Question

A group of statistics students decided to conduct a survey at their university to find the average (mean) amount of time students spent studying per week. They sampled 240 students and found a mean of 22.3 hours per week. Assuming a population standard deviation of six hours, what is the confidence interval at the 95% level of confidence? (Please use Student's t distribution (Appendix B.5) at infinite degrees of freedom for three decimal accuracy of the required z value.)

Answer 1

Answer: $(21.541,\ 23.059)$

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

Given : Sample size : $n=240$ , which is a large sample , so we apply z-test .

Sample mean : $\overline{x}=22.3$

Standard deviation : $\sigma= 6$

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.960$

Now, a confidence interval at the 95% level of confidence will be :-

$22.3\pm(1.960)\dfrac{6}{\sqrt{240}}\\\\\approx22.3\pm0.759\\\\=(22.3-0.759,\ 22.3+0.759)\\\\=(21.541,\ 23.059)$