Question

A particular paperback book is published in a choice of four different covers. A certain bookstore keeps copies of each cover on its racks. To test the hypothesis that sales are equally divided among the four choices, a random sample of 100 purchases is identified. a. If the resulting \(X^{2}\) value were \(6.4\), what conclusion would you reach when using a test with significance level .05? b. What conclusion would be appropriate at significance level \(.01\) if \(X^{2}=15.3 ?\) c. If there were six different covers rather than just four, what would you conclude if \(X^{2}=13.7\) and a test with \(\alpha=.05\) was used?

Short answer

a. At \(\alpha=.05\) significance level, we would not reject the null hypothesis that sales are equally divided among the four covers when \(X^{2}=6.4\). b. At \(\alpha=.01\) significance level, we would reject the null hypothesis meaning that sales are not likely equally divided among the four covers when \(X^{2}=15.3\). c. For six covers and at \(\alpha=.05\) significance level, we would reject the null hypothesis meaning that sales are not likely equally divided among the six covers when \(X^{2}=13.7\).

Step-by-step solution

Understanding chi-square test and significance level

The chi-square test is a statistical test which tells us if there's a significant difference between observed frequency and expected frequency. Here, we are using this test to analyze if the sales are equally divided among different covers. \nThe significance level indicates the probability of rejecting the null hypothesis if it is true. Small significance level indicates strong evidence against the null hypothesis.

Interpret the results for \(X^{2}=6.4\) with \(\alpha=.05\)

We need to compare the given chi-square statistic to the chi-square distribution critical value for 3 degrees of freedom (df = number of covers - 1 = 4 - 1 = 3) and a \(\alpha = .05\) significance level. Checking a chi-squared distribution table, we find our critical value to be approximately 7.81. Since \(6.4 < 7.81\), we do not reject the null hypothesis at this level. Thus, it seems sales are equally divided among the four covers.

Interpret the results for \(X^{2}=15.3\) with \(\alpha=.01\)

Similarly, we need to compare the given chi-square statistic to the chi-square distribution critical value for 3 degrees of freedom and a \(\alpha = .01\) significance level. Checking a chi-squared distribution table, we find our critical value to be approximately 11.34. Since \(15.3 > 11.34\), we reject the null hypothesis at this level. This means that it does not seem that sales are equally divided among the four covers.

Interpret the results for \(X^{2}=13.7\) with \(\alpha = .05\) for six covers

This time we need to compare the chi-square statistic to the chi-square distribution critical value for 5 degrees of freedom (df = number of covers -1 = 6 - 1 = 5) and a significance level of \(\alpha = .05\). Checking a chi-squared distribution table, we find our critical value to be approximately 11.07. Since \(13.7 > 11.07\), we reject the null hypothesis at this level. This means that it does not seem that sales are equally divided among the six covers.