Question

A study claims that the average home sale price in Kansas is less than the average home sale price in Missouri. The average sales price for 35 home sales in Kansas is $240,993 with a standard deviation of $25,875. The average home sales price for 37 homes in Missouri is $249,237 with a standard deviation of $27,110. At the (alpha) a= 0.10, is there enough evidence to reject the study’s claim?
Complete a full hypothesis test for the following. Include the hypotheses, critical

value(s), test value and graph, your decision to reject or not reject, and a summary of

the information​

Answer 1

Since $|z| = 1.301 \leq z_1 = 1.64$ null hypothesis on this data is not rejected.

Null and Alternative Hypothesis

Data;

• sample mean x1 = 240993
• population standard deviation σ1 = 25875
• sample size n1 = 35
• sample mean x2 = 249237
• population standard mean σ2 = 27110
• sample size n2 = 35
• significance level α = 0.10

H_o = μ1 = μ2

H_o = μ1 ≠ μ2

This is a two-tailed test and population standard deviation will be used.

Test Statistic

The z-stat is

$z = \frac{x_1 - x_2}{\sqrt{\frac{\sigma_1^2}{n_1}+ \frac{\sigma_1 ^2}{n_2} } } \\z=\frac{240993-249237}{\sqrt{\frac{25875^2}{35}+ \frac{27110^2}{35} } } \\z = \frac{-8244}{6334.6388} \\z = -1.301$

The significance level is  α = 0.1 and the critical value for a two-tailed test is

$Z_t = Z_1 - _\alpha _/_2 = 1.64$

Since $|z| = 1.301 \leq z_1 = 1.64$, we can assume that the null hypothesis is not rejected.

Learn more on null hypothesis here;

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