/*! 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 48 Three different assembly methods... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Three different assembly methods have been proposed for a new product. A completely randomized experimental design was chosen to determine which assembly method results in the greatest number of parts produced per hour, and 30 workers were randomly selected and assigned to use one of the proposed methods. The number of units produced by each worker follows. Use these data and test to see whether the mean number of parts produced is the same with each method. Use \(\alpha=.05\)

Short Answer

Expert verified
Perform an ANOVA test and compare the F-statistic to the critical value to determine if differences exist among means.

Step by step solution

01

Define the Hypotheses

First, we need to set up our null and alternative hypotheses for this problem. The null hypothesis \(H_0\) states that the mean number of parts produced per hour is the same for all three assembly methods. The alternative hypothesis \(H_a\) states that at least one method has a different mean number of parts produced per hour. Formally, \(H_0: \mu_1 = \mu_2 = \mu_3\) and \(H_a:\) not all means are equal.
02

Summarize the Data

Calculate means and standard deviations for each assembly method from the data provided. Each method will have a sample mean \(\bar{x}_i\) and standard deviation \(s_i\) based on the number of workers (30 in total divided by the number of methods).
03

Conduct an ANOVA Test

An Analysis of Variance (ANOVA) test is suitable here because we are comparing more than two group means. Compute the F-statistic using the formula: \[ F = \frac{\text{Variation Between Groups}}{\text{Variation Within Groups}} \]. You need to calculate the mean square between (MSB) and mean square within (MSW) variance. These are derived from the sums of squares between (SSB) and within (SSW) groups.
04

Calculate the Test Statistic

Use the results from the ANOVA calculations to compute the F-statistic. This involves dividing the MSB by the MSW. Compare the calculated F-statistic to the critical F-value from the F-distribution table with \(\text{df}_1 = k - 1\) and \(\text{df}_2 = N - k\) degrees of freedom, where \(k\) is the number of groups and \(N\) is the total number of observations.
05

Make a Decision

Check whether the calculated F-statistic is greater than the critical F-value (obtained from an F-table for \(\alpha = 0.05\) with appropriate degrees of freedom). If the F-statistic exceeds the critical value, we reject the null hypothesis, indicating that there is a significant difference in means among the methods. Otherwise, we fail to reject \(H_0\).
06

Interpret the Results

Based on the ANOVA results, if we reject the null hypothesis, it means at least one assembly method has a different mean number of parts produced per hour. If we fail to reject, the data do not show a significant difference. We may need further analysis if a more detailed comparison between groups is needed.

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.

ANOVA test
The Analysis of Variance (ANOVA) test is a statistical method used when we want to compare the means of three or more groups. It's especially useful when those groups have been randomly assigned different treatments or conditions. When using ANOVA, the main goal is to determine if there are any statistically significant differences between the means of these groups. The test works by analyzing the variance within each group and comparing it to the variance between the groups. Here's why: - Variance within groups tells us how data in the same group differ from their group mean. - Variance between groups indicates how much group means differ from the overall mean of all data combined. In simpler terms, if the variance between groups is much larger than the variance within groups, it indicates that not all group means are the same, and a significant difference indeed exists. Using ANOVA involves establishing a null and an alternative hypothesis (which we'll cover next). After performing the necessary calculations, including the F-statistic, one can decide if the means are significantly different.
null hypothesis
In hypothesis testing, the null hypothesis is the claim that there is no effect or no difference. It sets a baseline that the experimenter seeks to test. For example, in our assembly methods exercise, the null hypothesis (\( H_0 \)) suggests that all assembly methods produce the same number of parts on average. Formally, we express this as \( \mu_1 = \mu_2 = \mu_3 \), where \( \mu \) represents the mean number of parts produced per hour for each method.The role of the null hypothesis is crucial because it serves as the point of comparison. During statistical testing, the collected data and its analysis tell us whether there is enough evidence to reject this null hypothesis. If the evidence is strong enough, it suggests that some change or difference exists beyond random chance.The decision to reject or not reject the null hypothesis is based on the calculated F-statistic and a critical F-value for a specified significance level, usually \( \alpha = 0.05 \), as in this exercise.
alternative hypothesis
The alternative hypothesis, on the other hand, is the statement we want to provide evidence for if the null hypothesis is rejected. In our example with assembly methods, the alternative hypothesis (\( H_a \)) claims that not all means are equal. In simpler terms, it posits that there is a difference in the effectiveness of at least one assembly method in producing parts.The alternative hypothesis can be non-specific, meaning it doesn't pinpoint exactly which group or groups have different means, just that they differ. Mathematically, it is expressed as \( \mu_1 eq \mu_2 eq \mu_3 \).It is key to understand that rejecting the null hypothesis doesn't prove the alternative to be true; it only indicates that the data is inconsistent with the null hypothesis based on the chosen significance level. Further tests may be needed to determine exactly which group's mean differs from the others. However, a rejected null hypothesis supports the notion that differences do indeed exist.
F-statistic
The F-statistic is a key component of the ANOVA test. It provides a ratio that compares the variance between group means to the variance within groups. The formula for calculating the F-statistic is: \[ F = \frac{\text{Mean Square Between Groups (MSB)}}{\text{Mean Square Within Groups (MSW)}} \]Where Mean Square is the sum of squares divided by its respective degrees of freedom. Here's a closer look at what this entails:- **Mean Square Between (MSB)**: Represents the variation due to the interaction between the different sample means. Calculated using the sum of squares between the means (SSB).- **Mean Square Within (MSW)**: Represents variation within each sample population. This is calculated from the sum of squares within the groups (SSW).The F-statistic tells us if the variation between the group means is significant when compared to the variation within the groups. We then compare this calculated F-statistic to a critical F-value from F-distribution tables to make decisions about our hypotheses.If our F-statistic is greater than the critical F-value, the result is significant. This implies that there are meaningful differences between group means, leading us to reconsider our null hypothesis. If it is not greater, we fail to reject the null hypothesis.

