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. Her realtor friend informs them that the last 29 houses that sold in their neighborhood took an average time of 230 days to sell. The realtor also tells them that based on her prior experience, the population standard deviation is 35 days.
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.

Answer 1

Answer:

a. Assume that the population has a normal distribution.

b. The 90% confidence interval of the mean sale time for all homes in the neighborhood is between 219.31 days and 240.69 days.

Step-by-step explanation:

Question a:

We have to assume normality.

Question b:

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

$\alpha = \frac{1 - 0.9}{2} = 0.05$

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 $1 - 0.05 = 0.95$, so Z = 1.645.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

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

$M = 1.645\frac{35}{\sqrt{29}} = 10.69$

The lower end of the interval is the sample mean subtracted by M. So it is 230 - 10.69 = 219.31 days.

The upper end of the interval is the sample mean added to M. So it is 230 + 10.69 = 240.69 days.

The 90% confidence interval of the mean sale time for all homes in the neighborhood is between 219.31 days and 240.69 days.