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An independent random sampling design was used to compare the means of six treatments based on samples of four observations per treatment. The pooled estimator of \(\sigma^{2}\) is \(9.12,\) and the sample means follow: \(\bar{x}_{1}=101.6 \quad \bar{x}_{2}=98.4 \quad \bar{x}_{3}=112.3\) \(\bar{x}_{4}=92.9 \quad \bar{x}_{5}=104.2 \quad \bar{x}_{6}=113.8\) a. Give the value of \(\omega\) that you would use to make pairwise comparisons of the treatment means for $$ \alpha=.05 . $$ b. Rank the treatment means using pairwise comparisons.

Short Answer

Expert verified
Answer: The final ranking of the treatments is: Treatment 4, Treatment 2, Treatment 1, Treatment 5, Treatment 3, and Treatment 6.

Step by step solution

01

Understanding Tukey's HSD test

Tukey's HSD test is a popular method for performing multiple pairwise comparisons simultaneously. It takes into account the family-wise error rate and adjusts the significance level accordingly. The test statistic is based on the difference between the sample means and the estimation of the variance.
02

Calculate \(q\)-value

The first step to calculate the test statistic using Tukey's HSD test is finding the \(q\)-value. The value is obtained from the Studentized Range Distribution tables with \(k\) treatments and \(N-k\) degrees of freedom. In our case, we have six treatments (\(k=6\)) and 24 total observations, so we have \(N-k = 24-6 = 18\) degrees of freedom. For \(\alpha=0.05\), the \(q\)-value for \(k=6\) and \(\text{df}=18\) is approximately \(3.960\) (refer to a standard \(q\)-table).
03

Calculate the standard error

We now need to compute the standard error for the pairwise differences of the treatment means. To do this, we use the pooled estimator of \(\sigma^{2}\) and the number of observations per treatment: $$ \text{SE}_\text{pairwise} = \sqrt{\frac{2\cdot 9.12}{4}} = \sqrt{4.56} \approx 2.136 $$
04

Compute the critical value (\(\omega\))

Now, we can calculate the critical value (\(\omega\)) by multiplying the \(q\)-value obtained in Step 2 by the standard error calculated in Step 3: $$ \omega = 3.960 \cdot 2.136 \approx 8.46 $$
05

Perform pairwise comparisons and rank the treatments

With the critical value computed, we can now perform pairwise comparisons between the treatment means. If the absolute difference between any two sample means is greater than \(\omega\), we can conclude that the means are significantly different at the \(\alpha=0.05\) level. To rank the treatment means, we can simply order them in ascending or descending order. Based on pairwise comparisons (absolute differences between means), we can order the treatments as follows: 1. \(\bar{x}_{4}=92.9\) 2. \(\bar{x}_{2}=98.4\) 3. \(\bar{x}_{1}=101.6\) 4. \(\bar{x}_{5}=104.2\) 5. \(\bar{x}_{3}=112.3\) 6. \(\bar{x}_{6}=113.8\) So, the final ranking of the treatments would be: Treatment 4, Treatment 2, Treatment 1, Treatment 5, Treatment 3, and Treatment 6.

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Key Concepts

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

Pairwise Comparisons
Pairwise comparisons are a way to directly compare two sets of data against each other. When conducting statistical analyses, especially in experiments with multiple groups, researchers often want to determine how each group performs relative to all others. This is achieved by comparing each pair of groups or treatments.
The Tukey's HSD (Honestly Significant Difference) test is a popular method used to carry out these comparisons in a statistically rigorous way. It enables us to see if the difference in treatment effects between each pair is significant, considering the variability in the data. By calculating a critical value for comparison using Tukey's method, we can conclude if the observed differences are due to chance or a result of actual differences in treatment effects.
Each pair is measured for the absolute difference in means, and those differences are then compared to a calculated threshold (or critical value) to establish significance. If the difference exceeds this threshold, the difference is deemed significant.
Family-Wise Error Rate
The family-wise error rate (FWER) is the probability of making one or more false discoveries, or Type I errors, when performing multiple hypotheses tests. When you test many pairs in pairwise comparisons, you increase the chance of a Type I error because each test has a chance of giving a false positive.
Tukey's HSD method effectively controls for the family-wise error rate by adjusting the level at which differences are considered significant. This means that the overall error rate across all comparisons is maintained at the desired level, often set at 0.05 or 5%.
This adjustment is critical for maintaining the validity of test results, ensuring that conclusions about which treatment groups differ are more reliable. Without accounting for FWER, the likelihood of incorrectly claiming statistical significance would be unacceptably high, especially in experiments with many comparisons.
Standard Error
The standard error (SE) is a crucial concept in statistics that measures the variability or dispersion of a sample mean estimate from the true population mean. In the context of pairwise comparisons using Tukey's HSD test, the standard error provides a measure of the variability in the differences of treatment means.
The calculation of standard error for pairwise comparisons takes into account the pooled variance and the number of observations. The formula used is \[ \text{SE}_{\text{pairwise}} = \sqrt{\frac{2 \cdot \sigma^2}{n}} \]where \( \sigma^2 \) is the pooled variance and \( n \) is the number of observations per treatment.
A smaller standard error indicates that the sample mean is a more accurate reflection of the population mean, leading to a more reliable outcome when determining if treatment means significantly differ. In our solution, the calculated SE was used to determine the critical value needed for performing meaningful pairwise comparisons.
Treatment Means
Treatment means are the average outcomes measured across different treatment groups in a study. These means represent how each group performed based on the manipulations they were subjected to during the experiment. In our exercise, we had six treatment means ranging from 92.9 to 113.8.
The analysis of treatment means involves comparing these averages to discern patterns or differences between them. By using methods like Tukey's HSD test, we can objectively rank the treatments and identify which means are significantly different from one another.
Being able to analyze treatment means effectively helps in understanding the effects of different experimental conditions and can guide further research or practical applications. It is essential not only to compute these averages but also to assess them in context, considering the variability within each treatment and across them all.

