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.

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Answer:

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).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

[tex]n = 2305, \pi = \frac{339}{2305} = 0.147[/tex]

95% confidence level

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

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.147 - 1.96\sqrt{\frac{0.147*0.853}{2305}} = 0.133[/tex]

The upper limit of this interval is:

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

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).