/*! 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 26 There are four auto body shops i... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

There are four auto body shops in a community and all claim to promptly serve customers. To check if there is any difference in service, customers are randomly selected from each repair shop and their waiting times in days are recorded. The output from a statistical software package is: $$\begin{array}{|lclll|}\hline {\text { Summary }} \\ \text { Groups } & \text { Count } & \text { Sum } & \text { Average } & \text { Variance } \\\\\hline \text { Body Shop A } & 3 & 15.4 & 5.133333 & 0.323333 \\\\\text { Body Shop B } & 4 & 32 & 8 & 1.433333 \\\\\text { Body Shop C } & 5 & 25.2 & 5.04 & 0.748 \\\\\text { Body Shop D } & 4 & 25.9 & 6.475 & 0.595833 \\\\\hline\end{array}$$ $$\begin{array}{|lcrccc|}\hline {\text { ANOVA }} \\\\\hline \text { Source of Variation } & \text { SS } &\text { df } & \text { MS } & \text { F } & \text { p-value } \\\\\hline \text { Between Groups } & 23.37321 & 3 & 7.791069 & 9.612506 & 0.001632 \\\\\text { Within Groups } & 9.726167 & 12 & 0.810514 & & \\\\\text { Total } & 33.09938 & 15 & & & \\\\\hline\end{array}$$ Is there evidence to suggest a difference in the mean waiting times at the four body shops? Use the .05 significance level.

Short Answer

Expert verified
Yes, there is a significant difference in mean waiting times at the body shops.

Step by step solution

01

Understanding the Problem

The problem involves determining if there is a significant difference in the mean waiting times at four different auto body shops based on the provided ANOVA table with a significance level of \( \alpha = 0.05 \).
02

Identify the Hypotheses

The null hypothesis \( H_0 \) states that there is no difference in the mean waiting times among the four body shops. The alternative hypothesis \( H_a \) states that at least one shop has a different mean waiting time.
03

Examine the ANOVA Table

The ANOVA table provides the calculated F-value as 9.612506 and the p-value as 0.001632 for the test comparing the means of the four groups.
04

Decision Rule

If the p-value is less than the significance level (\( \alpha = 0.05 \)), we reject the null hypothesis. Otherwise, we do not reject the null hypothesis.
05

Conclusion

The p-value (0.001632) is less than the significance level (0.05), therefore, we reject the null hypothesis. There is enough statistical evidence to suggest a difference in the mean waiting times at the four body shops.

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.

Hypothesis Testing
Hypothesis testing is a fundamental part of statistics and is used to determine whether there is enough evidence in data to infer that a condition is true for a whole population. Imagine you want to test if all auto body shops deliver similar waiting times. This can be formulated into a hypothesis test.

We start by defining two hypotheses based on the research question. The null hypothesis, denoted as \( H_0 \), typically states that there is no effect or difference. In this scenario, \( H_0 \) is that the waiting times at all body shops are the same. The alternative hypothesis, \( H_a \), states the opposite, that at least one body shop's waiting time is different.

To test these hypotheses, we collect data, analyze it and assess whether the findings are consistent with the null hypothesis using statistical algorithms like ANOVA (Analysis of Variance). This serves as a pivotal part of making an informed decision. If the evidence suggests rejection of \( H_0 \), we conclude there are differences in waiting times across the shops.
P-Value
The p-value is a crucial metric in hypothesis testing, giving us a measure of the strength of evidence against the null hypothesis. Think of it as indicating the probability of observing the results we obtained, or more extreme, assuming that the null hypothesis is true.

In the example with the body shops, the p-value was calculated to be 0.001632. A very small p-value like this suggests that if \( H_0 \) were true (i.e., there is no difference in waiting times), obtaining such an extreme result would be highly unlikely.

Since the p-value is significantly low, it indicates a strong evidence against the null hypothesis. In practical scenarios, a low p-value implies that the observed data could not easily occur by chance, advocating for the acceptance of the alternative hypothesis.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold set by the researcher to decide when to reject the null hypothesis. It represents the probability of making a Type I error, which is when the null hypothesis is wrongly rejected.

In our body shop example, we use a significance level of 0.05. This means we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.
  • If our p-value is less than 0.05, we reject the null hypothesis, suggesting a significant difference in waiting times.
  • If the p-value is greater than or equal to 0.05, we do not reject the null hypothesis.

Significance levels must be chosen before conducting the test to avoid biases. They help in maintaining objectivity and reliability in statistical analysis.
F-Value
The F-value is a statistic used in ANOVA tests to determine if the means between multiple groups are significantly different. Essentially, it compares the variance between the groups to the variance within the groups.

