A poll showed that 48 out of 120 randomly chosen graduates of California medical schools last year intended to specialize in family practice. What is the width of a 90 percent confidence interval for the proportion that plan to specialize in family practice?
A)± .0447
B)± .0736
C)± .0876
D)± .0894

Question

Quizplus · Accepted Answer

To calculate the width of a 90% confidence interval for the proportion, we use the formula for the margin of error in a proportion: $ME = Z * \sqrt{\frac{p(1-p)}{n}}$, where $Z$ is the Z-score corresponding to the desired confidence level (for 90%, $Z \approx 1.645$), $p$ is the sample proportion, and $n$ is the sample size. Here, $p = \frac{48}{120} = 0.4$, and $n = 120$. Plugging these values into the formula gives $ME = 1.645 * \sqrt{\frac{0.4(0.6)}{120}} \approx 0.0736$. The width of the confidence interval is twice the margin of error, but since the question asks for the margin of error itself (which is half the width), the correct answer is ±0.0736.

A Poll Showed That 48 Out of 120 Randomly Chosen