5. A teacher wishes to compare two different groups of students with respect to their mean time to complete a standardized test. The time required is determined for each group. The data summary is given below. Test the claim at = 0.10, that there is no difference in variance. Give the critical region, test statistic value, and conclusion for the F test.

Question

n1 = 60 s1 = 11
n2 = 120 s2 = 17
= 0.10
· State the null and alternate hypotheses
· Determine which test statistic applies, and calculate it
· Determine the critical region
· State your decision: Should the null hypothesis be rejected?

Answer 1

Your data:

Sample 1
Sample size = 60
Variance = 121 (variance is standard deviation squared)
df = n - 1 = 59 (df = degrees of freedom)

Sample 2
Sample size = 120
Variance = 289
df = n - 1 = 119

Determine your test statistic from the appropriate formula using variance.

Critical value using an F-distribution at 0.10 level of significance with the above information is... (check an F-distribution table for this value).

If the test statistic exceeds the critical value from the table, the null will be rejected in favor of the alternative hypothesis. (The null would state that the ratio of the two variances is less than or equal to 1, and the alternative hypothesis would state that the ratio of the two variances is greater than 1.) If the test statistic does not exceed the critical value from the table, the null will not be rejected.

I hope this helps.