/*! 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 18 Work hours and education, Part I... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 18

Work hours and education, Part II. The General Social Survey described in Exercise 4.16 included random samples from two groups: US residents with a college degree and US residents without a college degree. For the 505 sampled US residents with a college degree, the average number of hours worked each week was 41.8 hours with a standard deviation of 15.1 hours. For those 667 without a degree, the mean was 39.4 hours with a standard deviation of 15.1 hours. Conduct a hypothesis test to check for a difference in the average number of hours worked for the two groups.

Short Answer

Expert verified
There is a significant difference in average work hours between those with and without a degree.

Step by step solution

01

Define the Hypotheses

We want to test if there is a significant difference in the number of hours worked per week between those with a college degree and those without. Let \( \mu_1 \) be the mean hours worked by those with a college degree and \( \mu_2 \) be the mean for those without. The null hypothesis (\( H_0 \)) is that there is no difference in the means: \( H_0: \mu_1 = \mu_2 \). The alternative hypothesis (\( H_a \)) is that there is a difference: \( H_a: \mu_1 eq \mu_2 \).
02

Determine the Significance Level

Select a significance level, commonly \( \alpha = 0.05 \), for the hypothesis test. This means we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.
03

Identify the Test Statistic

We will use a two-sample t-test to compare the means of the two independent samples. The formula for the test statistic is: \[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] where \( \bar{x}_1 = 41.8 \), \( s_1 = 15.1 \), and \( n_1 = 505 \) are the sample mean, standard deviation, and size for the college degree group, and \( \bar{x}_2 = 39.4 \), \( s_2 = 15.1 \), and \( n_2 = 667 \) for the group without a degree.
04

Calculate the Test Statistic

Substitute the values into the test statistic formula:\[ t = \frac{(41.8 - 39.4)}{\sqrt{\frac{15.1^2}{505} + \frac{15.1^2}{667}}} \] Calculate \( t \):\[ t = \frac{2.4}{\sqrt{\frac{228.01}{505} + \frac{228.01}{667}}} \] \[ t = \frac{2.4}{\sqrt{0.4515 + 0.3418}} \] \[ t = \frac{2.4}{\sqrt{0.7933}} \] \[ t = \frac{2.4}{0.89} \] \[ t \approx 2.70 \]
05

Find the Critical Value and Make a Decision

For a two-tailed test at \( \alpha = 0.05 \), compute the degrees of freedom using the formula for two-sample t-tests, which we approximate using the smaller sample size minus one, \( n_{min} - 1 = 505 - 1 = 504 \). Using a t-distribution table or calculator, the critical value for two-tailed \( \alpha = 0.05 \), \( df = 504 \) is about \( t_{critical} \approx \pm 1.96 \). Since \( |t| \approx 2.70 \) is greater than \( 1.96 \), we reject the null hypothesis.

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 91影视!

Key Concepts

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

Two-sample t-test
A two-sample t-test is a statistical method used to determine whether there is a significant difference between the means of two independent groups. It's commonly applied in scenarios where researchers want to compare the performance, effectiveness, or other characteristics across two distinct groups. In this particular exercise, the goal is to compare the average work hours of individuals with and without a college degree.

To perform a two-sample t-test, we calculate the test statistic using the formula:\[t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]where:
  • \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means
  • \(s_1\) and \(s_2\) are the standard deviations
  • \(n_1\) and \(n_2\) are the sample sizes
The formula calculates how many standard deviations the difference in sample means is from zero, which allows us to test if this difference is statistically significant.
Statistical Significance
Statistical significance is a cornerstone of hypothesis testing and indicates whether a result is likely due to a specific effect rather than random chance.

When conducting a hypothesis test, we choose a significance level, denoted by \(\alpha\), that represents the probability of rejecting the null hypothesis when it is true. Common values for \(\alpha\) are 0.05 or 0.01.

In this example, an \(\alpha\) level of 0.05 was used. This means there is a 5% risk of concluding there is a difference in work hours between those with and without a college degree when, in fact, there is not. To determine significance, the computed test statistic is compared against a critical value derived from the t-distribution. If the absolute value of the test statistic exceeds the critical value, the result is considered statistically significant, prompting the rejection of the null hypothesis.
Null and Alternative Hypothesis
Hypotheses are fundamental components in statistical testing, helping to frame questions and guide decision-making. The null hypothesis, denoted as \(H_0\), assumes no effect or no difference between groups. In this case, the null hypothesis (\[H_0: \mu_1 = \mu_2\]) proposes that there is no difference in mean working hours between those with and without a college degree.

