Respuesta :
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 [tex]\bar x[/tex] and [tex]\bar y[/tex] be the sample means of two sets of data of size [tex]n_x[/tex] and [tex]n_y[/tex] respectively. We assume that the distribution's of x and y are:
[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]
So on this case the best answer would be :
b. Student-t with 48 degrees of freedom
