When the two independent small samples of sizes \(n_1\) and \(n_2\) with means \(\overline{x_1}\) and \(\overline{x_2}\) respectively are drawn from populations with identical population variances, the test-statistic is computed as
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(t=\displaystyle\frac{\overline{x_1}-\overline{x_2}}{S_p\sqrt{\displaystyle\frac{1}{n_1}+\frac{1}{n_2}}}\), where \(S_p\) is pooled standard deviation.
The two-sample t-test compares the difference of the sample means with its standard error. With equal population variances, the two sample variances are combined into the pooled standard deviation
\( \displaystyle S_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}} \)
For independent samples the variances of the two means add, so the standard error of \(\overline{x_1} - \overline{x_2}\) is
\( \displaystyle S_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}} \)
Hence
\( \displaystyle t = \frac{\overline{x_1} - \overline{x_2}}{S_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \)
with \(n_1 + n_2 - 2\) degrees of freedom.
Why the other options are incorrect
Option (2) \(t=\displaystyle\frac{\overline{x_1}-\overline{x_2}}{S_p\sqrt{\displaystyle\frac{1}{n_1}-\frac{1}{n_2}}}\), where \(S_p\) is pooled standard deviation. — Incorrect. The variances add, so the root must contain \(\frac{1}{n_1} + \frac{1}{n_2}\); with a minus sign the root would be 0 for equal sample sizes.
Option (3) \(t=\displaystyle\frac{\overline{x_1}+\overline{x_2}}{S_p\sqrt{\displaystyle\frac{1}{n_1}-\frac{1}{n_2}}}\), where \(S_p\) is pooled standard deviation. — Incorrect. The test uses the difference of the means, not their sum, and the root has the wrong sign.
Option (4) \(t=\displaystyle\frac{\overline{x_1}+\overline{x_2}}{\sqrt{\displaystyle\frac{1}{n_1}-\frac{1}{n_2}}}\), where \(S_p\) is pooled standard deviation. — Incorrect. Besides the sum of the means and the minus sign, the factor \(S_p\) is missing from the denominator.
Note: This topic was not in the CUET 2026 syllabus.