Question

A class survey in a large class for first-year college students asked, “About how many minutes do you study on a typical weeknight?” The mean response of the 463 students was = 118 minutes with a standard deviation of 65 minutes in the population of all first-year students at this university. Create a 99% confidence interval for the mean study time of all first-year students. Assume all necessary assumptions and conditions apply given the large sample size. Interpret your interval in context.

Answer 1

Answer:  (110.22, 125.78)

Step-by-step explanation:

The confidence interval for the population mean is given by :-

$\mu\ \pm z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size = 463

$\mu=118\text{ minutes}$

$\sigma=65\text{ minutes}$

Significance level : $\alpha=1-0.99=0.01$

Critical value : $z_{\alpha/2}=z_{0.005}=\pm2.576$

We assume that the population is normally distributed.

Now, the 90% confidence interval for the population mean will be :-

$118\ \pm\ 2.576\times\dfrac{65}{\sqrt{463}} \\\\\approx118\pm7.78=(118-7.78\ ,\ 118+7.78)=(110.22,\ 125.78)$

Hence, 99% confidence interval for the mean study time of all first-year students = (110.22, 125.78)