Question

In a random sample of 2,305 college students, 339 reported getting 8 or more hours of sleep per night. Create a 95% confidence interval for the proportion of college students who get 8 or more hours of sleep per night. Use a TI-83, TI-83 plus, or TI-84 calculator, rounding your answers to three decimal places.

Answer 1

The 95% confidence interval for the proportion of college students who get 8 or more hours of sleep per night is (0.133, 0.161).

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{(\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 = 2305, \pi = (339)/(2305) = 0.147`

95% confidence level

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)} = 0.147 - 1.96\sqrt{(0.147*0.853)/(2305)} = 0.133`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.147 + 1.96\sqrt{(0.147*0.853)/(2305)} = 0.161`

The 95% confidence interval for the proportion of college students who get 8 or more hours of sleep per night is (0.133, 0.161).