Answer: [tex](336.49,\ 355.51)[/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= 314 [/tex] , which is a large sample , so we apply z-test .
Sample mean : [tex]\overline{x}=346 [/tex]
Standard deviation : [tex]\sigma= 86[/tex]
Significance level : [tex]\alpha=1-0.95=0.05[/tex]
Critical value : [tex]z_{\alpha/2}=1.96[/tex]
Now, a confidence interval at the 95% level of confidence will be :-
[tex]346\pm(1.96)\dfrac{86}{\sqrt{314}}\\\\\approx346\pm9.51\\\\=(346-9.51,\ 346+9.51)\\\\=(336.49,\ 355.51)[/tex]
The 96% confidence interval estimate for the mean monthly rent of all unmarried BYU students in winter 2018 is (336.488, 355.512)
The given parameters are:
[tex]\mu = 346[/tex] --- mean
[tex]\sigma = 86[/tex] ---- the standard deviation
[tex]n=314[/tex] --- sample size
For 96% confidence interval, the critical value (z) is 1.96
The confidence interval is calculated using:
[tex]CI = \mu \± z \times \frac{\sigma}{\sqrt n}[/tex]
So, we have:
[tex]CI = 346 \± 1.96 \times \frac{86}{\sqrt {314}}[/tex]
[tex]CI = 346 \± 9.512[/tex]
Split
[tex]CI = (346 - 9.512,346 + 9.512)[/tex]
[tex]CI = (336.488,\ 355.512)[/tex]
Hence, the required confidence interval estimate is (336.488, 355.512)
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