You are testing the claim that the mean GPA of night students is different from the mean GPA of day students. You sample 35 night students, and the sample mean GPA is 2.72 with a standard deviation of 0.54. You sample 60 day students, and the sample mean GPA is 3.19 with a standard deviation of 0.32. Test the claim using a 1% level of significance. Assume the population standard deviations are unequal and that GPAs are normally distributed. Give answer to at least 4 decimal places.

Question

What are the correct hypotheses?

H0:
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=
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H1:
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?

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Based on the hypotheses, find the following:

Test Statistic =

Critical Values =
(Just enter the positive CV.)

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

Answer 1

To test the claim that the mean GPA of night students is different from the mean GPA of day students, we will conduct a two-sample t-test for the difference between means. Let's define the hypotheses and then calculate the test statistic.

Hypotheses

• Null Hypothesis (H0): \( \mu_{\text{night}} = \mu_{\text{day}} \)
• Alternative Hypothesis (H1): \( \mu_{\text{night}} \neq \mu_{\text{day}} \)

Given Data

• Night Students:
  • Sample Size (\( n_1 \)) = 35
  • Sample Mean (\( \bar{x}_1 \)) = 2.72
  • Sample Standard Deviation (\( s_1 \)) = 0.54
• Day Students:
  • Sample Size (\( n_2 \)) = 60
  • Sample Mean (\( \bar{x}_2 \)) = 3.19
  • Sample Standard Deviation (\( s_2 \)) = 0.32

Test Statistic Calculation

We will use the formula for the t-test statistic for two independent samples:

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

Substituting the values:

\[ t = \frac{2.72 - 3.19}{\sqrt{\frac{0.54^2}{35} + \frac{0.32^2}{60}}} \]

Calculating the denominator:

1. Calculate \( s_1^2 \) and \( s_2^2 \):

\[ s_1^2 = 0.54^2 = 0.2916 \] \[ s_2^2 = 0.32^2 = 0.1024 \]

2. Substitute into the formula for standard error:

\[ SE = \sqrt{\frac{0.2916}{35} + \frac{0.1024}{60}} = \sqrt{0.008336 + 0.0017067} = \sqrt{0.0100427} \approx 0.1002 \]

Now substituting back into the t-statistic formula:

\[ t = \frac{2.72 - 3.19}{0.1002} = \frac{-0.47}{0.1002} \approx -4.69 \]

Critical Values

With a significance level of \( \alpha = 0.01 \) and degrees of freedom calculated using Welch’s approximation:

\[ df \approx \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]

Calculating degrees of freedom:

Let \( a = \frac{0.2916}{35} \approx 0.008336 \) and \( b = \frac{0.1024}{60} \approx 0.0017067 \):

\[ df \approx \frac{(0.008336 + 0.0017067)^2}{\frac{(0.008336)^2}{34} + \frac{(0.0017067)^2}{59}} \approx \frac{0.0001019235}{0.00000209309 + 0.00000008174} \approx 50.434 \]

Round down to \( df \approx 50 \).

Now, using a t-table or calculator, the critical values for \( t \) at \( df = 50 \) and \( \alpha = 0.01 \) for a two-tailed test is \( t_{critical} \approx 2.576 \).

Summary

• Test Statistic: \( t \approx -4.69 \)
• Critical Value: \( 2.576 \)

Decision Rule

If the absolute value of the test statistic exceeds the critical value, we reject the null hypothesis.

Since \( |-4.69| > 2.576 \), we reject the null hypothesis.

Conclusion

The correct summary would be: We reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the mean GPA of night students is different from the mean GPA of day students.