Question

A family is relocating from St. Louis, Missouri, to California. Due to an increasing inventory of houses in St. Louis, it is taking longer than before to sell a house. The wife is concerned and wants to know when it is optimal to put their house on the market. They ask their realtor friend for help and she informs them that the last 26 houses that sold in their neighborhood took an average time of 218 days to sell. The realtor also tells them that based on her prior experience, the population standard deviation is 72 days. Use Table 1.
a.
What assumption regarding the population is necessary for making an interval estimate of the population mean?

Assume that the population has a normal distribution.
Assume that the population has a uniform distribution.
b.
Construct a 90% confidence interval of the mean sale time for all homes in the neighborhood. (Round your intermediate calculations to 4 decimal places, "z" value and final answers to 2 decimal places.)

Answer 1

Answer:

a) Assume that the population has a normal distribution.

And the reason is because this distribution is tabulated and is possible to find quantiles for a confidence level given. And for the uniform distribution we don't have somthing like this.

b) $218- 1.6449 \frac{72}{\sqrt{26}}= 194.77$

$218+ 1.6449 \frac{72}{\sqrt{26}}= 241.23$

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".

$\bar X =218$ represent the sample mean

$\mu$ population mean (variable of interest)  

$\sigma=72$ represent the population standard deviation  

n=26 represent the sample size  

Calculate the confidence interval

Part a

For this case the correct answer would be:

Assume that the population has a normal distribution.

And the reason is because this distribution is tabulated and is possible to find quantiles for a confidence level given. And for the uniform distribution we don't have somthing like this.

Part b

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

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)  

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.6449$

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

$218- 1.6449 \frac{72}{\sqrt{26}}= 194.77$

$218+ 1.6449 \frac{72}{\sqrt{26}}= 241.23$