Question

Two independent samples of sizes 20 and 30 are randomly selected from two normally distributed populations. Assume that the population variances are unknown but equal. In order to test the difference between the population means, , the sampling distribution of the sample mean difference, , is
a. normal. b. Student-t with 48 degrees of freedom. c. Student-t with 50 degrees of freedom. d. None of these choices.

Answer 1

Answer:

b. Student-t with 48 degrees of freedom

Step-by-step explanation:

For this case we need to use a Two Sample t Test: equal variances.

Assumptions

When running a two-sample equal-variance t-test, the basic assumptions are "that the distributions of the two  populations are normal, and that the variances of the two distributions are the same".

Let $\bar x$ and $\bar y$ be the sample means of two sets of data of size $n_x$ and $n_y$ respectively. We assume that the distribution's of x and y are:

$x \sim N(\mu_x ,\sigma_x =\sigma)$

$y \sim N(\mu_y ,\sigma_y=\sigma)$

Both are normally distributed but without the variance equal for both populations.

The system of hypothesis can be:

Null hypothesis: $\mu_x =\mu_y$

Alternative hypothesis: $\mu_x \neq \mu_y$

We can define the following random variable:

$t=\frac{(\bar x -\bar y)-(\mu_x -\mu_y)}{s\sqrt{\frac{1}{n_x}+\frac{1}{n_y}}}$

The random variable t is distributed $t \sim t_{n_x +n_y -2}$, with the degrees of freedom $df=n_x +n_y -2= 20+30-2=48$

And the pooled variance can be founded with the following formula:

$s^2=\frac{(n_x -1)s_x^2 +(n_y-1)s_y^2}{n_x +n_y -2}$

So on this case the best answer would be :

b. Student-t with 48 degrees of freedom