Question

In a survey of 1000 large corporations, 260 said that, given a choice between a job candidate who smokes and an equally qualified nonsmoker, the nonsmoker would be preferred. Construct a 95% confidence interval for the proportion of corporations preferring a nonsmoking candidate.

Answer 1

The 95% confidence interval for the proportion of corporations preferring a nonsmoking candidate is (0.2328, 0.2872)

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 = 1000, p = (260)/(1000) = 0.26`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.26 - 1.96\sqrt{(0.26*0.74)/(1000)} = 0.2328`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.26 + 1.96\sqrt{(0.26*0.74)/(1000)} = 0.2872`

The 95% confidence interval for the proportion of corporations preferring a nonsmoking candidate is (0.2328, 0.2872)