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.)

Accepted Solution

Answer: [tex](21.541,\ 23.059)[/tex]Step-by-step explanation:The confidence interval for population mean is given by :-[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]Given : Sample size : [tex]n=240[/tex] , which is a large sample , so we apply z-test .Sample mean : [tex]\overline{x}=22.3[/tex]Standard deviation : [tex]\sigma= 6[/tex]Significance level : [tex]\alpha=1-0.95=0.05[/tex]Critical value : [tex]z_{\alpha/2}=1.960[/tex]Now, a confidence interval at the 95% level of confidence will be :-[tex]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)[/tex]