To test the claim that the mean GPA of night students is greater than the mean GPA of day students, we can use a two-sample t-test for the difference in means.
Step 1: Define the parameters.
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Sample size of night students (\( n_1 = 30 \))
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Sample mean GPA of night students (\( \bar{x}_1 = 2.66 \))
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Standard deviation of night students (\( s_1 = 0.98 \))
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Sample size of day students (\( n_2 = 60 \))
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Sample mean GPA of day students (\( \bar{x}_2 = 2.32 \))
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Standard deviation of day students (\( s_2 = 0.54 \))
Step 2: Calculate the standard error (SE) of the difference in means.
\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]
Calculating \( s_1^2 \) and \( s_2^2 \):
\[ s_1^2 = 0.98^2 = 0.9604 \] \[ s_2^2 = 0.54^2 = 0.2916 \]
Now, substituting the values:
\[ SE = \sqrt{\frac{0.9604}{30} + \frac{0.2916}{60}} \]
Calculating each part:
\[ \frac{0.9604}{30} = 0.03201333 \] \[ \frac{0.2916}{60} = 0.00486 \]
Now, combine these:
\[ SE = \sqrt{0.03201333 + 0.00486} = \sqrt{0.03687333} \approx 0.6072 \]
Step 3: Calculate the test statistic (t).
The formula for the t-statistic when comparing two means is given by:
\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} \]
Substituting the values:
\[ t = \frac{2.66 - 2.32}{0.6072} = \frac{0.34}{0.6072} \approx 0.56 \]
Step 4: Round the test statistic to two decimal places:
\[ t \approx 0.56 \]
Therefore, the test statistic is:
\[ \boxed{0.56} \]