Take this test as you would take a test in class.For each exercise, perform the steps below.

d. Find the appropriate standardized test statistic.


A real estate agency says that the mean home sales price in Olathe, Kansas, is greater than in Rolla, Missouri. The mean home sales price for 39 homes in Olathe is \$392,453. Assume the population standard deviation is \$224,902. The mean home sales price for 38 homes in Rolla is \$285,787. Assume the population standard deviation is \$330,578. At α=0.05, is there enough evidence to support the agency’s claim? (Adapted from Realtor.com)

Verified step by step guidance

1. Identify the hypotheses for the test. Since the agency claims that the mean home sales price in Olathe is greater than in Rolla, set the null hypothesis as \(H_0: \mu_{Olathe} \leq \mu_{Rolla}\) and the alternative hypothesis as \(H_a: \mu_{Olathe} > \mu_{Rolla}\).
2. Determine the type of test to use. Because population standard deviations are known and sample sizes are given, use a two-sample z-test for the difference of means.
3. Calculate the standardized test statistic using the formula for the difference between two means with known population standard deviations: \[ Z = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \] Here, \(\bar{X}_1\) and \(\bar{X}_2\) are the sample means, \(\sigma_1\) and \(\sigma_2\) are the population standard deviations, and \(n_1\) and \(n_2\) are the sample sizes. Since under the null hypothesis \(\mu_1 - \mu_2 = 0\), this simplifies the numerator.
4. Substitute the given values into the formula: - \(\bar{X}_1 = 392,453\), \(\sigma_1 = 224,902\), \(n_1 = 39\) (Olathe) - \(\bar{X}_2 = 285,787\), \(\sigma_2 = 330,578\), \(n_2 = 38\) (Rolla) Calculate the denominator by finding the square root of the sum of the variances divided by their respective sample sizes.
5. Compute the value of the test statistic \(Z\) using the substituted values. This \(Z\) value will then be compared to the critical value from the standard normal distribution at \(\alpha = 0.05\) for a one-tailed test to decide whether to reject the null hypothesis.

Key Concepts

Hypothesis Testing

Hypothesis testing is a statistical method used to decide whether there is enough evidence to support a specific claim about a population parameter. It involves formulating a null hypothesis (no effect or difference) and an alternative hypothesis (the claim), then using sample data to determine if the null can be rejected at a given significance level (α).

Two-Sample Z-Test for Means

A two-sample z-test compares the means of two independent populations when population standard deviations are known. It calculates a standardized test statistic to assess if the difference between sample means is statistically significant, assuming normality or large sample sizes for approximation.

Significance Level and Critical Value

The significance level (α) is the threshold probability for rejecting the null hypothesis, commonly set at 0.05. The critical value corresponds to α and defines the rejection region for the test statistic. If the test statistic exceeds this critical value, the null hypothesis is rejected in favor of the alternative.