A random sample of 130 mortgages in the state of Maryland was randomly selected. From this sample, 14 were found to be delinquent on their current payment. The 98% confidence interval for the proportion based on this sample is ________.
A) [0.044, 0.171]
B) [0.036, 0.180]
C) [0.029, 0.188]
D) [0.015, 0.201]

Question

Quizplus · Accepted Answer

To calculate the 98% confidence interval for the proportion, we use the formula for the confidence interval for a proportion: $\hat{p} \pm Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $Z_{\alpha/2}$ is the Z-score corresponding to the confidence level, and $n$ is the sample size. Here, $\hat{p} = \frac{14}{130}$, $n = 130$, and for a 98% confidence level, $Z_{\alpha/2} \approx 2.33$. Plugging these values into the formula gives a confidence interval of approximately [0.044, 0.171].

A Random Sample of 130 Mortgages in the State of Maryland