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

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 $\pi$, and a confidence level of $1-\alpha$, 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:

$n = 2305, \pi = \frac{339}{2305} = 0.147$

95% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.147 - 1.96\sqrt{\frac{0.147*0.853}{2305}} = 0.133$

The upper limit of this interval is:

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

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).