/*! 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 24 Clark Heter is an industrial eng... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 24

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the afternoon shift than on the day shift. A sample of 54 day-shift workers showed that the mean number of units produced was \(345,\) with a standard deviation of 21 . A sample of 60 afternoon-shift workers showed that the mean number of units produced was \(351,\) with a standard deviation of 28 units. At the .05 significance level, is the number of units produced on the afternoon shift larger?

Short Answer

Expert verified
The mean units produced on the afternoon shift is not significantly larger.

Step by step solution

01

State the Hypotheses

We begin by stating the null and alternative hypotheses. The null hypothesis (\( H_0 \)) is that there is no difference in the mean number of units produced between the afternoon and day shifts, i.e., \( \mu_1 = \mu_2 \). The alternative hypothesis (\( H_a \)) is that the mean number of units produced on the afternoon shift is greater than on the day shift, i.e., \( \mu_1 > \mu_2 \).
02

Gather the Data

We have the following data for the day and afternoon shifts:- Day shift: \( n_1 = 54 \), \( \bar{x}_1 = 345 \), \( s_1 = 21 \).- Afternoon shift: \( n_2 = 60 \), \( \bar{x}_2 = 351 \), \( s_2 = 28 \).
03

Calculate the Test Statistic

Since we have two samples, we use the formula for the two-sample t-test. The test statistic is given by:\[t = \frac{(\bar{x}_2 - \bar{x}_1) - 0}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}\]Plugging in the values:\[t = \frac{(351 - 345) - 0}{\sqrt{\frac{21^2}{54} + \frac{28^2}{60}}}\]Calculating the denominator first:\[\sqrt{\frac{21^2}{54} + \frac{28^2}{60}} = \sqrt{\frac{441}{54} + \frac{784}{60}} = \sqrt{8.1667 + 13.0667} = \sqrt{21.2334} \approx 4.607\]Then the test statistic:\[t = \frac{6}{4.607} \approx 1.302\]
04

Determine the Critical Value

At a significance level of 0.05 for a one-tailed test with degrees of freedom calculated approximately using the formula:\[df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}\]Calculating the degrees of freedom:\[df \approx \frac{21.2334^2}{\frac{(\frac{441}{54})^2}{53} + \frac{(\frac{784}{60})^2}{59}} \approx 110.6\]Using a large sample approximation, we use the t-value from critical t-table for df = 110 and \(\alpha = 0.05\), which is approximately 1.645.
05

Make the Decision

Compare the calculated test statistic (1.302) with the critical value (1.645): Since 1.302 < 1.645, we fail to reject the null hypothesis.
06

Conclusion

Based on our calculation, at the 0.05 significance level, we do not have enough evidence to conclude that the number of units produced on the afternoon shift is significantly larger than the day shift.

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.

Hypothesis Testing
In statistics, hypothesis testing is a fundamental method for making decisions. It starts with an assumption or educated guess about a population parameter, known as a hypothesis. In this context, there are two types of hypotheses:
  • **Null Hypothesis ( \( H_0 \) )**: Represents the default or original assumption, stating that there is no effect or difference in the population. For example, "there is no difference in the mean number of units produced between the afternoon and day shifts," which can be mathematically expressed as \( \mu_1 = \mu_2 \) .
  • **Alternative Hypothesis ( \( H_a \) )**: Contradicts the null hypothesis and represents what the researcher aims to demonstrate. In this case, it's that the afternoon shift produces more units on average than the day shift, noted as \( \mu_1 > \mu_2 \) .
Once the hypotheses are set, the test utilizes statistical evidence from sample data to decide whether to reject the null hypothesis. The main goal is to determine whether there is enough statistical evidence to support the alternative hypothesis.
Significance Level
The significance level, often denoted by alpha ( \( \alpha \) ), is a crucial component in hypothesis testing. It represents the probability threshold at which the null hypothesis will be rejected. Commonly used significance levels are 0.05, 0.01, and 0.10.
  • **0.05 Significance Level**: Suggests that there is a 5% risk of concluding that a difference exists when there is none. This level is typically considered a fair balance between rigor and feasibility for many experiments.
When conducting a two-sample t-test, if the probability of the observed result is less than the significance level ( \( \alpha \) ), the null hypothesis is rejected. This means there's strong enough evidence to support the alternative hypothesis. In the given problem, a significance level of 0.05 was chosen, indicating the researcher was willing to accept a 5% chance of being wrong when claiming that more units are produced on the afternoon shift.
Degrees of Freedom
Degrees of freedom ( \( df \) ) is a statistical concept used to approximate the number of independent values or quantities which can vary freely in a calculation. In a two-sample t-test, degrees of freedom is essential to determine the critical value from the t-distribution.
  • **Formula for Degrees of Freedom**: It's calculated using the formula for pooled variance which considers the sample sizes and standard deviations of both samples.
  • **Importance**: A higher degrees of freedom usually equates to a more reliable estimate because it implies a larger sample size, which is closer to the actual population distribution.
In the exercise, we calculated that degrees of freedom were approximately 110.6. For practicality and ease, especially with large sample sizes, the t-distribution table is often used with the nearest whole number or a large approximation to find the critical t-value. This helps determine whether the observed data lies within the expected distribution should the null hypothesis be true.

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

The research department at the home office of New Hampshire Insurance conducts ongoing research on the causes of automobile accidents, the characteristics of the drivers, and so on. A random sample of 400 policies written on single persons revealed 120 had at least one accident in the previous three-year period. Similarly, a sample of 600 policies written on married persons revealed that 150 had been in at least one accident. At the . 05 significance level, is there a significant difference in the proportions of single and married persons having an accident during a three-year period?

The null and alternate hypotheses are: $$\begin{array}{l}H_{0}: \mu_{1}=\mu_{2} \\\H_{1}: \mu_{1} \neq \mu_{2}\end{array}$$ A random sample of 10 observations from one population revealed a sample mean of 23 and a sample deviation of \(4 .\) A random sample of 8 observations from another population revealed a sample mean of 26 and a sample standard deviation of \(5 .\) At the .05 significance level, is there a difference between the population means?

The management of Discount Furniture, a chain of discount furniture stores in the Northeast, designed an incentive plan for salespeople. To evaluate this innovative plan, 12 salespeople were selected at random, and their weekly incomes before and after the plan were recorded. Was there a significant increase in the typical salesperson's weekly income due to the innovative incentive plan? Use the .05 significance level. Estimate the \(p\) -value, and interpret it.

A recent study compared the time spent together by single- and dual-earner couples. According to the records kept by the wives during the study, the mean amount of time spent together watching television among the single-earner couples was 61 minutes per day, with a standard deviation of 15.5 minutes. For the dual-earner couples, the mean number of minutes spent watching television was 48.4 minutes, with a standard deviation of 18.1 minutes. At the .01 significance level, can we conclude that the single-earner couples on average spend more time watching television together? There were 15 single-earner and 12 dual-earner couples studied.

The null and alternate hypotheses are: $$\begin{array}{l}H_{0}: \pi_{1}=\pi_{2} \\\H_{1}: \pi_{1} \neq \pi_{2}\end{array}$$ A sample of 200 observations from the first population indicated that \(X_{1}\) is \(170 .\) A sample of 150 observations from the second population revealed \(X_{2}\) to be \(110 .\) Use the .05 significance level to test the hypothesis. a. State the decision rule. b. Compute the pooled proportion. c. Compute the value of the test statistic. d. What is your decision regarding the null hypothesis?
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

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.