Answer:
Explanation:
To determine if the professor should be surprised, we can use statistical hypothesis testing. The null hypothesis (\(H_0\)) is that the proportion of students who engage in binge drinking is equal to the reported national average of 41%. The alternative hypothesis (\(H_1\)) is that the proportion is different.
Given:
- Sample proportion (\(p\)): 111/287 ≈ 0.3861 (the proportion of students in the sample who admitted to binge drinking).
- Population proportion under the null hypothesis (\(p_0\)): 0.41 (the reported national average).
We can use the Z-test to compare the sample proportion to the population proportion.
\[ Z = \frac{p - p_0}{\sqrt{\frac{p_0 \cdot (1 - p_0)}{n}}} \]
Where:
- \( p \) is the sample proportion.
- \( p_0 \) is the population proportion under the null hypothesis.
- \( n \) is the sample size.
\[ Z = \frac{0.3861 - 0.41}{\sqrt{\frac{0.41 \cdot (1 - 0.41)}{287}}} \]
Calculate \( Z \) to determine the number of standard deviations away from the mean the sample proportion is.
Now, the professor can check the Z-value against a critical value (usually 1.96 for a 95% confidence interval) or use a p-value to determine statistical significance.
- If \( |Z| > 1.96 \), he may be surprised, and the result is statistically significant at the 0.05 level.
- If \( |Z| \leq 1.96 \), he may not be surprised, and the result is not statistically significant at the 0.05 level.
The standard deviation for him being surprised or not surprised depends on the critical value or p-value calculated during the hypothesis test.