Question

Suppose Sora is arguing with her mother, Jane. Jane believes that Sora eats mostly candy and doesn't eat enough fruits and vegetables, but Sora claims that she eats candy, fruits, and vegetables equally often. To test Sora's claim, Jane carefully monitors Sora's eating habits for several weeks. She records the number of times that Sora eats fruits, vegetables, and candy.

Jane conducts a chi-square test for goodness-of-fit to test Sora's claim. Jane's results are summarized in the table
Type of food Fruits Vegetables Candy
Observed 62 54 84
Test proportion 0.333 0.333 0.333
Expected 66.667 66.667 66.667
Contribution to chi-square 0.327 2.407 4.506
Chi-square statistic: 7.2400
Degrees of freedom: 2
What is the p-value for Jane's chi-square test for goodness-of-fit? Round your answer to three decimal places.

Answer 1

0.027

Explanation:

The p-value for Jane's chi-square test for goodness of fit is computed by using value of test statistic and degree of freedom.

Using Excel function CHISQ.DIST.RT(7.24,2) for computing p-value for chi-square goodness of fit test. 7.24 is the value of test statistic and 2 is degree of freedom. The resultant value is 0.026783. By rounding off to three decimal places the required p-value is 0.027.