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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.

Sagot :

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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

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

This means that [tex]n = 918, \pi = \frac{521}{918} = 0.5675[/tex]

90% confidence level

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

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

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

[tex]M = 1.645\sqrt{\frac{0.5675*0.4325}{918}} = 0.0269[/tex]

The margin of error is 0.0269.

C) construct the confidence interval.

[tex]\pi \pm M[/tex]

So

[tex]\pi - M = 0.5675 - 0.0269 = 0.5406[/tex]

[tex]\pi + M = 0.5675 + 0.0269 = 0.5944[/tex]

The confidence interval is (0.5406, 0.5944).