Consider the following hypothesis test:
$H_0:μ_1-μ_2=0$
$H_a:μ_1-μ_2≠0$
The following results are from independent sample taken from two populations:
|
Sample 1 |
Sample 2 |
|
$n_1 = 35$ |
$n_2 = 40$ |
|
$x_1 = 13.6$ |
$x_2 = 10.1$ |
|
$S_1 = 5.2$ |
$S_2 = 8.5$ |
What is the degrees of freedom for the t-distribution?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 65
Given, $D_0 = 0, n_1 = 35, n_2 = 40, x_1 = 13.6, x_2 = 10.1, S_1 = 5.2$ and $S_2 = 8.5$
$df=\frac{\left( \frac{S_1^2}{n_1} + \frac{S_2^2}{n_2} \right)^2}{\frac{1}{n_1 - 1}\left( \frac{S_1^2}{n_1} \right)^2 + \frac{1}{n_2 - 1}\left( \frac{S_2^2}{n_2} \right)^2}$
$=\frac{\left(\frac{(5.2)^2}{35} + \frac{(8.5)^2}{40}\right)^2}{\frac{1}{35-1}\left(\frac{(5.2)^2}{35}\right)^2+\frac{1}{40-1}\left(\frac{(8.5)^2}{40}\right)^2}$
$=\frac{(0.77251+1.80625)^2}{\frac{1}{34}(0.77251)^2+\frac{1}{39}(1.80625)^2}=\frac{6.6503}{\frac{1}{34}(0.5969)+\frac{1}{39}(3.2625)}$
$=\frac{6.6503}{0.0176+0.0837}=\frac{6.6503}{0.1013}=65.64$
So $df=65$