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Most popular questions from this chapter

The sample means corresponding to populations 1 and 2 in Exercise 11.4 are \(\bar{x}_{1}=88.0\) and \(\bar{x}_{2}=83.9\) a. Find a \(90 \%\) confidence interval for \(\mu_{1}\). b. Find a \(90 \%\) confidence interval for the difference \(\left(\mu_{1}-\mu_{2}\right)\)

An experiment was conducted to compare the glare characteristics of four types of automobile rearview mirrors. Forty drivers were randomly selected to participate in the experiment. Each driver was exposed to the glare produced by a headlight located 30 feet behind the rear window of the experimental automobile. The driver then rated the glare produced by the rearview mirror on a scale of 1 (low) to 10 (high). Each of the four mirrors was tested by each driver; the mirrors were assigned to a driver in random order. An analysis of variance of the data produced this ANOVA table: $$ \begin{array}{lcc} \text { Source } & d f & \text { SS } & \text { MS } \\ \hline \text { Mirrors } & 46.98 & \\ \text { Drivers } & & 8.42 \\ \text { Error } & & & \\ \hline \text { Total } & 638.61 & \end{array} $$ a. Fill in the blanks in the ANOVA table. b. Do the data present sufficient evidence to indicate differences in the mean glare ratings of the four rearview mirrors? Calculate the approximate \(p\) -value and use it to make your decision. c. Do the data present sufficient evidence to indicate that the level of glare perceived by the drivers varied from driver to driver? Use the \(p\) -value approach. d. Based on the results of part b, what are the practical implications of this experiment for the manufacturers of the rearview mirrors?

Four chemical plants, producing the same product and owned by the same company, discharge effluents into streams in the vicinity of their locations. To check on the extent of the pollution created by the effluents and to determine whether this varies from plant to plant, the company collected random samples of liquid waste, five specimens for each of the four plants. The data are shown in the table: $$ \begin{array}{llllll} \text { Plant } & {\text { Polluting Effluents }} {\text { (Ib/gal of waste) }} \\ \hline \text { A } && 1.65 & 1.72 & 1.50 & 1.37 & 1.60 \\ \text { B } & &1.70 & 1.85 & 1.46 & 2.05 & 1.80 \\ \text { C } && 1.40 & 1.75 & 1.38 & 1.65 & 1.55 \\ \text { D } && 2.10 & 1.95 & 1.65 & 1.88 & 2.00 \end{array} $$ a. Do the data provide sufficient evidence to indicate a difference in the mean amounts of effluents discharged by the four plants? b. If the maximum mean discharge of effluents is 1.5 lb/gal, do the data provide sufficient evidence to indicate that the limit is exceeded at plant \(\mathrm{A} ?\) c. Estimate the difference in the mean discharge of effluents between plants \(\mathrm{A}\) and \(\mathrm{D},\) using a \(95 \%\) confidence interval.

The data that follow are observations collected from an experiment that compared four treatments, \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D},\) within each of three blocks, using a randomized block design. $$ \begin{array}{lrrrrrr} &&&{\text { Treatment }} \\ \hline \text { Block } & \text { A } & \text { B } & \text { C } & \text { D } & \text { Total } \\ \hline 1 & 6 & 10 & 8 & 9 & 33 \\ 2 & 4 & 9 & 5 & 7 & 25 \\ 3 & 12 & 15 & 14 & 14 & 55 \\ \hline \text { Total } & 22 & 34 & 27 & 30 & 113 \end{array} $$ a. Do the data present sufficient evidence to indicate differences among the treatment means? Test using $$ \alpha=.05 . $$ b. Do the data present sufficient evidence to indicate differences among the block means? Test using \(\alpha=.05 .\) c. Rank the four treatment means using Tukey's method of paired comparisons with \(\alpha=.01\) d. Find a \(95 \%\) confidence interval for the difference in means for treatments \(\mathrm{A}\) and \(\mathrm{B}\). e. Does it appear that the use of a randomized block design for this experiment was justified? Explain.

Refer to Exercise \(11.28 .\) Find a \(95 \%\) confidence interval for the difference between a pair of treatment means \(\mathrm{A}\) and \(\mathrm{B}\) if \(\bar{x}_{\mathrm{A}}=21.9\) and \(\bar{x}_{\mathrm{B}}=24.2\).

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