Welch’s t Test

Mean comparison is the central theme of many classical statistical procedures. The well-known independent-sample t test is often used to test the equality of two means from independent populations with equal variances, whereas Welch’s t test is generally preferred when the variances are not equal.

Let $y_{11},y_{21},\dots ,y_{n_{1}1}$ and $y_{12},y_{22},\dots ,y_{n_{2}2}$, be two independent random samples from two populations with means (or expected values) ${\mu }_j=E\left(y_{ij}\right)$ and variances ${\sigma }_j^2=\text{Var}(y_{ij}),j=1,2.$ The sample counterparts of µj and ${\sigma }_j^2$ are ${\overline{y}}_j=\frac{1}{n_j}{\displaystyle \sum _{i=1}^{n_j}{y}_{ij}}$, and $s_j^2=\frac{1}{n_j-1}{\displaystyle \sum _{i=1}^{n_j}{(y_{ij}-{\overline{y}}_j)}^2},$ respectively. Because ${\overline{y}}_j$ is an unbiased estimator of µj, any difference between µ1 and µ2 should be reflected by ${\overline{y}}_1-{\overline{y}}_2.$ Of course, even when the null hypothesis

$$H_0:{\mu }_1={\mu }_2$$

holds, ${\overline{y}}_1-{\overline{y}}_2$ does not hold in general because of the sampling error, which is characterized by the variance

$${\nu }^2=\text{Var}({\overline{y}}_1-{\overline{y}}_2)=\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}.$$

Notice that, when ${\sigma }_1^2={\sigma }_2^2={\sigma }^2,{\nu }^2$ will be ...