Question

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. $$\begin{array}{lrr} & \text { Whiting } & \text { Franklin } \\\\\hline \text { Mean age } & 62.1 \text { years } & 55.6 \text { years } \\\\\text { Standard deviation } & 5.4 \text { years } & 3.9 \text { years }\end{array}$$

Short answer

Reject the null hypothesis; Whiting's houses are statistically older.

Step-by-step solution

Define the Hypotheses

Formulate the null and alternative hypotheses. The null hypothesis (^0) states that there is no difference in the average age of houses between Whiting and Franklin. The alternative hypothesis (^a) suggests that houses in Whiting are older. Thus: \[ \begin{align*} H_0: \mu_{\text{W}} - \mu_{\text{F}} \leq 0 \ H_a: \mu_{\text{W}} - \mu_{\text{F}} > 0 \end{align*} \] where \(\mu_{\text{W}}\) and \(\mu_{\text{F}}\) are the mean ages of houses in Whiting and Franklin, respectively.

Calculate the Test Statistic

The formula for the test statistic for two independent means is: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where \(\bar{x}_1 = 62.1\), \(s_1 = 5.4\), \(n_1 = 20\) for Whiting and \(\bar{x}_2 = 55.6\), \(s_2 = 3.9\), \(n_2 = 20\) for Franklin. Calculate: \[ t = \frac{62.1 - 55.6}{\sqrt{\frac{5.4^2}{20} + \frac{3.9^2}{20}}} \] \[ t = \frac{6.5}{\sqrt{\frac{29.16}{20} + \frac{15.21}{20}}} \] \[ t = \frac{6.5}{\sqrt{2.58 + 0.76}} \] \[ t \approx \frac{6.5}{1.88} \approx 3.46 \]

Determine the Degrees of Freedom

Use the formula for degrees of freedom for two independent samples, which is approximated as follows in this case: \[ df = \left\lfloor\frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1 - 1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2 - 1}}\right\rfloor \] Calculate this: \[ df \approx \frac{\left(\frac{29.16}{20} + \frac{15.21}{20}\right)^2}{\frac{\left(\frac{29.16}{20}\right)^2}{19} + \frac{\left(\frac{15.21}{20}\right)^2}{19}} \] \[ df \approx \frac{11.09}{0.053 + 0.016} \approx 36.44 \] Thus, round down to use \( df \approx 36 \).

Find the Critical Value and P-value

For \(\alpha = 0.05\) and \(\text{df} = 36\), use a t-table or calculator to find the critical t-value for a one-tailed test. The critical value \(t_{\alpha, df}\) is approximately 1.686. Now, find the P-value using the calculated t-score (\(t = 3.46\)). Looking at a t-distribution table or using software, the P-value is much less than 0.05.

Decision and Conclusion

Since the calculated t-statistic is greater than the critical t-value (3.46 > 1.686) and the P-value is less than 0.05, we reject the null hypothesis. There is sufficient evidence to conclude that the houses in Whiting are older than those in Franklin.

Key concepts

Null Hypothesis

In hypothesis testing, the null hypothesis is a statement that there is no effect or no difference. It acts as a starting point for the test and assumes that any observed difference is due to random chance.
The goal is to test whether the data contradicts this assumption.

• For the given exercise, the null hypothesis (\( H_0 \)) claims that there is no difference in the average age of houses between Whiting and Franklin, or that the houses in Whiting are not older than those in Franklin.
• This is mathematically expressed as: \[ H_0: \mu_{\text{W}} - \mu_{\text{F}} \leq 0 \]

By maintaining the null hypothesis, researchers start with the assumption that any difference is accidental. This standard approach helps in creating a baseline for comparison.

Alternative Hypothesis

The alternative hypothesis is what the researcher aims to support. It proposes a specific effect or difference that stands in contrast to the null hypothesis.
When evidence is strong enough to reject the null, the alternative hypothesis is considered plausible.

• In our context, the alternative hypothesis (\( H_a \)) argues that houses in Whiting are older than those in Franklin.
• It is mathematically stated as: \[ H_a: \mu_{\text{W}} - \mu_{\text{F}} > 0 \]

This hypothesis is what researchers hope to prove with their data. It suggests a real, non-random effect based on the statistical evidence collected.

P-value

The P-value is a key component in hypothesis testing. It measures the strength of evidence against the null hypothesis by providing the probability of observing the data, or something more extreme, assuming the null is true.

• A smaller P-value indicates stronger evidence against the null hypothesis.
• In this exercise, the calculated P-value is much less than 0.05, suggesting strong evidence against the null hypothesis.

When a P-value is less than a chosen significance level (\( \alpha \)), it advises rejecting the null hypothesis. For \( \alpha = 0.05 \), rejecting it denotes a less than 5% probability of observing such data if the null hypothesis were true.

Degrees of Freedom

Degrees of Freedom (df) relate to the number of independent pieces of information available to estimate a parameter. They inform the shape of statistical distributions, such as the t-distribution in this case.

• For two independent sample t-tests, df is determined using an approximation formula. This helps in assessing the validity of test results.
• In our problem, the calculated df was approximately 36.44, which was rounded down to 36 for practical use.

Degrees of freedom are critical for interpreting the t-test results. They help in finding the critical t-value to decide whether the test statistic is significant enough to reject the null hypothesis.