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.
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Calculate the variances:
- \( s_1^2 = (0.32)^2 = 0.1024 \)
- \( s_2^2 = (0.53)^2 = 0.2809 \)
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Compute the standard error of the difference in means: \[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{0.1024}{35} + \frac{0.2809}{40}} \]
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Calculate the individual components:
- \( \frac{0.1024}{35} \approx 0.002926 \)
- \( \frac{0.2809}{40} \approx 0.0070225 \)
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Summing these: \[ SE \approx \sqrt{0.002926 + 0.0070225} \approx \sqrt{0.0099485} \approx 0.0997 \]
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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: \[ \boxed{2.51} \]