Question

Ages of Homes Whiting, Indiana, leads the "Top 100 Cities with the Oldest Houses" list with the average age of houses being 66.4 years. Farther down the list resides Franklin, Pennsylvania, with an average house age of 59.4 years. Researchers selected a random sample of 20 houses in each city and obtained the following statistics. At \(\alpha=0.05,\) can it be concluded that the houses in Whiting are older? Use the \(P\) -value method. $$ \begin{array}{ccc}{} & {\text { Whiting }} & {\text { Franklin }} \\ \hline \text { Mean age } & {62.1 \text { years }} & {55.6 \text { years }} \\\ {\text { Standard deviation }} & {5.4 \text { years }} & {3.9 \text { years }}\end{array} $$

Short answer

The houses in Whiting are statistically older than those in Franklin at \( \alpha = 0.05 \).

Step-by-step solution

Define Hypotheses

First, establish the null and alternative hypotheses for the test. The null hypothesis (\( H_0 \)) states that the mean age of houses in Whiting is less than or equal to the mean age in Franklin. In contrast, the alternative hypothesis (\( H_a \)) posits that the mean age of houses in Whiting is greater than in Franklin. Thus, we have: \( H_0: \mu_1 \leq \mu_2 \) and \( H_a: \mu_1 > \mu_2 \), where \( \mu_1 \) is the mean age of houses in Whiting, and \( \mu_2 \) is the mean age in Franklin.

Analyze Sample Information

From the problem, we know that the sample sizes are both 20: \( n_1 = 20 \) and \( n_2 = 20 \). The sample means are \( \bar{x}_1 = 62.1 \) and \( \bar{x}_2 = 55.6 \). The sample standard deviations are \( s_1 = 5.4 \) and \( s_2 = 3.9 \) years, respectively.

Calculate the Test Statistic

Use the formula for the t-test statistic for independent samples: \[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\left(\frac{s_1^2}{n_1}\right) + \left(\frac{s_2^2}{n_2}\right)}} \]. Substituting the given values: \[ t = \frac{(62.1 - 55.6)}{\sqrt{\left(\frac{5.4^2}{20}\right) + \left(\frac{3.9^2}{20}\right)}} \]. Calculate this value to find the test statistic.

Find the Critical Value and Conclusion

The degrees of freedom (as per the formula for a t-test for independent samples with approximate calculation) are \( df = 34 \) (using a conservative approach as we're using equal sample sizes, \( df = n_1 + n_2 - 2 \)). Looking up the critical t-value for a one-tailed test at \( \alpha = 0.05 \) and \( df = 34 \), we find it to be approximately 1.690. If our calculated t-value is greater than this critical value, we reject the null hypothesis.

Key concepts

Understanding the T-Test

A t-test is a statistical method used to determine if there is a significant difference between the means of two groups. It helps us decide if the differences we observe in sample data reflect true differences in the populations we're studying or if they're simply due to random chance.
A t-test is particularly useful when dealing with small sample sizes. For our exercise, we are comparing the average age of houses in two cities: Whiting and Franklin, using a sample of 20 houses from each city. This scenario calls for an independent t-test because the samples are distinct and do not involve repeated measures from the same group.

• It provides a way to test the hypothesis and conclude whether the mean difference is statistically significant.
• Requires identifying the null and alternative hypotheses, calculating the test statistic, and comparing it with a critical value or using the P-value.

By conducting a t-test, we ultimately aim to determine if we have enough evidence to say one city's houses are indeed older on average than the other's.

Null Hypothesis in Hypothesis Testing

The null hypothesis, often denoted as \( H_0 \), is a statement suggesting that there is no effect or difference in the situation being tested. It's the position that one seeks to validate or refute through statistical analysis.
In the context of this problem, the null hypothesis states that the mean age of houses in Whiting is less than or equal to the mean age of houses in Franklin. That means: \( H_0: \mu_1 \leq \mu_2 \).
The null hypothesis serves as a baseline or starting point of any statistical test. By comparing it to an alternative hypothesis, researchers can assess whether any observed differences are meaningful or just due to natural variability. Rejection or acceptance of the null hypothesis will depend on the test results and statistical significance.

Alternative Hypothesis in Hypothesis Testing

The alternative hypothesis, represented as \( H_a \), is what you want to prove or validate through your test. It's the statement that shows the expected effect or difference, opposing the null hypothesis.
In our exercise, the alternative hypothesis claims that the mean age of houses in Whiting is greater than the mean age in Franklin: \( H_a: \mu_1 > \mu_2 \).
This hypothesis suggests a specific direction of difference, implying a one-tailed test. Here, we are only interested in knowing if Whiting's houses are older, not if they are the same or newer compared to Franklin's.

• If test results support the alternative hypothesis, it suggests a real difference exists beyond natural sampling variation.
• By validating the alternative hypothesis, we counter the default position (null hypothesis) with statistical evidence.

This focus on proving something contrary to the null hypothesis provides statistical significance to our findings.

Understanding the P-Value Method

The P-value method is used in hypothesis testing to help determine the strength of the results. A P-value measures the probability of obtaining a result at least as extreme as the one observed during the test, assuming that the null hypothesis is true.
The smaller the P-value, the stronger the evidence against the null hypothesis. Within this exercise, we compare the calculated P-value to our alpha level (significance level), which in this case is 0.05. Important things to consider:

• If the P-value is less than or equal to the alpha level, we reject the null hypothesis, indicating significant evidence that Whiting's houses are older.
• If the P-value is greater than the alpha level, we fail to reject the null hypothesis, indicating insufficient evidence to support our claim.

Using the P-value method provides a more intuitive way to interpret how significant the test results are and helps to make informed decisions based on the evidence at hand.