u a l i t y i n q u e s t i o n 11. 0 1 0 0 0 1» 5 2 1 0 2 0 0 0 1 0 1 0 0 0 2 3 5 0 1 1 5i 1 0 1 0 97 CHAPTEfi V SUMMARY, CONCLUSIONS AND RECOMMENDATIONS T h i s chapter c o n t a i n s a b r i e f review of the problem, the f i n d i n g s and the c o n c l u s i o n s , l i m i t a t i o n s o f the study, i m p l i c a t i o n s f o r e d u c a t i o n a l p r a c t i c e , and recommendations f o r f u t u r e r e s e a r c h . Review of the Problem The purpose o f t h i s study i s to determine the r e l a t i o n s h i p between the l e v e l of c o n s e r v a t i o n of volume and the degree to which s i x t h grade students l e a r n the volume a l g o r i t h m of a cuboid "V = L x W x H". The problem i s a conseguence o f an apparent discrepancy between the present s c h o o l programs - and the c o g n i t i v e theory of Piaget concerning the time t o i n t r o d u c e the volume a l g o r i t h m of a cuboid. Two widely used textbook s e r i e s i n B r i t i s h Columbia i n t r o d u c e the al g o r i t h m "V = L x W x H" i n grade 5 (age 10) ( D i l l e y e t a l . , 1974 and E i c h o l z e t a l . , 1974). Another s e r i e s i n t r o d u c e s the a l g o r i t h m f o r m a l l y i n grade 4 (age- 9) ( E l l i o t e t a l . , 1974) and uses i t i n f o r m a l l y i n grade 3 (age 8 ) . Most proponents of P i a g e t ' s theory would disagree with 98 such e a r l y i n t r o d u c t i o n of the a l g o r i t h m and c l a i m t h a t most c h i l d r e n do not develop the necessary c o g n i t i v e a b i l i t i e s f o r l e a r n i n g i t before grade 6 (age 11). Piaget (1960) h i m s e l f , f o r example, holds t h a t " i t i s not u n t i l stage IV (formal o p e r a t i o n a l ) t h a t c h i l d r e n understand how they can a r r i v e a t an area or volume simply by m u l t i p l y i n g boundary edges" (p. 408). P i a g e t (1960) c o n s i d e r s the concept of c o n s e r v a t i o n to be necessary f o r any meaningful computation i n both area and volume: ... C h i l d r e n a t t a i n a c e r t a i n kind of c o n s e r v a t i o n of area £and volume], based on the p r i m i t i v e c o n c e p t i o n o f area (and volume) as t h a t which i s bounded by l i n e s (or f a c e s ) . That understanding comes long before the a b i l i t y to c a l c u l a t e areas and volumes by mathematical m u l t i p l i c a t i o n , i n v o l v i n g r e l a t i o n s between u n i t s of d i f f e r e n t powers ... (Piaget, 1960, p. 355) On the other hand, many educators b e l i e v e t h a t " a c q u i s i t i o n of f o r m a l s c i e n t i f i c reasoning may be f a r more dependent on s p e c i f i c i n s t r u c t i o n a l experiences and f a r l e s s dependent on g e n e r a l maturation than hypothesized by I n h e l d e r and P i a g e t (1960)" ( S i e g l e r and A t l a s , 1976, p. 368). Graves (1972, p. 223), f o r example, c o n s i d e r e d e d u c a t i o n and experience to be necessary f o r volume c o n s e r v a t i o n . L o v e l l (197 1, p. 179) went f u r t h e r to suggest t h a t even seven and e i g h t year o l d s (grade 2 and 3) can l e a r n how to use the a l g o r i t h m "V = L x W x H" i n order to c a l c u l a t e the volume. There seems to be a d i s c r e p a n c y between the present s c h o o l programs and the theory of P i a g e t . Some of the present programs i n t r o d u c e the volume a l g o r i t h m of a cuboid as e a r l y as grade 3. This p o s i t i o n seems to be backed by some educators who c l a i m 99 that s c i e n t i f i c r e a s o n i n g i s acre dependent on t r a i n i n g and i n s t r u c t i o n than maturation. P i a g e t and h i s proponents seem to argue t h a t c o n s e r v a t i o n of volume i s a p r e r e q u i s i t e f o r any-meaningful c a l c u l a t i o n of volume. However, one ought not n e c e s s a r i l y t o d e l a y the i n t r o d u c t i o n of the volume a l g o r i t h m "V = L x W x H" u n t i l a l l students conserve volume.. S t u d i e s have i n d i c a t e d t h a t the m a j o r i t y of a d u l t s do not conserve volume ( E l k i n d , 1962; Towler and Wheatley, 1971; Graves, 1972),. In such a predicament t h e r e seems to be a need f o r r e s e a r c h i n order t o j u s t i f y our present s c h o o l c u r r i c u l u m or suggest i t s m o d i f i c a t i o n . T h i s need has been acknowledged by such educators as DeVault .who advocates t h a t " i t seems reasonable ... to a s s e r t t h a t the s t u d i e s most l i k e l y t o produce u s e f u l r e s u l t s f o r c u r r i c u l u m work would be experimental s t u d i e s (DeVault, 1966, p. 639)." F i n d i n g s and C o n c l u s i o n s The aims of t h i s study which were s t a t e d i n Chapter 1 w i l l be r e s t a t e d i n t h i s s e c t i o n . A l s o a summary o f the f i n d i n g s r e l a t e d t o each of the aims w i l l be reported. Aim 1.. To determine the v a r i o u s degrees to which c o n s e r v e r s , p a r t i a l c o n s e r v e r s , and nonconservers of volume l e a r n the volume a l g o r i t h m of a cuboid "V = L x W x H". F i n d i n g s . The r e s u l t s of the p o s t t e s t showed a s i g n i f i c a n t (p < 0.05) s u p e r i o r i t y of the c o n s e r v e r s group over the p a r t i a l 100 c o n s e r v e r s group i n the Volume Achievement P o s t t e s t . No s i g n i f i c a n t d i f f e r e n c e was found between any o t h e r c o n s e r v a t i o n groups at the 0.05 l e v e l . There was no s i g n i f i c a n t d i f f e r e n c e found i n Volume Achievement R e t e n t i o n Test scores between c o n s e r v a t i o n groups at the 0.05 l e v e l . Aim 2. To determine the degree of e f f e c t i v e n e s s f o r each of the two t e a c h i n g methods on l e a r n i n g the volume a l g o r i t h m f o r a cuboid. F i n d i n g s . The r e s u l t s showed a s i g n i f i c a n t s u p e r i o r i t y of volume treatment group over m u l t i p l i c a t i o n treatment group (p < 0.01) and over c o n t r o l treatment group (p < 0.01) on the Volume Achievement P o s t t e s t . No s i g n i f i c a n t d i f f e r e n c e was found between m u l t i p l i c a t i o n treatment group and c o n t r o l treatment group a t the 0.05 l e v e l . S i m i l a r l y , the r e s u l t s of the r e t e n t i o n t e s t showed a s i g n i f i c a n t s u p e r i o r i t y of volume treatment group over m u l t i p l i c a t i o n treatment group (p < 0.01) and over c o n t r o l treatment group (p < 0.01); no s i g n i f i c a n t d i f f e r e n c e was found between m u l t i p l i c a t i o n treatment group and c o n t r o l treatment group a t the 0.05 l e v e l . Aim 3. To determine the e f f e c t of l e a r n i n g the volume alg o r i t h m of a cuboid on the t r a n s i t i o n from one volume c o n s e r v a t i o n l e v e l t e another. F i n d i n g s . The t r a n s i t i o n from a lower to a higher l e v e l of c o n s e r v a t i o n between the p r e t e s t and the p o s t e s t was found to be independent of volume achievement s c o r e s a t the 0.05 l e v e l . S i m i l a r l y , the t r a n s i t i o n from a lower to a higher l e v e l of 101 c o n s e r v a t i o n between the p r e t e s t and the r e t e n t i o n t e s t was found t o be independent c f volume achievement s c o r e s a t the 0.05 l e v e l . L i k e w i s e , the t r a n s i t i o n from a h i g h e r t o a lower l e v e l of c o n s e r v a t i o n was found t o be independent of volume achievement s c o r e s between the p r e t e s t and t h e p o s t t e s t and between the p r e t e s t and the r e t e n t i o n t e s t at the 0.05 l e v e l . Even though the t r a n s i t i o n from one c o n s e r v a t i o n l e v e l to another was found to be independent of volume achievement sc o r e s t h e r e was a g e n e r a l improvement i n the s u b j e c t s * c o n s e r v a t i o n l e v e l s . ftim,4. To determine the r e l a t i o n s h i p between sex and the l e v e l s of c o n s e r v a t i o n of volume. F i n d i n g s . The Volume C o n s e r v a t i o n P r e t e s t r e v e a l e d t h a t the i n i t i a l l e v e l of c o n s e r v a t i o n of males was found t o be s i g n i f i c a n t l y (p < 0.01) b e t t e r than t h a t of the females.. Out of a t o t a l 150 s t u d e n t s , 76 were males and 74 were females. The nonconservers were 33 males and 45 females, the p a r t i a l c o n s e r v e r s were 13 males and 9 females and the co n s e r v e r s were 30 males and 20 females. Rim,5. To determine the r e l a t i o n s h i p between sex and the degree of l e a r n i n g the volume a l g o r i t h m f o r a cuboid. F i n d i n g s . The Volume achievement P o s t t e s t s c o r e s were not found t o be s i g n i f i c a n t l y (p < 0.05) c o r r e l a t e d t o sex. S i m i l a r l y , the Volume achievement E e t e n t i o n T e s t scores were not found s i g n i f i c a n t l y c o r r e l a t e d with sex a t the 0.05 l e v e l . aim 6. To determine the r e l a t i o n s h i p between mathematics 1 102 achievement and the l e v e l s of c o n s e r v a t i o n of volume. F i n d i n g s . The i n i t i a l l e v e l o f c o n s e r v a t i o n was found to be independent of the mathematics achievement scores of the p r e t e s t as measured by the computation s e c t i o n of SAT. Aim 7. To determine the r e l a t i o n s h i p between mathematics achievement and the degree of l e a r n i n g the volume algorithm f o r a cuboid. F i n d i n g s . The Volume Achievement P o s t t e s t s c o r e s were found to be s i g n i f i c a n t l y (r = 0.35, p < 0.001) c o r r e l a t e d to the mathematics achievement p r e t e s t s c o r e s measured by SAT. S i m i l a r l y , Volume Achievement Retention T e s t s c o r e s were found to be s i g n i f i c a n t l y (r =0.37, p < 0.001) c o r r e l a t e d t o the p r e t e s t of SAT s c o r e s . Summary of C o n c l u s i o n s and D i s c u s s i o n Importance of Conservation L e v e l s . This study was an attempt t o determine the r e l a t i o n s h i p between the l e v e l o f c o n s e r v a t i o n of volume and the degree to which s i x t h grade students l e a r n the volume a l g o r i t h m . ; The onl y s i g n i f i c a n t r e s u l t i n t h i s connection was t h a t the con s e r v e r s scored higher than the p a r t i a l c o n s e r v e r s on the Volume Achievement P o s t t e s t . The conservers d i d not score s i g n i f i c a n t l y h i g h e r than the nonconservers on the p o s t t e s t ; p a r t i a l c o n s e r v e r s scored lower, though not s i g n i f i c a n t l y lower, than the nonconservers on the p o s t t e s t . On the r e t e n t i o n t e s t , the s c o r e s o f the c o n s e r v a t i o n groups on the Volume Achievement Test d i d not d i f f e r 103 s i g n i f i c a n t l y . Furthermore* students of the volume treatment group scored 65% on each of the Volume Achievement P o s t t e s t and Volume Achievement Retention Test. C h i l d r e n i n grade 6 seem to be able to apply the volume a l g o r i t h m f o r a cuboid, at the computation and comprehension l e v e l s , r e g a r d l e s s o f t h e i r c o n s e r v a t i o n l e v e l . So f a r as students who have reached the 6th grade l e v e l are concerned, i t appears, t h a t c o n s e r v a t i o n l e v e l i s not an important f a c t o r i n l e a r n i n g the volume a l g o r i t h m as d e f i n e d i n t h i s study. I t i s p o s s i b l e , although t h i s study has no data to support i t , that c o n s e r v a t i o n l e v e l might l i k e w i s e be r e l a t i v e l y unimportant as a f a c t o r t h a t i n f l u e n c e s s u c c e s s f u l l e a r n i n g of the volume a l g o r i t h m by students i n , say, grades 4 and 5. I f t h i s be so, then the present s c h o o l programs which do present the volume a l g o r i t h m i n those grades may not be unreasonable. So long as the c r i t e r i o n f o r r e a s o n a b i l i t y i s l e a r n a b i l i t y , the present study does not support the i d e a t h a t the i n t r o d u c t i o n of the volume a l g o r i t h m should not take place before the l e a r n e r s have become co n s e r v e r s of volume. However, t h e r e may be f a c t o r s ether than volume c o n s e r v a t i o n which would a l s o i n f l u e n c e t h e l e a r n a b i l i t y of the volume a l g o r i t h m . , The need f o r f u r t h e r r e s e a r c h , r e g a r d i n g the time to i n t r o d u c e the a l g o r i t h m , w i l l be d i s c u s s e d l a t e r i n t h i s c h a p t e r . E f f e c t of Treatments. S u b j e c t s who were i n the volume treatment d i d s i g n i f i c a n t l y b e t t e r , on the Volume Achievement P o s t t e s t and Volume Achievement Retention T e s t , than those s u b j e c t s who were i n each of the other treatments; the s u b j e c t s 104 i n the o t h e r two treatments d i d not d i f f e r s i g n i f i c a n t l y i n volume performance. The s u b j e c t s who were i n the m u l t i p l i c a t i o n treatment d i d s i g n i f i c a n t l y b e t t e r , on the M u l t i p l i c a t i o n Achievement P o s t t e s t , than ' those who were i n the volume treatment and the c o n t r o l treatment; those who were i n the volume treatment and those who were i n the c o n t r o l treatment d i d not d i f f e r s i g n i f i c a n t l y . T h i s i n d i c a t e d t h a t at the p o s t t e s t l e v e l the m u l t i p l i c a t i o n treatment was s u c c e s s f u l . That i s , the students had l e a r n e d the m u l t i p l i c a t i o n m a t e r i a l which was taught. In s h o r t , i t may be concluded t h a t , at the p o s t t e s t l e v e l , the s u b j e c t s who were i n the m u l t i p l i c a t i o n treatment l e a r n e d the m u l t i p l i c a t i o n m a t e r i a l but the s u b j e c t s who were i n the volume treatment d i d s i g n i f i c a n t l y b e t t e r than those who were i n the m u l t i p l i c a t i o n treatment, on the Volume Achievement P o s t t e s t . The c o n c l u s i o n s mentioned above seem t o suggest t h a t the volume treatment i s b e t t e r than the m u l t i p l i c a t i o n treatment i n t e a c h i n g s i x t h graders the volume a l g o r i t h m of a cuboid. The volume treatment i n c l u d e d a c t i v i t i e s f o r determining the volume of cuboids by b u i l d i n g them with cubes and c o u n t i n g the number of cubes; t h i s method l a t e r used the a l g o r i t h m "V = L x W x H" f o r computing the volume of a cuboid. The m u l t i p l i c a t i o n treatment c o n s i s t e d mainly of studying the e f f e c t of v a r y i n g f a c t o r s on t h e i r product and v a r y i n g f a c t o r s when t h e i r product i s c onstant; t h i s task was supplemented by a b r i e f a p p l i c a t i o n to the volume a l g o r i t h m "V = L x W x H. H 105 The r e s u l t s of t h i s study do not, t h e r e f o r e , support the c o n j e c t u r e made i n Chapter 2 t h a t students who are p r o f i c i e n t i n v a r y i n g f a c t o r s of a f i x e d product can r a p i d l y p r e d i c t and determine the volumes or dimensions of cuboids. On the c o n t r a r y , the r e s u l t s seem to support P i a g e t ' s c l a i m t h a t " i t i s one t h i n g to m u l t i p l y two numbers tog e t h e r and q u i t e another to m u l t i p l y two l e n g t h s o r three l e n g t h s and understand t h a t t h e i r product i s an area or a volume ... (Piaget e t a l . , 1960, p. 408)." T r a n s i t i o n between C o n s e r v a t i o n L e v e l s . R e s u l t s o f the study r e v e a l e d t h a t t h e r e was, g e n e r a l l y , an improvement o f the students* c o n s e r v a t i o n l e v e l s r e g a r d l e s s of t h e i r volume achievement scores or t h e i r treatments. The t r a n s i t i o n from a lower to a higher l e v e l of c o n s e r v a t i o n between the p r e t e s t and each of the p o s t t e s t and r e t e n t i o n t e s t was found independent of volume achievement sc o r e s and o f treatments. I t appears t h a t the improvement of the s u b j e c t s ' c o n s e r v a t i o n l e v e l was i n f l u e n c e d by some f a c t o r ( s ) other than treatments and volume achievement scores. P o s s i b l e f a c t o r s c o u l d have been growth, peer i n f l u e n c e , t e s t i n f l u e n c e ( s e n s i t i z a t i o n ) and 'Hawthorne e f f e c t ' . Growth i s suspected t o have been a f a c t o r because the experiment l a s t e d about two months durin g which s u b j e c t s i n grade 6, e s p e c i a l l y those who were "on the doorstep" of c o n s e r v a t i o n , could have developed from one stage of c o g n i t i v e development to the next. U n c o n t r o l l a b l e students' d i s c u s s i o n s (peer i n f l u e n c e ) o f the Volume Co n s e r v a t i o n Test o u t s i d e the classroom c o u l d have 106 i n f l u e n c e d the r e s u l t s of the p o s t t e s t and the r e t e n t i o n . t e s t s i n c e these t e s t s were i d e n t i c a l t o the p r e t e s t . The t e s t i t s e l f c o u l d have i n f l u e n c e d some s u b j e c t s to thi n k s e r i o u s l y about c o n s e r v a t i o n tasks and t o c o r r e c t t h e i r own e r r o r s i n l a t e r t e s t s . F i n a l l y , the development