Question

Keisha wants to estimate the percentage of managers at her company that hold an MBA. She surveys 320 managers and finds that 70 hold an MBA. Find the margin of error for the confidence interval for the population proportion with a 90% confidence level.

Answer 1

Answer:

The margin of error is of 0.038 = 3.8%.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 320, \pi = \frac{70}{320} = 0.21875$

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$M = 1.645\sqrt{\frac{0.21875(1-0.21875)}{320}}$

$M = 0.038$

The margin of error is of 0.038 = 3.8%.