e-Statistics > 4480-5480 Probability and Statistics II

Comparison of Two Means

Data are collected from two groups, say ``Group 1'' and ``Group 2,'' concerning with how Group 1 and Group 2 differ in terms of their respective population means $ \mu_1$ and $ \mu_2$. Then the hypothesis test can be stated with the alternative hypothesis

$ H_A:\hspace{0.05in}\mu_1$ $ \mu_2$

The test statistic is calculated from the sample means $ \bar{X} =$ and $ \bar{Y} =$ and the sample standard deviations $ S_1 =$ and $ S_2 =$ from Group 1 and Group 2 with the respective sample sizes $ n =$ and $ m =$

In general procedure the variance estimate of $ (\bar{X}-\bar{Y})$ is given by $ S_{\bar{X}-\bar{Y}}^2 = \frac{S_1^2}{n} + \frac{S_2^2}{m}$ with the sample variances $ S_1^2$ and $ S_2^2$ of Group 1 and 2. Then the test statistic $ T = \frac{\bar{X} - \bar{Y}}{S_{\bar{X}-\bar{Y}}}$ is likely observed around zero, positive, or negative, under the respective null hypothesis `` $ \mu_1 = \mu_2$,'' `` $ \mu_1 \ge \mu_2$,'' or `` $ \mu_1 \le \mu_2$.'' The opposite of such an observation is expressed by the p-value smaller than $ \alpha$, and it suggests an evidence against the null hypothesis $ H_0$.

When it is reasonable to assume that the two population variances $ \sigma_1^2$ and $ \sigma_2^2$ of Group 1 and 2 are equal, the variance estimate is given by $ S_{\bar{X}-\bar{Y}}^2
= \left(\frac{1}{n} + \frac{1}{m}\right) S_p^2$ via pooled sample variance $ S_{p}^2 = \frac{(n-1)S_x^2 + (m-1)S_y^2}{n+m-2}$.

If the null hypothesis is rejected, it would be preferable to construct the confidence interval for the population mean difference $ \mu_1 - \mu_2$.

$ \displaystyle
\left(\bar{X} - \bar{Y}
- t_{\alpha/2,df} S_{\bar{X}-\bar{Y}},\:
\bar{X} - \bar{Y}
+ t_{\alpha/2,df} S _{\bar{X}-\bar{Y}}
\right) =$ ( , )

Here the choices of confidence level $ (1-\alpha)$ are 90%, 95%, or 99%.