The alternative hypothesis, \(H_a\), is what you aim to support through testing. It posits that there is a difference between the groups. Here, the alternative hypothesis (\[H_a: \mu_1 eq \mu_2\]) suggests that the average work hours differ between the two sets of residents.

The decision on which hypothesis to support is determined based on the statistical analysis outcome. If the data provides strong evidence against \(H_0\), it is rejected in favor of \(H_a\). In our exercise, the calculated test statistic was greater than the critical value, allowing us to conclude that the difference in average working hours is indeed significant.

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

Does the Paleo diet work? The Paleo diet allows only for foods that humans typically consumed over the last 2.5 million years, excluding those agriculture-type foods that arose during the last 10,000 years or so. Researchers randomly divided 500 volunteers into two equal-sized groups. One group spent 6 months on the Paleo diet. The other group received a pamphlet about controlling portion sizes. Randomized treatment assignment was performed, and at the beginning of the study, the average difference in weights between the two groups was about 0. After the study, the Paleo group had lost on average 7 pounds with a standard deviation of 20 pounds while the control group had lost on average 5 pounds with a standard deviation of 12 pounds. (a) The \(95 \%\) confidence interval for the difference between the two population parameters (Paleo \- control) is given as (-0.891,4.891) . Interpret this interval in the context of the data. (b) Based on this confidence interval, do the data provide convincing evidence that the Paleo diet is more effective for weight loss than the pamphlet (control)? Explain your reasoning. (c) Without explicitly performing the hypothesis test, do you think that if the Paleo group had lost 8 instead of 7 pounds on average, and everything else was the same, the results would then indicate a significant difference between the treatment and control groups? Explain your reasoning.

Weight gain during pregnancy. In \(2004,\) the state of North Carolina released to the public a large data set containing information on births recorded in this state. This data set has been of interest to medical researchers who are studying the relationship between habits and practices of expectant mothers and the birth of their children. The following histograms show the distributions of weight gain during pregnancy by 867 younger moms (less than 35 years old) and 133 mature moms ( 35 years old and over) who have been randomly sampled from this large data set. The average weight gain of younger moms is 30.56 pounds, with a standard deviation of 14.35 pounds, and the average weight gain of mature moms is 28.79 pounds, with a standard deviation of 13.48 pounds. Calculate a \(95 \%\) confidence interval for the difference between the average weight gain of younger and mature moms. Also comment on whether or not this interval provides strong evidence that there is a significant difference between the two population means.

\(t^{*}\) vs. \(z^{*}\). For a given confidence level, \(t_{d f}^{*}\) is larger than \(z^{*}\). Explain how \(t_{d f}^{*}\) being slightly larger than \(z^{*}\) affects the width of the confidence interval.

True or false, Part III. Determine if the following statements are true or false, and explain your reasoning for statements you identify as false. If the null hypothesis that the means of four groups are all the same is rejected using ANOVA at a \(5 \%\) significance level, then ... (a) we can then conclude that all the means are different from one another. (b) the standardized variability between groups is higher than the standardized variability within groups. (c) the pairwise analysis will identify at least one pair of means that are significantly different. (d) the appropriate \(\alpha\) to be used in pairwise comparisons is \(0.05 / 4=0.0125\) since there are four groups.

Coffee, depression, and physical activity. Caffeine is the world's most widely used stimulant, with approximately \(80 \%\) consumed in the form of coffee. Participants in a study investigating the relationship between coffee consumption and exercise were asked to report the number of hours they spent per week on moderate (e.g., brisk walking) and vigorous (e.g., strenuous sports and jogging) exercise. Based on these data the researchers estimated the total hours of metabolic equivalent tasks (MET) per week, a value always greater than \(0 .\) The table below gives summary statistics of MET for women in this study based on the amount of coffee consumed. 3 (a) Write the hypotheses for evaluating if the average physical activity level varies among the different levels of coffee consumption. (b) Check conditions and describe any assumptions you must make to proceed with the test. (c) Below is part of the output associated with this test. Fill in the empty cells. (d) What is the conclusion of the test?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
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.