For a Poisson distribution with parameter $\lambda$, the probability mass function is $P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$.

Given $P(X = 1) = P(X = 3)$:

$$\frac{e^{-\lambda} \lambda^1}{1!} = \frac{e^{-\lambda} \lambda^3}{3!}$$ $$\lambda = \frac{\lambda^3}{6} \implies 6\lambda = \lambda^3 \implies \lambda^2 = 6 \implies \lambda = \sqrt{6} \approx 2.4495$$

The mode of a Poisson distribution when $\lambda$ is non-integral is the integer part of $\lambda$:

$$\text{Mode} = [\lambda] = [\sqrt{6}] = [2.4495] = 2$$

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

$$t = \frac{\bar{X}_1 - \bar{X}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$$

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

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

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:

$$t = \frac{r\sqrt{n - 2}}{\sqrt{1 - r^2}}$$

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.