Verizon Wireless would like to estimate the proportion of households that use cell phones for their phone service without a land line. A random sample of 150 households was selected and 48 relied strictly on cell phones for their service. The margin of error for a 90% confidence interval for the proportion based on this sample is ------.
A)0.0561
B)0.0627
C)0.0333
D)0.0475

Question

Quizplus · Accepted Answer

The margin of error (ME) for a proportion is calculated using the formula: $ME = z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $n$ is the sample size, and $z$ is the z-score corresponding to the confidence level. Here, $\hat{p} = \frac{48}{150}$, $n = 150$, and for a 90% confidence interval, the z-score is approximately 1.645. Plugging these values into the formula gives $ME = 1.645 \times \sqrt{\frac{(48/150)(1-48/150)}{150}} \approx 0.0627$.

Verizon Wireless Would Like to Estimate the Proportion of Households