Question

A standardized test is given to a sixth grade class. Historically, the test scores have had a standard deviation of 21. The superintendent believes that the standard deviation of performance may have recently changed. She randomly sampled 30 students and found a mean of 160 with a standard deviation of 21.3846. Is there evidence that the standard deviation of test scores has increased at the α=0.025 level? Assume the population is normally distributed.

Step 1 of 5: State the hypotheses in terms of the standard deviation. Round the standard deviation to four decimal places when necessary.

Ho:

Ha:

Step 2 of 5: Determine the critical value(s) of the test statistic. If the test is two-tailed, separate the values with a comma. Round your answer to three decimal places.

Step 3 of 5: Determine the value of the test statistic. Round your answer to three decimal places.

Step 4 of 5: Make the decision.

Reject Null Hypothesis

Fail to Reject Null Hypothesis

Step 5 of 5: What is the conclusion?

There is sufficient evidence to support the superintendent's claim.

There is not sufficient evidence to support the superintendent's claim.

Answer 1

Solution:

Given: Historically, the test scores have had a standard deviation of 21. Thus

We have to test if there is evidence that the standard deviation of test scores has increased at the α=0.025 level.

Thus this is right tailed test.

Sample Size = n = 30

Sample Standard Deviation = 21.3846

Step 1 of 5: State the hypotheses in terms of the standard deviation.

Step 2 of 5: Determine the critical value(s) of the test statistic.

We use Chi-square distribution to find critical value.

Since this is right tailed test , look in Chi-square table for df = n - 1 = 30 - 1 = 29 and right tail area = 0.025 and find corresponding Chi-square critical value.

Chi-square critical value = 45.722

Step 3 of 5: Determine the value of the test statistic.

Step 4 of 5: Make the decision.

Reject null hypothesis H0, if Chi square test statistic > Chi-square critical value = 45.722, otherwise we fail to reject H0.

Since Chi square test statistic = < Chi-square critical value = 45.722, we fail to reject null hypothesis H0.

Thus correct option is:

Fail to Reject Null Hypothesis

Step 5 of 5: What is the conclusion?

There is not sufficient evidence to support the superintendent's claim.