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A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, [tex]n = 913[/tex] and [tex]x = 556[/tex] who said "yes." Use a 90% confidence level.

Answer :

The poll's margin of error is 0.0414, the confidence interval is 0, and the best point forecast for the population proportion is p = 0.6157. (0.5743, 0.6571)

What is a poll?

In order to represent the opinions of a population, opinion polls typically involve a series of questions and an attempt to extrapolate broad generalizations in ratio or within confidence intervals. A person who conducts polls is known as a pollster.

We have been given;

n = 916 people participated in the poll.

x = 564 is the number of people who said yes.

The formula would yield the population proportion;

p = x/n

p = 564/916

p = 0.6157

The formula for the margin of error is;

E = (z_α/2)√(p^(1 - p^)/n)

Where;

(z_α/2) is the critical value at the given confidence level.

The critical value z_α/2 for a confidence level of 99% from tables is: 2.576.

So,

E = 2.576√(0.6157(1 - 0.6157)/916)

E = 0.0414

The formula for the confidence interval is;

CI = p^ ± (z_α/2)√(p^(1 - p^)/n)

From B it can be known that

(z_α/2)√(p^(1 - p^)/n) = 0.0414

So,

CI = 0.6157 ± 0.0414

CI = (0.6157 - 0.0414), (0.6157 + 0.0414)

CI = (0.5743, 0.6571)

Thus, the answer for parts a, b, and c is p = 0.6157, E = 0.0414, and CI = (0.5743, 0.6571) respectively.

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Your question seems incomplete, the missing part of the question is:

(a) Find the best point estimate of the population proportion p.

​(b) Identify the value of the margin of error E.

​(c) Construct the confidence interval.

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