Mathematics, 10.03.2020 08:27 cyndiann2002

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.

Answers: 3

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Answers

Answer from: kvekrcurf

Using the t-distribution, it is found that the correct option is:

a. The null hypothesis is rejected.

At the null hypothesis, it is tested if the mean time that families in Gotham have been living in their current homes is less than families in Metropolis, that is:

$H_0: \mu_G - \mu_H \geq 0$

At the alternative hypothesis, it is tested if it is less, that is:

$H_0: \mu_G - \mu_H < 0$

The standard errors are:

$s_G = \sqrt{\frac{900}{100}} = 3$

$s_M = \sqrt{\frac{105}{150}} = 0.8367$

The distribution of the difference is given by:

$\overline{x} = \mu_G - \mu_H = 35 - 50 = -15$

$s = \sqrt{s_G^2 + s_M^2} = \sqrt{3^2 + 0.8367^2} = 3.1145$

The t-test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{-15 - 0}{3.1145}$

t = -4.82

The p-value is found using a left-tailed test, as we are testing if the mean is less than a value, with t = -4.82 and 150 + 100 - 2 = 248 df.

Using a t-distribution calculator, this p-value is of approximately 0, hence the null hypothesis is rejected and option a is correct.

A similar problem is given at link

A real estate company is interested in testing whether the mean time that families in Gotham have be

Answer from: shambrayperkins

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:

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