Question

a real estate agency says that the mean home sales price in casper, wyoming is the same as in cheyenne, wyoming. the mean home sales price for 35 homes in casper, wyoming is $349,237. assume the population standard deviation is $152,867. the mean home sales price for 41 homes in cheyenne, wyoming is $435,244. assume the population standard deviation is $137,829. at α = 0.05, is there enough evidence to reject the agency’s claim? complete parts (a) through (e).

a. the mean home sales price in casper, wyoming is the same as in cheyenne, wyoming.

b. the mean home sales price in casper, wyoming is greater than in cheyenne, wyoming.

c. the mean home sales price in casper, wyoming is not the same as in cheyenne, wyoming.

d. the mean home sales price in casper, wyoming is less than in cheyenne, wyoming.

let casper, wyoming be sample 1 and let cheyenne, wyoming be sample 2. identify h₀ and hₐ.

h₀: μ₁ = μ₂

hₐ: μ₁ ≠ μ₂

(b) find the critical value(s) and identify the rejection region. select the correct choice below and fill in the answer box within your choice.
(round to two decimal places as needed.)

a. the critical values are z₀ = ±

b. the critical value is z₀ =

Explanation:

Step1: Determine Test Type

This is a two - tailed z - test for the difference between two population means. The significance level \(\alpha = 0.05\). For a two - tailed test, we split the \(\alpha\) into two equal parts, \(\alpha/2=0.025\) in each tail.

Step2: Find Critical Z - Values

We need to find the \(z\) - values such that the area in each tail is \(0.025\). Looking up in the standard normal distribution table (or using a calculator with a normal - distribution function), the \(z\) - value corresponding to an area of \(0.975\) (since \(1 - 0.025=0.975\)) to the left of the \(z\) - value is \(z = 1.96\). For a two - tailed test, the critical values are \(z_{0}=\pm1.96\).

Answer:

A. The critical values are \(z_{0}=\pm1.96\)