Question

Consider independent simple random samples that are taken to test the difference between the means of two populations. The variances of the populations are unknown but are assumed to be equal. The sample sizes of each population are n1?

Answer 1

Answer:

d. t distribution with df = 80

Step-by-step explanation:

Assuming this problem:

Consider independent simple random samples that are taken to test the difference between the means of two populations. The variances of the populations are unknown, but are assumed to be equal. The sample sizes of each population are n1 = 37 and n2 = 45. The appropriate distribution to use is the:

a. t distribution with df = 82.

b. t distribution with df = 81.

c. t distribution with df = 41.

d. t distribution with df = 80

Solution to the problem

When we have two independent samples from two normal distributions with equal variances we are assuming that  

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

This last one is an unbiased estimator of the common variance $\sigma^2$

So on this case the degrees of freedom are given by:

$df= 37+45-2=80$

And the best answer is:

d. t distribution with df = 80