Question

A survey of 1000 air travelers1 found that 60 % prefer a window seat. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is StartItalic UpperWord S E EndItalic equals 0.015.

Use a normal distribution to find a 90 % confidence interval for the proportion of air travelers who prefer a window seat.


Round your answers to three decimal places.


The 90% confidence interval is _____________to ____________

Answer 1

The 90% confidence interval is 0.575 to 0.625.

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:

`n = 1000, \pi = 0.60`

90% confidence interval

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`z = 1.645`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.60 - 1.645\sqrt{(0.60*0.40)/(1000)} = 0.575`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.60 + 1.645\sqrt{(0.60*0.40)/(1000)} {119}} = 0.625`

The 90% confidence interval is 0.575 to 0.625.