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

Answer:

The 90% confidence interval is 0.575 to 0.625.

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 interval $1-\alpha$, 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:

$n = 1000, \pi = 0.60$

90% confidence interval

So $\alpha = 0.1$, z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$, so $z = 1.645$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

The 90% confidence interval is 0.575 to 0.625.