of the students from one c o n s e r v a t i o n l e v e l to the next could have been p a r t i a l l y a t t r i b u t e d to the f a c t t h a t they were chosen f o r the experiment (Hawthorne e f f e c t ) and consequently t o the i n f l u e n c e of f e e l i n g s p e c i a l and worthy. E f f e c t of_Hathematics- Achievement. Re s t i l t s -of the study r e v e a l e d t h a t volume achievement s c o r e s on the p o s t t e s t and the r e t e n t i o n t e s t were c o r r e l a t e d to mathematics achievement sc o r e s measured by the computation s e c t i o n of SAT. The i n i t i a l l e v e l o f c o n s e r v a t i o n of volume was found tp be independent o f the mathematics achievement s c o r e s measured by the computation s e c t i o n of SAT. The above-mentioned r e s u l t s seem t o suggest t h a t a competency i n mathematics computation may i n d i c a t e a competency i n volume achievement or v i c e versa. Furthermore, the mathematics achievement score and volume c o n s e r v a t i o n l e v e l seem t o be independent. E f f e c t of Sex. The e f f e c t of se^c on the degree of l e a r n i n g the volume a l g o r i t h m of a cuboid and on the i n i t i a l l e v e l of c o n s e r v a t i o n of volume was examined. The degree of l e a r n i n g the volume a l g o r i t h m of a cuboid, a t the p o s t t e s t and r e t e n t i o n t e s t l e v e l s was not found to be r e l a t e d t o sex. On the other hand, the males were found t o have a s i g n i f i c a n t l y (p < 0.01) 107 higher i n i t i a l l e v e l of c o n s e r v a t i o n than the females. The above-mentioned s u p e r i o r i t y of males over females i n volume c o n s e r v a t i o n has been a l s o r e p o r t e d by other r e s e a r c h e r s such as Graves (1972), E l k i n d (1961-b and 1962) and Wheatley (1971). The s u p e r i o r i t y of the males to the females i n the i n i t i a l l e v e l of c o n s e r v a t i o n c o u l d be a t t r i b u t e d t o the more a c t i v e p a r t i c i p a t i o n of males i n p r a c t i c a l e x p e r i e n c e s i n v o l v i n g m a n i p u l a t i v e s k i l l s . ( P r i c e - W i l l i a m s e t a l . , 1969 and Grave, 1972). L i m i t a t i o n s of the Study There are s e v e r a l l i m i t a t i o n s of the study. Some of these l i m i t a t i o n s are r e l a t e d t o the type of s u b j e c t s chosen f o r the experiment while other l i m i t a t i o n s are conseguences of the procedures of the study. The s u b j e c t s of t h i s study were s i x t h grade students of a suburban area. T h e i r average f a m i l y income was s l i g h t l y h i g h e r than the average income of the g r e a t e r m e t r o p o l i t a n area. The s u b j e c t s c o n s i s t e d o r i g i n a l l y of 171 students but when those who missed any t e s t or treatment day were e l i m i n a t e d , the f i n a l sample was reduced to 105 s t u d e n t s . There were 57 nonconservers, 32, conservers and only 16 p a r t i a l c o n s e r v e r s . Any g e n e r a l i z a t i o n based upon the r e s u l t s and c o n c l u s i o n s of t h i s study i s l i m i t e d to t h i s or to a s i m i l a r p o p u l a t i o n o f s t u d e n t s . On the other hand, the s u b j e c t s used i n t h i s study were grade 6 students who f o l l o w e d a mathematics program 108 t y p i c a l of those used i n North America, They were of ages and socioeconomic s t a t u s j s i m i l a r t o those of suburban grade 6 s t u d e n t s i n North America. These s u b j e c t s seem, t h e r e f o r e , to be r e p r e s e n t a t i v e of the p o p u l a t i o n of suburban North American grade 6 students and f i n d i n g s c o u l d be g e n e r a l i z e d t o t h i s l a r g e r p o p u l a t i o n . The l i m i t a t i o n s r e l a t i n g t o the procedures i n v o l v e c o n s t r a i n t s due to the treatments and c o n s t r a i n t s r e l a t i n g to the t e s t s . One of the experimental treatments, the volume treatment, was more comprehensive and r e q u i r e d more students' a c t i v e involvement than i s normal i n elementary classrooms. The other experimental treatment, the m u l t i p l i c a t i o n treatment, was d i f f e r e n t from u s u a l s c h o o l approaches i n i t s emphasis on m u l t i p l i c a t i o n s k i l l s i n v o l v e d i n the l e a r n i n g of the a l g o r i t h m "V = L x W x H." Furthermore, only one v e r s i o n of each of the Volume Achievement T e s t , Volume Conservation Test and M u l t i p l i c a t i o n Achievement T e s t was used at d i f f e r e n t stages of the experiment. The c l a s s i f i c a t i o n of s u b j e c t s i n t o volume c o n s e r v a t i o n l e v e l s was achieved using a judgement-based volume c o n s e r v a t i o n t e s t . The g e n e r a l i z a t i o n s of t h i s study are l i m i t e d by the treatments and t e s t instruments used. 109 I m p l i c a t i o n s f o r E d u c a t i o n a l P r a c t i c e I m p l i c a t i o n 1. The students of the volume treatment had an adjusted mean score of 65% on both the Volume Achievement P o s t t e s t and Volume Achievement E e t e n t i o n T e s t . T h i s seems to i n d i c a t e t h a t students i n grade 6 are capable of l e a r n i n g the volume a l g o r i t h m f o r a cuboid "V = L x W x H." The volume achievement s c o r e s of such students do not seem to be a f f e c t e d by t h e i r l e v e l s of c o n s e r v a t i o n of volume. S i n c e the c o n s e r v a t i o n l e v e l d i d not seem to be an important f a c t o r i n l e a r n i n g the volume al g o r i t h m , u s i n g c o n s e r v a t i o n as a c r i t e r i o n , the present s c h o o l programs t h a t i n t r o d u c e the a l g o r i t h m p r i o r to grade 6 are not proven unreasonable. This study does not, t h e r e f o r e , suggest the delay of i n t r o d u c i n g the volume a l g o r i t h m f o r a cuboid "V = L x W x H." T h i s i s not to say t h a t the p r e v a l e n t s c h o o l p r a c t i c e s are j u s t i f i e d with r e s p e c t to the theory of P i a g e t . The s e c t i o n on f u t u r e r e s e a r c h o u t l i n e s p o s s i b l e ways f o r pursuing the matter of j u s t i f i c a t i o n of the s c h o o l programs. I m p l i c a t i o n 2. The r e s u l t s and c o n c l u s i o n s of t h i s study i n d i c a t e t h a t the a c t i v i t y - o r i e n t e d volume treatment was s u c c e s s f u l . T h i s treatment was based on determining the volume of cuboids by b u i l d i n g them with cubes and counting the number of cubes; l a t e r the a l g o r i t h m "V = L x W x H" was used f o r computing the volume of a cuboid,. I t seems, t h e r e f o r e , a p p r o p r i a t e to teach the volume a l g o r i t h m of a cuboid u s i n g an a c t i v i t y o r i e n t e d method. 110 I m p l i c a t i o n 3 , The mathematics computation s c o r e s o f SAT were found t o be p o s i t i v e l y c o r r e l a t e d with volume achievement s c o r e s . T h i s seems t o suggest t h a t a competency i n mathematics computation may i n d i c a t e a competency i n volume achievenent or v i c e v e r s a . I m p l i c a t i o n 4. Females seem to be as capable as males i n l e a r n i n g the volume a l g o r i t h m f o r a cuboid "V = L x W x H." However, the s u p e r i o r i t y of the males to the females i n the i n i t i a l l e v e l of c o n s e r v a t i o n c o u l d be a t t r i b u t e d t o the more a c t i v e p a r t i c i p a t i o n of males i n p r a c t i c a l e x periences i n v o l v i n g manipulative s k i l l s . A c t i v i t y - o r i e n t e d programs i n the t e a c h i n g of volume concepts may be b e n e f i c i a l t o the a c q u i s i t i o n of volume c o n s e r v a t i o n by females. Recommendations f o r Future Besearch The purpose of t h i s study was to determine the r e l a t i o n s h i p between the l e v e l of c o n s e r v a t i o n of volume and the deqree to which students l e a r n the volume a l g o r i t h m f o r a cuboid "V = L x W x H." On the b a s i s of the f i n d i n g s , c o n c l u s i o n s and i m p l i c a t i o n s of the study, f u r t h e r r e s e a r c h i s needed on t h i s t o p i c . I t i s recommended t h a t the experiment be r e p l i c a t e d on a l a r g e r sample. , The sample of t h i s study c o n s i s t e d of 105 students only 16 of which were p a r t i a l c o n s e r v e r s . A l a r g e r sample might i n f l u e n c e the r e s u l t s found i n t h i s study. 111 I t i s a l s o recommended t h a t the experiment be r e p l i c a t e d on s u b j e c t s i n a grade lower than s i x . The independence of c o n s e r v a t i o n l e v e l and volume achievement c o u l d have been i n f l u e n c e d by the high grade l e v e l chosen. Students i n grade 6 c o u l d have developed l e a r n i n g h a b i t s t h a t were very e f f e c t i v e , or those s t u d e n t s c o u l d have been "on the doorstep" of volume c o n s e r v a t i o n and became co n s e r v e r s e a r l y i n the experiment. The c h o i c e of a lower grade and consequently a lower l e v e l of development might r e v e a l the importance of volume c o n s e r v a t i o n l e v e l more c l e a r l y . A f u r t h e r recommendation i s t o conduct a study, i n which volume l e a r n i n g at h i g h e r c o g n i t i v e l e v e l s than computation and comprehension i s i n v e s t i g a t e d with r e s p e c t t o a c o r r e l a t i o n with volume c o n s e r v a t i o n . . I t i s a l s o recommended t h a t a volume c o n s e r v a t i o n p o s t t e s t and a volume c o n s e r v a t i o n r e t e n t i o n t e s t be developed which are p a r a l l e l forms o f , but not i d e n t i c a l t o , the volume c o n s e r v a t i o n pretest,. The i d e n t i c a l c o n s e r v a t i o n t e s t s given i n the p r e t e s t , p o s t t e s t and r e t e n t i o n t e s t could have allowed a g r e a t e r peer i n f l u e n c e or s e n s i t i z a t i o n e f f e c t than i f they were not i d e n t i c a l . I t i s a l s o recommended t h a t the f o l l o w i n g o b s e r v a t i o n s which were made i n the Post Hoc Q u a l i t a t i v e Analyses be i n v e s t i g a t e d f u r t h e r . 1. I t was noted t h a t t h e r e was a g e n e r a l p rogress i n the c o n s e r v a t i o n l e v e l s of s u b j e c t s i n a l l groups. However, i n cases of r e g r e s s i o n , the p a r t i a l conservers who were given 112 experiences i n volume a c t i v i t i e s seem to have regr e s s e d the most.,Further i n v e s t i g a t i o n , which i n c l u d e s i n t e r v i e w i n g of s u b j e c t s , i s needed f o r cases where r e g r e s s i o n occurs. Such i n v e s t i g a t i o n may r e v e a l the e f f e c t of volume a c t i v i t i e s on the s t a b i l i t y of the c o n s e r v a t i o n l e v e l of s u b j e c t s , p a r t i c u l a r l y the p a r t i a l c o n servers. 2. Students were asked, i n question 11 of the Volume Conservation T e s t , t o w r i t e reasons f o r t h e i r responses. Those responses d i d not seem t o provide s u f f i c i e n t v a l i d i t y i n f o r m a t i o n f o r the f i n a l judgement about the t e s t used. A c o n s i d e r a b l e number c f students gave an i n c o r r e c t response and a reason which might support the c o r r e c t response while other students gave a c o r r e c t response and a reason r e l a t e d t o weight. F u r t h e r i n v e s t i g a t i o n i s needed t o v a l i d a t e . the a s s e s s i n g of volume c o n s e r v a t i o n . I t i s recommended t h a t methodological s t u d i e s be undertaken t o determine the r e l a t i o n s h i p between nonverbal and i n t e r r o g a t i o n methods of a s s e s s i n g c o n s e r v a t i o n of volume. 3. Question 12 of the Volume Co n s e r v a t i o n T e s t concerned two unegual volumes where st u d e n t s were asked t o write reasons f o r t h e i r responses. I t was noted t h a t the number of p a r t i a l c o n s e r v e r s and c o n s e r v e r s , who answered q u e s t i o n 12 c o r r e c t l y , i n c r e a s e d i n the p o s t t e s t and decreased i n the r e t e n t i o n t e s t . The number of nonconservers who answered q u e s t i o n 12 c o r r e c t l y seemed t o have improved s t e a d i l y i n the p r e t e s t , p o s t t e s t and r e t e n t i o n t e s t . On the other hand, most c o r r e c t responses to q u e s t i o n 12 were supported by e x p l i c i t reasons. F u r t h e r 113 r e s e a r c h i s needed f o r q u e s t i o n 12 i n p a r t i c u l a r and cases of c o n s e r v a t i o n of i n e q u a l i t y of volume i n g e n e r a l . F i n a l l y , i n t h i s study the r e l a t i o n s h i p between volume c o n s e r v a t i o n and the degree of l e a r n i n g a volume a l g o r i t h m i n v o l v i n g m u l t i p l i c a t i o n s k i l l s was i n v e s t i g a t e d . I t i s recommended t h a t s i m i l a r r e s e a r c h be conducted to determine the r e l a t i o n s h i p between v a r i o u s c o n s e r v a t i o n tasks and the degree of l e a r n i n g other a l g o r i t h m s i n v o l v i n g m u l t i p l i c a t i o n s k i l l s i n the elementary s c h o o l . For example, rese a r c h may be designed t o i n v e s t i g a t e the r e l a t i o n s h i p between area c o n s e r v a t i o n and l e a r n i n g the a l g o r i t h m f o r the area of a r e c t a n g u l a r r e g i o n , "A = L x W", t h a t i s , area equals l e n g t h times width. In summary, f u r t h e r r e s e a r c h i s needed before the p r e v a l e n t s c h o o l p r a c t i c e of i n t r o d u c i n g the volume a l g o r i t h m f o r a cuboid (V = L x W x H) i n grades e a r l i e r than grade 6 can be j u s t i f i e d on the b a s i s of the c o g n i t i v e theory of P i a g e t . 114 REFERENCES A r l i n , P. K. The a p p l i c a t i o n of P i a g e t i a n theory to i n s t r u c t i o n a l d e c i s i o n s (Eeport No. 78:1, E d u c a t i o n a l Research I n s t i t u t e o f B r i t i s h Columbia). B. C. , 1977. Bat-Haee, M. A. C o n s e r v a t i o a of mass, weight, and volume i n i n t e r m e d i a t e grades. 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J . , & Anderson, L. Number c o n s e r v a t i o n : The r o l e of r e v e r s i b i l i t y , a d d i t i o n - s u b t r a c t i o n , and m i s l e a d i n g cues,. Child,, Developmenent, 1967, 38, 425-442. 35-39. 122 Winer, B. J . S t a t i s t i c a l p r i n c i p l e s i n experimental design. New York: McGraw-Hill Book Company, 1971. W o h l w i l l , J . F. On e s s a i d 1 a p p r e n t i s s a g e dans l e domaine de l a c o n s e r v a t i o n du nombre. Etudes d ,'epistomologie Genetigue, 1959, 9, 125-135. W o h l w i l l , J . F., & Lowe R. C. Experimental a n a l y s i s of the development of the c o n s e r v a t i o n of number.. C h i l d Development. 1962, 33, 153-167. appendix a DESCRIPTION OF THE INSTRUCTIONAL UNITS 124 Treatment ft Lesson 1 B e h a v i o r a l O b j e c t i v e s : 1. Given two models of cuboids (closed boxes or s o l i d blocks) the volume of which d i f f e r m a c r o s c o p i c a l l y , the students w i l l be a b l e to s t a t e which one has the g r e a t e r volume. 2. Given f i v e models of cuboids (closed boxes or s o l i d blocks) the volume of any two of which d i f f e r m a c r o s c o p i c a l l y , the students w i l l be able t o order the f i v e models by volume. 3. Given two c l o s e d boxes, the volumes o f which do not n e c e s s a r i l y d i f f e r m a c r o s c o p i c a l l y , and g i v e n a s e t of decimetre cubes to be used as u n i t s , the students w i l l be a b l e to b u i l d models congruent to the c l o s e d boxes, and thereby to s t a t e the volume of each box. 4. Given f i v e c l o s e d boxes, the volume of any two of which do not n e c e s s a r i l y d i f f e r m a c r o s c o p i c a l l y , and given a s e t of decimetre cubes to be used as u n i t s , the students w i l l be able t o b u i l d models congruent to the c l o s e d boxes, and thereby t o order the f i v e boxes by volume. 5. Given a p i c t u r e of p o l y h e d r a l model b u i l t from u n i t cubes, some of which may not be v i s i b l e , and given a set of u n i t cubes, the students w i l l be a b l e to b u i l d the p i c t u r e d model and s t a t e i t s volume. O u t l i n e : 1. D i r e c t comparison of o b j e c t s . 2. D i r e c t o r d e r i n g of o b j e c t s . 3. I n d i r e c t comparison of c l o s e d boxes. 3.1. Need f o r u n i t s ; volume i s the number of u n i t cubes. 3.2. - Non-standard u n i t s : D i s c u s s i o n . 125 4. Standard u n i t s : m3, dm3 and cm 3. 5. I n d i r e c t o r d e r i n g of c l o s e d boxes. 6. Volume of p o l y h e d r a l models b u i l t from u n i t cubes. 7. Worksheet. M a t e r i a l s : 1. Cardboard boxes. 2. 1 inch-cubes.. 3. Some decimetre cubes and centimetre cubes. 4. A poster o f p o l y h e d r a l models b u i l t from u n i t cubes. 5. A cm r u l e r . A c t i v i t i e s : Give each student 25 inch-cubes ( r e f e r t o t h e s e cubes as simply "cubes" and not as " i n c h - c u b e s " ) . Ask the students to l a y the b l o c k s a s i d e because they w i l l be used l a t e r i n the p e r i o d . 