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.

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

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 1000, p = \frac{260}{1000} = 0.26[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.26 - 1.96\sqrt{\frac{0.26*0.74}{1000}} = 0.2328[/tex]

The upper limit of this interval is:

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

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