A real estate company is interested in testing whether the mean time that families in Gotham have been living in their current h

Question

25 days ago

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A real estate company is interested in testing whether the mean time that families in Gotham have been living in their current h

omes is less than families in Metropolis. Assume that the two population variances are equal. A random sample of 100 families from Gotham and a random sample of 150 families in Metropolis yield the following data on length of residence in current homes. Gotham: XG = 35 months, SG2 = 900 Metropolis: XM= 50 months, SM2 = 105
Which of the following represents the result of the relevant hypothesis test?

a. The null hypothesis is rejected.
b. The alternative hypothesis is rejected.
c. The null hypothesis is not rejected.
d. Insufficient information exists on which to make a decision.

Mathematics

1 answer:

Inessa [11.2K] · Answer 1 · 2026-03-11T19:44:17+03:00

Answer:

Correct choice: a. The null hypothesis is rejected.

Step-by-step explanation:

Utilizing a statistical hypothesis test for comparing means can help ascertain if the average duration that families in Gotham have lived in their homes is shorter than that of families in Metropolis.

The hypothesis is:

H₀: The average duration of residence for families in Gotham is not shorter than that of families in Metropolis, meaning μ₁ ≥ μ₂.

Hₐ: The average duration of residence for families in Gotham is shorter than that of families in Metropolis, meaning μ₁ < μ₂.

Provided:

$\bar X_{G}=35\\\bar X_{M}=50\\S_{G}^{2}=900\\S_{M}^{2}=105\\n_{G}=100\\n_{M}=150$

Assuming the significance level of the test is α = 0.10.

The calculated statistic is:

$t=\frac{\bar X_{G}-\bar X_{M}}{s_{p}\sqrt{\frac{1}{n_{G}}+\frac{1}{n_{M}}}}$

Here, s $_{p}$ refers to the pooled standard deviation.

Calculate s $_{p}$ as follows:

$s_{p} = \sqrt{\frac{(n_{G}-1)S_{G}^{2}+(n_{M}-1)S_{M}^{2}}{n_{G}+n_{M}-2}}=\sqrt{\frac{(100-1)900+(150-1)105}{100+150-2}}=20.55$

Determine the value of the test statistic as follows:

$t=\frac{\bar X_{G}-\bar X_{M}}{s_{p}\sqrt{\frac{1}{n_{G}}+\frac{1}{n_{M}}}}=\frac{35-50}{20.55\sqrt{\frac{1}{100}+\frac{1}{150}}}=-5.654$

The result for the test statistic is -5.654.

The critical value for the test is, $t_{\alpha, (n_{G}+n_{M}-2)}=t_{0.10,(100+150-2)}=t_{0.10,248}=-1.282$

*Refer to a t-table for the necessary value.

*The negative value denotes that this is a left-tailed test.

*Since the table lacks an entry for 248 degrees of freedom, use the closest larger value available.

Decision rule:

If the test statistic is lower than the critical value, the null hypothesis is dismissed.

Test statistic = -5.654 < Critical t = -1.282.

Therefore, the null hypothesis is rejected.