/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Commuting Times for College Stud... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

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

Step by step solution

01

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., \(\mu_1 eq \mu_2\).
02

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 \(n_2 = 30\).
03

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:\[ Z \approx \frac{5.7}{1.80099} \approx 3.16 \]
04

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.
05

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.
06

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For Exercises 2 through \(12,\) perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Reducing Errors in Spelling A ninth-grade teacher wishes to see if a new spelling program will reduce the spelling errors in his students' writing. The number of spelling errors made by the students in a five-page report before the program is shown. Then the number of spelling errors made by students in a five-page report after the program is shown. At \(\alpha=0.05,\) did the program work? $$ \begin{array}{lllllll}{\text { Before }} & {8} & {3} & {10} & {5} & {9} & {11} & {12} \\ \hline \text { After } & {6} & {4} & {8} & {1} & {4} & {7} & {11}\end{array} $$

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Bullying Bullying is a problem at any age but especially for students aged 12 to 18 . A study showed that \(7.2 \%\) of all students in this age bracket reported that bullied at school during the past six months with 6 th grade having the highest incidence at \(13.9 \%\) and 12 th grade the lowest at \(2.2 \%\). To see if there is a difference between public and private schools, 200 students were randomly selected from each. At the 0.05 level of significance, can a difference be concluded? $$ \begin{array}{ccc}{} & {\text { Private }} & {\text { Public }} \\ \hline \text { Sample size } & {200} & {200} \\ {\text { No. bullied }} & {13} & {16}\end{array} $$

For these exercises, perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified and assume the variances are unequal. Tax-Exempt Properties A tax collector wishes to see if the mean values of the tax-exempt properties are different for two cities. The values of the tax- exempt properties for the two random samples are shown. The data are given in millions of dollars. At \(\alpha=0.05,\) is there enough evidence to support the tax collector's claim that the means are different? $$ \begin{array}{cccc|cccc}{} & {\text { City } \mathbf{A}} & {} & {} & {} & {\text { City } \mathbf{B}} & {} & {} \\ \hline 113 & {22} & {14} & {8} & {82} & {11} & {5} & {15} \\ {25} & {23} & {23} & {30} & {295} & {50} & {12} & {9} \\ {44} & {11} & {19} & {7} & {12} & {68} & {81} & {2} \\ {31} & {19} & {5} & {2} & {} & {20} & {16} & {4} & {5}\end{array} $$

Noise Levels in Hospitals The mean noise level of 20 randomly selected areas designated as "casualty doors", was \(63.1 \mathrm{dBA}\), and the sample standard deviation is \(4.1 \mathrm{dBA}\). The mean noise level for 24 randomly selected areas designated as operating theaters was \(56.3 \mathrm{dBA}\), and the sample standard deviation was \(7.5 \mathrm{dBA}\). At \(\alpha=0.05,\) can it be concluded that there is a difference in the means?

ACT Scores A random survey of 1000 students nationwide showed a mean ACT score of 21.4 . Ohio was not used. A survey of 500 randomly selected Ohio scores showed a mean of 20.8 . If the population standard deviation is \(3,\) can we conclude that Ohio is below the national average? Use \(\alpha=0.05 .\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.