Question

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.

Answer 1

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).