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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 12.1.1b
Chapter 12, Problem 12.1.1b

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 measurements (mm) for small, midsize, large, and SUV car sizes from crash tests.


Anova


b. If the objective is to test the claim that the four car sizes have the same mean chest compression, why is the method referred to as analysis of variance?

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1
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 chest compression using a 0.05 significance level. The method used is analysis of variance (ANOVA).
Step 2: Recognize why the method is called analysis of variance. ANOVA is used to compare means of multiple groups by analyzing the variability within each group and between groups. It determines if the observed differences in sample means are statistically significant.
Step 3: Organize the data. The table provides chest compression measurements for four car sizes. Each row corresponds to a car size, and each column represents a measurement. Calculate the mean and variance for each group (Small, Midsize, Large, SUV).
Step 4: Perform ANOVA. Compute the between-group variance (how much the group means differ from the overall mean) and the within-group variance (how much individual measurements differ within each group). Use these variances to calculate the F-statistic.
Step 5: Compare the F-statistic to the critical value at the 0.05 significance level. If the F-statistic exceeds the critical value, reject the null hypothesis that all group means are equal. Otherwise, fail to reject the null hypothesis.

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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 compare the means of three or more groups to determine if at least one group mean is significantly different from the others. It assesses the impact of one or more factors by comparing the variance within groups to the variance between groups. In this context, it helps to test the claim that different car sizes have the same mean amount of chest compression.
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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, and it serves as the default assumption in hypothesis testing. In this case, the null hypothesis would state that the mean chest compression measurements for small, midsize, large, and SUV cars are equal. The goal of the ANOVA test is to determine whether there is enough evidence to reject this null hypothesis.
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Step 1: Write Hypotheses

Significance Level

The significance level, often denoted as alpha (α), is the threshold for determining whether a result is statistically significant. In this scenario, a significance level of 0.05 indicates that there is a 5% risk of concluding that a difference exists when there is none. If the p-value obtained from the ANOVA test is less than 0.05, it suggests that the means of the car sizes are significantly different.
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Step 4: State Conclusion Example 4
Related Practice
Textbook Question

Interaction


a. What is an interaction between two factors?


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Textbook Question

Interaction


b. In general, when using two-way analysis of variance, if we find that there is an interaction effect, how does that affect the procedure?


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Textbook Question

Transformations of Data Example 1 illustrated the use of two-way ANOVA to analyze the sample data in Table 12-3. How are the results affected in each of the following cases?


b. Each sample value is multiplied by the same nonzero constant.

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Textbook Question

Bonferroni Test Shown below are weights (kg) of poplar trees obtained from trees planted in a rich and moist region. The trees were given different treatments identified in the table below. The data are from a study conducted by researchers at Pennsylvania State University and were provided by Minitab, Inc. Also shown are partial results from using the Bonferroni test with the sample data.

b. What do the displayed Bonferroni SPSS results tell us?

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Textbook Question

In Exercises 1–5, refer to the following list of numbers of years that deceased U.S. presidents, popes, and British monarchs lived after their inauguration, election, or coronation, respectively. (As of this writing, the last president is George H. W. Bush, the last pope is John Paul II, and the last British monarch is George VI.) Assume that the data are samples from larger populations.


[Image]


Exploring the Data Include appropriate units in all answers.


d. Are there any obvious outliers?

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Textbook Question

c. Shown below is an interaction graph constructed from the data in Exercise 1. What does the graph suggest?

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