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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. (

Use The Sample Data And Confidence Level Given Below To Complete Parts A Through DA Research Institute Poll Asked Respondents If They Felt Vulnerable To Identit class=

Sagot :

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.

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