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Assisted-Living Facility Rent.Costs are rising for all kinds of medical care. The mean monthly rent at assisted-living facilities was reported to have increased 17% over the last five years to $3486 (the Wall Street Journal, October 27, 2012). Assume this cost estimate is based on a sample of 120 facilities and, from past studies, it can be assumed that the population standard deviation is s = $650. a. Develop a 90% confidence interval estimate of the population mean monthly rent.
Answer:
[tex]CI: 3388.39<X<3583.61[/tex]
Step-by-step explanation:
Sample Size n=120
Mean \=x =3486
Standard Deviation \sigma=650
Confidence interval CI=0.9
Therefore
Level of sig [tex]\alpha=0.1[/tex]
Therfore
The Critical Value from table is
Z_c=1.645
Generally the equation for Standard error is mathematically given by
[tex]S.E=\frac{\sigma}{\sqrt{n}}[/tex]
[tex]S.E=\frac{650}{\sqrt{120}}[/tex]
[tex]S.E=59.3366[/tex]
Generally the equation for Margin error is mathematically given by
[tex]M.E= = Z_c * SE[/tex]
[tex]M.E=1.65 * 59.34[/tex]
[tex]M.E= 97.61[/tex]
Therefore
[tex]CI= \=x \pm M.E[/tex]
[tex]CI= 3486 \pm 97.61[/tex]
Lower limit
[tex]LL= \=x-M.E=3486-97.6087[/tex]
[tex]LL= 3388.39[/tex]
Upper limit:
[tex]UL= \=x+E=3486+97.6087[/tex]
[tex]UL= 3583.61[/tex]
Therefore The 90% confidence interval estimate of the population mean monthly rent.
[tex]CI: 3388.39<X<3583.61[/tex]
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