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Adapting the academic motivation scale for use in pre-tertiary mathematics classrooms

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Abstract

The Academic Motivation Scale (ams) is a comprehensive and widely used instrument for assessing motivation based on the self-determination theory. Currently, no such comprehensive instrument exists to assess the different domains of motivation (stipulated by the self-determination theory) in mathematics education at the pre-tertiary level (grades 11 and 12) in Asia. This study adapted the ams for this use and assessed the properties of the adapted instrument with 1610 students from Singapore. Exploratory and confirmatory factor analyses indicated a five-factor structure for the modified instrument (the three original ams intrinsic subscales collapsed into a single factor). Additionally, the modified instrument exhibited good internal consistency (mean α = .88), and satisfactory test-retest reliability over a 1-month interval (mean r xx = .73). The validity of the modified ams was further demonstrated through correlational analyses among scores on its subscales, and with scores on other instruments measuring mathematics attitudes, anxiety and achievement.

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Appendices

Appendix 1

Marking scheme for sample item in achievement test

Method 1 for (i)

\( \begin{array}{l}x=2 \sin \theta \hfill \\ {}\frac{\mathrm{d}x}{\mathrm{d}\theta }=2 \cos \theta \hfill \end{array} \)

B1 – differentiate substitution correctly

\( \begin{array}{l}x=2 \sin \theta =\sqrt{3}\Rightarrow \theta ={ \sin}^{-1}\frac{\sqrt{3}}{2}=\frac{\uppi}{3}\hfill \\ {}x=2 \sin \theta =1\Rightarrow \theta ={ \sin}^{-1}\frac{1}{2}=\frac{\uppi}{6}\hfill \end{array} \)

B1 – correct limits

B1

\( \begin{array}{l}{\displaystyle {\int}_1^{\sqrt{3}}\sqrt{4-{x}^2}\mathrm{d}x}={\displaystyle {\int}_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\sqrt{4-4{ \sin}^2\theta }\ 2 \cos \theta\ \mathrm{d}}\theta \hfill \\ {}={\displaystyle {\int}_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}2\sqrt{1-{ \sin}^2\theta }\ 2 \cos \theta\ \mathrm{d}}\theta \hfill \\ {}={\displaystyle {\int}_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}4{ \cos}^2\theta\ \mathrm{d}}\theta \hfill \\ {}={\displaystyle {\int}_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\ 2\left( \cos 2\theta +1\right)\ \mathrm{d}}\theta \hfill \\ {}={\left[ \sin 2\theta +2\theta \right]}_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\hfill \end{array} \)

M1– substituting for x and dx

A1 – k cos2 θ

B1 – 2 cos2 θ = cos 2θ + 1

B1– integrate cos2 θ correctly to obtain \( \frac{ \sin 2\theta }{2}+\theta \)

\( =\left(\frac{\sqrt{3}}{2}+\frac{2\uppi}{3}\right)-\left(\frac{\sqrt{3}}{2}+\frac{\uppi}{3}\right)=\frac{\uppi}{3} \)

A1 – correct answer (AG)

Method 2 for (i)

\( \begin{array}{l}x=2 \sin \theta \hfill \\ {}\frac{\mathrm{d}x}{\mathrm{d}\theta }=2 \cos \theta \hfill \end{array} \)

B1 – differentiate substitution correctly

\( \begin{array}{l}{\displaystyle \int \sqrt{4-{x}^2}\mathrm{d}x}={\displaystyle \int \sqrt{4-4{ \sin}^2\theta }\ 2 \cos \theta\ \mathrm{d}}\theta \hfill \\ {}={\displaystyle \int 2\sqrt{1-{ \sin}^2\theta }\ 2 \cos \theta\ \mathrm{d}}\theta \hfill \\ {}={\displaystyle \int 4{ \cos}^2\theta\ \mathrm{d}}\theta \hfill \\ {}={\displaystyle \int\ 2\left( \cos 2\theta +1\right)\ \mathrm{d}}\theta \hfill \\ {}= \sin 2\theta +2\theta \hfill \end{array} \)

M1

A1 – k cos2 θ

B1 – 2 cos2 θ = cos 2θ + 1

B1– integrate cos2 θ correctly to obtain \( \frac{ \sin 2\theta }{2}+\theta \)

\( \begin{array}{c}\hfill \begin{array}{l}=2 \sin \theta \cos \theta +2\theta \hfill \\ {}=2\left(\frac{x}{2}\right)\sqrt{1-{\left(\frac{x}{2}\right)}^2}+2{ \sin}^{-1}\left(\frac{x}{2}\right)\hfill \end{array}\hfill \\ {}\hfill \hfill \end{array} \)

M1 – convert back from θ to x correctly

A1 – \( 2\left(\frac{x}{2}\right)\sqrt{1-{\left(\frac{x}{2}\right)}^2}+2{ \sin}^{-1}\left(\frac{x}{2}\right) \)

