In a two-sample independent t-test with pooled variance:

Where the pooled sample variance $s_p^2$ is estimated using sums of squares from both samples:

Because two sample means (parameters) are estimated, 2 degrees of freedom are lost from the total $(n_1 + n_2)$ observations, leaving $(n_1 + n_2 - 2)$ degrees of freedom.

To test the null hypothesis $H_0: \rho = 0$ (that the population correlation coefficient is zero) against $H_1: \rho \ne 0$, the Student's t-test statistic is:

This statistic follows a Student's t-distribution with $(n - 2)$ degrees of freedom, because two parameters (the means of $X$ and $Y$) are estimated from the bivariate sample.

For dependent/paired bivariate samples $(X_i, Y_i)$ of size $n$, the differences $d_i = X_i - Y_i$ are computed. The test statistic is:

where $\bar{d} = \frac{1}{n}\sum d_i$ and $s_d^2 = \frac{1}{n-1}\sum (d_i - \bar{d})^2$.
This follows Student's t-distribution with $(n - 1)$ degrees of freedom. Hence, Option 2 is correct.

While the asymptotic (large-sample) mean of Pearson's chi-square test for an $r \times s$ contingency table equals its degrees of freedom $(r - 1)(s - 1)$, J.B.S. Haldane (1939) derived the exact mean under conditional multinomial distribution with fixed marginal totals as:

Notice that as sample size $N \to \infty$, $\frac{N}{N-1} \to 1$, recovering the standard asymptotic mean $(r - 1)(s - 1)$. Thus, Option 3 is correct.