a. The best point of estimate of the population of portion p is given by the formula:
[tex]p^{\prime}=\frac{x}{n}[/tex]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:
[tex]p^{\prime}=\frac{545}{1047}=0.521[/tex]b. The value of the margin of error E is given by the following formula:
[tex]E=(z_{\alpha/2})\cdot(\sqrt[]{\frac{p^{\prime}q^{\prime}}{n}})[/tex]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:
[tex]\begin{gathered} E=1.96\cdot\sqrt[]{\frac{0.521\cdot(1-0.521)}{1047}} \\ E=1.96\cdot\sqrt[]{\frac{0.521\cdot0.479}{1047}} \\ E=1.96\cdot\sqrt[]{\frac{0.2496}{1047}} \\ E=1.96\cdot\sqrt[]{0.0002} \\ E=1.96\cdot0.0154 \\ E=0.0303 \end{gathered}[/tex]c. The confidence interval is then:
[tex]\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}[/tex]d. We estimate with a 95% confidence that between 49% and 55.1% of the people felt vulnerable to identity theft.
