# Question

In trials of an experimental Internet-based method of learning statistics, pre-tests and post- tests were given to two groups: traditional instruction (22 students) and Internet-based (17 students). Pre-test scores were not significantly different. On the post-test, the first group (traditional instruction) had a mean score of 8.64 with a standard deviation of 1.88, while the second group (experimental instruction) had a mean score of 8.82 with a standard deviation of 1.70.

(a) Construct a 90 percent confidence interval for the true difference of the means assuming equal variances. Show all work clearly.

(b) Repeat, using the assumption of unequal variances with either Welch's formula for d.f. or the quick rule for degrees of freedom.

(c) Did the assumption about variances change the conclusion?

(d) Construct separate confidence intervals for each mean. Do they overlap?

(a) Construct a 90 percent confidence interval for the true difference of the means assuming equal variances. Show all work clearly.

(b) Repeat, using the assumption of unequal variances with either Welch's formula for d.f. or the quick rule for degrees of freedom.

(c) Did the assumption about variances change the conclusion?

(d) Construct separate confidence intervals for each mean. Do they overlap?

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