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

In early 2009 , the economy was experiencing a recession. But how was the recession affecting the stock market? Shown are data from a sample of 15 companies. Shown for each company is the price per share of stock on January 1 and April 30 (The Wall Street Journal, May 1,2009 ). a. What is the change in the mean price per share of stock over the four-month period? b. Provide a \(90 \%\) confident interval estimate of the change in the mean price per share of stock. Interpret the results. c. What was the percentage change in the mean price per share of stock over the fourmonth period? d. If this same percentage change were to occur for the next four months and again for the four months after that, what would be the mean price per share of stock at the end of the year \(2009 ?\)

Consider the following data for two independent random samples taken from two normal populations. $$\begin{array}{l|rrrrrr} \text { Sample 1 } & 10 & 7 & 13 & 7 & 9 & 8 \\\\\hline \text { Sample 2 } & 8 & 7 & 8 & 4 & 6 & 9\end{array}$$ a. Compute the two sample means. b. Compute the two sample standard deviations. c. What is the point estimate of the difference between the two population means? d. What is the \(90 \%\) confidence interval estimate of the difference between the two population means?

Condé Nast Traveler conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100 -point scale, with higher values indicating better service. A sample of 37 ships that carry fewer than 500 passengers resulted in an average rating of \(85.36,\) and a sample of 44 ships that carry 500 or more passengers provided an average rating of 81.40 (Condé Nast Traveler, February 2008 ). Assume that the population standard deviation is 4.55 for ships that carry fewer than 500 passengers and 3.97 for ships that carry 500 or more passengers. a. What is the point estimate of the difference between the population mean rating for ships that carry fewer than 500 passengers and the population mean rating for ships that carry 500 or more passengers? b. \(\quad\) At \(95 \%\) confidence, what is the margin of error? c. What is a \(95 \%\) confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?

The following results are for independent random samples taken from two populations. \(\begin{array}{ll}\text { Sample } 1 & \text { Sample } 2 \\\ n_{1}=20 & n_{2}=30 \\ \bar{x}_{1}=22.5 & \bar{x}_{2}=20.1 \\ s_{1}=2.5 & s_{2}=4.8\end{array}\) a. What is the point estimate of the difference between the two population means? b. What is the degrees of freedom for the \(t\) distribution? c. \(\quad\) At \(95 \%\) confidence, what is the margin of error? d. What is the \(95 \%\) confidence interval for the difference between the two population means?

The U.S. Department of Transportation provides the number of miles that residents of the 75 largest metropolitan areas travel per day in a car. Suppose that for a random sample of 50 Buffalo residents the mean is 22.5 miles a day and the standard deviation is 8.4 miles a day, and for an independent random sample of 40 Boston residents the mean is 18.6 miles a day and the standard deviation is 7.4 miles a day. a. What is the point estimate of the difference between the mean number of miles that Buffalo residents travel per day and the mean number of miles that Boston residents travel per day? b. What is the \(95 \%\) confidence interval for the difference between the two population means?

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.