Question

A university is comparing the grade point average averages of biology majors with the grade point averages of engineering majors. 25 students from each major are randomly selected. The mean and standard deviation for each sample are shown in the table.
The university wants to test if there is a significant difference in the GPAs for students in two majors. How is the appropriate test statistic calculated?



Biology Majors - (Sample Mean 3.22 , Sample Standard Deviation 0.05)

Engineering Majors - (Sample Mean 3.17 , Sample Standard Deviation 0.03).

Answer 1

Considering the hypothesis tested, using the t-distribution, as we have the standard deviation for the sample, it is found that the adequate test statistic is given by:

$t = \frac{(3.22 - 3.17) - 0}{\sqrt{\frac{0.05^2 + 0.03^2}{25}}}$

What are the hypothesis tested?

At the null hypothesis, we test if there is no difference, that is, the subtraction of their means is 0, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, we test if there is a difference, that is, the subtraction of their means is not 0, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

What is the distribution of the differences?

For Biology Majors, we have that:

$\mu_1 = 3.22, \sigma_1 = 0.05, n_1 = 25, s_1 = \frac{0.05}{\sqrt{25}}$

For Engineering Majors, we have that:

$\mu_2 = 3.17, \sigma_2 = 0.03, n_2 = 25, s_2 = \frac{0.03}{\sqrt{25}}$

Then, for the distribution of differences, we have that:

$\overline{x} = \mu_1 - \mu_2 = 3.22 - 3.17$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{\left(\frac{0.05}{\sqrt{25}}\right)^2 + \left(\frac{0.03}{\sqrt{25}}\right)^2} = \sqrt{\frac{0.05^2 + 0.03^2}{25}}$

What is the test statistic?

It is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{(3.22 - 3.17) - 0}{\sqrt{\frac{0.05^2 + 0.03^2}{25}}}$

To learn more about the t-distribution, you can take a look at brainly.com/question/13873630