1. (2 min) D i r e c t comparison of o b j e c t s : D i s p l a y the two c l o s e d cardboard boxes A and B of s i z e s 6 cm X 4 cm X 2 cm and 40 cm X 20 cm X 10 cm r e s p e c t i v e l y . Ask the s t u d e n t s to guess which i s b i g g e r , which occupies more space and which has the g r e a t e r volume. Conclude t h a t box B i s bigger than box A and t h a t any one of the f o l l o w i n g sentences d e s c r i b e s t h i s f a c t . a. Box B i s bigger than box A b. Eox B takes up more room than box A c. Box B occupies more space than box A d. Eox B has a l a r g e r volume than box A. 2. (3 min) D i r e c t o r d e r i n g of o b j e c t s : D i s p l a y , i n t h i s order, the f i v e c l o s e d boxes F, G, H, I , and J of s i z e s 2 dm3, 4 dm3, 16 dm3, 1 dm3 and 10 dm3 r e s p e c t i v e l y . Ask the students t o help to order the boxes from l a r g e s t to s m a l l e s t . Allow time f o r responses then order the boxes by comparing any two boxes and then p u t t i n g a t h i r d i n i t s proper p o s i t i o n between the f i r s t two and so on. 126 3. (5 min) I n d i r e c t comparison of c l o s e d boxes: 3. 1 • Need for, u n i t s : D i s p l a y a c l o s e d cardboard box i n each of two d i s t a n t l o c a t i o n s of the classroom (use box P and box Q) and ask the students t o compare the volumes of the boxes without moving them. Lead the d i s c u s s i o n i n order t o conclude t h a t p e r c e p t i o n may be d e c e i v i n g ; determine and compare the volumes using the f o l l o w i n g a c t i v i t i e s : a. Use u n i t s (smaller boxes provided) to b u i l d next to each box a cuboid with the same shape and volume as t h a t of the box. b. Count the number of u n i t s and write the volumes of the boxes on the board. (Volume of P = 7 u n i t s , Volume of Q = 8 u n i t s ) . c. Compare the volumes of the boxes using the numbers of u n i t s found i n b. S t r e s s t h a t i n t h i s p r o c e s s , a_vy u n i t s of the same s i z e c o u l d be used, but the same u n i t s must be used throughout. 3-2. Non-standard u n i t s : -Discussion,: Encourage i n d i v i d u a l students to suggest items which c o u l d be used as u n i t s . D i s c u s s with the students the - f e a s i b i l i t y and convenience of some of the suggested u n i t s . 4. (5 min) Standard u n i t s : dm3 and cm 3: Ask the students about the common u n i t s f o r measuring l e n g t h ( d e s i r e d answer: m, dm, cm, . . . ) . I f necessary use a cm r u l e r t o measure l e n g t h . Lead the d i s c u s s i o n i n order to conclude that m3, dm3, cm 3 are c o n s i s t e n t with the u n i t s we u s u a l l y use to measure l e n g t h . Show cubes of volume 1 dm3 and 1 cm 3. Emphasize t h a t these volumes are the u s u a l metric u n i t s used i n i n d u s t r y and commerce. Use decimetre cubes t o b u i l d a congruent shape and to measure the volume of box B (4 dm X 2 dm X 1 dm) and use c e n t i m e t r e cubes to b u i l d a congruent shape and to measure the volume of box A (6 cm X 4 cm X 2 cm). Write on the board statements such as "Volume of box B = 8 dm3" and "Volume of box A = 48 cm 3." 5. (5 min) I n d i r e c t o r d e r i n g of c l o s e d boxes: D i s p l a y three c l o s e d boxes ( I , F, K) and ask the students to suggest how to order the boxes from l a r g e s t to s m a l l e s t u s i n g the 1 dm3 blocks and t h e method d e s c r i b e d i n s e c t i o n 3. 1 above. Determine the volume of each box and order the boxes a c c o r d i n g l y . 127 6. (5 min) Volume of p o l y h e d r a l models b u i l t from u n i t cubes: D i s p l a y a pos t e r (#1.1) of two arrangements of cubes. The f i r s t of the arrangements i n c l u d e s one block "absent" although i t may appear to be "present." The second of the arrangements i n c l u d e s one block " p r e s e n t " although i t may appear to be "absent." Ask the students t o use the cubes (inch-cubes) i n order to b u i l d cuboids e x a c t l y l i k e the ones p i c t u r e d i n the p o s t e r and to s t a t e the volumes i n terms of the cubes. 7. (10 min) Worksheet: D i s p l a y boxes X, Y, W and Z i n one area of the classroom ( s t a t i o n #1). D i s p l a y a l s o boxes #1 and #2 as w e l l as 15 decimetre cub-es i n another area ( s t a t i o n #2). Give each of the students a copy of the attached worksheet and e x p l a i n i t to them. D i v i d e the p u p i l s i n t o 3 groups. L e t two o f the groups each be of about 1/4 the c l a s s and the t h i r d of about 1/2 of the c l a s s . Ask each of the s m a l l e r groups to s t a r t on s t a t i o n #1 or #2 and the l a r g e r group on the seat work. I n s t r u c t the students i n the s m a l l e r groups to f i n i s h the work at t h e i r own s t a t i o n , then, with your a p p r o v a l , to move to the other s t a t i o n and f i n i s h the work t h e r e , and then to r e t u r n to t h e i r s e a t s f o r the seat work. S i m i l a r l y i n s t r u c t the students i n the l a r g e r group to f i n i s h t h e i r s e a t work f i r s t then wait f o r your app r o v a l i n moving to the s t a t i o n s . Send students from t h i s l a r g e r group to the s t a t i o n s as they f i n i s h t h e i r s e a t work and as space and order at the s t a t i o n s allow. At the end of the p e r i o d c o l l e c t the worksheets. 128 12S Hasc: l a s t : F i r s t : l e s s o n A l - Worksheet ( S t a t i o n #1) 1. Examine the two fccxes l e t t e r e d 1 and i . i h i c h o c c u p i e s l e s s space, ¥ or X? which has the s i a l l e r volume I or X? 2. Examine the f o u r boxes l e t t e r e d W, X, I, and Z; then order thea f r c a l a r g e s t t c s . a l l e s t and re c o r d year answers below. ( l a r g e s t ) , , (smallest) ( S t a t i o n 12) 3. B u i l d s t a c k s c f cubes s i m i l a r t o these bcres and answer the f o l l o w i n g : The volume c f fccx 11 i s : cubes. The vclune cf fccx #2 i s : cubes. (Seat work) 4. f o r each of t i e f i g u r e s below use the cubes to b u i l d a ao d e l . Ccunt the nuaber c f cubes and record the vcluae. F i g u r e A F i g u r e B F i g u r e C Vcluae = cubes Voluae = cubes Voluse = cubes 5 . 1 i s t 1, E, and C i n order (Largest) (Smallest) 130 Treatment A Lesson 2 B e h a v i o r a l O b j e c t i v e s : 1. Given a cuboid or a diagram of a c u b o i d , which i s completely p a r t i t i o n e d or p a r t i a l l y p a r t i t i o n e d i n t o u n i t cubes, the students w i l l be a b l e to b u i l d a l a y e r and determine the volume of the l a y e r , the number of l a y e r s and the t o t a l volume. 2. Given a diagram of a n o n - p a r t i t i o n e d cuboid, the dimensions shown e i t h e r by numerals or by the f a c t of i t s edges being marked i n u n i t segments, the students w i l l be able to determine the volume of a l a y e r , the number of l a y e r s and the t o t a l volume o f the cuboid. O u t l i n e : 1. F a l l o w up of the worksheet from the p r e v i o u s l e s s o n . 2. Volume of p a r t i t i o n e d and n o n ^ p a r t i t i o n e d cuboids. 3. A l g o r i t h m f o r the volume of a cuboid "V = L X W X H." 4. A p p l i c a t i o n of the volume a l g o r i t h m t o cuboids and diagrams of cuboids. 5. . Worksheet. M a t e r i a l s : 1. Three cardboard boxes. 2. Some decimetre cubes. 3. A p o s t e r o f : a p a r t i t i o n e d c u b o i d , a p a r t i a l l y p a r t i t i o n e d cuboid and a n o n - p a r t i t i o n e d cuboid,. 131 a c t i v i t i e s : 1. (3 min) Fellow up o f the,worksheet from the p r e v i o u s  j e s s o n : Give each student h i s c o r r e c t e d worksheet from the previous l e s s o n . E x p l a i n t h a t i n order t o compare volumes one can count the cubes and compare the numbers o b t a i n e d . For i l l u s t r a t i o n count the cubes of the shapes i n #4 as each student f o l l o w s on h i s own worksheet, r e p o r t the volumes of the shapes and s t a t e t h e i r o rder. 2. (7 min) Volume of p a r t i t i o n e d and n o n - p a r t i t j o n e d cuboids: Give each student 25 inch-cubes. D i s p l a y a poster (#2.1) of a p a r t i t i o n e d cuboid of dimensions 2, 2 and 5. Ask the students to b u i l d with the cubes a cuboid e x a c t l y l i k e the one p i c t u r e d i n the p o s t e r , t o count the number of b l o c k s and to s t a t e the volume of the cuboid. Draw on the board a diagram of a n o n - p a r t i t i o n e d cuboid of dimensions L = 4, W = 3 and H = 2. Write these dimensions along the edges. Ask the students t o guess the number of cubes necessary to b u i l d the cuboid. P a r t i t i o n the top l a y e r then the r e s t of the c u b o i d . Ask the s t u d e n t s to determine the volume by using the b l o c k s to b u i l d a model then count the number of b l o c k s used. 3. (12 min) Algorithm f o r the volume of a cuboid _ I = _ 1 _ J ___ D i s p l a y a c l o s e d cardboard box (H) whose dimensions are L = 4 dm, W = 2 dm and H = 2 dm. D i s p l a y a l s o about 20 decimetre cubes. Discuss with the students how one can determine the volume of the box using the a v a i l a b l e cubes. Lead the a c t i v i t i e s u s i n g the cubes i n order t o determine: a. The number of decimetre cubes along the l e n g t h of the bottom l a y e r . Write on the board "L(length) = 4 dm" i . e . / 4 cubes f i t along the length, of the bottom l a y e r . b. The number o f decimetre cubes along the width o f the bottom l a y e r . Write oh the board "W (width) = 2 dm" c. B u i l d a l a y e r and conclude t h a t the number of b l o c k s t h a t can f i t i n the bottom l a y e r i s given by L X W. / d. The number of decimetre cubes along the h e i g h t of the box. Write on the board "H (height) = 2 dm" e. B u i l d a shape congruent t o the box and conclude t h a t the t o t a l volume i s the volume cf one' l a y e r (L X W) 132 m u l t i p l i e d by the number of l a y e r s (H) i . e . , V = L 2 W X E. D i s p l a y another cardboard box (R) whose dimensions are 3 dm, 3 dm and 2 dm. D i s p l a y a l s o about 10 decimetre cubes. Discuss with the students how one can determine the volume of the box using the a v a i l a b l e cubes, knowing t h a t there are not enough cubes to b u i l d a shape congruent t o the box or even b u i l d ' a shape congruent t o a l a y e r . Develop again the a l g o r i t h m "V = L X W X H" by f o l l o w i n g steps a to e of the p r e v i o u s a c t i v i t y but without a c t u a l l y b u i l d i n g a shape s i m i l a r to the box. Test the a l g o r i t h m "V = L X W X H" u s i n g a d i f f e r e n t cardboard box (#1) of dimensions 4 dm, 3 dm and 1 dm: f . F i n d the l e n g t h , width and height of the box i n decimetres and m u l t i p l y them. g. B u i l d a shape s i m i l a r to the box with decimetre cubes and count the number o f cubes. h. Compare the r e s u l t s i n f* and g. • 4. (6 min) A p p l i c a t i o n of the, volume a l g o r i t h m t o diagrams of cuboids: Present a p o s t e r (#2.2) of diagrams of cuboids and apply the a l g o r i t h m as d i r e c t e d . a. Refer to the p a r t i t i o n e d cuboid on the poster. Without a c t u a l l y using the b l o c k s develop again the a l g o r i t h m "V = L X W X H" by v e r b a l l y f o l l o w i n g steps a to e o f the p r e v i o u s a c t i v i t y . b. Refer to the p a r t i a l l y p a r t i t i o n e d and the non-p a r t i t i o n e d c uboids. In each case determine the volume of each l a y e r , the number o f l a y e r s and the t o t a l volume..Use the a l g o r i t h m "V = L X W X H" to compute the volume. Compare the two answers. 5. , (7 min) Worksheet: Give each of the students a copy of the attached worksheet and e x p l a i n i t to them. Go around and help them to complete i t . C o l l e c t the worksheets at the end of the p e r i o d . 133 P o s t e r s of Lesson A2 No. 2.2 131 Dane: l a s t : F i r s t ; lesson K2 - Worksheet 1, Fina the vclune of each of the f i g u r e s below, the cubes to b u i l d mcdels of the f i g u r e s ) . (you Bay use Figure C Volane of a l a y e r = , Huafcer of l a y e r s = _ Voluae = cubes 2. vclome of the fcettes l a y e r Number c f l a y e r s = • Yolune = Figure F ?oluae of a l a y e r = Hunber of l a y e r s = foluae = cubes 3. I = 1 -H = V = 4. Vclnae of the box = 135 Treatment , A Lesson 3 B e h a v i o r a l o b j e c t i v e s : 1. Given e i t h e r a diagram of a n o n - p a r t i t i o n e d cuboid with i t s dimensions marked, or a word d e s c r i p t i o n of the dimensions of a cuboid without a diagram, the students w i l l be a b l e to use the volume a l g o r i t h m i n order to determine the volume of the cuboi d . 2. Given a diagram of cuboids touching s i d e by s i d e and a l l of the r e q u i r e d dimensions, the students w i l l be able t o use the volume a l g o r i t h m i n " order t o determine the t o t a l volume of the cuboids. 3. Given a diagram of a cuboid to which t h e r e are attached h a l f cubes ( r e c t a n g u l a r p a r a l l e l e p i p e d s or t r i a n g u l a r prisms) the students w i l l be able t o determine the t o t a l volume. 4. Given a diagram of a p a r t i t i o n e d cuboid which i s p a r t i a l l y covered, the s t u d e n t s w i l l be able to determine the t o t a l volume of the cuboi d . O u t l i n e : 1. Follow up of the worksheet from the p r e v i o u s l e s s o n . 2. A p p l i c a t i o n of the volume al g o r i t h m t o the f o l l o w i n g c a s e s : a. Word d e s c r i p t i o n of cuboids. b. Cuboids touching s i d e by s i d e . C. Diagrams of cuboids with some u n i t cubes att a c h e d or removed. d. Attachments of h a l f cubes to cuboids. e. Diagrams of p a r t i a l l y covered cuboids. 3. Worksheet. 136 1. P o s t e r s of cuboids t o u c h i n g s i d e by s i d e , cubo,ids with some u n i t s attached or removed and attachments of h a l f cubes to cuboids. A c t i v i t i e s , : 1. (7 min) Follow up o f the worksheet from the p r e v i o u s  l e s s e n : Give each student h i s c o r r e c t e d worksheet from the p r e v i o u s l e s s o n and e x p l a i n t h a t "V = L X W X H" , h e l p s us to compute the volume (the number of cubes needed to b u i l d a s i m i l a r shape) of each of the shapes drawn on the worksheet. For i l l u s t r a t i o n e x p l a i n , as each student f o l l o w s on h i s own worksheet, t h a t i n #2 the volume of the top l a y e r (of L = 4 and W = 2) i s L X W = 4 X 2 = 8 , t h a t there are 6 l a y e r s and t h a t the volume i s the number of cubes i n one l a y e r (8) m u l t i p l i e d by the number of l a y e r s (6) i . e . , V = 4 X 2 X 6 = 48. 2. (18 min) A p p l i c a t i o n of the volume al g o r i t h m to the f o l l o w i n g cases: a. Word d e s c r i p t i o n of cuboid: Write on the board a v e r b a l d e s c r i p t i o n of the dimensions of a r e c t a n g u l a r box (L = 6 dm, W = 2 dm, H = 5 dm). S u b s t i t u t e the given numbers f o r L, W, and H i n "V - L X W X H" and compute the volume. Draw a diagram on the board and i l l u s t r a t e t h a t "L X W" g i v e s the number o f b l o c k s i n one l a y e r and L X W X H i s the t o t a l volume. b. Cuboids touching s i d e by s i d e : Present the p o s t e r (#3.1) of two cuboids touching s i d e by s i d e . Ask students t o compute the volume of each cuboid and add to determine the t o t a l volume. c. Diagrams of cuboids with some u n i t cubes att a c h e d or removed: Present the poster (#3.2) of cuboids with some u n i t cubes attached or removed. Lead the students to compute the volume of the cuboid as i f there were nothing attached or removed. Determine the volume of the u n i t b l o c k s to be added or s u b t r a c t e d . Add or s u b t r a c t to determine the volume. d. Attachments of h a l f cubes to cuboids: 137 Present the poster (#3.3) of the diagrams of the attachments of h a l f cubes to cuboids. In each case ask the students to determine the volume of the c u b o i d , the volume of the attached h a l f cubes and add to determine the t o t a l volume. €» Diagrams of p a r t i a l l y covered cuboids: Present the poster (#3.4) of a p a r t i a l l y covered cuboid (4X3X6). P o i n t out t h a t the b l o c k s of the top l a y e r are shown and there i s a t o t a l of 6 l a y e r s . Determine the l e n g t h , width and h e i g h t ' of the cuboid and use the a l g o r i t h m t o compute the volume. Count the number of b l o c k s i n the top l a y e r (12) and conclude t h a t each of the 6 l a y e r s has 12 b l o c k s . Determine the t o t a l volume (6x12). Compare the two answers. 3. (10 min) Worksheet: Give each o f the students a copy of the attached worksheet and e x p l a i n i t to them. Go around and help them t o complete i t . C o l l e c t the worksheets a t the end of the p e r i o d . 138 Pesters of Lessen ft3 No- 3-1 Ho. 3.2 139 Ha&e: L a s t : F i r s t : Lesson_A3_;,Worksheet 1. A box has a length of 14, a width of 11 and a height of 3. What i s the volume of the box? _ _ What i s the volume k a e t a l block and a h a l f of a d i f f e r e n t block on top Bhat i s the volume What i s the volume 5. A p i l e of cubes and h a l f cubes 6 What i s the volume wmmi A p i l e of cubes p a r t i a l l y covered that i s the volume 140 Treatment A Lesson 4 B e h a v i o r a l O b j e c t i v e : 1. Given a diagram or a word d e s c r i p t i o n of a cuboid o f known dimensions and given a proposed a d d i t i v e or m u l t i p l i c a t i v e dimensional t r a n s f o r m a t i o n , the students w i l l be able t o s t a t e the volume of the cuboid t h a t would r e s u l t a f t e r the t r a n s f o r m a t i o n . O u t l i n e : 1. Follow up of the worksheet from the p r e v i o u s l e s s o n . 