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

Answer:

a) $884-2.426\frac{50}{\sqrt{40}}=864.82$    

$884+2.426\frac{50}{\sqrt{40}}=903.179$    

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

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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):

$884-2.426\frac{50}{\sqrt{40}}=864.82$    

$884+2.426\frac{50}{\sqrt{40}}=903.179$    

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