Question

after a product recall, a company solicits feedback from a random sample of 100 customers regarding their opinion about the company since the recall. customers were asked, \do you approve, disapprove, or have no opinion about the company?\ the public relations department is concerned that the opinions may be equally distributed across these three options. the survey reveals that 38 customers approve, 45 disapprove, and 17 have no opinion. they decide to carry out a test of the following hypotheses: ( h_{0} ): customer opinion is equally distributed across the three options. ( h_{a} ): customer opinion is not equally distributed across the three options. the chi - square test statistic is ( chi^{2}=7.77 ) and the ( p ) - value is between 0.02 and 0.025. because the ( p ) - value is less than 0.05 the public relations department rejects the null hypothesis. because the null hypothesis is rejected, they would like to know which term contributed the significance of the test. complete a follow - up analysis. which term contributed the most to the significance of this test? were there more or fewer people with this opinion than expected? approve, more than expected approve, fewer than expected disapprove, more than expected disapprove, fewer than expected

Explanation:

Step1: Calculate expected frequency

Under \(H_0\), since there are \(n = 100\) customers and \(k=3\) categories, the expected frequency \(E=\frac{n}{k}=\frac{100}{3}\approx33.33\) for each category.

Step2: Calculate \((O - E)^2/E\) for each category

• For "approve": \(O = 38\), \(\frac{(38 - 33.33)^2}{33.33}=\frac{(4.67)^2}{33.33}=\frac{21.8089}{33.33}\approx0.654\)
• For "disapprove": \(O = 45\), \(\frac{(45 - 33.33)^2}{33.33}=\frac{(11.67)^2}{33.33}=\frac{136.1889}{33.33}\approx4.086\)
• For "no opinion": \(O = 17\), \(\frac{(17 - 33.33)^2}{33.33}=\frac{(- 16.33)^2}{33.33}=\frac{266.6689}{33.33}\approx7.99\)

Answer:

disapprove, more than expected