Question

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

Answer 1

Answer:

Correct option: a. The null hypothesis is rejected.

Step-by-step explanation:

A statistical hypothesis test for difference between two means can be used to determine whether the mean time that families in Gotham have been living in their current homes is less than families in Metropolis.

The hypothesis is:

H₀: The mean time that families in Gotham in their current homes is not less than families in Metropolis,i.e. μ₁ ≥ μ₂.

Hₐ: The mean time that families in Gotham in their current homes is less than families in Metropolis,i.e. μ₁ < μ₂.

Given:

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

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

The test statistic is:

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

Here s$_{p}$ = pooled standard deviation.

Compute the value of s$_{p}$ as follows:

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

Compute the test statistic value 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 test statistic value is -5.654.

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

*Use a t-table for the value.

*The negative sign is because of the test is left-tailed.

*Since the table does not consist the value for the degrees of freedom 248 use the next highest value.

Decision rule:

If the test statistic value is less than the critical value then the null hypothesis is rejected.

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

Thus, the null hypothesis is rejected.