Respuesta :
Answer:
[tex]t=\frac{(\bar x -\bar y)-(\mu_x -\mu_y)}{s\sqrt{\frac{1}{n_x}+\frac{1}{n_y}}}[/tex]
The random variable t is distributed [tex]t \sim t_{n_x +n_y -2}[/tex], with the degrees of freedom [tex]df=n_x +n_y -2= 20+30-2=48[/tex]
And the pooled variance can be founded with the following formula:
[tex]s^2=\frac{(n_x -1)s_x^2 +(n_y-1)s_y^2}{n_x +n_y -2}[/tex]
The degrees of freedom are given by:
[tex] df = 20 +25-2 = 43[/tex]
So then the best answer for this case:
c. t-distributed with 43 degrees of freedom
Step-by-step explanation:
When we use a 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".
[tex]x \sim N(\mu_x ,\sigma_x =\sigma)[/tex]
[tex]y \sim N(\mu_y ,\sigma_y=\sigma)[/tex]
Both are normally distributed but without the variance equal for both populations.
The system of hypothesis can be:
Null hypothesis: [tex]\mu_x =\mu_y[/tex]
Alternative hypothesis: [tex]\mu_x \neq \mu_y [/tex]
We can define the following random variable:
[tex]t=\frac{(\bar x -\bar y)-(\mu_x -\mu_y)}{s\sqrt{\frac{1}{n_x}+\frac{1}{n_y}}}[/tex]
The random variable t is distributed [tex]t \sim t_{n_x +n_y -2}[/tex], with the degrees of freedom [tex]df=n_x +n_y -2= 20+30-2=48[/tex]
And the pooled variance can be founded with the following formula:
[tex]s^2=\frac{(n_x -1)s_x^2 +(n_y-1)s_y^2}{n_x +n_y -2}[/tex]
The degrees of freedom are given by:
[tex] df = 20 +25-2 = 43[/tex]
c. t-distributed with 43 degrees of freedom
