10.51 A statistics teacher compared the marks of her morning class and her afternoon class on the same quiz. The average score of the 42 students in the morning was 83.52 with a standard deviation of 7.81, while
the mean score of the 46 afternoon students was 79.44 with a standard deviation of 6.62. (whose
test would you use to compare means? (b) State the assumptions to see if there was a difference.
(c) Obtain a test statistic and a p-value. (d) Why is there a difference between morning and
afternoon notes? Explain. (e) Is it reasonable to assume equal variances? (f) Carry out a
test for variances equal to á = 0.05, clearly showing all the steps.
could you please help explain the steps to this problem Try a two sample test.
Hypothesis:
Ho: µ1 = µ2
Ha: µ1 is not equal to µ2
This is a two-tailed test because the problem is simply asking if there is a difference (no specific direction in the alternative hypothesis).
The P value is the actual level of the test statistic (using a z or t array, depending on the test you are using).
A test involving two variances (the standard deviation is the square root of the variance) would have these assumptions: 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 square.
Complete the calculation. Using an F-distribution table at the 0.05 significance level, determine the critical or cut-off value to reject the null value. (You will need to use degrees of freedom to determine the value.) If the test statistic exceeds the critical value in the table, the null value is rejected in favor of the alternative hypothesis. If the test statistic does not exceed the critical value in the array, the null value is not rejected.
I hope these few tips will help you get started.