Question

In a survey of 1,200 airline​ travelers, 820 responded that the airline fee that is most unreasonable is additional charges to redeem​ points/miles. Construct a​ 95% confidence interval estimate for the population proportion of airline travelers who think that the airline fee that is most unreasonable is additional charges to redeem​ points/miles.

1. A​ 95% confidence interval estimate for the population proportion __________? ​(Round to four decimal places as​ needed.)

Answer 1

A​ 95% confidence interval estimate for the population proportion is (0.6570, 0.7096).

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence interval
`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:

In a survey of 1,200 airline​ travelers, 820 responded that the airline fee that is most unreasonable is additional charges to redeem​ points/miles. This means that
`n = 1200, \pi = (820)/(1200) = 0.6833`

95% confidence interval

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)} = 0.6833 - 1.96\sqrt{(0.6833*0.3167)/(1200)} = 0.6570`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.6833 + 1.96\sqrt{(0.6833*0.3167)/(1200)} = 0.7096`

A​ 95% confidence interval estimate for the population proportion is (0.6570, 0.7096).