THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Suppose two food preservatives are extensively tested and determined safe for use in meats.A processor wants to compare the preservatives for their effects on retarding spoilage.Suppose 16 cuts of fresh meat are treated with preservative A and another 12 cuts of meat are treated with preservative B.The number of hours until spoilage begins is recorded for each of the 28 cuts of meat.The results are summarized in the table below.

-The value of the test statistic for determining if there is a difference in the population variances for preservatives A and B is equal to:
A)0.784
B)1.625
C)1.275
D)1.129

Question

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:
Suppose two food preservatives are extensively tested and determined safe for use in meats.A processor wants to compare the preservatives for their effects on retarding spoilage.Suppose 16 cuts of fresh meat are treated with preservative A and another 12 cuts of meat are treated with preservative B.The number of hours until spoilage begins is recorded for each of the 28 cuts of meat.The results are summarized in the table below.
 
-The value of the test statistic for determining if there is a difference in the population variances for preservatives A and B is equal to:
A)0.784 
B)1.625 
C)1.275 
D)1.129

Quizplus · Accepted Answer

To determine if there is a difference in population variances, we use the F-test. The value of the test statistic for this problem is F = S^2A/S^2B = 11.25/6.93 = 1.625. We compare this value to the F-distribution with 15 degrees of freedom for the numerator (16-1) and 11 degrees of freedom for the denominator (12-1). At a significance level of 0.05, the critical value for the F-distribution with these degrees of freedom is 2.46. Since our calculated F-value (1.625) is smaller than the critical value (2.46), we fail to reject the null hypothesis that the population variances are equal. Therefore, the best choice is B.