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)`

Explanation:

The confidence interval for population mean is given by :-

`\overline{x}\pm z_(\alpha/2)(\sigma)/(√(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)(6)/(√(240))\\\\\approx22.3\pm0.759\\\\=(22.3-0.759,\ 22.3+0.759)\\\\=(21.541,\ 23.059)`