2. A p p l i c a t i o n of the volume a l g o r i t h m t o cuboids with proposed dimensional t r a n s f o r m a t i o n s . 3. Summary of g e n e r a l i z a t i o n s . 4. Worksheet. M a t e r i a l s : 1. & poster of a p a r t i t i o n e d cuboid. 141 A c t i v i t i e s : 1. (5 min) Fellow up of the worksheet from the p r e v i o u s l e s s o n : 1 Give each student h i s c o r r e c t e d worksheet from the prev i o u s l e s s o n and e x p l a i n t h a t i n #2 f o r example, t h e r e are 2 h a l f - c u b e s attached t o a r e c t a n g u l a r p i l e of cubes of L = 3, W = 3 and H = 2. The volume o f the 2 h a l f - c u b e s i s 1, the volume o f the p i l e i s 3 X 3 X 2 = 18 (using V = L X W X H) and the t o t a l volume i s t h e r e f o r e 18 + 1 = 19. S i m i l a r l y f o r #6, e x p l a i n t h a t the volume of the block i s 5 X 3 X 4 = 60 (using V = L X W x H), the volume o f the h a l f block on top i s (3 X 2 X 1)/2 = 3 and the t o t a l volume i s t h e r e f o r e 60 + 3 = 63. 2. (18 min) A p p l i c a t i o n o f the volume a l g o r i t h m to cuboids  with proposed dimensional t r a n s f o r m a t i o n s : Write on the board "V = L X W X H", d i s p l a y a poster (#4.1) of a p a r t i t i o n e d cuboid of dimensions 6, 4 and 10 and r e p l a c e L, W, and H by the numbers 6, 4 and 10. Compute the volume (240) and r e p l a c e V by 240. Apply the f o l l o w i n g changes t o the f a c t o r s (L, W, H) and observe the changes i n the product (volume). Encourage the students t o s t a t e and t e s t c o n j e c t u r e s about the e f f e c t on the volume when the dimensions are changed. 2.1. A d d i t i v e (or m u l t i p l i c a t i v e ) i n c r e a s e - i n one, two or three of the dimensions produces a d d i t i v e (or m u l t i p l i c a t i v e )  i n c r e a s e i n the volume. 2.1.1. A d d i t i v e i n c r e a s e i n one, two or three of the dimensions: Ask the students to use the a l g o r i t h m "V = L X W X H" to c a l c u l a t e the volume of the cuboid i f i t s length i n c r e a s e s to 7 u n i t s . . W r i t e "7 X 4 X 10 = 280" underneath "6 X 4 X 10 =240." Continue by asking the students t o p r e d i c t what happens to the volume (240) i f the width i n c r e a s e s o r the hei g h t i n c r e a s e s . Allow time f o r responses, write on the board the two sentences "6 X 5 X 10 = " and " 6 X 4 X 1 1 = _", d i v i d e the students i n t o two groups and ask each group t o complete one of the statements. S o l i c i t the answers (300,264) and complete the statements w r i t t e n on the board. Ask the students t o p r e d i c t what happens t o the volume (240) i f any one of the dimensions i n c r e a s e s . When c o n j e c t u r e s are made t e s t them using examples such as "8 X 4 X 10 = 320" and "6 X 4 X 12 = 288." Lead the d i s c u s s i o n i n order to conclude that the volume i n c r e a s e s i f any one of the dimensions 142 i n c r e a s e s . S i m i l a r l y , ask the students t o p r e d i c t what happens to the volume (240) i f two of the dimensions i n c r e a s e . Allow time f o r responses and t e s t them u s i n g examples such as "7 X 5 X 10 = 350" and "6 X 6 X 10 = 360." Lead the d i s c u s s i o n i n order t o conclude t h a t the volume i n c r e a s e s i f two o f the dimensions i n c r e a s e . S i m i l a r l y , l e a d the d i s c u s s i o n i n order t o conclude t h a t the volume i n c r e a s e s i f a l l dimensions i n c r e a s e . Ose "7 X 5 X 11 = 385" f o r i l l u s t r a t i o n . 2.1.2. M u l t i p l i c a t i v e - i n c r e a s e i n one or two of the dimensions: Ask the students t o p r e d i c t what happens to the volume (240) i f any one of the dimensions i s m u l t i p l i e d by "2" (doubled). Allow time f o r responses and ask t h r e e groups t o t e s t them using "12 X 4 X 10 = .__", "6 X 8 X 10 = _ _ _ " and "6 X 4 X 20 = " (answer: 480),. Conclude t h a t the volume i s m u l t i p l i e d by 2 (doubled). Continue by asking the students to p r e d i c t what happens to the volume (240) i f any one o f the dimensions i s m u l t i p l i e d by 3 ( t r i p l e d ) or 4. T e s t the c o n j e c t u r e s using: 6 X (3 X 4) X 10 = 720 = (3 X 240) and 6 X 4 X (4 X 10) = 960 = (4 X 240). Lead the d i s c u s s i o n i n order t o conclude t h a t the volume i s m u l t i p l i e d by 2 (doubled), 3 ( t r i p l e d ) , ... i f any one of the dimensions i s m u l t i p l i e d by 2 (doubled), 3 ( t r i p l e d ) , ... L i k e w i s e , ask the students to p r e d i c t what happens to the volume (240) i f each of two dimensions i s m u l t i p l i e d by 2; then i f one i s m u l t i p i e d by 2 and another by 3, e t c . Ose examples such as the ones w r i t t e n below to e s t a b l i s h t h at i f one dimension i s m u l t i p l i e d by a whole number and another dimension i s a l s o m u l t i p l i e d by a whole number, the volume w i l l be m u l t i p l i e d by the product of the two numbers. (5 X 6) X (2 X 4) X 10 = (5 X 2) X 240 30 X 8 X 10 = 2400 6 X (3 X 4) X (2 X 10) = (3 X 2) X 240 6 X 12 X 20 = 1440 2.2» A d d i t i v e (or m u l t i p l i c a t i v e ) d e c r e a s e : i n one, two or  th r e e of the dimensions, produces a d d i t i v e (or m u l t i p l i c a t i v e )  decrease i n the, volume,. Model the d i s c u s s i o n of t h i s s e c t i o n a f t e r the d i s c u s s i o n of the p r e v i o u s s e c t i o n (2.1). For each o f the g e n e r a l i z a t i o n s 143 below ( l e t t e r e d a, b, ...) ask the students t o p r e d i c t what happens to the volume (240) i f the proposed changes i n the dimensions are a p p l i e d , t e s t the students* p r e d i c t i o n s u s i n g the g i v e n examples and conclude the g e n e r a l i z a t i o n . 2.2.1, Ad d i t i v e , decrease, i n one,, two or t h r e e of the dimensions: (move q u i c k l y through t h i s section) a. The volume decreases i f any, one of the dimensions decreases. Examples: 5 X 4 X 10 = 200; 6 X 3 X 10 = 180. b. The volume decreases i f any two of the dimensions decrease. Examples: 5 X 3 X 10 = 150; 5 X 4 X 3 = 60 c. The volume decreases i f a l l dimensions decrease. Example: 4 X 3 X 8 = 96 2.2.2. M u l t i p l i c a t i v e decrease i n one,or two of the dimensions: a. The volume w i l l be d i v i d e d by 2 (halved), 3, ... i f any one of the dimensions i s d i v i d e d by 2, 3, ... Examples: 6 X (4/2) X 10 = 240/2 6 1 2 X 10 = 120 6 X 4 X (10/5) = 240/5 6 X 4 X 2 = 48 b. I f one o f the dimensions i s d i v i d e d by a whole number and another dimension i s d i v i d e d by a whole number, the volume w i l l be d i v i d e d by the product of the two numbers. Examples: (6/2) X (4/2) X 10 = 240/(2 X 2) = 240/4 3 X 2 X 10 = 60 2.3. A d d i t i v e i n c r e a s e i n one.pr tuo of the dimensions and a d d i t i v e decrease i n one, or two can produce any one of the  f o l l o w i n g : a. A d d i t i v e i n c r e a s e i n the volume b. A d d i t i v e decrease i n the volume c. No change i n the volume Ask the students whether they can p r e d i c t what happens to the volume i f one of the dimensions i n c r e a s e s and another decreases. Allow time f o r p r e d i c t i o n s and ask t h r e e groups t o t e s t them using the f o l l o w i n g examples: 144 6 X 5 X 9 = 270 (the volume i n c r e a s e s ) 7 X 4 X 3 = 84 (the volume decreases) 8 X 3 X 10 = 240 (the volume does not change) Conclude t h a t the volume may i n c r e a s e , decrease or stay the same i f one dimension i n c r e a s e s and another decreases; each example has to be examined i n d i v i d u a l l y . 3. (2 min) Summary of g e n e r a l i z a t i o n s ; Summarize the g e n e r a l i z a t i o n s made i n t h i s l e s s o n by aski n g t h i s s e r i e s of gue s t i o n s and encourage students to answer them: i . What happens to the volume i f we i n c r e a s e one, two or three of the dimensions? i i . What happens to the volume i f we decrease one, two or th r e e of the dimensions? i i i . What happens to the volume i f we i n c r e a s e one o f the dimensions and decrease another? i v . a. What happens t o the volume i f we double only one dimension? b. What happens to the volume i f we double two dimensions? c. What happens to the volume i f we double a l l t h r e e dimensions? v. What happens to the volume i f we double one dimension and halve another dimension? 4. (10 min) worksheet,: Give each o f the students a copy of the attached worksheet and e x p l a i n i t to them. Go around and help them t o complete i t . C o l l e c t the worksheets a t the end of the p e r i o d . Poster of Lesson 146 Kane:Last; F i r s t ; fresscn a4-MorKsheet Cosplete the f o l l o w i n g : 1. A p l a s t i c hex. I f ve increased tfce width of t h i s box by three u n i t s and the length and the height stayed tke sane, then i n the new bcx what would be the l e n g t h ? i i a t h ? Height? Volune? 2. 3-II a e t a l bcx. I f we decreased the height by 3 un i t s and the t i e length by 2 u n i t s but the width stayed tfce sane, then i n the new bcx what would be the length? l i d t h ? Height? Ycluae? 1 wooden bcx. I f we decreased the length by 1 u n i t and we increased the height by 2 u n i t s but the width stayed the sa«e, then i n the new bcx what would be the length? l i d t h ? _ Height? _ • Vcluae? 4. A r e c t a n g u l a r p i l e of cubes. I f we t r i p l e d the h e i g h t o f t h i s p i l e and the l e n g t h and the width stayed the same, then i n the new p i l e what would be the Length? width? Height? Volume? 5. A r e c t a n g u l a r p i l e of cubes. I f we doubled the l e n g t h and halved the width but the h e i g h t stayed the same, then i n the new p i l e what would be the Length? Width? Height? Volume? 6- A cardboard box f u l l of cubes. I f we m u l t i p l i e d the height by 3 and d i v i d e d the l e n g t h by 6 and the width by 2, then i n the new box what would be the Length? Width? Height? Volume? 148 Treatment B Lesson 1 B e h a v i o r a l O b j e c t i v e s : a, b and n denote n a t u r a l numbers. 1. Given a m u l t i p l i c a t i o n equation with the same t h r e e f a c t o r s on each s i d e of the equal s i g n , such t h a t the f a c t o r s on both s i d e s are not i n the same order and t h a t one or two f a c t o r s of one s i d e are missing, the students w i l l be a b l e to s t a t e the missing f a c t o r s . 2. Given an a s s e r t i o n of the form a X b > a X c, a X b < a X c , c X a > b X a o r c X a < b X a such t h a t a i s known, but only one of b or c i s known, the students w i l l be able t o a s c r i b e c o r r e c t l i m i t s t o the range o f value s f o r the unknown number. 3. Given an a s s e r t i o n of the form a X b X n > a X c X n o r n X c X a > n X b X a such t h a t n and a are known and only one of b or c i s known, the students w i l l able to a s c r i b e c o r r e c t l i m i t s t o the range of valu e s f o r the unknown number t o make.the a s s e r t i o n t r u e . O u t l i n e : 1. Review of the commutative and a s s o c i a t i v e p r i n c i p l e s . 2. P r e s c r i b i n g the range f o r the missing f a c t o r i n an i n e q u a l i t y i n v o l v i n g two f a c t o r s a t each of i t s s i d e s . 3. P r e s c r i b i n g the range f o r the missing f a c t o r i n an i n e q u a l i t y i n v o l v i n g three f a c t o r s a t each of i t s s i d e s 4. Worksheet. M a t e r i a l s : 1. ft poster of 3 X 5 g r i d . 149 A c t i v i t i e s : 1.a. (5 min) Review of th__.Commutative p r i n c i p l e : Present the poster (#1.1) of 3 X 5 g r i d and ask the students to t e l l the number of rows (3) and the number of squares per row (5). Ask the students to give a m u l t i p l i c a t i o n sentence to d e s c r i b e the t o t a l number of squares (3 X 5 = 15). Write " 3 X 5 = 15" on the board. D e c l a r e t h a t the numbers 3 and 5 are c a l l e d the f a c t o r s while 15 i s c a l l e d the product. Turn the p o s t e r 90 degrees and ask s i m i l a r q u e s t i o n s to the ones i n the previous paragraph. Conclude t h a t the sentence d e s c r i b i n g the t o t a l number of squares i s now " 5 X 3 = 15". Write " 5 X 3 = 15" on the board underneath " 3 X 5 = 15".. Lead the d i s c u s s i o n t o i l l u s t r a t e t h a t both "3 X 5" and "5 X 3" d e s c r i b e the same number of squares and are t h e r e f o r e e q u i v a l e n t . Write the statement "3 X 5 = 5 X 3" on the board. Ask the students t o make a g e n e r a l i z a t i o n about the order of the f a c t o r s based on the statement "3 X 5 = 5 X 3". . Allow time f o r responses and emphasize the commutative p r i n c i p l e (without mentioning the term "commutative") i . e . , the order o f m u l t i p l y i n g the f a c t o r s does not a f f e c t the product. Test t h i s p r i n c i p l e using 3 X 4 and 4 X 3 by determining the answer to 3 X 4 and 4 X 3., Confirm the commutative p r i n c i p l e by a s k i n g the s tudents t o compute the answer (180) to each si d e of the sentence 12 X 15 = 15 X 12. 1. b. (5 min) Review of the a s s o c i a t i v e p r i n c i p l e : Write " 2 X 3 X 4 " on the board and ask the students i f i t makes any d i f f e r e n c e i n choosing the numbers t o be m u l t i p l i e d f i r s t , 2 X 3 or 3 X 4 . Then write ( 2 X 3 ) X 4 = _____ and 2 X ( 3 X 4 ) = on the board. Ask the students to help i n the s t e p s f o r f i n d i n g the answers and conclude t h a t one may m u l t i p l y the f i r s t two f a c t o r s f i r s t or the l a s t two f i r s t . Write on the board (2 X 3) X 4 = 2 X (3 X 4)..Ask the students to make a g e n e r a l i z a t i o n about the order of m u l t i p l y i n g the f a c t o r s based on the sentence w r i t t e n on the board. Allow time f o r responses and emphasize the a s s o c i a t i v e p r i n c i p l e (without mentioning the term " a s s o c i a t i v e " ) f o r the product of any t h r e e f a c t o r s i . e . , the grouping of the f a c t o r s does not a f f e c t the product..Test and confirm t h i s p r i n c i p l e using "5 X 3 X 8". Write the f o l l o w i n g statements on the board and ask the stqdents to f i n d the numbers which complete the statements. X 8 X 1 1 = 8 X 6 X 1 1 ; 7 X 1 2 X 4 = _____ X 7 X 2. (10 min) Prescribing..the range f o r the missing f a c t o r  i n an i n e q u a l i t y i n v o l v i n g two f a c t o r s at each of i t s s i d e s . Write on the board "3 X [ J > 3 X 5 " and f o l l o w the 150 d i s c u s s i o n below i n order to l e a d the students to p r e s c r i b e whole numbers that can f i t i n the £ ]. T: I f we r e p l a c e £ J by 10, w i l l i t be t r u e ? (Response: "Yes". Write 10 on the board underneath [ J) T: Is t h e r e a number g r e a t e r than 10 that would make i t t r u e ? I f so, t e l l me such a number? (Write p u p i l s * responses i n ascending o r d e r . Leave spaces f o r c o n t i n g e n c i e s ) T: What i f we r e p l a c e £ ] by 1, w i l l i t be tru e ? (Response: "No",. Ask how we t e l l ) T: Are t h e r e any numbers l e s s than 10 t h a t would make i t tr u e ? I f so, which numbers? (Write them with the sequence i n order) T: What i s the l e a s t number we have been able t o r e p l a c e £ ] by, so f a r ? Does t h a t mean t h a t 6 i s the l e a s t whole number we can use? Can anyone name a s m a l l e r whole number than 6 t o make i t t r u e ? Why not? T: (Point out the "gap" i n the sequence on the board which probably l o o k s l i k e t h i s : Numbers t h a t make i t t r u e : 6, 8, 10, 64, 100) Are there any numbers between these two ( p o i n t to 6 and 8) t h a t make i t t r u e ? How about here? (point t o 10 and 64) e t c . T: Could we p o s s i b l y l i s t a l l of the numbers t h a t make i t t r u e ? Why not? T: So without a c t u a l l y l i s t i n g them a l l , what co u l d we wri t e t h a t would c l e a r l y i n d i c a t e a l l of the numbers t h a t make i t true? Could t h i s do? (write on the board "Any number g r e a t e r than 5" o r "n > 5") S i m i l a r l y l e t the students p r e s c r i b e answers to the f o l l o w i n g statements. 3 X 8 < [ J X 8 12 X 15 > i ] X 12 3. (5 min) P r e s c r i b i n g , the..range f o r the missing f a c t o r i n  an i n e q u a l i t y i n v o l v i n g three f a c t o r s at each of i t s s i d e s : Write the statement " 2 X 3 X 6 < 2 X [ ] X 6" on the board and f o l l o w the steps below i n order to l e a d the stu d e n t s to p r e s c r i b e numbers t h a t can f i t i n the £ ], (Answer: "Any number g r e a t e r than 3" or "n > 3"). 151 3.1. Determine a number, say 10, t h a t makes the statement t r u e . . 3.2. Determine a number, say 2, t h a t makes the statement f a l s e . 3.3. E s t a b l i s h a lower bound; F i n d a l l the numbers l e s s than 10 which make the statement t r u e . 3.4. Determine a l a r g e r number than 10, say 50, which makes the statement t r u e . 