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 σ = $650 . a) Develop a 90% confidence interval estimate of the population mean monthly rent.

b) Develop a 95% confidence interval estimate of the population mean monthly rent.

c) Develop a 99% confidence interval estimate of the population mean monthly rent.

d) What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? Explain.

Given that 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

Sample size n =120

Population std dev = [tex]\sigma = 650[/tex]

Since sigma is known we can use Z critical values for confidence interval

a) 90% confidence interval estimate of the population mean monthly rent

=Mean±1.645*std error

=[tex]3486+/- \frac{650}{\sqrt{120} } *1.645\\\\=(3476.239, 3496.761)[/tex]

b) 95% confidence interval estimate of the population mean monthly rent.

==Mean±1.96*std error

=[tex]3486+/- \frac{650}{\sqrt{120} }*1.96\\=(3474.37, 3497.63)[/tex]

c) 99% confidence interval estimate of the population mean monthly rent.

=Mean±2.58*std error

=[tex]3486+/- \frac{650}{\sqrt{120} }*2.58\\=(3470.691, 3501.309)[/tex]

d) Width increases as confidence level increases. This is reasonable because margin of error increases due to increase in critical value.