Question

PEI Real Estate company believes that their average house price in 2020 of $290,000 is higher than the mean price of all houses sold in 2019 of $270,000. Assuming that their estimate was based on the first 40 sales in 2020 and the population standard deviation of $70,000, test this hypothesis at the 95% confidence interval.

1. State the hypotheses based on a one-tailed test.

2. What is the level of significance?

3. Select a test statistic.  

4. Formulate the decision rule. Sketch this on a graph.

5. Calculate the test value.

6. What is the decision?

7. Does your conclusion change if the 90% confidence interval is selected? Explain.

Answer 1

Here claim is that mean is greater than 270000

As null hypothesis always have equality sign, hypothesis here is

vs

Here CI we need to use is 95%, so

As populations tandard deviation is known, so we will use z statistics

The z-critical value for a right-tailed test, for a significance level of α=0.05 is

zc​=1.64

Graphically

As test statistics falls in the rejection region, we reject the null hypothesis

Hence we have sufficient evidence to support the claim that mean is greater than 270000

The z-critical value for a right-tailed test, for a significance level of α=0.1 is

zc​=1.28

Graphically

As test statistics falls in this rejection region so our conclusion will not change.