Question

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. 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 .

Answer 1

Complete Question

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:

`CI: 3388.39<X<3583.61`

Explanation:

Sample Size n=120

Mean \=x =3486

Standard Deviation \sigma=650

Confidence interval CI=0.9

Therefore

Level of sig
`\alpha=0.1`

Therfore

The Critical Value from table is

Z_c=1.645

Generally the equation for Standard error is mathematically given by

`S.E=(\sigma)/(√(n))`

`S.E=(650)/(√(120))`

`S.E=59.3366`

Generally the equation for Margin error is mathematically given by

`M.E= = Z_c * SE`

`M.E=1.65 * 59.34`

`M.E= 97.61`

Therefore

`CI= \=x \pm M.E`

`CI= 3486 \pm 97.61`

Lower limit

`LL= \=x-M.E=3486-97.6087`

`LL= 3388.39`

Upper limit:

`UL= \=x+E=3486+97.6087`

`UL= 3583.61`

Therefore The 90% confidence interval estimate of the population mean monthly rent.

`CI: 3388.39<X<3583.61`