Affective determinants of mathematical problem posing: the case of Chinese Miao students

Abstract

Students’ affective characteristics have been confirmed to shape their mathematics learning outcomes, including problem-solving performance and mathematics achievement. However, it remains unclear whether affect influences student mathematical problem posing - a process closely related to mathematical problem solving. Drawn from the expectancy-value theory (EVT), this study examined the relationship between students’ affective factors (self-concept, intrinsic value, and test anxiety) and their mathematical problem posing performance (complexity, quantity, and accuracy). Structural equation models were employed to analyze the data of 302 Chinese Miao students. The results showed that self-concept had a positive association with the complexity and accuracy of the problems posed. Intrinsic value was positively related to the complexity and quantity of the problems posed. Conversely, test anxiety negatively predicted the complexity of the problems. Our findings provide quantitative evidence of the significant influence of students’ affective characteristics on their problem posing performance and offer a better understanding of the problem-posing ability of Chinese minority students. Moreover, this study validates an instrument measuring student affect in mathematical problem posing based on EVT.

Appendices

Appendix 1

Table 2 The example responses of the three problem-posing tasks

Table 3 The items in mathematical problem-posing affect scales and factor loadings

Appendix 2. Examples of coding process

1.1 Example 1:

As shown in Fig. 4, the Point A, B, and C are three points on ⨀O.

Question (a): Prove: ∠AOB = 2∠ACB.

Question (b): A cow is tied to a tree on a grassland with a two-meter-long rope. How large area of grass can the cow eat?

Question (a): The hypothesized proving steps are presented as follows:

Point D is the intersection point of ⨀O and line OC.

Reasoning steps:

Step 1: ∵ A, B and C are three points on ⨀O.

∴ AO = OC = OB.

Step 2: ∴ ∠OAC = ∠OCA; ∠OBC = ∠OCB.

Step 3: and ∵ ∠AOD = ∠OAC + ∠OCA; ∠BOD = ∠OBC + ∠BCO.

∴ ∠AOB = ∠AOD +∠BOD = (∠OAC + ∠OCA) + (∠OBC + ∠BCO) = 2∠OCA + 2∠OCB = 2(∠OCA + ∠OCB) = 2∠ACB.

Question (b):

Reasoning step: we consider abstracting “grassland problem” (real situation) into “area of a circle” (mathematical model) as one reasoning step.

Computational step: The area of grass = The area of a circle with a radius of 2 meters = π ∙ r² = 4π m².

In this example, student posed two solvable questions. Thus, quantity and accuracy of this task were coded as two and 100%, respectively. Three and two steps are needed to solve Question (a) and (b), respectively; thus, based on the most complex problem student posed in this task (i.e., Question a), the complexity was coded as three.

1.2 Example 2:

As shown in Fig. 5, ⨀O is the inscribed circle of △ABC, Point E and F are the tangent points of ⨀O on line AC and AB, respectively.

(a) Prove: △ AEO ≅ △ AFO.

(b) Find the area of △ AEO.

Question (a): The hypothesized proving steps are presented as follows:

Step1: ∵ ⨀O is the inscribed circle of △ ABC, O is the center of ⨀O.

∴ AO is the angle bisector of ∠BAC (∠EAO = ∠OAF).

Step 2: ∵ Point E and F are the tangent points of ⨀O on sides AC and AB of △ABC.

∴ OE ⊥ AC; OF ⊥ AB (∠AEO = ∠AFO = 90°).

Step3: and ∵ Rt △ AEO and Rt △ AFO have the common hypotenuse AO.

∴ △ AEO ≅ △ AFO.

Question (b): this is an unsolvable problem as there is no sufficient information to calculate the area of the triangle.

In this example, student posed two questions, only one of them is solvable. Thus, quantity and accuracy of this task were coded as one and 50%, respectively. Three reasoning steps are needed to prove the statement in Question (a), thus the complexity was coded as three.

Example 3:

As shown in Fig. 6, there are four right triangles in square ABCD. Point E, F, G, H are the right-angled vertices of the four right triangles and locate on AH, BE, CF and DG, respectively.

Question: Given AB = 2 cm, ∠ABE = 30°, find the area of △ADH.

Reasoning steps:

Step1: ∵ Point E and H are the right-angled vertices of △ABE and △AHD, respectively. Quadrilateral ABCD is square. AB = 2 cm.

∴ ∠BAD = ∠AEB = ∠AHD = 90°; AD = AB = 2 cm.

Step 2: and ∵ ∠ABE = 30°.

∴ ∠DAH = ∠BAD – ∠BAE = ∠BAD – (180° – ∠AEB – ∠ABE) = 30°.

Computational steps:

Step 1: $DH=\mathit{\sin}\angle DAH\times AD=1\ cm; AH=\mathit{\cos}\angle DAH\times AD=\sqrt{3}\; cm.$

Step 2: ${S}_{\varDelta ADH}=\frac{1}{2}\ AH\times DH=\frac{\sqrt{3}}{2}\ {\mathrm{cm}}^2.$

In this example, student posed one solvable question. Thus, quantity and accuracy of this task were coded as one and 100%, respectively. Two reasoning steps and two computational steps are needed to solve this problem; thus, the complexity was coded as four.