Question

Use the sample data and confidence level given below to complete parts (a) through (d)A research institute poll asked respondents if they felt vulnerable to identity theft. In the poll, n = 1047 and x = 545 whosaid "yes." Use a 95% 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. (

Answer 1

a. The best point of estimate of the population of portion p is given by the formula:

`p^(\prime)=(x)/(n)`

where x is the number of successes x=545 and n is the sample n=1047.

Replace these values in the formula and find p:

`p^(\prime)=(545)/(1047)=0.521`

b. The value of the margin of error E is given by the following formula:

`E=(z_(\alpha/2))\cdot(\sqrt[]{(p^(\prime)q^(\prime))/(n)})`

Where z is the z-score at the alfa divided by 2, q'=1-p'.

As the confidence level is 95%=0.95, then alfa is 1-0.95=0.05, and alfa/2=0.025

The z-score at 0.025 is 1.96.

Replace the known values in the formula and solve for E:

`\begin{gathered} E=1.96\cdot\sqrt[]{(0.521\cdot(1-0.521))/(1047)} \\ E=1.96\cdot\sqrt[]{(0.521\cdot0.479)/(1047)} \\ E=1.96\cdot\sqrt[]{(0.2496)/(1047)} \\ E=1.96\cdot\sqrt[]{0.0002} \\ E=1.96\cdot0.0154 \\ E=0.0303 \end{gathered}`

c. The confidence interval is then:

`\begin{gathered} (p^(\prime)-E,p^(\prime)+E)=(0.521-0.0303,0.521+0.0303) \\ \text{Confidence interval=}(0.490,0.551) \end{gathered}`

d. We estimate with a 95% confidence that between 49% and 55.1% of the people felt vulnerable to identity theft.