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Chapter 27: Problem 27

As part of an ANOVA that compares three treatments, you carry out Tukey pairwise tests at the overall \(5 \%\) significance level. The Tukey tests find that \(\mu_{1}\) is significantly different from \(\mu_{3}\) but that the other two comparisons show no significant difference. You can be \(95 \%\) confident that a. \(\mu_{1} \neq \mu_{3}\) and \(\mu_{1}=\mu_{2}\) and \(\mu_{2}=\mu_{3}\). b. just \(\mu_{1} \neq \mu_{3}\); there is not enough evidence to draw conclusions about the other pairs of means. c. \(\mu_{1}=\mu_{2}\) and \(\mu_{2}=\mu_{3}\) and this implies that it must also be true that \(\mu_{1}=\mu_{3}\).

Short Answer

Expert verified
Option b: Just \( \mu_{1} \neq \mu_{3} \); no conclusions about other pairs.

Step by step solution

01

Understand the Tukey Test Findings

The Tukey pairwise test results show that there is a statistically significant difference between means \( \mu_{1} \) and \( \mu_{3} \). However, there is no significant difference found between other pairs of means (\( \mu_{1} \) and \( \mu_{2} \), \( \mu_{2} \) and \( \mu_{3} \)).
02

Interpret 95% Confidence

The 95% confidence level indicates that we are confident in the results of the tests within a certain margin of error, affirming the significance of the difference between \( \mu_{1} \) and \( \mu_{3} \). For means where significant differences are not found (\( \mu_{1} \) and \( \mu_{2} \), \( \mu_{2} \) and \( \mu_{3} \)), the test lacks sufficient evidence to declare them significantly different.
03

Evaluate the Options

Option (a) claims equality of pairs except \( \mu_{1} eq \mu_{3} \); however, equality is not concluded for non-significant results. Option (c) mistakenly implies \( \mu_{1} = \mu_{3} \) from non-significant results, contradicting Tukey's findings. Option (b) aligns with the Tukey test results: \( \mu_{1} eq \mu_{3} \), without drawing unfounded conclusions about other pairs.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

ANOVA
The analysis of variance (ANOVA) is a statistical method used to compare means across multiple groups. Imagine you have three different treatments, let's call them Group 1, Group 2, and Group 3. ANOVA allows you to determine if there are any statistically significant differences between the means of these groups.

This is done by analyzing the variance within each group and between the groups. If the variance between groups is significantly greater than the variance within groups, ANOVA tells us that there is evidence to suggest that the group means are not all equal.

In our exercise, ANOVA sets the stage for further analysis with the Tukey Test, comparing each pairwise mean difference to find where those differences lie.
pairwise comparisons
Following an ANOVA test, if results suggest that there are differences amongst group means, pairwise comparisons can be employed to further probe these differences. Specifically, the Tukey Test is frequently used for this purpose.

The Tukey Test is a method of making all possible comparisons between pairs of means, called pairwise comparisons. It controls for the inflation of Type I error that occurs when multiple pairwise tests are performed.
  • Each pair of means, such as Group 1 vs Group 3, is compared.
  • It helps determine precisely which pairs, if any, are statistically different.
  • In the exercise, the Tukey results showed that Group 1 and Group 3 are significantly different.
These comparisons provide a clearer picture of exactly where the differences between groups lie.
significance level
The significance level is a threshold we set to determine how unlikely a result must be under the null hypothesis for it to be considered statistically significant. Commonly set at 5% (or 0.05), the significance level in an ANOVA context tells us the probability of observable differences being due only to chance.

It is vital for interpreting the results of an analysis:
  • If the tests yield a p-value less than 0.05, we reject the null hypothesis, suggesting the difference in means is significant.
  • This level of significance was used in the exercise to establish that the difference between the means of Group 1 and Group 3 is real and not due to random variation.
Understanding the significance level allows researchers to gauge the reliability of their findings.
confidence interval
The confidence interval provides a range of values which is likely to contain the true population parameter with a certain degree of confidence. Usually set at 95%, as in the exercise, it indicates how confident we are that a range contains the true mean difference.

Here's why it matters:
  • A 95% confidence interval means that if we were to repeat the experiment 100 times, the interval would capture the true mean difference 95 times out of 100.
  • In the exercise, a 95% interval supports that the mean difference between Group 1 and Group 3 is not only statistically significant but also falls within the calculated interval.
  • Confidence intervals help in understanding the precision of the estimate, offering more information than a simple significance test alone.
This helps educators and students alike understand not just whether differences exist, but how pronounced they might be.

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Most popular questions from this chapter

Which of the following would be evidence that ANOVA inference is not safe to use? a. The sample sizes are less than four. b. The largest sample variance is more than twice as large as the smallest sample variance. c. The largest sample standard deviation is more than twice as large as the smallest sample standard deviation.

What Music Will You Play? People often match their behavior to their social environment. One study of this idea first established that the type of music most preferred by Black college students is R\&B and that Whites' most preferred music is rock. Will students hosting a small group of other students choose music that matches the racial composition of the people attending? Assign 90 Black business students at random to three equal-sized groups. Do the same for 96 White students. Each student sees a picture of the people he or she will host. Group 1 sees six Blacks, Group 2 sees three Whites and three Blacks, and Group 3 sees six Whites. Ask how likely the host is to play the type of music preferred by the other race. Use ANOVA to compare the three groups to see whether the racial mix of the gathering affects the choice of music. \(\frac{8}{1}\) a. For the White subjects, \(F=16.48\). What are the degrees of freedom? b. For the Black subjects, \(F=2.47\). What are the degrees of freedom?

Perceived Exertion while Exercising. Can the introduction of pleasant sensory stimuli lead to a more pleasant exercise environment and decrease perceived exertion during a four-minute stepping task? Forty-three students from a southeastern university were assigned at random to three conditions: "taste," in which participants inserted a lemon-flavored mouth guard during the task; "placebo," in which participants inserted a non-flavored mouth guard; and "control," in which no mouth guard was used. Twelve students were assigned to the taste group, 15 to the placebo group, and 16 to the control group. Ratings of perceived exertion (RPE) scores were measured on a standard 15-point scale ranging from 6 (very, very light) to 20 (exhausted). 14

The purpose of analysis of variance is to compare a. the variances of several populations. b. the standard deviations several populations. c. the means of several populations.

Using Tablets in Class. Does the use of technology, such as tablets, in the classroom affect student learning? To explore this, a researcher examined the NAEP mathematics scores for a representative sample of 150,100 eighth-graders. Each student was classified by how frequently he or she had used a tablet in the classroom (never, in fewer that half of all classes, in about half of all classes, in more than half of all classes, or in all classes). Here are the mean NAEP mathematics scores for the five categories of frequency of tablet use in class: 2 An analysis of variance finds \(F=10,297\), with \(P<0.0001\). a. What are the null and alternative hypotheses for the ANOVA \(F\) test? Be sure to explain what means the test compares. b. Based on the sample means and the \(F\) test, what do you conclude?
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