Problem

You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 25 night students, and the sample mean GPA is 2.8 with a standard deviation of 0.75 . You sample 60 day students, and the sample mean GPA is 2.44 with a standard deviation of 0.52 . Test the claim using a $10 \%$ level of significance. Assume the population standard deviations are unequal and that GPAs are normally distributed. Give answer to at least 4 decimal places.

What are the correct hypotheses?
$\mathrm{H}_{0}:$ Select an answer $\hat{\imath}=$ Select an answer $\hat{\imath}$

Based on the hypotheses, find the following:

Test Statistic $=$ Critical Values $= \pm \square$ (Just enter the positive CV.)

The correct decision is to
Select an answer
The correct summary would be:
Select an answer that the mean GPA of night students is different from the mean GPA of day students.

Answer

Expert–verified
Hide Steps
Answer

\(\boxed{\text{Final Answer:}}\) There is enough evidence to support the claim that the mean GPA of night students is different from the mean GPA of day students.

Steps

Step 1 :Given values are: sample size of night students \(n1 = 25\), sample mean GPA of night students \(x1 = 2.8\), standard deviation of night students \(s1 = 0.75\), sample size of day students \(n2 = 60\), sample mean GPA of day students \(x2 = 2.44\), standard deviation of day students \(s2 = 0.52\), and level of significance \(\alpha = 0.10\).

Step 2 :The null hypothesis \(H_0\) is that the mean GPA of night students is equal to the mean GPA of day students, i.e., \(\mu_{1} = \mu_{2}\). The alternative hypothesis \(H_a\) is that the mean GPA of night students is not equal to the mean GPA of day students, i.e., \(\mu_{1} \neq \mu_{2}\).

Step 3 :Calculate the test statistic using the formula: \(t_{stat} = \frac{x1 - x2}{\sqrt{(s1^2/n1) + (s2^2/n2)}}\). Substituting the given values, we get \(t_{stat} = 2.1906\).

Step 4 :Calculate the degrees of freedom using the formula: \(df = n1 + n2 - 2\). Substituting the given values, we get \(df = 83\).

Step 5 :Calculate the critical value using the t-distribution table with \(df = 83\) and \(\alpha = 0.10\). The critical value is \(t_{critical} = 1.6634\).

Step 6 :Since the test statistic \(t_{stat} = 2.1906\) is greater than the critical value \(t_{critical} = 1.6634\), we reject the null hypothesis \(H_0\).

Step 7 :\(\boxed{\text{Final Answer:}}\) There is enough evidence to support the claim that the mean GPA of night students is different from the mean GPA of day students.

link_gpt