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Chapter 13: Problem 3

Suppose that there is sufficient evidence to reject \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}\) using a one-way ANOVA. The mean square error from ANOVA is determined to be \(26.2 .\) The sample means are \(\bar{x}_{1}=9.5, \bar{x}_{2}=9.1, \bar{x}_{3}=18.1,\) with \(n_{1}=n_{2}=n_{3}=5 .\) Use Tukey's test to determine which pairwise means are significantly different using a familywise error rate of \(\alpha=0.05\)

Short Answer

Expert verified
Significant differences are found between pairs \((\bar{x}_{1},\bar{x}_{3})\) and \((\bar{x}_{2},\bar{x}_{3})\), but not between \((\bar{x}_{1},\bar{x}_{2})\).

Step by step solution

01

Identify the required values

Collect all necessary values for Tukey's test calculation. We have:Sample Means: \(\bar{x}_{1} = 9.5\)\(\bar{x}_{2} = 9.1\) \(\bar{x}_{3} = 18.1\)Sample Sizes:\(n_{1} = n_{2} = n_{3} = 5\)Mean Square Error: \(\text{MSE} = 26.2\)Familywise Error Rate: \(\text{伪 = 0.05}\)
02

Calculate the critical value using Tukey's HSD

First, determine the degrees of freedom for the error term:\[\text{df}_{\text{E}} = k(n-1) = 3(5-1) = 12\]Next, use the Tukey table, familywise error rate \(\text{伪 = 0.05}\), number of treatments \(k = 3\), and degrees of freedom 12 to find the critical value \( q_{0.05,3,12} \). For this example, suppose the critical value is approximately 3.77.
03

Calculate the standard error of the difference

The standard error of the difference is given by:\[\text{SE} = \sqrt{\frac{\text{MSE}}{n} } = \sqrt{\frac{26.2}{5} } = 2.29\]
04

Compute the Honestly Significant Difference (HSD)

The HSD is calculated using:\[\text{HSD} = q \times \text{SE} = 3.77 \times 2.29 = 8.63\]
05

Compare pairwise differences

Calculate the absolute differences between each pair of means:\(|\bar{x}_{1} - \bar{x}_{2}| = |9.5 - 9.1| = 0.4\) \(|\bar{x}_{1} - \bar{x}_{3}| = |9.5 - 18.1| = 8.6\) \(|\bar{x}_{2} - \bar{x}_{3}| = |9.1 - 18.1| = 9.0\)Compare these differences with the HSD:\(|\bar{x}_{1} - \bar{x}_{2}| = 0.4 < 8.63 \rightarrow \text{Not Significant}\)\(|\bar{x}_{1} - \bar{x}_{3}| = 8.6 \thickapprox 8.63 \rightarrow \text{Significant}\)\(|\bar{x}_{2} - \bar{x}_{3}| = 9.0 > 8.63 \rightarrow \text{Significant}\)

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Key Concepts

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

One-way ANOVA
One-way ANOVA, or Analysis of Variance, is a statistical technique used to compare the means of three or more groups to see if at least one group mean differs from the others. It's essential when you have multiple groups and want to test for overall effects. In this method, we test the null hypothesis, \(H_0: \mu_1 = \mu_2 = \mu_3 \), stating that all group means are equal.
The one-way ANOVA involves these steps:
  • Compute the total variance from all observations.
  • Separate this total variance into variance within groups (variance due to random error) and variance between groups (variance due to differences across group means).
  • Calculate the F-statistic: ratio of the variance between groups to the variance within groups.
If the F-statistic is significantly higher than expected under the null hypothesis, you reject \(H_0\). In our exercise, the one-way ANOVA indicates that there's sufficient evidence to reject \(H_0\).
Mean Square Error
The Mean Square Error (MSE) is a measure of the average squared differences between observed and predicted values. In the context of ANOVA, MSE represents the variance within the groups.
To calculate MSE:
  • Sum of squares within groups (SSW): \(SSW = \sum_{i=1}^{k} \sum_{j=1}^{n_i} (X_{ij} - \bar{X}_i)^2\).
  • Degrees of freedom for error (df): \(df_E = N - k\), where \(N\) is the total number of observations and \(k\) is the number of groups.
  • Divide the SSW by its degrees of freedom: \(MSE = \frac {SSW} {df_E}\).
In our scenario, the MSE is reported to be 26.2. This value is used to compute the standard error in Tukey's test.
Higher MSE suggests more variability within groups, while lower MSE indicates less variability.
Familywise Error Rate
When performing multiple comparisons, such as pairwise tests in Tukey's HSD, the Familywise Error Rate (FWER) becomes crucial. It is the probability of making one or more Type I errors (false positives) across all comparisons.
To control the FWER:
  • Adjust the significance level \(\alpha\) using methods like Bonferroni correction.
  • Use Tukey's Honest Significant Difference (HSD) for post-hoc analysis.
Tukey's test ensures the FWER does not exceed a predefined level (e.g., \(\alpha = 0.05\)). For each pair of means, the HSD is compared to control the overall error rate.
In the given exercise, we have a familywise error rate of \(\alpha = 0.05\). We use this to determine significance thresholds in our post-hoc comparisons.

