Question

As part of their business promotional package, the Milwaukee Chamber of Commerce would like an estimate of the mean cost per month to lease a one-bedroom apartment. The mean cost per month for a random sample of 40 apartments currently available for lease was $884. The standard deviation of the sample was $50. a. Develop a 98% confidence interval for the population mean. b. Would it be reasonable to conclude that the population mean is $950 per month

Answer 1

a)
`884-2.426(50)/(√(40))=864.82`

`884+2.426(50)/(√(40))=903.179`

So on this case the 98% confidence interval would be given by (864.82;903.179)

b) For this case the upper bound of the confidence interval is lower than 950 so we don't have a significant evidence to conclude that the true mean is 950 per month

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=884` represent the sample mean

`\mu` population mean (variable of interest)

s=50 represent the sample standard deviation

n=40 represent the sample size

Solution to the problem

Part a

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=40-1=39`

Since the Confidence is 0.98 or 98%, the value of
`\alpha=0.02` and
`\alpha/2 =0.01`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,39)".And we see that
`t_(\alpha/2)=2.426`

Now we have everything in order to replace into formula (1):

`884-2.426(50)/(√(40))=864.82`

`884+2.426(50)/(√(40))=903.179`

So on this case the 98% confidence interval would be given by (864.82;903.179)

Part b

For this case the upper bound of the confidence interval is lower than 950 so we don't have a significant evidence to conclude that the true mean is 950 per month