Question

Russell is doing some research before buying his first house. He is looking at two different areas of the city, and he wants to know if there is a significant difference between the mean prices of homes in the two areas. For the 33 homes he samples in the first area, the mean home price is $168,300. Public records indicate that home prices in the first area have a population standard deviation of $37,825. For the 32 homes he samples in the second area, the mean home price is $181,900. Again, public records show that home prices in the second area have a population standard deviation of $25,070. Let Population 1 be homes in the first area and Population 2 be homes in the second area. Construct a 95% confidence interval for the true difference between the mean home prices in the two areas.

Answer 1

Answer:

The 95% confidence interval for the true difference between the mean home prices in the two areas is (-$29156.52, $1956.52).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

First area:

33 homes, mean of $168,300, standard deviation of $37,825. Thus:

$\mu_1 = 168300$

$s_1 = \frac{37825}{\sqrt{33}} = 6584.5$

Second area:

33 homes, mean of $181,900, standard deviation of $25,070. Thus:

$\mu_2 = 1819000$

$s_2 = \frac{25070}{\sqrt{32}} = 4431.8$

Distribution of the difference:

$\mu = \mu_1 - \mu_2 = 168300 - 181900 = -13600$

$s = \sqrt{s_1^2+s_2^2} = \sqt{6584.5^2 + 4431.8^2} = 7937$

Confidence interval:

$\mu \pm zs$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.  

The lower bound of the interval is:

$\mu - zs = -13600 - 1.96*7937 = -29156.52$

The upper bound of the interval is:

$\mu + zs = -13600 + 1.96*7937 = 1956.52$

The 95% confidence interval for the true difference between the mean home prices in the two areas is (-$29156.52, $1956.52).