In the example, the F-value is calculated to be 9.612506. A high F-value indicates that there is more variability between the group means than within the groups, suggesting a potential difference among them.
  • F-value is computed from the data and compared against a critical value from F-distribution tables, which depend on the chosen significance level \( \alpha \) and degrees of freedom.
  • If the calculated F-value is larger than the critical value, it implies a significant difference between group means, leading to the rejection of the null hypothesis.
The F-value provides a formal assessment in ANOVA tests, thereby confirming whether the variations in group means are statistically 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

When only two treatments are involved, ANOVA and the Student \(t\) test (Chapter 10 ) result in the same conclusions. Also, \(t^{2}=\) F. As an example, suppose that 14 randomly selected students were divided into two groups, one consisting of 6 students and the other of 8 . One group was taught using a combination of lecture and programmed instruction, the other using a combination of lecture and television. At the end of the course, each group was given a 50-item test. The following is a list of the number correct for each of the two groups. $$\begin{array}{|cc|}\hline \begin{array}{l}\text { Lecture and } \\\\\text { Programmed } \\\\\text { Instruction }\end{array} & \begin{array}{c}\text { Lecture and } \\\\\text { Television }\end{array} \\\\\hline19 & 32 \\\17 & 28 \\\23 & 31 \\\22 & 26 \\\17 & 23 \\\16 & 24 \\\& 27 \\\& 25 \\\\\hline\end{array}$$ a. Using analysis of variance techniques, test \(H_{0}\) that the two mean test scores are equal; \(\alpha=.05 .\) b. Using the \(t\) test from Chapter 10 , compute \(t\). c. Interpret the results.

The following is sample information. Test the hypothesis that the treatment means are equal. Use the .05 significance level. $$\begin{array}{|rcc|}\hline \text { Treatment } 1 & \text { Treatment } 2 & \text { Treatment } 3 \\\\\hline 8 & 3 & 3 \\\6 & 2 & 4 \\\10 & 4 & 5 \\\9 & 3 & 4 \\\\\hline\end{array}$$ a. State the null hypothesis and the alternate hypothesis. b. What is the decision rule? c. Compute SST, SSE, and SS total. d. Complete an ANOVA table. e. State your decision regarding the null hypothesis.

Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random sample of six hourly periods is chosen for each assembly line and the number of components produced during these periods for each line is recorded. The output from a statistical software package is: $$\begin{array}{|lclll|}\hline {\text { Summary }} \\\\\hline \text { Groups } & \text { Count } & \text { Sum } & \text { Average } & \text { Variance } \\\\\hline \text { Line A } & 6 & 250 & 41.66667 & 0.266667 \\\\\text { Line B } & 6 & 260 & 43.33333 & 0.666667 \\\\\text { Line C } & 6 & 249 & 41.5 & 0.7 \\\\\hline\end{array}$$ $$\begin{array}{|lccccc|}\hline{\text { ANOVA }} \\\\\hline \text { Source of Variation } & \mathbf{S S} & l{d f} & \text { MS } & {F} &{p} \text { -value } \\\\\hline \text { Between Groups } & 12.33333 & 2 & 6.166667 & 11.32653 & 0.001005 \\\\\text { Within Groups } & 8.166667 & 15 & 0.544444 & & \\\\\text { Total } & 20.5 & 17 & & & \\\\\hline\end{array}$$ a. Use a .01 level of significance to test if there is a difference in the mean production of the three assembly lines. b. Develop a 99 percent confidence interval for the difference in the means between Line \(\mathrm{B}\) and Line \(\mathrm{C}\)

The fuel efficiencies for a sample of 27 compact, midsize, and large cars are entered into a statistical software package. Analysis of variance is used to investigate if there is a difference in the mean mileage of the three cars. What do you conclude? Use the .01 significance level. $$\begin{array}{|lcccc|}\hline {\text { Summary }} \\\\\hline \text { Groups } & \text { Count } & \text { Sum } & \text { Average } & \text { Variance } \\\\\hline \text { Compact } & 12 & 268.3 & 22.35833 & 9.388106 \\\\\text { Midsize } & 9 & 172.4 & 19.15556 & 7.315278 \\\\\text { Large } & 6 & 100.5 & 16.75 & 7.303 \\\\\hline\end{array}$$ Additional results are shown below. $$\begin{array}{|lccccc|}\hline {\text { ANOVA }} \\\\\hline \text { Source of Variation } & \text { SS } & \text { df } & \text { MS } & {F} & {p} \text { -value } \\\\\hline \text { Between Groups } & 136.4803 & 2 & 68.24014 & 8.258752 & 0.001866 \\\\\text { Within Groups } & 198.3064 & 24 & 8.262766 & & \\\\\text { Total } & 334.7867 & 26 & & & \\\\\hline\end{array}$$

The City of Maumee comprises four districts. Chief of police Andy North wants to determine whether there is a difference in the mean number of crimes committed among the four districts. He recorded the number of crimes reported in each district for a sample of six days. At the .05 significance level, can the chief of police conclude there is a difference in the mean number of crimes? $$\begin{array}{|cccc|}\hline {\text { Number of Crimes }} \\\\\hline \text { Rec Center } & \text { Key Street } & \text { Monclova } & \text { Whitehouse } \\\\\hline 13 & 21 & 12 & 16 \\\15 & 13 & 14 & 17 \\\14 & 18 & 15 & 18 \\\15 & 19 & 13 & 15 \\\14 & 18 & 12 & 20 \\\15 & 19 & 15 & 18 \\\\\hline\end{array}$$

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.