To construct a confidence interval for the population proportion \( p \), we can use the formula:
\[ \hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
Where:
- \( \hat{p} \) is the sample proportion
- \( z_{\alpha/2} \) is the z-score corresponding to the desired confidence level
- \( n \) is the sample size
Step 1: Calculate sample proportion \( \hat{p} \)
From the data provided:
- Number of students who own a car = 120
- Total number of students surveyed = 319
\[ \hat{p} = \frac{120}{319} \approx 0.376 \]
Step 2: Find the z-score for 99% confidence
For a 99% confidence level, the z-score corresponding to \( \alpha/2 = 0.005 \) (since it's a two-tailed test) is approximately \( 2.576 \).
Step 3: Calculate the standard error
\[ \text{Standard Error} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.376(1-0.376)}{319}} \approx \sqrt{\frac{0.376 \times 0.624}{319}} = \sqrt{\frac{0.234624}{319}} \approx \sqrt{0.000734} \approx 0.0271 \]
Step 4: Calculate the margin of error
\[ \text{Margin of Error} = z_{\alpha/2} \times \text{Standard Error} = 2.576 \times 0.0271 \approx 0.0699 \approx 0.070 \]
Step 5: Construct the confidence interval
\[ \text{Confidence Interval} = \hat{p} \pm \text{Margin of Error} \]
\[ \text{Lower limit} = 0.376 - 0.070 \approx 0.306 \]
\[ \text{Upper limit} = 0.376 + 0.070 \approx 0.446 \]
Final Result
Thus, the 99% confidence interval for the true proportion of college students who own a car is:
\[ 0.306 < p < 0.446 \]
The correct answer is C) 0.306 < p < 0.446.
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