Hence,

\( \begin{array}{l}{\displaystyle {\int}_1^{\sqrt{3}}\sqrt{4-{x}^2}\mathrm{d}x}={\left[2\left(\frac{x}{2}\right)\sqrt{1-{\left(\frac{x}{2}\right)}^2}+2{ \sin}^{-1}\left(\frac{x}{2}\right)\right]}_1^{\sqrt{3}}\hfill \\ {}=\left(\frac{\sqrt{3}}{2}+\frac{2\uppi}{3}\right)-\left(\frac{\sqrt{3}}{2}+\frac{2\uppi}{6}\right)=\frac{\uppi}{3}\hfill \end{array} \)

A1 – correct answer (AG)

(ii) When x = 1, (1)2 + y 2 = 4

\( \Rightarrow y=\sqrt{3} \) (since y is positive)

When \( x=1,y=\frac{\sqrt{3}}{1}=\sqrt{3} \)

When \( x=\sqrt{3},{\left(\sqrt{3}\right)}^2+{y}^2=4 \)

⇒ y = 1 (since y is positive)

When \( x=\sqrt{3},y=\frac{\sqrt{3}}{\sqrt{3}}=1 \)

Thus, two curves intersect at the points \( \left(1,\sqrt{3}\right) \) and \( \left(\sqrt{3},1\right) \).(Verified)

B1 – able to verify one of the points correctly

B1 – verify other point correctly

\( \mathrm{Area}={\displaystyle {\int}_1^{\sqrt{3}}\sqrt{4-{x}^2}\mathrm{d}x}-{\displaystyle {\int}_1^{\sqrt{3}}\frac{\sqrt{3}}{x}\mathrm{d}x} \)

B3 – correct expressions, correct limits

B2 – correct expressions, incorrect limits

\( =\frac{\uppi}{3}-{\left[\sqrt{3} \ln x\right]}_1^{\sqrt{3}} \)

√ M1– use (i)

B1 – \( \sqrt{3} \ln x \)

\( \begin{array}{l}=\frac{\uppi}{3}-\left(\sqrt{3} \ln \sqrt{3}-0\right)\hfill \\ {}=\frac{\uppi}{3}-\sqrt{3} \ln \sqrt{3}\hfill \end{array} \)

√A1 – \( \sqrt{3} \ln \sqrt{3} \) answer from previous part

A1 – correct answer

(iii) Volume = \( \pi {\displaystyle {\int}_1^{\sqrt{3}}\left(4-{y}^2-\frac{3}{y^2}\right)}\ \mathrm{d}y \)

B1 – correct expression ∫(4 − y 2) dy

B1 – correct expression \( {\displaystyle \int \frac{3}{y^2}}\ \mathrm{d}y \)

B1 – correct limits

B1 – \( {\displaystyle {\int}_1^{\sqrt{3}}\left(4-{y}^2-\frac{3}{y^2}\right)}\ \mathrm{d}y \)

= 0.262 π or 0.822

B2 – 0.262 π

[B1 if π omitted]

Notes. M = mark allocated for a correct method applied to appropriate numbers, A = mark allocated for accuracy and depends on M marks; M0 A1 is not possible, B = independent accuracy marks; A fully correct final answer may receive full marks without the need to check for method.

Appendix 2

Academic motivation scale

Directions:

In the answer sheet and using the scale below, shade the number that best indicates the extent to which each of the following items presently corresponds to one of the reasons why you spend time studying mathematics.

Why do you spend time studying mathematics?

Does not Correspond at all

Corresponds a little

Corresponds moderately

Corresponds a lot

Corresponds exactly

1

2

3

4

5

1. Honestly, I don’t know; I feel that it is a waste of time studying mathematics.

2. Because I want to show to others (e.g., teachers, family, friends) that I can do mathematics.

3. Because I want to show myself that I can do well in mathematics.

4. I am not sure; I don’t see how mathematics is of value to me.

5. Because without a good grade in mathematics, I will not be able to find a high-paying job later on.

6. Because I believe that mathematics will improve my work competence.

7. For the pleasure I experience when I discover new things in mathematics that I have never learnt before.

8. In order to obtain a more prestigious job later on.

9. To show myself that I am an intelligent person.

10. In order to have a better salary later on.

11. For the pleasure that I experience when I feel completely absorbed by what mathematicians have come up with.

12. Because what I learn in mathematics now will be useful for the course of my choice in university.

13. Because I want to feel the personal satisfaction of understanding mathematics.

14. Because studying mathematics will be useful for me in the future.

15. For the pleasure that I experience in broadening my knowledge about mathematics.

16. Because I want to have “the good life” later on.

17. I don’t know; I can’t understand what I am doing in mathematics.

18. Because I think that mathematics will help me better prepare for my future career.

19. I can’t see why I study mathematics and frankly, I couldn’t care less.

20. Because of the fact that when I do well in mathematics, I feel important.

21. For the pleasure that I experience when I learn how things in life work, because of mathematics.

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Lim, S.Y., Chapman, E. Adapting the academic motivation scale for use in pre-tertiary mathematics classrooms. Math Ed Res J 27, 331–357 (2015). https://doi.org/10.1007/s13394-014-0140-9

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