Question

Home Prices A real estate agent compares the selling prices of randomly selected homes in two municipalities in southwestern Pennsylvania to see if there is a difference. The results of the study are shown. Is there enough evidence to reject the claim that the average cost of a home in both locations is the same? Use \(\alpha=0.01 .\) $$ \begin{array}{cc}{\text { Scott }} & {\text { Ligonier }} \\ \hline \overline{X_{1}=\$ 93,430^{*}} & {\bar{X}_{2}=\$ 98,043^{*}} \\\ {\sigma_{1}=\$ 5602} & {\sigma_{2}=\$ 4731} \\ {n_{1}=35} & {n_{2}=40}\end{array} $$

Short answer

There is enough evidence at \(\alpha=0.01\) to reject the claim that the average home costs are the same in both locations.

Step-by-step solution

Understand the Hypotheses

Formulate the null and alternative hypotheses. The null hypothesis \(H_0\) is that the average costs of homes in both locations are the same, \(\mu_1 = \mu_2\). The alternative hypothesis \(H_1\) is that the average costs are different, \(\mu_1 eq \mu_2\).

Select the Appropriate Test Statistic

Because we have sample means, standard deviations, and sample sizes for both groups, use a two-sample z-test for the mean. The test statistic is calculated with the formula: \( z = \frac{\overline{X}_1 - \overline{X}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \).

Compute the Z-Statistic

Calculate the z-test statistic using the given values: \(\overline{X}_1 = 93430\), \(\overline{X}_2 = 98043\), \(\sigma_1 = 5602\), \(\sigma_2 = 4731\), \(n_1 = 35\), and \(n_2 = 40\). Substitute these into the formula: \[ z = \frac{93430 - 98043}{\sqrt{\frac{5602^2}{35} + \frac{4731^2}{40}}} \].

Calculate the Standard Error

First, calculate the individual components: \( \frac{5602^2}{35} = 896132.4 \) and \( \frac{4731^2}{40} = 559333.025 \). The combined standard error \(SE\) is \( \sqrt{896132.4 + 559333.025} = 1198.88 \).

Calculate the Test Statistic Value

Use the standard error to find the test statistic: \( z = \frac{93430 - 98043}{1198.88} \approx -3.84 \).

Determine the Critical Value

For \(\alpha = 0.01\) in a two-tailed test, the critical z-values are approximately \(-2.576\) and \(2.576\).

Make the Decision

Since the calculated \(z\) value of \(-3.84\) is less than \(-2.576\), it falls in the critical region. Thus, we reject the null hypothesis.

Key concepts

Hypothesis Testing

Hypothesis testing is a method used in statistics to determine whether there is enough evidence in a sample of data to infer that a certain condition is true for the entire population. In our exercise, we are examining if the average home price in two municipalities is the same or different. To achieve this, we start by formulating two hypotheses:

1. **Null Hypothesis (\(H_0\))**: This assumes there is no effect or difference. Here, it claims the average home prices in both locations are equal (\(\mu_1 = \mu_2\)).
2. **Alternative Hypothesis (\(H_1\))**: This is what you want to prove and claims there is a difference in average prices (\(\mu_1 eq \mu_2\)).

We conduct hypothesis testing by choosing a significance level (\(\alpha\)), which in this case is 0.01. This level represents a 1% risk of concluding that a difference exists when there is no actual difference.

Standard Error Calculation

The standard error is a critical component in hypothesis testing, as it measures the accuracy with which a sample represents a population. A lower standard error indicates more precise estimates.

In our context, since we have two independent samples, we calculate the standard error for the difference between the sample means. The formula used is:
\[SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\]where:

• \(\sigma_1\) and \(n_1\) are the standard deviation and sample size for the first population.
• \(\sigma_2\) and \(n_2\) are for the second population.

In the provided exercise, these figures lead to a standard error of 1198.88. This value indicates the variability in the difference between the two sample means.

Critical Z-Values

Critical z-values define the cutoff points for deciding whether to reject the null hypothesis. They are derived from the standard normal distribution and are crucial in two-tailed tests, like ours.

With a significance level (\(\alpha\)) of 0.01, the critical z-values for a two-tailed test are approximately -2.576 and 2.576. What this means in practice is that any calculated z-score falling below -2.576 or above 2.576 indicates a statistically significant result, allowing us to reject the null hypothesis.

In our example, the calculated z-statistic is -3.84, which is less than -2.576. This shows substantial evidence against the null hypothesis, suggesting that the average home costs in the two municipalities are indeed different.

Null and Alternative Hypotheses

In the realm of hypothesis testing, clearly defining the null and alternative hypotheses is an integral first step. These hypotheses form the foundation of your statistical test.

The **null hypothesis (\(H_0\))** posits that there is no difference or effect. In the housing price case, it suggests that the average prices in the two municipalities are the same (\(\mu_1 = \mu_2\)).

On the other hand, the **alternative hypothesis (\(H_1\))** asserts the opposite. It proposes that there is a measurable difference (\(\mu_1 eq \mu_2\)).

These hypotheses are tested using sample data. If you find enough evidence (like significant z-scores in our example), you can reject the null hypothesis and conclude that the average prices differ. If not, the null hypothesis stands, and you'd assert that the data does not show a significant difference in average home prices between Scott and Ligonier.