In a survey of n = 100 teachers, 50 preferred year-around school.Set the .95 confidence interval around this sample proportion.
A) .40 to .60
B) .45 to .55
C) .50 to .60
D) .55 to .65

Question

Quizplus · Accepted Answer

The confidence interval for a proportion is calculated using the formula $p \pm Z \sqrt{\frac{p(1-p)}{n}}$, where $p$ is the sample proportion, $Z$ is the Z-score corresponding to the confidence level (for 95% confidence, $Z \approx 1.96$), and $n$ is the sample size. Here, $p = 0.50$ and $n = 100$, so the margin of error is $1.96 \sqrt{\frac{0.50 \times 0.50}{100}} = 0.098$. Adding and subtracting this from the sample proportion gives an interval of $0.50 \pm 0.098$, which is $0.402$ to $0.598$, rounded to $0.40$ to $0.60$.

In a Survey of N = 100 Teachers, 50 Preferred