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In a survey of 623 adults, 95 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for the proportion of adults who regularly lie to people conducting surveys. Use a TI-83, TI-83 plus, or TI-84 calculator, rounding your answers to three decimal places.

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Answer:

The 99% confidence interval for the proportion of adults who regularly lie to people conducting surveys is (0.116, 0.19).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
\pi, and a confidence level of
1-\alpha, we have the following confidence interval of proportions.


\pi \pm z\sqrt{(\pi(1-\pi))/(n)}

In which

z is the zscore that has a pvalue of
1 - (\alpha)/(2).

For this problem, we have that:


n = 623, \pi = (95)/(623) = 0.153

99% confidence level

So
\alpha = 0.01, z is the value of Z that has a pvalue of
1 - (0.01)/(2) = 0.995, so
Z = 2.575.

The lower limit of this interval is:


\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.153 - 2.575\sqrt{(0.153*0.848)/(623)} = 0.116

The upper limit of this interval is:


\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.1525 + 2.575\sqrt{(0.153*0.845)/(623)} = 0.19

The 99% confidence interval for the proportion of adults who regularly lie to people conducting surveys is (0.116, 0.19).

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