Let us suppose that two independent random samples of sizes $n_1$ and $n_2$ has been drawn from the same normal population then degree of freedom of statistic t-distribution is.

\( n_1 + n_2 - 2 \)

The correct answer is Option (2) → \( n_1 + n_2 - 2 \)

Given two independent random samples of sizes n1 and n2 drawn from the same normal population.

The t-statistic for comparing two means uses the formula:

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

If population variances are assumed equal, the pooled variance is used:

$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1 + n_2 - 2}$

The degrees of freedom for the t-distribution is:

$df = n_1 + n_2 - 2$