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	2,671	2,755	2,500	2,489	2,446	2,376	2,327	2,450	2,834
Pinewood	2,659	2,340	2,813	2,775	2,734	2,420	2,457	2,535	3,449

(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 μ₁ is the mean of home sizes in Greenwood and μ₂ is the mean of home sizes in Pinewood.

(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.)

Answer 1

The provided sample means are shown below:

Also, the sample standard deviations are:

s1​=174.3588, s2​=330.7111

and the sample sizes are n1​=9 and n2​=9.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ1​ = μ2​

Ha: \μ1​ ≠ μ2​

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

The critical values are FL​=0.226 and FU​=4.433, and since F = 0.278, then the null hypothesis of equal variances is not rejected.

(2) Rejection Region

The significance level is α = 0.10, and the degrees of freedom are df = 16. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:

Hence, it is found that the critical value for this two-tailed test is tc​ = 1.745884, for α = 0.10 and df = 16.

The rejection region for this two-tailed test is R = { t : ∣t∣ > 1.745884}.

(3) Test Statistics

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

​​

(4) Decision about the null hypothesis

Since it is observed that ∣t∣ = 0.004 ≤ tc ​= 1.745884, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.9965, and since p = 0.9965 ≥ 0.10, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level.

We can conclude that there is no difference between the average sizes of homes in the two neighborhoods at the .10 significance level

d.

Assuming unequal variances.

s1​=174.3588, s2​=330.7111

n1​=9 and n2​=9

On solving, we get df = 12.1285.

Therefore, the p-value is calculated by 2 * P(|t| > -0.004) at df = 12.1285 is 0.9968735.

2*pt(q = -0.004,df = 12.1285,lower.tail = TRUE)

The p-value is 0.9968735