Let two independent random samples of sizes $n_1$ and $n_2$ respectively have been drawn from the same normal population. Let $\overline{X_1}$ and $\overline{X_2}$ be the means and let $s_1$ and $s_2$ be their standard deviations. In order to test whether the the two sample means $\overline{X_1}$ and $\overline{X_1}$ differ significantly or not, the t-test statistic is given by

$t=\frac{\overline{X_1}-\overline{X_1}}{S\sqrt{\frac{1}{n_1}-\frac{1}{n_2}}},S=\sqrt{\frac{n_1s_1^2+n_2s_2^2}{n_1+n_2-2}}$

The correct answer is Option (3) → $t=\frac{\overline{X_1}-\overline{X_1}}{S\sqrt{\frac{1}{n_1}-\frac{1}{n_2}}},S=\sqrt{\frac{n_1s_1^2+n_2s_2^2}{n_1+n_2-2}}$

For two independent samples from the same normal population, the t-test statistic to compare the means is:

$t = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

Where:

• $\bar{X}_1, \bar{X}_2$ = sample means
• $s_1, s_2$ = sample standard deviations
• $n_1, n_2$ = sample sizes

Answer: $t=\frac{\overline{X_1}-\overline{X_1}}{S\sqrt{\frac{1}{n_1}-\frac{1}{n_2}}},S=\sqrt{\frac{n_1s_1^2+n_2s_2^2}{n_1+n_2-2}}$