Question

A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n=1086 and x=571 who said "yes." Use a 99% confidence level. A. find the best point of estimate of the population of portion p.B. Identify the value of the margin of error E.E= round to four decimal places as needed.C. Construct the confidence interval._ < p <_ round to three decimal places.

Answer 1

Given:

$\begin{gathered} n=1086 \\ x=571 \end{gathered}$

To Determine: Using 99% confidence interval, the best point of estimate of the population of portion p

Solution

$\begin{gathered} Point-of-estimation(PE)=\frac{x}{n} \\ PE=\frac{571}{1086} \\ PE=0.5258 \end{gathered}$

The formula for finding the margin of error is

$MOE=\sqrt{\frac{0.5258\times(1-0.5258)}{1086}}\times z$$\begin{gathered} MOE=\sqrt{\frac{0.24933436}{1086}}\times z \\ MOE=0.01515\times z \end{gathered}$

The z score corresponding to 99% confidence interval from the table is 2.575

Therefore

$\begin{gathered} MOE=0.01515\times2.575 \\ MOE=0.039 \end{gathered}$

Hence

A. Point of estimation of the population of portion p is approximately 0.5258 (4 decimal places) OR 0.526(3 decimal places)

B. The margin of Error is apprimately 0.039