e-Statistics

t-Test

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 and from Group 1 and Group 2 with the respective sample sizes and

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 and 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 .

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%.