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

Answer:

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

Step-by-step 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.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 1000, p = \frac{260}{1000} = 0.26$

95% confidence level

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

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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