Question

Nine homes are chosen at random from real estate listings in two suburban neighborhoods, and the square footage of each home is noted in the following table.

Size of Homes in Two Subdivisions
Subdivision	Square Footage
Greenwood	2,780	2,710	2,404	2,622	2,413	2,888	2,378	2,609	2,650
Pinewood	2,484	2,356	2,453	2,759	2,631	2,672	2,373	2,511	3,093

(a) Choose the appropriate hypothesis to test if there is a difference between the average sizes of homes in the two neighborhoods at the .10 significance level. Assume ?₁ is the mean of home sizes in Greenwood and ?₂ is the mean of home sizes in Pinewood.

a. H₀: ?₁ – ?₂ = 0 vs. H₁: ?₁– ?₂ ? 0

b. H₀: ?₁ – ?₂ ? 0 vs. H₁: ?₁– ?₂ = 0

(b) Specify the decision rule with respect to the p-value.

Reject the null hypothesis if the p-value is (Click to select)greater thanless than 0.10.

(c) Find the test statistic t_calc. (A negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

(d) Assume unequal variances to find the p-value. (Use the quick rule to determine degrees of freedom. Round your answer to 4 decimal places.)

p-value            

(e) Make a decision.

We reject or do not reject the null hypothesis.

(f) State your conclusion.

We can or cannot conclude that there is a difference between the sizes of homes in the two neighborhoods.

Answer 1

(a)

Hypotheses are:

(b)

If p-value is less than 0.10 so we reject the null hypothesis.

(c)

Following is the output of descriptive statistics:

Descriptive statistics
	Greenwood	Pinewood
count	9	9
mean	2,606.00	2,592.44
sample standard deviation	177.78	231.30
sample variance	31,606.75	53,501.53
minimum	2378	2356
maximum	2888	3093
range	510	737

So we have

The test statistics will be

(d)

Since it is not given that variances are equal so degree of freedom of the test is

So df is 8.

The test is two tailed so p-value is: 0.8929

(e)

Fail to reject the null hypothesis.

(f)

We cannot conclude that there is a difference between the sizes of homes in the two neighborhoods.