Posted by **cindy** on Thursday, March 15, 2012 at 8:44pm.

10.51 A statistics professor compared the scores of her morning class and afternoon class on the same

quiz. The 42 morning students’ average score was 83.52 with a standard deviation of 7.81, while

the 46 afternoon students’ average score was 79.44 with a standard deviation of 6.62. (a) Which

test would you use to compare the means? (b) State the hypotheses to see if there was a difference.

(c) Obtain a test statistic and p-value. (d) Why might there be a difference between morning and

afternoon scores? Explain. (e) Is it reasonable to assume equal variances? (f) Carry out a formal

test for equal variances at á = .05, showing all steps clearly.

could you please help explain the steps to this problem

- statistics -
**MathGuru**, Friday, March 16, 2012 at 6:45pm
Try a two-sample test.

Hypothesis:

Ho: µ1 = µ2

Ha: µ1 does not equal µ2

This is a two-tailed test because the problem is just asking if there is a difference (no specific direction in the alternate hypothesis).

P-value is the actual level of the test statistic (using a z or t-table, depending on the test you use).

A test involving two variances (standard deviation is the square root of the variance) would include these hypotheses: Ho would be the ratio of the two variances less than or equal to 1 and Ha would be the ratio of the two variances greater than 1.

Test statistic = (7.81)^2/(6.62)^2 = ?

Note: ^2 means squared.

Finish the calculation. Using an F-distribution table at .05 level of significance, determine the critical or cutoff value to reject the null. (You will need to use degrees of freedom to determine the value.) If the test statistic exceeds the critical value from the table, the null is rejected in favor of the alternative hypothesis. If the test statistic does not exceed the critical value from the table, then the null is not rejected.

I hope these few hints will help get you started.

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