A real estate company would like to see if the average sale price of townhouses in a particular school district has changed in the last 12 months. A study conducted 12 months ago indicated that the average sale price of condominiums in this locality was $280,000. Data on recent sales were as follows:

Sample size n 36

Mean $294,365

Std ... Deviation $22,898

Required:

a. Write down the null and alternative hypotheses for this problem.

b. Specify the rejection region for conducting a two-tail test at a significance level of 5%.

c. Based on the data in the table, would you reject or fail to reject the null hypothesis for a 5% significance level?

Step-by-step explanation:

a) We would set up the hypothesis test.

For the null hypothesis,

H:0 µ = 280000

For the alternative hypothesis,

H1: µ ≠ 280000

This is a 2 tailed test

Since the population standard deviation is given, z score would be determined from the normal distribution table. The formula is

z = (x - µ) / (σ/√n)

Where

x = sample mean

µ = population mean

σ = population standard deviation

n = number of samples

From the information given,

µ = 280000

x = 294365

σ = 22898

n = 36

z = (294365 - 280000) / (22898/√36) = 3.76

b) Since α = 0.05, the critical value is determined from the normal distribution table.

For the left, α/2 = 0.05/2 = 0.025

The z score for an area to the left of 0.025 is - 1.96

For the right, α/2 = 1 - 0.025 = 0.975

The z score for an area to the right of 0.975 is 1.96

In order to reject the null hypothesis, the test statistic must be smaller than - 1.96 or greater than 1.96

Therefore, the rejection regions are area to the left of - 1.96 and to the right of 1.96 on the normal distribution curve.

The calculated test statistic is 3.76 for the right tail and - 3.76 for the left tail

c) Since - 3.76 1.96, we would reject the null hypothesis.