Question

Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p.

When 319 college students are randomly selected and surveyed, it is found that 120 own a car. Find a 99% confidence interval for the true proportion of all college students who own a car.

A) 0.323 < p < 0.429
B) 0.313 < p < 0.439
C) 0.306 < p < 0.446
D) 0.332 < p < 0.421

Answer 1

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**.