#### Confidence Intervals for a Differences in Means

**A Motivating Problem.**Are scores in men's and women's college basketball games similar? More specifically, we can ask if scores in the championship game of men's and women's NCAA basketball tournament are equal. We will take the viewpoint that each tournament is the realization of a random process and that one outcome of the random process is a total score in the championship game. We assume that all the tournaments are mutually independent and that the expected total scores in championship games do not change in time.**A model.**We represent these tournaments with two buckets filled with numbered balls, representing total scores in the championship games in men's and women's tournaments respectively. We take two independent samples from these buckets.-
**Data.**Here are summary statistics over 16 years.Tournament mean sd n men's 150.3 14.2 16 women's 150.3 14.2 16 Neither strong skewness nor extreme outliers are present in the data.

**Theory.**The sampling distribution of a difference in sample means from independent samples is approximately normal.mean(xbar

_{1}-xbar_{2}) = mu_{1}- mu_{2}.

SE(xbar_{1}-xbar_{2}) = sqrt(sigma_{1}^{2}/ n_{1}+ sigma_{2}^{2}/ n_{2})The shape is exactly normal if the populations are normal and approximately normal for nonnormal populations and sufficiently large sample sizes.

If population standard deviations are known, the test statistic

z = ((xbar

_{1}-xbar_{2}) - (mu_{1}-mu_{2})) / SEhas the standard normal distribution. If the population standard deviations (sigma

_{1}and sigma_{2}) are replaced with sample standard deviations, the test statistic has an approximate t distribution with degrees of freedom that can be estimated from the observed data.A common approach is to assume that the two population standard deviations are equal, but unknown. In this case, the standard error is

SE = sigma sqrt( 1/n

_{1}+ 1/n_{2})The common standard deviation sigma may be estimated by pooling the two sample standard deviations. The pooled estimate is the weighted average of the sample standard deviations with the weights proportional to the degrees of freedom.

s

_{p}= sqrt( ((n_{1}-1)s_{1}^{2}+ (n_{1}-1)s_{1}^{2}) / (n_{1}+ n_{1}- 2 ) )The test statistic

t = ((xbar

_{1}-xbar_{2}) - (mu_{1}-mu_{2})) / ( s_{p}sqrt( 1/n_{1}+ 1/n_{2}) )has a t distribution with n

_{1}+ n_{1}- 2 degrees of freedom.**Confidence Interval.**A confidence interval for the mean men's total score minus the mean women's total score follows the basic formula(estimate) ± (multiplier)(standard error)

The standard error is 18.8 sqrt(1/16 + 1/16) = 6.65. A t multiplier for the middle 95% of a t distribution with 30 degrees of freedom is t

^{*}= 2.04. The 95% confidence interval is:12.3 ± 13.6 or -1.3 to 25.9.

We are 95% confident that the difference in population mean total scores for men's tournaments minus women's tournaments is between a 1.3 point higher score for women to a 25.9 higher score for men.

**Hypothesis test.**We can formally test the hypothesis that the two mean scores are equal versus the alternative that they are different.H

_{0}: mu(men) = mu(women)

H_{0}: mu(men) not= mu(women)

The test statistic is t = (150.3 - 138.0) / (18.8 sqrt(1/16 + 1/16) ) = 1.85.

The p-value is twice the area to the right of 1.85 under a t distribution with 30 degrees of freedom. This area is between 0.05 and 0.10.

This result is not significant at the 5% level. There is marginal evidence that the mean score in men's championship games differs from women's.

- You are not responsible for statistical inferences about ratios of variances (6-4, 7-4).

Last modified: April 10, 2001

Bret Larget, larget@mathcs.duq.edu