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

B) The margin of error is 0.0269.

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

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

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

This means that
`n = 918, \pi = (521)/(918) = 0.5675`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

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

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`M = 1.645\sqrt{(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).