3.5. Determine two or th r e e numbers between 10 and 50 t h a t make the statement t r u e . 3.6. E s t a b l i s h t h a t i t i s impossible t o l i s t a l l the numbers t h a t make the statement t r u e . 3.7- Write a sentence t h a t p r e s c r i b e s a l l the numbers t h a t make the statement t r u e . ("Any number g r e a t e r than 3" or "n > 3") . S i m i l a r l y l e t the students p r e s c r i b e numbers t h a t f i t i n the I ] i n order to make the statement " 6 X 2 X 1 ] < 6 X 2 X 20" t r u e . 4. (10 min) Worksheet: Give each of the students a copy of the attached worksheet and e x p l a i n i t t o them. Move around and help them t o complete i t . At the end of the p e r i o d c o l l e c t the worksheets. 152 P o s t e r of Lesson Bl No- 1-1 Name: L a s t : F i r s t : Lesson B1 - Worksheet Complete the f o l l o w i n g : 1. 5 X 7 X 12 = 12 X 5 X 2. 9 X (14 X 10) = 10 X ( X ) 3. (23 X 7) X 13 = ( X 13) X 4. 7 X 9 > 7 X [ ] What number or numbers c o u l d go i n the _ ]? 5. 53 X 16 > 16 X [ ] What number o r numbers c o u l d go i n the £ ]? 6. 22 X 25 X 26 < 22 X £ ] X 26 What number or numbers co u l d go i n the £ ]? 7. 213 X 19 X [ ] < 213 X 19 X 6 What number o r numbers could go i n the £ ]? i n each of the f o l l o w i n g w r i t e : < # = or > i n the 9-8. 10. 11. 12. 9 X (8 X 13) 4 X(2 X 3) 2) X 3 X 13) X 9 154 Treatment B l e s s o n 2 B e h a v i o r a l O b j e c t i v e s : a, b, c and P denote n a t u r a l numbers. 1. Given an a s s e r t i o n of the form a X b = P where a, b and P are known, the students w i l l be able to d e s c r i b e the e f f e c t on P when any one of the f o l l o w i n g changes i s a p p l i e d a d d i t i v e l y or m u l t i p l i c a t i v e l y : i . I n c r e a s e i n a, or b, or both, i i . Decrease i n a, or b, or both. ' i i i . I n c r e a s e s i n a tog e t h e r with decrease i n b (or v i c e versa) . 2. Given an a s s e r t i o n of the form a X b X c = P the students w i l l be able to d e s c r i b e the e f f e c t on P when any one of the f o l l o w i n g changes i s a p p l i e d a d d i t i v e l y or m u l t i p l i c a t i v e l y t o the f a c t o r s . i . Increase i n a, b or c i i . Decrease i n a, b or c i i i . I ncrease i n one or two of the f a c t o r s a, b or c arid decrease i n two or one of the f a c t o r s a, b or c 1. Follow up of the worksheet from the p r e v i o u s l e s s o n . 2. E f f e c t on the product of two f a c t o r s when these f a c t o r s are changed a d d i t i v e l y o r m u l t i p l i c a t i v e l y . (Note: " a d d i t i v e l y " and " m u l t i p l i c a t i v e l y " w i l l subsume decrease as w e l l as increase.) 3- E f f e c t on the product of three f a c t o r s when these f a c t o r s are changed a d d i t i v e l y or m u l t i p l i c a t i v e l y . . ( N o t e : " a d d i t i v e l y " and " m u l t i p l i c a t i v e l y " w i l l subsume decrease as w e l l as increase.) 4. Worksheet 155 i M a t e r i a l s : (The a c t i v i t i e s of t h i s l e s s o n are mainly number manipulations and do not r e q u i r e p h y s i c a l m a t e r i a l s ) A c t i v i t i e s : 1. (4 min) Follow up of the worksheet from the p r e v i o u s  l e s s o n : Give each student h i s c o r r e c t e d worksheet from the previous l e s s o n and e x p l a i n t h a t i n #4, any number l e s s than 9 can qo i n the £ ] and make the statement t r u e . Any number q r e a t e r than or equal t o 9 w i l l make the statement f a l s e . S i m i l a r l y , i n #7 any number l e s s than 6 w i l l make the statement t r u e and any number g r e a t e r than or equal t o 6 w i l l make the statement f a l s e . 2. (20 min) E f f e c t , on.the product of two f a c t o r s when  these f a c t o r s are changed a d d i t i v e l y or , m u l t i p l i c a t i v e l y . , Write on the board the statement " 6 X 4 = 24". Apply the f o l l o w i n g changes t o the f a c t o r s (4 and 6) and observe the changes i n the product (24). Encourage the students t o s t a t e and t e s t c o n j e c t u r e s about the e f f e c t on the product when the f a c t o r s are changed. 2.1. A d d i t i v e (or m u l t i p l i c a t i v e ) i n c r e a s e i n one or two of the f a c t o r s , produces a d d i t i v e (or m u l t i p l i c a t i v e ) i n c r e a s e i n the product. 2.1.1. A d d i t i v e i n c r e a s e i n one or two of the f a c t o r s : Ask the students t o p r e d i c t what happens to the product i f the f a c t o r 6 i s r e p l a c e d by a l a r g e r number, say 7 or 8 ( d e s i r a b l e response: " I t i n c r e a s e s " ) . Allow time f o r responses and write on the board " 7 X 4 = 28" underneath " 6 X 4 = 24." Continue by asking the students t o p r e d i c t what happens to the product (24) i f any one of the f a c t o r s i n c r e a s e s . When c o n j e c t u r e s are made t e s t them using examples such as " 8 X 4 = 32" and " 6 X 7 = 42." Lead the d i s c u s s i o n i n order to conclude t h a t the product i n c r e a s e s i f any one of the f a c t o r s i n c r e a s e s . Write on the board 6 X 4 = 24 1 i I i n c r e a s e s s t a y s i n c r e a s e s the same S i m i l a r l y , ask the students t o p r e d i c t what happens to the product (24) i f both of the f a c t o r s i n c r e a s e . Allow time f o r 156 responses and t e s t them using examples such as " 7 X 5 = 35" and "10 X 6 = 60." Lead the d i s c u s s i o n i n order to conclude t h a t the product i n c r e a s e s i f both of the f a c t o r s i n c r e a s e . Write on the board 6 X 4 = 24 i i 1 i n c r e a s e s i n c r e a s e s i n c r e a s e s 2.1.2. M u l t i p l i c a t i v e i n c r e a s e i n one or two of the  f a c t o r s : Ask the students t o p r e d i c t what happens to the product (24) i f any one of the f a c t o r s i s m u l t i p l i e d by 2 (doubled). Allow time f o r responses and ask two groups t o t e s t them using "12 X 4 = (answer: 48)" and " 6 X 8 = (answer: 48)." Conclude t h a t the product i s m u l t i p l i e d by 2 (doubled). Continue by a s k i n g the students to p r e d i c t what happens to the product (24) i f any one of the f a c t o r s i s m u l t i p l i e d by 3 ( t r i p l e d ) or 4. Test the c o n j e c t u r e s u s i n g : 6 X 4 = 24 ( o r i g i n a l statement) 6 X 12 = 72 = 3 X 24 and 24 X 4 = S6 = 4 X 24. Lead the d i s c u s s i o n i n order to conclude t h a t the product i s m u l t i p l i e d by 2 (doubled), 3 ( t r i p l e d ) , ... i f any one of the f a c t o r s i s m u l t i p l i e d by 2 (doubled), 3 ( t r i p l e d ) , ... Write on the board 6 X 4 = 24 1 i 1 s t a y s i s i s the same m u l t i p l i e d m u l t i p l i e d by 3 by 3 L i k e w i s e , ask the s tudents t o p r e d i c t what happens to the product (24) i f one of the two f a c t o r s i s m u l t i p l i e d by 2 and the other by 3, e t c . Use examples such as the one w r i t t e n below to e s t a b l i s h t h a t i f one f a c t o r i s m u l t i p l i e d by a whole number and another by a whole number, the product w i l l be m u l t i p l i e d by the product of the two numbers. 6 X 4 = 24 ( o r i g i n a l statement) 30 X 8 = 240 = 10 X 24 2.2. A d d i t i v e (or m u l t i p l i c a t i v e ) decrease i n one or two  of the f a c t o r s produces a d d i t i v e (or m u l t i p l i c a t i v e ) „ decrease i n the product. (Move q u i c k l y through t h i s s e c t i o n ) Model t h e d i s c u s s i o n of t h i s s e c t i o n a f t e r the d i s c u s s i o n of the p r e v i o u s s e c t i o n (2.1). For each of the g e n e r a l i z a t i o n s below ( l e t t e r e d a, b, «..) ask the students t o p r e d i c t what 157 happens to the product (24) i f the proposed changes i n the f a c t o r s are a p p l i e d , t e s t the students' p r e d i c t i o n s using the giv e n examples and conclude the g e n e r a l i z a t i o n . 2.2.1. A d d i t i v e decrease i n one or two of the f a c t o r s : (move q u i c k l y through t h i s s e c t i o n ) a. The product decreases i f any one of the f a c t o r s decreases. Examples: 5 X 4 = 20; 6 X 3 = 18. Write on the board 6 X 4 = 24 1 i i s t a y s decreases decreases the same b. The product decreases i f both o f the f a c t o r s decrease. Example: 5 X 3 = 15. Write on the board 6 X 4 = 24 i i I decreases decreases decreases 2.2.2. M u l t i p l i c a t i v e . d e c r e a s e i n one, two or three of the  f a c t o r s : (move g u i c k l y through t h i s section) a. The product w i l l be d i v i d e d by 2 (halv e d ) , 3, ... i f any one of the f a c t o r s i s d i v i d e d by 2, 3, ... Example: 6 X 2 = 12 = 24/2 Write on the board '6 X 4 24 J i I s t a y s i s i s the sai ne d i v i d e d d i v i d e d by 2 by 2 B. The product (24) w i l l be- d i v i d e d by the product of two numbers i f one of the f a c t o r s i s d i v i d e d by one of the numbers and another f a c t o r i s d i v i d e d by the other number. Examples: 3 X 2 = 6 = 24/4 2 X 1 = 2 = 24/12 wr i t e on the board 158 6 3 . 4 i s d i v i d e d by 3 4 4 i s d i v i d e d by 4 24 4 i s d i v i d e d by 12 2.3- A d d i t i v e (or, m u l t i p l i c a t i v e ) i n c r e a s e i n one of the f a c t o r s and a d d i t i v e (or m u l t i p l i c a t i v e ) , decrease i n the other  f a c t o r can produce any one of the t f o l l o w i n g : a- A d d i t i v e (or m u l t i p l i c a t i v e ) i n c r e a s e i n the product b. A d d i t i v e (or m u l t i p l i c a t i v e ) decrease i n the product c. No change i n the,, product 2.3.1. A d d i t i v e i n c r e a s e i n one of the f a c t o r s and a d d i t i v e decrease i n the other f a c t o r : Ask the students whether they can p r e d i c t what happens to the product i f one of the f a c t o r s i n c r e a s e s and the other decreases. Allow time f o r p r e d i c t i o n s and ask three groups to t e s t them u s i n g the f o l l o w i n g examples: 6 X 4 = 24 ( o r i g i n a l statement) 5 X 9 = (45, the product i n c r e a s e s ) 7 X 3 = _____ (21, the product decreases) 8 X 3 = , (24, the product does not change) Conclude t h a t the product may i n c r e a s e , decrease or stay the same i f one f a c t o r i n c r e a s e s and the other decreases; each example has to be examined i n d i v i d u a l l y . Write on the board 6 X 4 = 24 i n c r e a s e s decreases i n c r e a s e s , decreases or s t a y s the same 3. (4 min) E f f e c t on,, the product of t h r e e f a c t o r s when  these f a c t o r s are changed a d d i t i v e l y or m u l t i p l i c a t i v e l y : (move q u i c k l y through t h i s section) Write of the board the statement "4 X 6 X 10 = 240" and ask the students whether they can p r e d i c t the change i n the product (240) of t h r e e f a c t o r s (4, 6, 10) when these f a c t o r s vary. Ask t h i s f o l l o w i n g s e r i e s of q u e s t i o n s and encourage students to answer them by making g e n e r a l i z a t i o n s . When necessary t e s t the g e n e r a l i z a t i o n s using numerical examples. i . What happens t o the product i f we i n c r e a s e one, two or thr e e of the f a c t o r s ? (answer: i t increases) i i . .What happens t o the product i f we decrease one, two or thr e e of the f a c t o r s ? (answer: i t decreases) 159 i i i . What happens t o the product i f we i n c r e a s e one of the f a c t o r s and decrease the other? (answer: we can't t e l l ; i t may i n c r e a s e , decrease o r stay the same. Each example has to be examined i n d i v i d u a l l y ) i v . a. What happens t o the product i f we double o n l y one f a c t o r ? b. What happens to the product i f we double two f a c t o r s ? c. What happens t o the product i f we double a l l t h r e e f a c t o r s ? (answers: i t w i l l be m u l t i p l i e d by 2, 4 or 8 ) . v. What happens to the product i f we m u l t i p l y one f a c t o r by 6 and d i v i d e another by 2? (answer: i t w i l l be m u l t i p l i e d by 6 and d i v i d e d by 2. 4. (7 min) Worksheet: Give each o f the students a copy of the attached worksheet and e x p l a i n i t t o them. Warn the students t h a t they may not have enough time to f i n i s h the work and ask them to do as much a-s they can. Move around and help them t o complete i t . At the end of the p e r i o d c o l l e c t the worksheets. 160 Name:Last: ; , F i r s t : . Lesson B2-Worksheet Complete the f o l l o w i n g : 1. 16 X 24 = 384 1 1 1 decreases s t a y s ? the same I f 16 were r e p l a c e d by a s m a l l e r number and 24 stayed the same, what should happen to the product? 2- 24 X 15 = 360 I . I i i n c r e a s e s i n c r e a s e s ? I f 24 were r e p l a c e d by a l a r g e r number and 15 were r e p l a c e d by a l a r g e r number, what should happen to the product? 3. 35 X 28 = 980 * 1 4 becomes becomes 3 ? twice times as as b i g b i g I f 35 were r e p l a c e d by a number twice as b i g and 28 were r e p l a c e d by a number three times as b i g , what should happen t o the product? 161 4. 17 X 9 X 14 2142 4 1 1 I s t a y s decreases decreases ? the same I f 9 were r e p l a c e d by a s m a l l e r number, 14 were r e p l a c e d by a s m a l l e r number and 17 stayed the same, what should happen t o the product? 5. 3 X 36 X 13 1404 i I I i become becomes becomes ? twice o n e - t h i r d 3 times as b ig as b i g as b i g I f 3 were r e p l a c e d by a number twice as b i g , 13 were r e p l a c e d by a number three times as b i g and 36 were r e p l a c e d by a number o n e - t h i r d as b i g , what should happen to the product? 162 Treatment B Lesson 3 B e h a v i o r a l O b j e c t i v e s : a, b, c and P denote n a t u r a l numbers. 1. Given an a s s e r t i o n o f the form a X b = P where a and b are known and given a c o n d i t i o n a l statement t h a t P remains f i x e d while one f a c t o r i s r e p l a c e d by one of i t s m u l t i p l e s or d i v i s o r s , the students w i l l be able t o a n t i c i p a t e a s u i t a b l e replacement f o r b. 2., Given an a s s e r t i o n of the form a X b X c = P such t h a t a l l v a r i a b l e s are known and given a c o n d i t i o n a l statement t h a t P remains f i x e d while one or two o f the f a c t o r s are r e p l a c e d by t h e i r m u l t i p l e s or d i v i s o r s , the students w i l l be a b l e to a n t i c i p a t e s u i t a b l e replacement f o r the remaining f a c t o r or f a c t o r s . O u t l i n e : 1. Follow up of the worksheet from the p r e v i o u s l e s s o n . 2. E f f e c t on one of two f a c t o r s when the other f a c t o r i s changed and the product i s f i x e d 3. E f f e c t on two (or one) of the three f a c t o r s when one (or two) of these f a c t o r s i s (are) changed and the product i s f i x e d . 4. Worksheet M a t e r i a l s : (The a c t i v i t i e s of t h i s l e s s o n are mainly number manipulations and do not r e q u i r e p h y s i c a l m a t e r i a l s ) 163 A c t i v i t i e s ; 1. (4 min) Fellow up o f the worksheet from t h e ^ p r e v i o u s l e s s o n : Give each student h i s c o r r e c t e d worksheet from the p r e v i o u s l e s s o n and e x p l a i n how one can o b t a i n the c o r r e c t answers to #2 and #5. For #2 w r i t e "24 X 15 = 360" on the board and e x p l a i n t h a t i f 24 and 15 were r e p l a c e d by l a r g e r numbers the product would i n c r e a s e . I l l u s t r a t e by w r i t i n g "30 X 20 = 600" underneath "24 X 15 = 360." S i m i l a r l y f o r #5, write 3 X 36 X 13 = 1404 then 6 X 12 X 39 = ••„-.• E x p l a i n t h a t the product w i l l be m u l t i p l i e d by 2 X 3 and d i v i d e d by 3. Therefore the number to f i l l i n the blank w i l l be 2 X 1404 = 2808. 2. (8 min) E f f e c t on one of two f a c t o r s when the other  f a c t o r i s changed and the product i s . f i x e d ; Write on the board the statement " 9 X 6 = 54" and underneath i t w r i t e "18 X I ] = 54" and t e l l the students t h a t i n the examples on the board, the product 54 was f i x e d while the f a c t o r 9 was doubled to 18. Ask the students to suggest a numbers t h a t can go i n the i_ ] to make the statement t r u e (answer: 3). Continue by t e l l i n g the students t h a t 9 was m u l t i p l i e d by 2 to get 18 and ask them to p r e d i c t what happened to 6 to get 3. Help the students conclude t h a t 6 was d i v i d e d by 2 s i n c e 9 was m u l t i p l i e d by 2. S i m i l a r l y w r i t e on the board " 5 X 8 = 40" and underneath i t write "10 X ( ] = 40". Ask the students to guess how 10 was obtained from 5 ( m u l t i p l i e d by 2 or doubled) and t o p r e d i c t what would happen t o 8 i n order to to get the same product 40 ( d i v i d e i t by 2 or halve i t ) . Allow time f o r responses and r e p l a c e "£ J" by "4". Write on the board " 4 X 6 = 24" and ask the students to p r e d i c t what happens t o one of the f a c t o r s i f the product (24) i s f i x e d and the other f a c t o r i s m u l t i p l i e d by 2. Test the students* p r e d i c t i o n s using "8 X J. ] = 24" and "J, ] X 12 = 24". Lead the students t o conclude t h a t i f the product i s f i x e d and one of the f a c t o r s i s m u l t i p l i e d by 2 (doubled) the other should be d i v i d e d by 2 (halved). S i m i l a r l y , r e f e r to "3 X 12 = 36" and ask the students to p r e d i c t what happens to one of the f a c t o r s i f the product i s f i x e d and one of the f a c t o r s i s d i v i d e d by 3. T e s t the s t u d e n t s ' p r e d i c t i o n s using "1 X £ ] = 36" and "£ ] X 4 = 36". Lead the students t o conclude t h a t i f the product i s f i x e d and one of the f a c t o r s i s d i v i d e d by 3 the other f a c t o r should be m u l t i p l i e d by 3. 