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Chapter 11: Problem 19

Suppose you wish to use Tukey's method of paired comparisons to rank a set of population means. In addition to the analysis of variance assumptions, what other property must the treatment means satisfy?

Short Answer

Expert verified
Answer: In addition to satisfying the ANOVA assumptions, the treatment means must also satisfy the property of additivity, meaning that the differences between the population means should be constant and the interaction effects should be negligible.

Step by step solution

01

Review Analysis of Variance (ANOVA) Assumptions.

The assumptions for ANOVA are: 1. The observations are independent. 2. The samples are drawn from normally distributed populations. 3. The population variances are equal (homoscedasticity).
02

Explain Tukey's Method of Paired Comparisons

Tukey's method of paired comparisons is a multiple comparison procedure used to determine whether there are any significant differences between the means of multiple groups (treatments). It is often used after an ANOVA test has been conducted and found to have a statistically significant result. The method involves comparing all possible pairs of group means in a pairwise fashion and adjusting the significance level to take into account the multiple comparisons being made.
03

Identify the Additional Property for Treatment Means in Tukey's Method

In addition to the three ANOVA assumptions mentioned in Step 1, for utilizing Tukey's method of paired comparisons, the treatment means must satisfy the following property: Additivity: The treatment effects must be additive. This means that the differences between the population means should be constant and the interaction effects should be negligible. In other words, the difference between any two treatment means should only be due to the treatments themselves and not due to any interaction with other variables or factors. Putting everything together, when using Tukey's method of paired comparisons to rank a set of population means, apart from satisfying the ANOVA assumptions, the treatment means must also satisfy the property of additivity.

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Most popular questions from this chapter

Refer to Exercise \(11.46 .\) The means of two of the factor-level combinations- say, \(\mathrm{A}_{1} \mathrm{~B}_{1}\) and \(\mathrm{A}_{2} \mathrm{~B}_{1}-\) are \(\bar{x}_{1}=8.3\) and \(\bar{x}_{2}=6.3,\) respectively. Find a \(95 \%\) confidence interval for the difference between the two corresponding population means.

A study was conducted to compare automobile gasoline mileage for three formulations of gasoline. A was a non-leaded 87 octane formulation, \(\mathrm{B}\) was a non-leaded 91 octane formulation, and \(\mathrm{C}\) was a non-leaded 87 octane formulation with \(15 \%\) ethanol. Four automobiles, all of the same make and model, were used in the experiment, and each formulation was tested in each automobile. Using each formulation in the same automobile has the effect of eliminating (blocking out) automobile-toautomobile variability. The data (in miles per gallon) follow. $$ \begin{array}{lcccc} && {\text { Automobile }} \\ \hline \text { Formulation } & 1 & 2 & 3 & 4 \\ \hline \mathrm{A} & 25.7 & 27.0 & 27.3 & 26.1 \\ \mathrm{~B} & 27.2 & 28.1 & 27.9 & 27.7 \\ \mathrm{C} & 26.1 & 27.5 & 26.8 & 27.8 \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in mean mileage per gallon for the three gasoline formulations? b. Is there evidence of a difference in mean mileage for the four automobiles? c. Suppose that prior to looking at the data, you had decided to compare the mean mileage per gallon for formulations A and B. Find a \(90 \%\) confidence interval for this difference. d. Use an appropriate method to identify the pairwise differences, if any, in the average mileages for the three formulations.

If the sample size for each treatment is \(n_{t}\) and if \(s^{2}\) is based on \(12 d f\), find \(\omega\) in these cases: a. \(\alpha=.05, k=4, n_{t}=5\) b. \(\alpha=.01, k=6, n_{t}=8\)

The analysis of variance \(F\) -test in Exercise 11.17 (and data set EX 1117 ) determined that there was indeed a difference in the average cost of lumber for the four states. The following information from Exercise 11.17 is given in the table: $$ \begin{array}{ll|lr} \hline \text { Sample Means } & \bar{x}_{1}=242.2 & \text { MSE } & 41.25 \\ & \bar{x}_{2}=214.8 & \text { Error } d f: & 16 \\ & \bar{x}_{3}=231.6 & n_{i}: & 5 \\ & \bar{x}_{4}=248.6 & k: & 4 \\ \hline \end{array} $$ Use Tukey's method for paired comparisons to determine which means differ significantly from the others at the \(\alpha=.01\) level.

An experiment was conducted to compare the effects of four different chemicals, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D},\) in producing water resistance in textiles. A strip of material, randomly selected from a bolt, was cut into four pieces, and the four pieces were randomly assigned to receive one of the four chemicals, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) or \(\mathrm{D} .\) This process was replicated three times, thus producing a randomized block design. The design, with moistureresistance measurements, is as shown in the figure (low readings indicate low moisture penetration). Analyze the experiment using a method appropriate for this randomized block design. Identify the blocks and treatments, and investigate any possible differences in treatment means. If any differences exist, use an appropriate method to specifically identify where the differences lie. What are the practical implications for the chemical producers? Has blocking been effective in this experiment? Present your results in the form of a report. $$$ \begin{aligned} &\text { Blocks (bolt samples) }\\\ &\begin{array}{|c|c|c|} \hline 1 & 2 & 3 \\ \hline \mathrm{C} & \mathrm{D} & \mathrm{B} \\ 9.9 & 13.4 & 12.7 \\ \mathrm{A} & \mathrm{B} & \mathrm{D} \\ 10.1 & 12.9 & 12.9 \\ \mathrm{B} & \mathrm{A} & \mathrm{C} \\ 11.4 & 12.2 & 11.4 \\ \mathrm{D}_{2} & \mathrm{C} & \mathrm{A} \\ 12.1 & 12.3 & 11.9 \end{array} \end{aligned} $$
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