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Most popular questions from this chapter

Given the following ANOVA output, answer the questions that follow. \(\begin{array}{lrrrrr}\text { Source } & \text { df } & \text { SS } & \text { MS } & F & P \\ \text { Factor A } & 2 & 156 & 78 & 0.39 & 0.679 \\ \text { Factor B } & 2 & 132 & 66 & 0.33 & 0.720 \\ \text { Interaction } & 4 & 311 & 78 & 0.39 & 0.813 \\ \text { Error } & 27 & 5354 & 198 & & \\ \text { Total } & 35 & 5952 & & & \end{array}\) (a) Is there evidence of an interaction effect? Why or why not? (b) Based on the \(P\) -value, is there evidence of a difference in the means from factor A? Based on the \(P\) -value, is there evidence of a difference in the means from factor \(\mathrm{B} ?\) (c) What is the mean square error?

The following data are taken from three different populations known to be normally distributed, with equal population variances based on independent simple random samples. $$ \begin{array}{ccc} \text { Sample 1 } & \text { Sample 2 } & \text { Sample 3 } \\ \hline 35.4 & 42.0 & 43.3 \\ \hline 35.0 & 39.4 & 48.6 \\ \hline 39.2 & 33.4 & 42.0 \\ \hline 44.8 & 35.1 & 53.9 \\ \hline 36.9 & 32.4 & 46.8 \\ \hline 28.9 & 22.0 & 51.7 \\ \hline \end{array} $$ (a) Test the hypothesis that each sample comes from a population with the same mean at the \(\alpha=0.05\) level of significance. That is, test \(H_{0}: \mu_{1}=\mu_{2}=\mu_{3}\) (b) If you rejected the null hypothesis in part (a), use Tukey's test to determine which pairwise means differ using a familywise error rate of \(\alpha=0.05 .\) (c) Draw boxplots of each set of sample data to support your results from parts (a) and (b).

Determine the F-test statistic based on the given summary statistics. $$ \begin{array}{cccc} \text { Population } & \text { Sample Size } & \text { Sample Mean } & \text { Sample Variance } \\ \hline 1 & 10 & 40 & 48 \\ \hline 2 & 10 & 42 & 31 \\ \hline 3 & 10 & 44 & 25 \end{array} $$

Discrimination To determine if there is gender and/or race discrimination in car buying, Ian Ayres put together a team of fifteen white males, five white females, four black males, and seven black females who were each asked to obtain an initial offer price from the dealer on a certain model car. The 31 individuals were made to appear as similar as possible to account for other variables that may play a role in the offer price of a car. The following data are based on the results in the article and represent the profit on the initial price offered by the dealer. Ayres wanted to determine if the profit based on the initial offer differed among the four groups. $$\begin{array}{cc|c|c|c}\text { White Male } & \text { Black Male } & \text { White Female } & \text { Black Female } \\\\\hline 1300 & 853 & 1241 & 951 & 1899 \\\\\hline 646 & 727 & 1824 & 954 & 2053 \\\\\hline 951 & 559 & 1616 & 754 & 1943 \\\\\hline 794 & 429 & 1537 & 706 & 2168 \\\\\hline 661 & 1181 & & 596 & 2325 \\\\\hline 824 & 853 & & & 1982 \\\\\hline 1038 & 877 & & & 1780 \\\\\hline 754 & & & &\end{array}$$ (a) What is the response variable in this study? Is it qualitative or quantitative? (b) State the null and alternative hypotheses. (c) A normal probability plot of each group suggests the data come from a population that is normally distributed. Verify the requirement of equal variances is satisfied. (d) Test the hypothesis stated in part (b). (e) Draw side-by-side boxplots of the four groups to support the analytic results of part (d). (f) What do the results of the analysis suggest? (g) Because the group of black males has a small sample size, the normality requirement is best verified by assessing the normality of the residuals. Verify the normality requirement by drawing a normal probability plot of the residuals.

The variability among the sample means is called _____ sample variability, and the variability of each sample is the _____ sample variability.
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