164 Write on the board " 9 X 8 = 72" and ask the students t o p r e d i c t what should happen to one of the f a c t o r s i f the product i s f i x e d and the other f a c t o r i s m u l t i p l i e d or d i v i d e d by a number. Allow time f o r responses and t e s t them u s i n g : 3 X £ ] = 72 (answer:3 X [24] = 72) 18 X [ J = 72 (answer: 18 X £ 4 ] = 72) £ ] X 72 = 72 (answer: £ (I J X 72 = 72) £ ] X 2 = 72 (answer: £ 36] X 2 = 72) Help the students g e n e r a l i z e t h a t i f the product i s f i x e d and one of the f a c t o r s i s m u l t i p l i e d (divided) by a number the other f a c t o r should be d i v i d e d ( m u l t i p l i e d ) by the same number. 3. (13 min) E f f e c t on..two (or one) of the three f a c t o r s when one (or two) of these f a c t o r s i s (are) changed m u l t i p l i c a t i v e l y and the product i s f i x e d . 3.1. M u l t i p l i c a t i v e increase, in, one of t h e ^ f a c t o r s and m u l t i p l i c a t i v e decrease i n another and v i c e versa (move q u i c k l y through t h i s s e c t i o n ) . Write of the board the statement "4 X 6 X 10 = 240" and underneath i t w r i t e "8 X £ ] X 10 = 240". T e l l the students t h a t the product 240 i s f i x e d and one of the f a c t o r s 10 i s a l s o f i x e d while another f a c t o r 4 was m u l t i p l i e d by 2 to become 8. Ask the students t o p r e d i c t what should happen to 6 i n order to f i n d the number (3) t h a t should go i n £ ] and make the statement t r u e . Test the st u d e n t s ' p r e d i c t i o n by r e p l a c i n g the sugqested number i n £ ] and complete the statement. Conclude t h a t 6 was d i v i d e d by 2. Refer to the statement "4 X 6 X 10 = 240" and ask the students to p r e d i c t what happens t o one of the f a c t o r s i f the product and one of the f a c t o r s are f i x e d while a f a c t o r i s m u l t i p l i e d by a number. Ask the students t o t e s t t h e i r p r e d i c t i o n s u s i n g : 12 X £ ] X 10 = 240 (answer: 12 X £ 2 ] X 10 = 240) £ ] X 6 X 20 = 240 (answer: £ 2 ] X 6 X 2 0 = 240) 4 X 1 X £ ] = 240 (answer: 4 X 1 X £60] - 240) Help the students to conclude t h a t i n t h i s case i f one f a c t o r i s m u l t i p l i e d by a number, the other should be d i v i d e d by the same number. S i m i l a r l y , ask the students t o p r e d i c t what happens t o one of the f a c t o r s i f another f a c t o r i s d i v i d e d by a number while the t h i r d f a c t o r and the product are f i x e d . Ask the students t o t e s t t h e i r p r e d i c t i o n s using 2 X £ ] X 10 = 240 (answer: 2 X £12] X 10 = 240) £ ] X 2 X 10 = 240 (answer: £12] X 2 X 10 = 240) 4 X 1 X £ ] = 240 (answer: 4 X 1 X £ 6 0 ] = 240) 165 3. 2 Increase i n two of the f a c t o r s and decrease i n one and v i c e versa B e f e r to "4 X 6 X 10 = 240" and t e l l the students t h a t the product 240 i s f i x e d while 6 i s m u l t i p l i e d by 2 and 10 i s m u l t i p l i e d by 2. , Write "£ J X 12 X 20 = 240" underneath "4 X 6 X 10 = 240" and ask the students to p r e d i c t what should happen t o the f a c t o r 4. Test the students' p r e d i c t i o n and l e a d the d i s c u s s i o n t o conclude t h a t s i n c e two f a c t o r s were m u l t i p l i e d by 2 each and the product i s f i x e d the t h i r d f a c t o r should be d i v i d e d by 2 X 2 or 4. S i m i l a r l y , ask the students to p r e d i c t what should happen to the f a c t o r 6 i f 10 was d i v i d e d by 5 and 4 by 2. Tes t the p r e d i c t i o n s and help the students to conclude t h a t the f a c t o r 6 should be m u l t i p l i e d by the product 5 X 2 (or 10). Test t h i s c o n c l u s i o n using 2 X [ ] X 2 = 240 (answer:2 X [ 6 0 ] X 2 = 240) Write "4 X 12 X 8 = 384" and t e l l the students t h a t the product 384 i s f i x e d and the f a c t o r 4 i s m u l t i p l i e d by 4 to get 16. Write "16 X [ ] X ( ) = 384" underneath "4 X 12 X 8 = 384." Ask the students to p r e d i c t what should happen t o 12 or to 8 or t o 12 and 8 i n order t o make the statement t r u e . T e s t the p r e d i c t i o n s and l e a d the students t o conclude t h a t s i n c e one of the f a c t o r s was m u l t i p l i e d by 4 any one of the f o l l o w i n g c o u l d be done: a. d i v i d e any other f a c t o r by 4 b. d i v i d e each of the f a c t o r s by 2. S i m i l a r l y , r e f e r t o "4 X 12 X 8 = 384" and t e l l the students t h a t the product 384 i s f i x e d and "the f a c t o r 12 i s d i v i d e d by 6 to get 2. Write ] X 2 X ( ) = 384" underneath "4 X 12 X 8 = 384." Ask the stu d e n t s t o p r e d i c t what should happen to 4 or to 8 or to 4 and 8 i n order t o make the statement t r u e . Test the p r e d i c t i o n s and lead the students to conclude t h a t s i n c e cne of the f a c t o r s was d i v i d e d by 6 any one of the f o l l o w i n g c o u l d be done a. m u l t i p l y any other f a c t o r by 6 b. m u l t i p l y one of the f a c t o r s by 2 and the other by 3. 4. (10 min) Worksheet Give each of the students a copy of the attached worksheet and e x p l a i n i t to them. Go around and help them t o complete i t . C o l l e c t the worksheets a t the end of t h e . p e r i o d . 166 Name: L a s t : , F i r s t : Lesson B3 - Worksheet Complete the f o l l o w i n g : 1. 8 X 5 = 40 £ J X 10 = 40 The product 40 stayed the same; 5 was r e p l a c e d by 10. What number should go i n the I ]? ___ 2. 9 X 15 = 135 27 X [ ] = 135 The product 135 stayed the same; 9 was re p l a c e d by 27. What number should go i n the £ J? 3. 34 X 12 X 9 = 3672 34 X £ ] X 36 = 3672 The product 3672 stayed the same; 34 stayed the same; 9 was r e p l a c e d by 36. What number should go i n the £ ]? 4. 11 X 15 X 22 = 3630 33 X £ ] X 22 = 3630 The product 3630 stayed the same; 22 stayed the same; 11 was r e p l a c e d by 33. What number should go i n the £ ]? 5. 26 X 45 = 1170 13 X £ j = 1170 I f the product 1170 s t a y s the same and 26 i s r e p l a c e d by 13, what number should r e p l a c e 45? _____ 6. 28 X 8 X 23 = 5152 £ ] X 32 X 23 = 5152 I f the product 5152 s t a y s the same, and 23 stay s the same and 8 i s r e p l a c e d by 32, what number should r e p l a c e 28? 7. 167 Treatment B Lesson 4 B e h a v i o r a l O b j e c t i v e s : 1. Given a cuboid or a diagram o f a cuboid, which i s non-p a r t i t i o n e d , p a r t i a l l y p a r t i t i o n e d or completely p a r t i t i o n e d i n t o u n i t cubes, the students w i l l be a b l e to use the volume al g o r i t h m "V = L X W X H" i n order to determine the volume 2. Given a diagram of a p a r t i t i o n e d cuboid which i s p a r t i a l l y covered the students w i l l be able to determine the t o t a l volume of the cuboid. 3. Given a diagram or a word d e s c r i p t i o n of a cuboid of known dimensions and given a proposed a d d i t i v e or m u l t i p l i c a t i v e dimensional t r a n s f o r m a t i o n the students w i l l be a b l e to s t a t e the volume of the cuboid t h a t would r e s u l t a f t e r the t r a n s f o r m a t i o n . O u t l i n e : 1. Fellow up of the worksheet from the p r e v i o u s l e s s o n . 2. C l a r i f i c a t i o n o f the concept of volume. 3. Algorithm f o r the volume of a cuboid "V = L X W X H." 4. A p p l i c a t i o n of the volume a l g o r i t h m t o p a r t i t i o n e d , p a r t i a l l y p a r t i t i o n e d and p a r t i a l l y covered cuboids.. 5. A p p l i c a t i o n of the volume a l g o r i t h m t o cuboids with proposed dimensional t r a n s f o r m a t i o n s . 6..Worksheet. M a t e r i a l s : 1. Cardboard boxes. 2. Decimetre cubes. 168 3. A p o s t e r of p o l y h e d r a l models b u i l t from u n i t cubes. 4. P o s t e r s of p a r t i t i o n e d , p a r t i a l l y p a r t i t i o n e d , non-p a r t i t i o n e d and p a r t i a l l y covered cuboids. A c t i v i t i e s : 1. (4 min) Fellow up of the worksheet from the p r e v i o u s  l e s s e n : Give each student h i s c o r r e c t e d worksheet from the p r e v i o u s l e s s o n and e x p l a i n how one can o b t a i n the answers to #5 and #7. For #5, e x p l a i n t h a t 26 was d i v i d e d by 2 i n order to get 13. T h e r e f o r e 45 should be m u l t i p l i e d by 2 i n order to get the same product, 1170. Check by w r i t i n g "13 X [ 9 0 ] = _" and computing the answer (1170). Write "1170" i n the blank. For #7, e x p l a i n that 18 was d i v i d e d by 3 to get 6, 32 was d i v i d e d by 2 to get 16. T h e r e f o r e 5 should be m u l t i p l i e d by 3 X 2 = 6 i n order to get the same product, 2880. Thus 5 should be m u l t i p l i e d by 30. Check by w r i t i n g 6 X [ 3 0 ] X 16 = ______ and computing the answer (2880). Write "2880" i n the blank. 2. (5 min) C l a r i f i c a t i o n of the concept of volume: D i s p l a y the twe c l o s e d cardboard boxes A and B of s i z e s 6 cm X 4 cm X 2 cm and 40 cm X 20 cm X 10 cm r e s p e c t i v e l y . Ask the s t u d e n t s to guess which i s b i g g e r , which occupies more space and which has the g r e a t e r volume. Conclude t h a t box B i s bigger than box A and t h a t any one of the f o l l o w i n g sentences d e s c r i b e s t h i s f a c t . a. Eox B i s bigger than box A b. Box B takes up more room than box A c. Eox B occupies more space than box A d. Box B has a l a r g e r volume than box A., D i s p l a y a c l o s e d cardboard box i n each of two d i s t a n t l o c a t i o n s of the classroom (use box #1 and box #2) and ask the students to compare the volumes of the boxes without moving them. Lead t h e d i s c u s s i o n i n order t o conclude t h a t p e r c e p t i o n may be d e c e i v i n g ; determine and compare the volumes u s i n g the f o l l o w i n g a c t i v i t i e s : a. Use decimetre cubes as u n i t s to b u i l d a cuboid next to each box with the same shape and volume as t h a t of the box. b. , Count the number of u n i t s and write the volumes of the boxes on the board. c. Compare the volumes of the boxes using the numbers of u n i t s found i n b. S t r e s s t h a t i n t h i s p r o c e s s , any u n i t s of the same s i z e 169 (i n t h i s case decimetre cubes) c o u l d be used, but the same u n i t s must be used throughout. 3. (5 min) A l g o r i t h m f o r the volume, of a cuboid _ _ _ _ _ _ _ _ J _ l ~ r ~ D i s p l a y a poster (#4.1) of a p a r t i t i o n e d cuboid whose dimensions are 6, 3, and 4. Remind the students t h a t the h e i g h t i s always the v e r t i c a l dimension, the l e n g t h i s the l o n g e r of the two h o r i z o n t a l dimensions and the width i s the s h o r t e r of the two h o r i z o n t a l dimensions. D i s c u s s with the students how one can determine the volume of the cuboid. Lead the a c t i v i t i e s i n order to determine: a. The number of cubes a l o n g the l e n g t h of the top l a y e r . Write on the hoard "L (length) = 6 " i . e., 6 cubes f i t along the l e n g t h of the top l a y e r . b. The number of cubes along the width of the top l a y e r . Write on the board "W (width) = 3 " c. Conclude t h a t the number of cubes t h a t can f i t i n the top l a y e r i s given by L X W. d. The number of cubes along the height of the box. Write on the board "H (height) = 4 " . T h i s i s the number of l a y e r s . e. Conclude t h a t the t o t a l volume i s the volume of one l a y e r (L X W) m u l t i p l i e d by the number of l a y e r s (H) i . e . , V = L X W X H. 4. (5 min) A p p l i c a t i o n of the volume, al g o r i t h m to the  f o l l o w i n g c a s e s : a. P a r t i a l l y p a r t i t i o n e d and, n o n - p a r t i t i o n e d , cuboids: Refer to the p a r t i a l l y p a r t i t i o n e d and the n o n - p a r t i t i o n e d cuboids on the p o s t e r (#4.2). In each case determine the volume of each l a y e r , the number of l a y e r s and the t o t a l volume. Ose the a l g o r i t h m "V = L X W X H" t o compute the volume. Compare the two answers., b. . P a r t i a l l y covered cuboids: Present the p o s t e r (#4.3) of a p a r t i a l l y covered cuboid (4X3X6). P o i n t out t h a t the b l o c k s of the top l a y e r are shown and there i s a t o t a l of 6 l a y e r s . Determine the l e n g t h , width and h e i g h t of the cuboid and use the a l g o r i t h m to compute the volume. Count the number of b l o c k s i n the top l a y e r (12) and conclude t h a t each o f the 6 l a y e r s has 12 b l o c k s . Determine the t o t a l volume (6x12). Compare the two answers. 170 5. (7 min) A p p l i c a t i o n of the, volume a l g o r i t h m t o cuboids with  proposed dimensional, t r a n s f o r m a t i o n s : Write on the board the statement 6 X 4 X 10 = 240. Review with the students how some of the changes i n the f a c t o r s (6,4,10) a f f e c t the product (240) by asking t h i s s e r i e s of qu e s t i o n s and encouraging the students t o make g e n e r a l i z a t i o n s . i^.What happens t o the product i f we i n c r e a s e one, two or thr e e of the f a c t o r s ? i i . What happens t o the product i f we decrease one, two or thr e e of the f a c t o r s ? i i i . a. What happens t o the product i f we double only one f a c t o r ? b. What happens t o the product i f we double two f a c t o r s ? c. What happens t o the product i f we double a l l three f a c t o r s ? i v . What happens t o the product i f we double one f a c t o r and halve another f a c t o r ? D i s p l a y a post e r (#4.4) of a p a r t i t i o n e d cuboid of dimensions 6, 4, and 10. Ask the students t o s t a t e the a l g o r i t h m f o r f i n d i n g the volume of the cuboid and write on the board V = L X W X H. Replace L, W, and H by 6, 4 and 10 r e s p e c t i v e l y and compute the volume (240)..Write on the board 6 X 4 X 10 = 240 underneath V = L X W X H. Ask the students t o p r e d i c t the changes i n the volume (240) when L (6), W (4), and H (10) vary. Help the stu d e n t s make the g e n e r a l i z a t i o n s by a s k i n g them t h i s s e r i e s o f question s . . i . What happens t o the volume i f we i n c r e a s e one, two or t h r e e of the dimensions? i i . What happens t o the.volume i f we decrease one, two or t h r e e of the dimensions? i i i . What happens to the volume i f we i n c r e a s e one of the dimensions and decrease another? i v . a. What happens to the volume i f we double only one dimension? b. What happens t o the volume i f we double two dimensions? c. What happens t o the volume i f we double a l l t h r e e dimensions? v. What happens t o the volume i f we double one dimension 171 and halve another dimension? 6. (9 min) Worksheet: Give each o f the students a copy of the attached worksheet and e x p l a i n i t to them. Go around and help them t o complete i t . C o l l e c t the worksheets a t the end of the p e r i o d . 172 Ecsters of Lessen E-t 173 3. A p i l e of cubes p a r t i a l l y covered. T o t a l volume = 4. i p l a s t i c .ox. I f we increased t i e width cf t h i s box by three u n i t s and the length and the height stayed the sane, then i n the new box what would be the Length? l i d t h ? . Height? Volute? 5. 1 wooden l e x . I f we decreased tfce length by 1 u n i t and we increased tfce height by 2 u n i t s but tfce width stayed the sane, then i n the new bcx what would be the l e n g t h ? l i d t h ? Height? Vcluae? 6. A r e c t a n g u l a r p i l e of cubes. I f we doubled tfce length and halved the widtfc but tfce height stayed the same, then i n the new p i l e what would be the length? l i d t h ? Height? Vcluae? 175 Treatment, C Lesson 1 B e h a v i o r a l O b j e c t i v e s : 1. Given a numeral i n Base 10 f o r a number n (n<10 s), the students w i l l be able to w r i t e the numeral i n expanded n o t a t i o n using e i t h e r of the forms i l l u s t r a t e d below: a. 324 = 3 hundreds • 2 tens • 4 ones b. 324 = 3 X 100 + 2 X 10 • 4 2. Given a numeral i n Base 10 f o r a number n (n<124) , the students w i l l be a b l e , with manipulative a i d s , t o convert the numeral to the e q u i v a l e n t numeral i n Base 5. O u t l i n e : 1. Review of Base 10 plac e value concepts 2. Bundling i n f i v e s and e x p r e s s i n g numbers i n Base 5 M a t e r i a l s : 1. P o p s i c l e s t i c k s 2. Twist t i e s 3. Base 10/Base 5 t a b l e 4. Ease 10 S Base 5 mats: a paper on one s i d e o f which t h e r e are three l a r g e columns with headings: t e n - t e n s , tens and ones; the other side has t h r e e l a r g e columns with headings: f i v e - f i v e s , f i v e s and ones. 176 A c t i v i t i e s : Give each student 45 p o p s i c l e s t i c k s , 5 t w i s t t i e s , a Base 10 and a Base 5 mat and a copy of the Base 10/Base 5 t a b l e i l l u s t r a t e d below. Ask the students t o l a y a s i d e the handouts, the s t i c k s and the t w i s t t i e s because they w i l l be used l a t e r i n the p e r i o d . | expanded form I short| | [form | f t e n - t e n s l t e n s l o n e s l , ,- - I I s h o r t , form | _ |f i v e - f i v e s expanded form f i v e s I onesj 1. (15 min) Review of Base 10 p l a c e value concepts: Write "111" on the board then ask the students t o read t h i s number. Give p u p i l s o p p o r t u n i t y to respond then ask about the r e l a t i o n s h i p of the "1" i n the tens place to the "1" i n the ones p l a c e and the "1" i n the hundreds p l a c e . Ose p o p s i c l e s t i c k s (1 s t i c k , 1 bundle of 10, and 1 bundle of 10 tens) i n order to i l l u s t r a t e t h a t * 1* i n the tens place means 10 times as much as the •1• i n the ones p l a c e ; the •1* i n the 100's p l a c e means 10 times as much as the '1* i n the 10«s p l a c e and 100 times as much as the '1' i n the ones p^ace. Conclude t h a t "111 = (100) + (10) +1". Write "444 = hundreds + . tens + ones" and then ask the c h i l d r e n to suggest numbers t h a t w i l l make the a s s e r t i o n t r u e . Review with the c h i l d r e n t h a t : 444 = 4 hundreds + 4 tens • 4 ones. 444 = (4 x 100) + (4 x 10) + 4. Repeat u s i n g "2795" i n order to conclude t h a t : 2795 = (2 x 1000) + (7 x 100) + (9 x 10) • 5. 