Question

You are testing the claim that the mean GPA of night students is greater than the mean GPA of day students.

You sample 35 night students, and the sample mean GPA is 2.36 with a standard deviation of 0.32

You sample 40 day students, and the sample mean GPA is 2.11 with a standard deviation of 0.53

Calculate the test statistic, rounded to 2 decimal places

Answer 1

To test the claim that the mean GPA of night students is greater than the mean GPA of day students, you can use a two-sample t-test for the means. The formula for the test statistic \( t \) is given by:

\[
t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\]

Where:
- \( \bar{x}_1 \) is the sample mean of the night students,
- \( \bar{x}_2 \) is the sample mean of the day students,
- \( s_1 \) is the standard deviation of the night students,
- \( s_2 \) is the standard deviation of the day students,
- \( n_1 \) is the sample size of the night students,
- \( n_2 \) is the sample size of the day students.

Given the information:
- \( \bar{x}_1 = 2.36 \)
- \( \bar{x}_2 = 2.11 \)
- \( s_1 = 0.32 \)
- \( s_2 = 0.53 \)
- \( n_1 = 35 \)
- \( n_2 = 40 \)

Now, we can calculate the test statistic step by step.

1. Calculate the variances:
- \( s_1^2 = (0.32)^2 = 0.1024 \)
- \( s_2^2 = (0.53)^2 = 0.2809 \)

2. Compute the standard error of the difference in means:
\[
SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{0.1024}{35} + \frac{0.2809}{40}}
\]

3. Calculate the individual components:
- \( \frac{0.1024}{35} \approx 0.002926 \)
- \( \frac{0.2809}{40} \approx 0.0070225 \)

4. Summing these:
\[
SE \approx \sqrt{0.002926 + 0.0070225} \approx \sqrt{0.0099485} \approx 0.0997
\]

5. Now calculate the test statistic \( t \):
\[
t = \frac{2.36 - 2.11}{0.0997} \approx \frac{0.25}{0.0997} \approx 2.51
\]

Finally, round the test statistic to two decimal places:
\[
\boxed{2.51}
\]