Question

Commuting Times for College Students The mean travel time to work for Americans is 25.3 minutes. An employment agency wanted to test the mean commuting times for college graduates and those with only some college. Thirty-five college graduates spent a mean time of 40.5 minutes commuting to work with a population variance of 67.24 . Thirty workers who had completed some college had a mean commuting time of 34.8 minutes with a population variance of \(39.69 .\) At the 0.05 level of significance, can a difference in means be concluded?

Short answer

Yes, there is a significant difference in commuting times at the 0.05 level of significance.

Step-by-step solution

State the Hypotheses

We want to determine if there is a significant difference between the mean commuting times of college graduates and those with some college education. The null hypothesis \(H_0\) states that there is no difference in commuting times, i.e., \(\mu_1 = \mu_2\). The alternative hypothesis \(H_a\) states that there is a difference, i.e., \(\mu_1 eq \mu_2\).

Determine the Test Statistic

We will use a two-sample Z-test for the difference of means since the population variances are known. The formula for the Z statistic is:\[ Z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]where \(\bar{x}_1 = 40.5\), \(\bar{x}_2 = 34.8\), \(\sigma_1^2 = 67.24\), \(\sigma_2^2 = 39.69\), \(n_1 = 35\), and \(n_2 = 30\).

Compute the Test Statistic

Substitute the given values into the Z formula:\[ Z = \frac{40.5 - 34.8}{\sqrt{\frac{67.24}{35} + \frac{39.69}{30}}} \]Calculate the denominator:\[ \sqrt{\frac{67.24}{35} + \frac{39.69}{30}} = \sqrt{1.9206 + 1.323} = \sqrt{3.2436} \approx 1.80099 \]Substitute back to find Z:\[ Z \approx \frac{5.7}{1.80099} \approx 3.16 \]

Determine the Critical Value

For a two-tailed test with a significance level of \(\alpha = 0.05\), the critical Z values are \(\pm 1.96\). This is the cutoff for determining significance.

Make a Decision

The calculated Z value is 3.16, which falls outside the range of \(-1.96\) to \(+1.96\). Therefore, we reject the null hypothesis.

Conclusion

There is enough statistical evidence at the 0.05 level of significance to conclude that there is a significant difference in the mean commuting times for college graduates and those with some college education.

Key concepts

Two-Sample Z-Test

In statistics, a two-sample Z-test is used to determine if there is a significant difference between the means of two independent groups. It is especially applicable when the population variances are known and the sample sizes are large enough to assume normal distribution of the sample means. In this exercise, the two groups are college graduates and those who have completed some college. To perform this test, we compare their commuting times using the formula: \[ Z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \] where \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(\sigma_1^2\) and \(\sigma_2^2\) are the population variances, and \(n_1\) and \(n_2\) are the sample sizes. The Z-test helps us assess if the observed difference between groups is due to random chance or if it's statistically significant.

Null Hypothesis

The null hypothesis, denoted as \(H_0\), is a statement that suggests there is no effect or no difference in the context of the test. It's the default assumption that any observed differences are due to random variability rather than a true effect. In our exercise, the null hypothesis is that there is no difference between the commuting times of college graduates and those with some college education, mathematically expressed as \(\mu_1 = \mu_2\). Establishing the null hypothesis is critical as the basis for statistical testing. When we conduct a Z-test, we are essentially trying to find evidence against the null hypothesis. If our data suggest a significant difference, we then reject \(H_0\). Otherwise, we do not reject \(H_0\) and conclude there's not enough evidence to say a difference exists.

Critical Value

The critical value is a threshold that helps decide whether to reject the null hypothesis. For two-tailed tests like the one in the exercise, the critical values are located at both extremes of the probability distribution, representing a chosen significance level's cut-off points. Here, at a significance level (\(\alpha\)) of 0.05, the critical Z values for a two-tailed test are \(\pm 1.96\). This means if the calculated Z-score falls beyond the range \([-1.96, +1.96]\), we have evidence to reject the null hypothesis, indicating a significant difference. Otherwise, if the Z-score falls within this range, we retain the null hypothesis.

Significance Level

The significance level, represented by \(\alpha\), is the probability of rejecting the null hypothesis when it is actually true. It's essentially a measure of the risk we are willing to take of making a Type I error, which is concluding that a difference exists when in reality, it doesn't. Commonly used significance levels include 0.01, 0.05, and 0.10. In the exercise, the significance level is set at 0.05, meaning there is a 5% chance of incorrectly rejecting the null hypothesis. By choosing a significance level before conducting the test, researchers can control the likelihood of a false positive, establishing a balance between sensitivity and specificity of the test. This helps ensure that any findings are not due to chance but reflect the true differences between the groups.