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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.)

User Burk
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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)

User Reilstein
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