2. (20 min) Bundling, i n ^ f i y e s and e x p r e s s i n g numbers i n  Base 5: T e l l the students t h a t a f t e r what they have j u s t done the f o l l o w i n g may seem to be r i d i c u l o u s l y easy. Give the s t u d e n t s the f o l l o w i n g i n s t r u c t i o n s and make sure t h a t each student completes the work. a. Count out 14 s t i c k s . b. Group the s t i c k s i n tens and ones. Use the t w i s t t i e s 177 to bundle each group of ten. c. P l a c e the bundles (1) and s t i c k s (4) i n the proper s e c t i o n s on the Base 10 mat. d. Enter i n the t a b l e on the l e f t what the bundles show, e. Enter i n the t a b l e on the l e f t the shor t form of what the bundles show. f . Unbundle the s t i c k s . g. Eegroup the s t i c k s i n f i v e s and ones u s i n g the t w i s t t i e s . h. Place the bundles (2) and the s t i c k s (4) i n the proper s e c t i o n s on the Base 5 mat. i . Enter i n the t a b l e on the r i g h t what the bundles show. j . Enter i n the t a b l e on the r i g h t the s h o r t form of what the bundles show. Eepeat the above s t e p s of i n s t r u c t i o n f o r " t h i r t y - t w o " and "twenty- s i x " ; i n step g say t h a t as soon as 5 bundles are made they should be bundled i n t o a l a r g e r bundle of f i v e - f i v e s . T e l l the students t h a t when they bundled i n f i v e s the w r i t t e n numbers were i n Base 5, j u s t as when they bundled i n tens the numbers were i n Base 10. Show the e q u i v a l e n c e of the numerals by using the numerals w r i t t e n on the t a b l e and w r i t i n g the f o l l o w i n g : 14 (Base 10) = 24 (Base 5) 22 (Base 10) = 42 (Base 5) 32 (Base 10) = 112 (Base 5) 26 (Base 10) = 101 (Base 5) 178 Treatment^ C Lesson 2 B e h a v i o r a l , O b j e c t i v e s : 1. Given a numeral i n Base 10 f o r a number n (n<124), the students w i l l be a b l e , with and without manipulative a i d s , to convert the numeral to the e q u i v a l e n t numeral i n Base 5. 2. Given a number no g r e a t e r than 124 (Base 10) which i s suggested by a given r e a l l i f e s i t u a t i o n , the st u d e n t s w i l l be a b l e to write the e q u i v a l e n t Base 5 numeral f o r t h a t given number. O u t l i n e : 1. Beview of s e c t i o n 2 of the previous l e s s o n 2. Counting i n Base 5 3. Converting numerals from Base 10 to Base 5 4. Worksheet M a t e r i a l s : 1. P o p s i c l e s t i c k s 2. Twist t i e s 3. Base 10/Base 5 t a b l e 4. Ease 10 S Base 5 mats: a paper on one s i d e of which t h e r e are three l a r g e columns with headings: t e n - t e n s , t e n s and ones; the other s i d e has three l a r g e columns with headings: f i v e - f i v e s , f i v e s and ones. A c t i v i t i e s : 1. (7 min) Beview of s e c t i o n 2 of the p r e v i o u s l e s s o n : Hold up 22 s t i c k s . Show the students t h a t 4 bundles of 5 179 s t i c k s each can be made and there w i l l be 2 s t i c k s l e f t over. Draw the Base 5 t a b l e on the board and write "4" under the " f i v e s " and "2" under the "ones". Then write on the board the statement "22 (Base 10) = 42 (Base 5)". 2. (8 min) Counting i n Base 5: Ask the students to enter "1", "2", "12" v e r t i c a l l y i n the l e f t s h o r t form s e c t i o n of the t a b l e . For each of these numerals write the e q u i v a l e n t Base 5 numeral. Emphasize t h a t i n Base 5, f i v e s must be grouped (bundled). Make a correspondence t h a t i n Base 5 t h e r e are ones, f i v e s , f i v e - f i v e s , ... j u s t as i n Base 10 there are ones, tens, ten^-tens, . . . A l s o p o i n t out th a t i n Base 5 the only d i g i t s needed are the f i v e d i g i t s "0, 1, 2, 3 and 4." Co n t r a s t with Base 10 i n which the ten d i g i t s "0, 1, 9" are needed. 3. (10 min) Converting numerals from Base 10 to Base 5: T e l l the students t o enter "38" i n the l e f t s h o r t form s e c t i o n of the t a b l e . Ask the students t o t h i n k of 38 s t i c k s and the number of bundles t h a t c o u l d be made. Ask i f 1 bundle of f i v e c o u l d be made (Yes), 2 bundles (Yes), 3 bundles (Yes), 4 bundles (Yes) and 5 bundles (Yes). Ask what we would do with 5 bundles of f i v e (Make 1 bundle o f f i v e - f i v e s o r twenty f i v e ) . Ask the students t o enter the number of bundles of twenty f i v e (1) i n the t a b l e . Continue by a s k i n g the stu d e n t s to c a l c u l a t e the number of s t i c k s t h a t would be l e f t over (13) , the number of bundles of f i v e t h a t c o u l d be made (2) and the number of s t i c k s l e f t over (3). Ask the students t o enter the numbers (2,3) r e p r e s e n t i n g the f i v e s and the ones i n the t a b l e and t o a l s o e n t e r the s h o r t form (123). Write on the board "38 (Base 10) = 123 (Base 5 ) . " Eepeat f o r "56" and conclude t h a t : 56 (Base 10) = 211 (Base 5) 4. (10 min) Worksheet: Give each of the students a copy of the attached worksheet and e x p l a i n i t to them. Go around and help them t o complete i t . At the end of the p e r i o d c o l l e c t the worksheets. 180 Name: L a s t : • : , F i r s t : . Lesson C2 - Worksheet Do each c a l c u l a t i o n and write your answer i n the space provided f o r i t . 1. 938 = 9 hundreds + t e n s + ones 2. 4207= ( x 1000) • ( ___ x 100) + (0 x 10) t 7 3. 279 = (2 x ) + (7 x ) + 5's | ones 4. 18 (Base 10) = = _____ (Base 5) I 25«s | 5»s | ones 5. 30 (Base 10) = — = • (Base 5) I I 25»s | 5«s | ones 6. 86 (Base 10) = = _______ (Base 5) I I 7. 57 (Base 10) = _________ (Base 5) 8. The Base 5 number f o r the number of days i n a week i s (Base 5) 9. The Base 5 number f o r the number of months i n a year i s (Base 5) 10. The Ease 5 number f o r the number of days i n January i s . (Base 5) 181 Treatment C Lesson 3 B e h a v i o r a l O b j e c t i v e s : 1. Given a numeral i n Base 5 f o r a number n £n<444(Base 5) J , the students w i l l be able t o convert the numeral to the e q u i v a l e n t numeral i n Base 10. O u t l i n e : 1. Follow up of the worksheet from the p r e v i o u s l e s s o n 2. Converting numerals from Base 5 t o Base 10 3. Worksheet M a t e r i a l s : (the a c t i v i t i e s of t h i s m a n ipulation and do not l e s s o n c o n s i s t mainly of r e q u i r e p h y s i c a l materials) number 182 A c t i v i t i e s : 1. (5 min) Fellow up of the worksheet from the, p r e v i o u s l e s s o n : Give each student h i s c o r r e c t e d worksheet from the prev i o u s l e s s o n and e x p l a i n how one can o b t a i n the answer to #6 and #10. For #6 e x p l a i n t h a t from 86 s t i c k s one can bundle 3 bundles of 25's, 2 bundles o f 5's and have 1 s t i c k l e f t . T h e refore "3", "2" and "1" should be w r i t t e n under "25«s", "5's" and "ones" and thus "321" should be w r i t t e n i n the space provided on the r i g h t . S i m i l a r l y , e x p l a i n t h a t i n #10 the 31 days i n January allow f o r w r i t i n g "1" under "25«s", "1" under "5«s" and "1" under "ones". The answer i s t h e r e f o r e 111 (Base 5). 2. (15 min) Converting numerals from Base 5 t o Base 10: Write the place values f o r Base 5 on the board as f o l l o w s : 1 I i I I Write a l s o "123 (Base 5)" on the board and ask the students what "123 (Base 5 ) " means (1 t w e n t y - f i v e , 2 f i v e s and 3 ones). Write "1", "2" and "3" i n the proper s e c t i o n s i n the place value t a b l e on the board. Then write on the board 123 (Base 5) = (1 X ) + (2 X ) + 3 and ask the students to s t a t e numbers t h a t w i l l make the a s s e r t i o n t r u e . Conclude t h a t 123 (Base 5) = (1 X 25) • ( 2 X 5 ) * 3 = 25 + 10 • 3 123 (Base 5) = 38 (Base 10). Likewise ask the students to s t a t e the us u a l way ( i n Base 10) of w r i t i n g the numbers i n the f o l l o w i n g e x e r c i s e s : 203 (Base 5) = (2 X ) + (0 X ) + 3 = (Base 10) 334 (Base 5) = (3 X ) + (3 X ) • 4 = - (Base 10) 3.,(15 min) Worksheet: Give each of the students a copy of the attached worksheet and e x p l a i n i t to them. Go around and help them t o complete i t . C o l l e c t the worksheets at the end of the p e r i o d . 25»s 183 Name: L a s t : ; F i r s t : Lesson C3 - Worksheet Do each c a l c u l a t i o n and write your answer i n the space provided f o r i t . 5»s | ones 1. 23 (Base 5) = = (Base 10) I 25*s | 5's | ones 2. 401 (Base 5) = = (Base 10) I I 3. 34 (Base 5) = (Base 10) 4. 332 (Base 5) = (Base 10) 5. 304 (Base 5) = . (Base 10) 6. 432 (Base 5) = (Base 10) 184 Treatment C Lesson 4 B e h a v i o r a l O b j e c t i v e s : 1. Given two numerals i n Base 5 which r e p r e s e n t two numbers the sum of which i s not g r e a t e r than 444(Base 5 ) , the students w i l l be able to c a l c u l a t e . the sum of the numbers and express i t by a numeral i n Base 5 without t r a n s l a t i n g the numerals i n t o Base 10. 2. Given two numbers i n Base 5, n e i t h e r of which i s g r e a t e r than 444(Base 5), the students w i l l be able to c a l c u l a t e the d i f f e r e n c e of the two numbers and express i t by a numeral i n Base 5 without t r a n s l a t i n g the numerals i n t o Ease 10. O u t l i n e : 1. Follow up of the worksheet from the p r e v i o u s l e s s o n 2. A d d i t i o n i n Base 5 with and without renaming 3. S u b t r a c t i o n i n Base 5 with and without renaming 4. Worksheet M a t e r i a l s ; 1. P o p s i c l e s t i c k s 2. Twist t i e s 3. Base 5 t a b l e 4. Ease 5 mat 185 A c t i v i t i e s : 1. (3 min) Follow up, of the worksheet from the p r e v i o u s  l e s s o n : Give each student h i s c o r r e c t e d worksheet from the previous l e s s o n and e x p l a i n how one can o b t a i n the c o r r e c t answer t o #6. E x p l a i n t h a t "4" means four 25's or 100, "3" means t h r e e f i v e s or 15 and "2" means two ones. Therefore "432" i n Base 5 i s e q u i v a l e n t to "117" i n Ease 10 (100 + 1 5 + 2 ) . 2. (12 min) A d d i t i o n i n Base 5 with and without renaming: 2.1. A d d i t i o n without renaming (using the s t i c k s ) : Give each student 40 p o p s i c l e s t i c k s , 5 t w i s t t i e s , a Base 5 mat and a copy of the Base 5 t a b l e i l l u s t r a t e d below. s h o r t | expanded form I form | j — - _ f i v e - f i v e s l f i v e s I ones I I I I I I I I I I I I I I I I I I I I I I I I I Hold up i n one hand one bundle of f i v e p o p s i c l e s t i c k s and 3 s t i c k s and ask the students t o use s t i c k s and t w i s t t i e s i n order t o i s o l a t e a s i m i l a r amount and p l a c e the bundle and the s t i c k s i n the proper p l a c e s on the mat. Ask the students to reco r d i n the Ease 5 t a b l e the expanded form of the numeral (13) r e p r e s e n t i n g a l l the s t i c k s on the mat. S i m i l a r l y , h o l d up i n the other hand two bundles of f i v e s t i c k s and one s t i c k . Ask the students to i s o l a t e a s i m i l a r amount, place the s t i c k s on the mat and record i n the same t a b l e underneath "13" the numeral (21) r e p r e s e n t i n g a l l the s t i c k s j u s t placed on the mat. Ask the students t o draw a h o r i z o n t a l l i n e underneath "21" and add the two numbers step by step as f o l l o w s : a. J o i n the s t i c k s (3 and 1) on the mat i n order t o get "4." Becord the number of s t i c k s (4) i n the proper plac e i n the t a b l e . b. J o i n the bundles (1 and 2) and r e c o r d the number' of bundles (3) i n the proper p l a c e i n the t a b l e . 2.2. A d d i t i o n with renaming (using the sticks),.,: 1 86 Repeat a c t i v i t i e s s i m i l a r to the ones d e s c r i b e d i n the above s e c t i o n (2.1) using 2 bundles and 4 s t i c k s (24) i n one hand and 1 bundle and 3 s t i c k s (13) i n the other. Remind the students t h a t i n Base 5 we have to bundle by f i v e s i f we can. Ask the students t o add 24 and 13 step by step as f o l l o w s : a. J o i n the s t i c k s (4 and 3) on the mat, make a bundle of f i v e s t i c k s and place i t with the bundles on the mat. Record i n the t a b l e the r e s u l t i n g number of s t i c k s ( 2). b. J o i n the bundles (1, 2 and 1) on the mat and r e c o r d i n the t a b l e the r e s u l t i n g number of bundles. S i m i l a r l y , c o n s i d e r 3 bundles of f i v e s and 2 s t i c k s (32) and 3 bundles of f i v e s and 4 s t i c k s (34). ask the students to place the s t i c k s on the mat, rec o r d the numerals and add step by step as f o l l o w s : a. J o i n the s t i c k s (2 and 4) on the mat, make a bundle of f i v e s t i c k s and place i t with the bundles on the mat. Record i n the t a b l e the r e s u l t i n g number of s t i c k s (1). , b. J o i n the bundles (1, 3 and 3) on the mat, bundle f i v e of them i n t o a bundle of f i v e - f i v e s and r e c o r d i n the t a b l e the r e s u l t i n g number of f i v e s (2). c. Record the number of bundles of f i v e - f i v e s (1) on the t a b l e . 2.3. a d d i t i o n without u s i n g t h e ^ s t i c k s ; Write on the board 31 (Base 5) + 2 (Base 5) (Base 5) Ask the students to copy the numerals to the s h o r t form s e c t i o n on the Base 5 t a b l e and add. Help the students t o c a l c u l a t e the number of ones and f i v e s r e s u l t i n g from a d d i t i o n without a c t u a l l y m a n i p u l a t i n g the s t i c k s . S i m i l a r l y , h e lp the students perform the a d d i t i o n without the s t i c k s f o r both of the f o l l o w i n g e x e r c i s e s : 23 (Base 5) 124 (Base 5) + 4 (Base 5) + 133 (Base 5) (Base 5) _____ (Base 5) 3 (12 min) S u b t r a c t i o n i n Base 5 with and without 187 renaming; 3.1. S u b t r a c t i o n without renaming (using the s t i c k s ) : Hold up th r e e bundles of f i v e s and two s t i c k s and ask the students to i s o l a t e a s i m i l a r amount, place the s t i c k s on the mat and r e c o r d i n the Base 5 t a b l e the expanded form of the numeral (32) r e p r e s e n t i n g a l l the s t i c k s . Ask the students to f o l l o w each o f the steps below i n order t o s u b t r a c t (take away, remove) 22 from the number represented i n the t a b l e . a. Record "22" i n the t a b l e underneath "32" and draw a h o r i z o n t a l l i n e underneath "22." b. Remove 2 s t i c k s and r e c o r d the number o f the r e s u l t i n g s t i c k s (0). c. Remove two bundles and record the number of the r e s u l t i n g bundles (1). 3.2. S u b t r a c t i o n with renaming (using the s t i c k s ) : Repeat a c t i v i t i e s s i m i l a r to the one d e s c r i b e d i n the above s e c t i o n (3.1) using 4 bundles, 2 s t i c k s and s u b t r a c t 2 bundles, 4 s t i c k s . Ask the students to c a r r y out the s u b t r a c t i o n (take away) by f o l l o w i n g the steps below: a. Place 4 bundles and two s t i c k s i n the proper columns on the mat. Record "42" i n the expanded form s e c t i o n o f the t a b l e . Write "24" underneath "42" and draw a h o r i z o n t a l l i n e underneath "24." b. In the ones column on the mat, "2 take away 4: can't do. " c. Regroup 1 bundle i n t o s t i c k s i n order to have s u f f i c i e n t amount of s t i c k s t o allow the "take away" o f 4. Pl a c e the s t i c k s i n the proper column on the mat. d. Remove 4 s t i c k s and rec o r d the r e s u l t i n g number (3) on the t a b l e . e. Remove 2 bundles and r e c o r d the r e s u l t i n g number (1) on the t a b l e . S i m i l a r l y , c o n s i d e r 1 bundle of f i v e - f i v e s , 1 bundle of f i v e s and 2 s t i c k s (112). Ask the students t o place the bundles and s t i c k s on the mat and rec o r d "112" i n the expanded form of the t a b l e . Ask the students to rec o r d "34" on the t a b l e underneath "112", draw a h o r i z o n t a l l i n e underneath "34" and s u b t r a c t (take away) by f o l l o w i n g the steps below: a. In the ones column on the mat, "2 take away 4: can't do. " 188 b. Regroup 1 bundle i n t o s t i c k s i n order to have s u f f i c i e n t amount of s t i c k s t o allow the "take away" of 4,. Pl a c e the s t i c k s i n the proper column on the mat. c. Remove 4 s t i c k s and r e c o r d the r e s u l t i n g number (3) on the t a b l e . d. In the f i v e s column on the mat, "0 take away 3: can't do. " e. Regroup the bundle of f i v e - f i v e s i n order to get s u f f i c i e n t amount of f i v e s to allow the "take away.'