Answer to Question #125935 in Statistics and Probability for Nur Syamimi

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 than the last 26 houses that sol 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.
a. What assumption regarding the population is necessary for making an interval estimate for the population mean? (2 Marks)
b. Construct the 90% confidence interval for the mean sale time for all homes in the neighborhood. (4 Marks)

Answer 1

a) The confidence interval is exact for normal populations and is approximately correct for large samples from non-normal populations. 1. For small samples, say, of size less than 15, the z-interval procedure should be used only when the variable under consideration is normally distributed or very close to being so.

b)The provided sample mean is  X = 218 and the population standard deviation is σ=72. The size of the sample is n = 26 and the required confidence level is 10%.

Based on the provided information,Using z table the critical z-value for α=0.1 is zc​=1.645.

Formula to find confidence interval:

$[\overline{X}-\frac{Zc*\sigma }{\sqrt{n}},\overline{X}+\frac{Zc*\sigma }{\sqrt{n}}] ; [218-\frac{1.645*72 }{\sqrt{26}},218+\frac{1.645*72 }{\sqrt{26}}] =(218−23.226,218+23.226) = (194.774, 241.226)$

which completes the calculation.