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Ch. 12 - Analysis of Variance
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
All textbooksTriola 14th EditionCh. 12 - Analysis of VarianceProblem 12.1.2
Chapter 12, Problem 12.1.2

In Exercises 1–4, use the following listed measured amounts of chest compression (mm) from car crash tests (from Data Set 35 “Car Data” in Appendix B). Also shown are the SPSS results from analysis of variance. Assume that we plan to use a 0.05 significance level to test the claim that the different car sizes have the same mean amount of chest compression.


Table showing chest compression data (mm) for small, midsize, large cars, and SUVs from crash tests.


Why Not Test Two at a Time? Refer to the sample data given in Exercise 1. If we want to test for equality of the four means, why don’t we use the methods of Section 9-2 “Two Means: Independent Samples” for the following six separate hypothesis tests?


Image showing six null hypotheses comparing means: μ1=μ2, μ1=μ3, μ1=μ4, μ2=μ3, μ2=μ4, μ3=μ4.

Verified step by step guidance
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Step 1: Understand the problem. The goal is to test the claim that the four car sizes (Small, Midsize, Large, SUV) have the same mean amount of chest compression using a significance level of 0.05. The question asks why we don't use six separate hypothesis tests for pairwise comparisons of the means.
Step 2: Recognize the issue with multiple hypothesis tests. Performing six separate tests for pairwise comparisons (as shown in the hypotheses H₀: μ₁ = μ₂, H₀: μ₁ = μ₃, etc.) increases the risk of Type I error. This means the probability of incorrectly rejecting a true null hypothesis increases with the number of tests conducted.
Step 3: Introduce the concept of ANOVA (Analysis of Variance). ANOVA is a statistical method designed to test for equality of means across multiple groups simultaneously. It controls the overall Type I error rate and is more efficient than conducting multiple pairwise tests.
Step 4: Explain the mechanics of ANOVA. ANOVA compares the variability between group means to the variability within groups. If the between-group variability is significantly larger than the within-group variability, we reject the null hypothesis that all group means are equal. The test statistic is based on the F-distribution.
Step 5: Highlight the advantage of ANOVA. By using ANOVA, we can test the equality of all four means in a single test, avoiding the need for six separate tests and reducing the risk of inflated Type I error. If ANOVA indicates significant differences, post-hoc tests (like Tukey's HSD) can be used to identify which specific means differ.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Analysis of Variance (ANOVA)

ANOVA is a statistical method used to test differences between two or more group means. It helps determine if at least one group mean is significantly different from the others, which is essential when comparing multiple groups, such as different car sizes in this case. By using ANOVA, we can avoid the increased risk of Type I error that occurs when conducting multiple t-tests.
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Variance & Standard Deviation of Discrete Random Variables

Null Hypothesis

The null hypothesis is a statement that there is no effect or no difference, serving as a starting point for statistical testing. In this context, the null hypothesis posits that the mean chest compression is the same across all car sizes. Testing this hypothesis allows researchers to determine if observed differences in means are statistically significant or due to random chance.
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Step 1: Write Hypotheses

Significance Level

The significance level, often denoted as alpha (α), is the threshold for determining whether to reject the null hypothesis. A common significance level is 0.05, indicating a 5% risk of concluding that a difference exists when there is none. In this scenario, using a 0.05 significance level means that if the p-value from the ANOVA test is less than 0.05, we would reject the null hypothesis and conclude that at least one car size has a different mean chest compression.
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Step 4: State Conclusion Example 4
Related Practice
Textbook Question

Sitting Heights The sitting height of a person is the vertical distance between the sitting surface and the top of the head. The following table lists sitting heights (mm) of randomly selected U.S. Army personnel collected as part of the ANSUR II study. Using the data with a 0.05 significance level, what do you conclude? Are the results as you would expect?

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Textbook Question

In Exercises 5–16, use analysis of variance for the indicated test.


Triathlon Times Jeff Parent is a statistics instructor who participates in triathlons. Listed below are times (in minutes and seconds) he recorded while riding a bicycle for five stages through each mile of a 3-mile loop. Use a 0.05 significance level to test the claim that it takes the same time to ride each of the miles. Does one of the miles appear to have a hill?

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Textbook Question

Distance Between Pupils The following table lists distances (mm) between pupils of randomly selected U.S. Army personnel collected as part of the ANSUR II study. Results from two-way analysis of variance are also shown. Use the displayed results and use a 0.05 significance level. What do you conclude? Are the results as you would expect?

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Textbook Question

Pancake Experiment Listed below are ratings of pancakes made by experts (based on data from Minitab). Different pancakes were made with and without a supplement and with different amounts of whey. The results from two-way analysis of variance are shown. Use the displayed results and a 0.05 significance level. What do you conclude?

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Textbook Question

Balanced Design Does the table given in Exercise 1 constitute a balanced design? Why or why not?

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Textbook Question

One-Way ANOVA In general, what is one-way analysis of variance used for?

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