Question

The chamber of commerce of a Florida Gulf Coast community advertises that area residential property is available at a mean cost of $125,000 or less per lot. Suppose a sample of 32 properties provided a sample mean of $130,000 per lot and a sample standard deviation of $12,500.
(a) Use a 0.05 level of significance to test the validity of the advertising claim.
(b) State the hypotheses.
(c) What is the p-value?
(d) What is your conclusion (use a level of significance of 0.05)?
(e) What is the interpretation of your conclusion?

Answer 1

Answer:

a) In the step-by-step explanation.

b)  The null and alternative hypothesis are:

$H_0: \mu=125000\\\\H_a:\mu> 125000$

c) P-value = 0.015

d) The conclusion, at a level of significance of 0.05, is that there is enough evidence to support the claim that that area residential property is available at a mean cost significantly higher than $125,000 per lot.

e) The claim of the chamber of commerce is false. We have statistical evidence against that claim and we can conclude that the mean prices are significantly higher than $125,000.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim, expressed in the alternative hypothesis, is that area residential property is available at a mean cost significantly higher than $125,000 per lot.

Then, the null and alternative hypothesis are:

$H_0: \mu=125000\\\\H_a:\mu> 125000$

The significance level is 0.05.

The sample has a size n=32.

The sample mean is M=130000.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12500.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12500}{\sqrt{32}}=2209.71$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{130000-125000}{2209.71}=\dfrac{5000}{2209.71}=2.26$

The degrees of freedom for this sample size are:

$df=n-1=32-1=31$

This test is a right-tailed test, with 31 degrees of freedom and t=2.26, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t>2.26)=0.015$

As the P-value (0.015) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that that area residential property is available at a mean cost significantly higher than $125,000 per lot.