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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=1003 and x = 519 who said "yes." Use a 99% confidence level.Click the icon to view a table of z scores.a) Find the best point estimate of the population proportion p.(Round to three decimal places as needed.)

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Solution

Part a

p=x/n = 519/1003= 0.517

Part b

The margin of error is given by:

[tex]ME=z_{\frac{\alpha}{2}}\cdot\sqrt[]{\frac{p(1-p)}{n}}[/tex]

At 99 % confidence level the z value is z= 2.576

Replacing we got:

[tex]ME=2.576\cdot\sqrt[]{\frac{0.517(1-0.517)}{1003}}=0.0407[/tex]

Rounded it would 0.041

Part c

For this case we can find the confidence interval on this way:

[tex]0.517\pm0.041=(0.476;0.558)[/tex]

Part d

For this case the correct answer is:

c. One has 99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion

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