Question

A student wants to see if the correct answers to multiple choice problems are evenly distributed. She heard a rumor that if you don't know the answer, you should always pick C. In a sample of 94 multiple-choice questions from prior tests and quizzes, the distribution of correct answers are given in the table below. In all of these questions, there were four options {A, B, C, D}.

	Correct Answers (n=94)n=94)
	A	B	C	D
Count	22	13	27	32

Test the claim that correct answers for all multiple-choice questions are not evenly distributed. Test this claim at the 0.05 significance level.

(a) Find the test statistic.

(b) Find the critical value.

(c) Is there sufficient data to support the claim?

Answer 1

Chi-Square Goodness of Fit test

Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H0: p1 =0.25, p2 =0.25, p3​=0.25, p4​=0.25 Ha​: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. Degrees of Freedom The number of degrees of freedom is df=n-1=4-1=3 (1) Test Statistics The Chi-Squared statistic is computed as follows: (2)Critical Value and Rejection Region Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df=n-1=4-1=3, so the critical value becomes 7.8147. Then the rejection region for this test is R={χ2:χ2>7.8147}. P-value The P-value is the probability that a chi-square statistic having 3 degrees of freedom is more extreme than 7.8147. The p-value is p=Pr(χ2>8.383)=0.0387 (3) The decision about the null hypothesis Since it is observed that χ2=8.383>χ2_crit​=7.8147, it is then concluded that the null hypothesis is rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis i.e. correct answers for all multiple-choice questions are not evenly distributed, at the α=0.05 significance level.