Answer:
Step-by-step explanation:
Hello!
The objective is to compare if the average time that families live in Gotham and Metropolis, the real state company thinks the average time is less in Gotham than in Metropolis. There are two populations of interest and therefore two variables:
X₁: Time a family has lived in Gotham.
n₁= 100 families
X[bar]₁= 35 months
S₁= 900 days
X₂: Time a family has lived in Metropolis.
n₂= 150 families
X[bar]₂= 50 months
S₂= 1050 days
You are studying the population means of both variables, so you have to work using the distribution of the sample means:
If:
X[bar]₁~N(μ₁;σ₁²/n₁)
X[bar]₂~N(μ₂;σ₂²/n₂)
We can say that the difference between these both distributions will have the following distribution:
(X[bar]₁-X[bar]₂)~N(μ₁-μ₂;σ₁²/n₁+σ₂²/n₂)
The population variance for the difference between the two sample means is:
σ₁²/n₁+σ₂²/n₂
Since both σ₁² = σ₂² = σ² we can say that:
σ² (1/n₁+1/n₂)To study the difference between two normal populations with unknown bu equal population variances the distribution to use is:
[tex]Z= \frac{(Xbar_1-Xbar_2)-(Mu_1-Mu_2)}{Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~N(0;1)[/tex]
The estimation of the standard error of the difference is:
[tex]Sa*\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }[/tex]
[tex]Sa^2= \frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} = \frac{(99*900^2)+(149*1050^2)}{100+150-2}= 985735.8871[/tex]
Sa= 992.8423≅ 992.84
[tex]992.84*\sqrt{\frac{1}{100} +\frac{1}{150} }= 128.1751= 128.18[/tex]
The standard error of the difference between the two sample means is 128.18.
I hope this helps!