! P l a c e the f i v e s i n the proper column on the mat. f . Remove 3 f i v e s and r e c o r d the r e s u l t i n g number of f i v e s (2) on the t a b l e . 3.3. S u b t r a c t i o n without using the s t i c k s : Write on the board Ask the students to copy the numerals t o the s h o r t form s e c t i o n on the Base 5 t a b l e and s u b t r a c t . Help the students to c a l c u l a t e the number of ones and f i v e s r e s u l t i n g from s u b t r a c t i o n without a c t u a l l y manipulating the s t i c k s . S i m i l a r l y , h e lp the st u d e n t s perform the s u b t r a c t i o n without the s t i c k s f o r both o f the f o l l o w i n g e x e r c i s e s : 44 (Base 5) - 32 (Base 5) (Base 5) 23 (Base 5) 4 (Base 5) 6. 340 (Base 5) - 243 (Base 5) (Base 5) (Base 5) 4. (8 min) Worksheet: Give each of the students a copy of the attached worksheet and e x p l a i n i t to them. Go around and help them to complete i t . C o l l e c t the worksheets at the end of the pe r i o d . 189 Name: L a s t : . ._, F i r s t : Lesson C4 - Worksheet Do each c a l c u l a t i o n and write your answer i n the space p r o v i d e d f o r i t . 1. 21 (Base 5) 2. 34 (Base 5) 3. 213 (Base 5) + 4 (Base 5) + 3 (Base 5) + 144 (Base 5) ._,_(Base 5) .__(Base 5) (Base 5) 4. 34 (Base 5) 5. 23 (Base 5) 6. 340 (Base 5) - 32 (Base 5) - 4 (Base 5) - 143 (Base 5) (Base 5) (Base 5) __(Base 5) 7. 143 (Base 5) • 31 (Base 5) = (Base 5) 8. . 201 (Base 5) - 32 (Base 5) = (Base 5) 9. 244 (Ease 5) +'40 (Base 5) = .__ (Base 5) 10,. 300 (Base 5) - 223 (Base 5) = (Base 5) Appendix B ITEM STATISTICS 191 Table Bl Volume Achievement P r e t e s t Item S t a t i s t i c s Item Percent P o i n t - b i s e r i a l Item Percent P o i n t - b i s e r i a l Number C o r r e c t C o e f f i c i e n t Number C o r r e c t C o e f f i c i e n t 1 80.7 0. 35 15 4.1 0.47 2 73. 1 0.51 16 19.9 0.69 3 52.0 0.58 17 11.7 0.49 4 28. 1 0.66 18 35. 1 0.64 5 8.8 0. 49 19 18. 1 0.74 6 22.8 0.68 20 11. 1 0.74 7 51. 5 0.53 21 13.5 0.74 8 25. 1 0.72 22 19.3 0.68 9 13. 5 0.69 23 19.3 0.68 10 27.5 0.66 24 9.9 0.67 11 24.0 0.67 25 11.7 0.68 12 32.7 0.65 26 21.6 0. 58 13 8. 2 0.65 27 45.6 0.54 14 14.6 0.63 Table B2 Volume Conservation P r e t e s t Item S t a t i s t i c s Item Percent P o i n t - b i s e r i a l Item Percent P o i n t - b i s e r i a l Number C o r r e c t C o e f f i c i e n t Number C o r r e c t C o e f f i c i e n t 1 72.5 0.67 7 57.3 0.78 2 76.6 0.75 8 66. 1 0.76 3 35.7 0.52 - 9 56. 1 0.72 4 56. 1 0.62 10 56. 1 0.71 5 58.5 0.65 11 42. 1 0.46 6 53.8 0.73 192 Table B3 Volume Achievement P o s t t e s t Item S t a t i s t i c s Item Percent P o i n t - b i s e r i a l Item Percent P o i n t - b i s e r i a l Number C o r r e c t C o e f f i c i e n t Number C o r r e c t C o e f f i c i e n t 1 78.4 0.50 15 8.2 0.38 2 74. 9 0.61 16 42.1 0.77 3 62.6 0.66 17 17.0 0. 35 4 53. 2 0.72 18 58.5 0.79 5 27.9 0.63 19 39.8 0.65 6 41. 5 0.64 20 32.7 0.73 ' 7 60.8 0.60 21 36.3 0.66 8 57. 9 0. 76 22 40.9 0.67 9 48*5 0.73 23 33.9 0.62 10 54. 4 0. 73 24 32.2 0.66 11 4 3-3 0.70 25 30.4 0. 65 12 51.5 0.75 26 41.5 0.62 13 23.4 0.64 27 56.7 0.61 14 43.3 0.74 Table B4 Volume Conservation P o s t t e s t Item S t a t i s t i c s Item Percent P o i n t - b i s e r i a l Item Percent P o i n t - b i s e r i a l Number C o r r e c t C o e f f i c i e n t Number C o r r e c t C o e f f i c i e n t 1 86.0 0.65 7 63.7 0.72 2 87. 1 0.75 8 76.0 0.75 3 59.6 0.60 9 7 3. 1 0.72 4 66.7 0. 47 10 71.9 0.70 5 74.3 0.56 1 1 56.1 0.46 6 6 7.3 0.70 193 Table B5 M u l t i p l i c a t i o n P o s t t e s t Item S t a t i s t i c s Item Percent P o i n t - b i s e r i a l Item Percent P o i n t - b i s e r i a l Number C o r r e c t C o e f f i c i e n t Number C o r r e c t C o e f f i c i e n t 1 85.4 0-58 11 71.3 0.58 2 80.7 0.59 12 63.7 0. 58 3 79.5 0.47 13 33.9 0.53 4 49.7 0.60 14 22.8 0.37 5 50. 9 0.65 15 19.9 0.42 6 33-3 0.62 16 67.3 0. 59 7 8.2 0.34 17 71.3 0.58 8 43.3 0.56 18 23.4 0.46 9 36.6 0.67 19 22.8 0.42 10 84.8 0.58 20 23.2 •0. 36 Table B6 Volume Achievement Betention Test Item S t a t i s t i c s Item Percent P o i n t - b i s e r i a l Item Percent P o i n t - b i s e r i a l Number C o r r e c t C o e f f i c i e n t Number C o r r e c t C o e f f i c i e n t 1 87. 1 0.39 15 12.9 0.41 2 82.5 0.54 16 54.4 0.71 3 . 74.3 0.57 17 22.8 0.41 4 59.6 0.62 18 60.2 0.71 5 29.2 0.59 19 46.2 0.60 6 46.8 0.65 20 36.8 0.65 7 63. 2 0. 58 21 42. 1 0.70 8 59. 1 0.73 22 45.6 0.73 9 49. 1 0.78 23 40.4 0.70 10 59. 1 0.68 24 34.5 0. 68 11 45.6 0.70 25 37.4 0.69 12 56.1 0.67 26 38.6 0.51 13 24. 6 0.58 27 6 3. 2 0.55 14 46.2 0.69 194 Table B7 Volume Conservation R e t e n t i o n Test Item S t a t i s t i c s Item Number Percent C o r r e c t P o i n t - b i s e r i a l C o e f f i c i e n t Item Number Percent C o r r e c t P o i n t - b i s e r i a l C o e f f i c i e n t 1 90.1 0.54 7 71.9 0.77 2 91.2 0, 59 8 82.5 0.70 3 68.4 0.60 9 76.0 0.72 4 81.9 0.54 10 75. 4 0.60 5 79.5 0.60 11 59.1 0. 33 6 71.9 0.65 Table B8 M u l t i p l i c a t i o n R etention Test Item S t a t i s t i c s Item Percent P o i n t - b i s e r i a l Item Percent P o i n t - b i s e r i a l Number C o r r e c t C o e f f i c i e n t Number C o r r e c t C o e f f i c i e n t 1 93.0 0.50 11 77.8 0. 40 2 91. 2 0. 46 12 66.7 0. 44 3 80.1 0.35 13 28.7 0.35 4 47.4 0.56 14 19.3 0.35 5 56. 1 0.55 15 22.8 0. 35 6 30.4 0. 55 16 71.3 0.50 7 7.6 0.34 17 73.7 0.47 8 50. 9 0.55 18 22.2 0.50 9 36.6 0.53 19 26.3 0. 44 10 88.9 0.44 20 14.6 0.40 appendix C EAW DATa Codes: 1 = Nonconserver, 2 = P a r t i a l Conserver, 3 = Conserver Pre = P r e t e s t , Post = Post T e s t , Ret = Retention Test Vol = Volume, Mul = M u l t i p l i c a t i o n , &ch = Achievement SAT = S t a n f o r d Achievement T e s t , Tr = Treatment V = Volume, M = M u l t i p l i c a t i o n , C = C o n t r o l C o n s e r v a t i o n V o l Vol Mul V o l Mul Subject L e v e l Ach Ach Ach Ach Ach Number Pre Post Ret Tr SAT Pre Post Post Ret Ret 002 3 2 2 V 30 4 18 10 17 14 004 2 1 1 V 33 4 12 5 8 9 006 1 1 1 M 29 4 2 10 2 8 007 1 2 1 C 25 1 4 7 5 7 008 3 3 3 M 31 3 0 15 1 11 014 3 3 3 C 40 16 23 15 23 13 015 2 3 3 c 37 3 1 9 3 7 016 1 3 1 M 31 5 6 9 15 • 8 022 1 1 3 C 32 5 6 8 6 7 023 1 1 1 V 26 17 18 13 18 14 024 3 3 3 V 30 8 15 8 7 12 027 1 2 1 M 35 4 10 8 7 11 031 1 1 1 M 43 4 8 14 8 9 032 1 3 3 V 34 14 21 4 21 13 033 3 2 2 M 39 12 21 14 24 8 034 2 2 3 V 25 7 15 3 19 6 036 1 2 3 c 28 3 0 7 3 7 037 1 1 1 c 32 3 2 11 ii 13 03 8 3 1 2 V 38 1 12 10 12 8 039 2 2 2 M 42 11 17 14 16 15 041 2 1 2 c 31 6 11 6 12 10 042 1 1 1 V 21 3 10 8 9 8 043 3 3 3 M 39 11 12 12 11 11 044 1 1 1 c 31 3 2 4 5 10 045 1 1 3 M 21 6 3 6 4 5 047 3 3 3 V 29 4 18 11 21 10 051 3 3 3 c 35 20 25 9. 22 15 052 1 1 1 V 39 9 21 13 25 12 053 1 2 2 V 34 7 25 6 23 9 056 1 1 1 c 40 2 1 8 1 10 057 2 2 3 c 18 2 0 0 0 5 058 1 1 2 M 26 4 19 10 18 11 059 3 3 3 V 40 9 19 12 23 10 061 1 1 1 V 28 7 13 11 23 12 06 2 1 1 1 V 35 2 5 3 10 9 063 1 3 1 V 43 18 22 12 25 13 06 5 2 2 2 M 40 3 3 14 1 8 066 1 1 3 c 37 2 2 5 3 13 067 2 1 1 V 33 5 6 8 5 9 06 9 3 3 3 c 39 2 5 13 4 12 070 1 1 1 c 28 3 2 6 4 12 071 1 3 3 M 30 2 2 8 6 8 197 074 3 3 3 M 39 13 25 14 25 11 075 1 1 1 C 35 7 7 10 12 12 076 1 3 3 M 35 9 6 6 15 12 077 1 1 3 C 38 3 4 5 7 4 07 8 3 3 V 36 18 23 11 22 13 079 1 1 M 23 3 1 13 2 6 080 1 1 1 C 36 3 7 14 3 10 082 3 3 V 35 19 26 16 23 15 083 1 1 3 V 33 4 24 4 19 8 084 1 3 3 V 20 5 12 7 8 8 085 1 1 1 c 43 3 3 11 1 14 087 1 1 2 c 44 12 26 19 26 15 088 1 1 1 c 37 1 13 9 10 7 089 1 1 1 h 27 10 12 6 1f* 5 090 1 1 1 V 30 3 16 9 9 7 091 1 1 1 M 33 2 10 12 14 13 092 1 3 3 M 38 7 14 15 12 15 093 1 1 1 V 38 5 17 11 21 9 094 1 1 B 42 23 25 19 26 19 095 1 1 3 C 41 3 3 8 11 13 096 1 1 2 V 28 2 20 12 17 12 097 1 1 1 c 41 4 8 15 15 14 098 1 2 V 44 23 25 16 26 18 099 1 1 1 M 3 6 6 9 14 20 16 100 1 1 2 V 36 7 22 10 20 9 101 1 1 1 V 41 5 22 12 24 13 102 1 1 1 H 39 7 24 14 23 14 104 3 3 M 44 7 20 18 22 18 10 5 3 3 M 39 24 24 19 25 9 109 1 3 3 C 27 0 0 9 7 7 110 1 1 1 C 37 16 20 12 20 12 111 1 1 1 V 42 14 24 18 23 17 113 1 3 3 H 39 21 21 14 22 17 114 1 3 3 M 28 9 22 16 17 15 119 1 3 3 M 18 10 12 9 15 9 120 1 2 1 C 20 5 7 15 1 9 121 1 1 1 V 23 3 17 9 6 9 122 3 3 c 44 26 26 19 26 19 124 1 1 3 V 30 11 18 8 23 10 126 1 3 3 M 29 11 3 9 10 6 127 2 3 3 V 31 4 22 8 15 6 128 3 2 3 M 31 5 14 11 23 12 129 1 1 3 V 28 1 9 13 24 10 130 3 3 3 & 35 4 19 15 13 11 131 3 3 3 M 42 18 22 18 23 14 133 2 2 1 _ 31 8 3 10 10 11 136 3 3 3 V 35 18 16 4 20 12 137 1 3 3 c 38 10 14 15 15 9 140 3 3 3 M 39 20 24 18 23 13 142 3 3 3 C 34 14 14 12 17 12 143 3 3 3 M 35 13 18 15 23 16 145 2 2 2 M 36 4 1 11 2 14 148 3 3 3 M 45 8 23 18 6 15 150 3 1 3 C 28 6 11 11 7 8 198 152 3 3 3 v 36 15 17 9 20 9 •154 3 3 3 V 33 4 16 10 18 10 157 3 3 3 V 33 2 14 7 13 5 160 3 3 3 V 33 17 25 11 21 11 161 3 3 3 M 29 6 18 15 9 11 164 2 3 3 C 36 12 14 12 19 11 165 3 3 3 M 36 6 10 17 10 13 169 1 2 3 M 36 16 20 11 22 15 170 3 3 3 M 36 9 4 11 17 11 Number of s u b j e c t s = 105 Appendix D ONAJDSTED DESCRIPTIVE STATISTICS 200 Table D1 Unadjusted Means, Standard D e v i a t i o n s , and Group S i z e s of Volume Achievement P o s t t e s t Scores f o r Treatments by Conservation L e v e l s (Maximum Score = 27) Co n s e r v a t i o n Treatments L e v e l Volume M u l t i p l i c a t i o n C o n t r o l T o t a l Non-conservers 17.68 12 .28 6.55 11.75 (5.63) (7 .27) (6.87) (6.58) 19 18 20 57 P a r t i a l c o n s ervers 17. 17 8 .33 6.50 11. 19 (7.41) (10. 17) (7.05) (8. 36) 6 6 4 16 Conservers 17.82 16 .93 17.33 17.31 (4.29) (7 .54) (8.59) (6.62) 11 15 6 32 T o t a l 17.64 13 .00 8.70 13. 36 (5.52) (7 .82) (7.24) (6.86) 36 39 30 105 Regression c o e f f i c i e n t = 0.66 Table D2 Unadjusted Means, Standard D e v i a t i o n s , and Group S i z e s of Volume Achievement Retention Test Scores f o r Treatments by Cons e r v a t i o n L e v e l s (Maximum Score = 27) Conservation .Treatments L e v e l Volume M u l t i p l i c a t i o n C o n t r o l T o t a l Non-conservers 18.32 13 .56 7.81 13. 13 (6.51) (6 .34) (6.90) (6.59) 19 18 20 57 P a r t i a l - c o n s e r v e r s 15.83 9 -50 8.50 11.62 (8.13) (10.00) (8.66) (8. 96) 6 6 4 16 Conservers 17.73 17 .00 16.50 17. 16 (5.08) (7 .96) (9-0 5) (7.17) 11 15 6 32 T o t a l 17.72 14 .26 9.63 14. 12 (6.34) (7 .53) (7.56) (7. 13) 36 39 30 105 Regression c o e f f i c i e n t = 0.67 Appendix TESTS Volume Conservation Test Page 1 (on vhite) Page 2 (on pink) 203 204 Page 5 (on green) Page 6 (on pink) Before After Before After 206 Page 9 (on goldenrod) "1 . j Before After Page 10 (on green) s i p . ' • '- " * * ' ' ' I Before After 207 208 Name: L a s t : F i r s t : School: Volume Achievement Test Height= Length-Width = 210 A rectangular p i l e of cubes. Volume of the top layer = Number of layers = A rectangular p i l e of cubes. Volume of a layer = Total volume = A glass box with some cubes i n i t . Volume of the box = A cardboard box f u l l of cubes. Volume of the bottom layer = flumber of layers = A box i s 10 units long, 5 units wide and 2 units high. What i s the volume of the box? A rectangular block of wood. Volume = . 211 212 A rectangular p i l e of cubes p a r t i a l l y covered. Total volume = A p i l e of cubes. 9 layers are covered. Total volume = A p i l e of cubes. If we removed the top layer, what would be the volume of the part l e f t ? A p i l e of cubes. If we removed both of the shaded portions, what would be the. volume of the part l e f t ? A p l a s t i c box. I f we increased the width of t h i s box by 1 unit but the length and height stayed the same, what would be the volume of the new box? A metal box. If we doubled the length of t h i s box but the width and the height stayed the same, what would be the volume of the new box? 213 19. A rectangular p i l e of cubes. If we halved the length and doubled the width but the height stayed the same, what would be the volume of the new shape? 20. A rectangular p i l e of cubes. If we doubled the length, t r i p l e d the height and halved the width, what would be the volume of the new shape? 21. A wooden box. If we decreased the width by 1 unit and we increased the height by 2 units but the length stayed the same, what would be the volume of the new shape? 22. A metal box. If we decreased the width by 2 u n i t s , decreased the length by 6 units and increased the height by 8 units, what would be the volume of the new shape? 23. A b r i c k has a height of 5 units, a length of 6 units and a volume of 120 u n i t s . What i s i t s width? 24. A rectangular p i l e of cubes p a r t i a l l y covered. Total volume = 42 cubes. How many layers are there altogether? Naae: Last: , F i r s t : 5-lt^p_,j,cati 1on yest A. |n each of w,.the_follpwjnq w r i t e your answer i n the__ox__t_the r i g h t . 1. (23 x 17) x 36 = 17 x ([ ] x 23) Ihat nuaber should go i n the £ j? 2. 13 x (5 x 19) = (£ J x ) x 5 i h a t two numbers should go i n the [ ] and i n the ? 3. 6 X 6 = 9 X £ ] What number should go i n the £ J? 4. 15 X [ ] = 30 X 25 i h a t number should go i n the [ 3-5. 24 X 11 = £ 3 X 33 i h a t nuaber should go i n the £ J? 6. 12 X 15 X 22 = 36 X £ 3 X 22 ihat nuaber should go i n the £ }? 7. 41 X 50 X 7 = [ 3 X 5 X 35 What number should go i n the £ 3? 6. 18 X 24 = 432 I f 18 i s replaced by 9 and i f the product, 432, stays the same, then 24 Bust be replaced by what number? 9. 9 X 13 X 21 = 2457 I f 9 i s replaced oy 27 and i f the product, 2457, stays the same, then 21 aust be replaced by what nuaber? 215 B. JJL_each of. the f o l l o w i n g w r i t e the l e t t e r corrasppndinq-to the c o r r e c t answer i n _ t h e box, ,aj fehe r i g h t * 10- 18 X 27 = 486 I f 27 i s replaced by a l a r g e r number and i f 18 stays the saae, what happens to the product, 486? a. I t becomes s m a l l e r b. I t becomes l a r g e r c. I t stays the same 11- £ } X 14 ( 9 X 14 i h a t number (or numbers) could go i n the £ J? a. Any number greater than 9 b. Any number l e s s than 9 c. Any nuober greater than 14 d. Any number l e s s than 14 e. 9 12. £ 3 X 27 ^> 41 X 27 What number (or numbers) could go i n the [ ]? a. Any number l e s s than 27 b. Any number greater than 27 c. 27 d. Any number l e s s than 41 e. Any number greater than 41 13. 38 X 53 = 2014 I f 38 i s replaced by a number twice as b i g and i f 53 i s replaced by a number three times as b i g , what happens to the product, 2014? a. I t increases to f i v e times as much b. I t increases to s i x t i a e s as much c. I t decreases to one h a l f as much d. I t decreases t o one f i f t h as much e. I t changes but i t i s impossible to know how ouch i t increases or decreases. • 14. 72 X 35 = 2520 I f 72 i s replaced by a number h a l f as b i q and i f 35 i s replaced by a nuaber one f i f t h as b i q , what happens to the product, 2520? a. I t decreases to one seventh as much b. I t increases t o ten times as much c. I t decreases to one tenth as much d. I t in c r e a s e s to seven times as auch e. I t changes but i t i s i a p o s s i b l e to know how •uch i t increas e s or decreases. 15. 40 X 24 = 960 I f 40 i s replaced by a number one fourth as b i q and i f 24 i s replaced by a number twice as b i q , what happens t o the product, 960? a. I t increas e s to s i x times as auch b. I t decreases t o one s i x t h as much c. I t in c r e a s e s to twice as much d. I t decreases t o one h a l f as auch e. I t changes but i t i s i a p o s s i b l e to know how auch i t i n c r e a s e s or decreases. 16. 12 X 25 X 18 <^ 12 X [ J X 18 Vhat number (or numbers) could go i n the [ ]? a. any number l e s s than 25 b. Any nuaber greater than 25 c. 12 d. Any nuaber greater than 12 e. Any nuaber l e s s than 12 17. 14 X 9 1 1 3 = 1638 I f 9 i s replaced by a s m a l l e r nuaber and i f 13 i s replaced by a s m a l l e r number and i f 14 stays the saae, what happens to the product, 1638? a. I t decreases b. I t i n c r e a s e s c. I t stays the saae d. I t changes but i t i s impossible to know whether i t increas e s or decreases e. I t could increase or i t could decrease or i t could stay the saae 18. 21 X 18 X 25 = 9450 I f 18 i s replaced by a nuaber h a l f as big and i f 21 i s replaced by a number one t h i r d as b i g and i f 25 stays the same, what happens to the product, 9450? a. I t increas e s to s i x times as auch b. I t in c r e a s e s to f i v e times as much c. I t decreases to one f i f t h as much d. I t decreases to one s i x t h as much e. I t changes but i t i s impossible t o know hov much i t i n c r e a s e s or decreases. 19. 4 X 27 X 17 = 1836 I f 4 i s replaced by a number twice as big and i f 17 i s replaced by a number three times as b i g and i f 27 i s replaced by a number one t h i r d as b i g , what happens to the product, 1836? a. I t increas e s to three times as much b. I t decreases to one t h i r d as ouch c. I t increas e s to twice as much d. I t decreases to one h a l f as much e. I t changes but i t i s impossible to know how much i t increases or decreases. 20. 16 X 154 X 2 = 4928 I f 16 i s replaced by a number twice as biq and i f 2 i s replaced by a number seven times as b i q and i f the product, 4928, stays the same, vhat happens to 154? a. I t increases t o nine times as much b. I t in c r e a s e s to fourteen times as much c. I t decreases t o one ni n t h as much d. I t decreases to one fourteenth as much e. I t changes but i t i s impossible to know how auch i t increases or decreases. ***************** End of Test **************** 

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