Question

A research institute poll asked respondents if they felt vulnerable to identity theft. In the​ poll, n=918 and x=521 who said​ "yes." Use a 90% confidence level.
B) Identify the value of the margin of error E.
C) construct the confidence interval.

Answer 1

Answer:

B) The margin of error is 0.0269.

C) The confidence interval is (0.5406, 0.5944).

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 $1 - \frac{\alpha}{2}$.

The margin of error is:

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

In the​ poll, n=918 and x=521 who said​ "yes."

This means that $n = 918, \pi = \frac{521}{918} = 0.5675$

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 $Z = 1.645$.

B) Identify the value of the margin of error E.

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

$M = 1.645\sqrt{\frac{0.5675*0.4325}{918}} = 0.0269$

The margin of error is 0.0269.

C) construct the confidence interval.

$\pi \pm M$

So

$\pi - M = 0.5675 - 0.0269 = 0.5406$

$\pi + M = 0.5675 + 0.0269 = 0.5944$

The confidence interval is (0.5406, 0.5944).