Question

An apartment complex developer is considering building apartments in College Town, but first wants to do a market study. A sample was selected of monthly rent values ($) for 80 studio apartments in College Town with sample mean of 498.76. (Based on past experience, the developer assumes a known value of s = $65 for the population standard deviation.)

a. Develop a 98% confidence interval for the mean monthly rent for all studio apartments in this city.
b. Suppose the apartment developer wants a 98% confidence interval estimate of the population mean with a margin of error of E = $15. What sample size is needed?

Answer 1

a. The 98% confidence interval for the mean monthly rent for all studio apartments in this city is between $481.85 and $515.67

b. A sample size of 102 is needed.

Explanation:

Question a:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.98)/(2) = 0.01`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.01 = 0.99`, so Z = 2.327.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 2.327(65)/(√(80)) = 16.91`

The lower end of the interval is the sample mean subtracted by M. So it is 498.76 - 16.91 = $481.85

The upper end of the interval is the sample mean added to M. So it is 498.76 + 16.91 = $515.67

The 98% confidence interval for the mean monthly rent for all studio apartments in this city is between $481.85 and $515.67

Question b:

This is n for which M = 15. So

`M = z(\sigma)/(√(n))`

`15 = 2.327(65)/(√(n))`

`15√(n) = 2.327*65`

`√(n) = (2.327*65)/(15)`

`(√(n))^2 = ((2.327*65)/(15))^2`

`n = 101.68`

Rounding up:

A sample size of 102 is needed.