Exercises: testing more than two groups

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1 TESTING YOUR UNDERSTANDING 1 Exercises: testing more than two groups 1 Testing your understanding Question 1. True or False? 1. The null hypothesis when performing an analysis of variance is that there are differences between group means; however, no prediction is made concerning where the differences lie. False, the omnibus null hypothesis is that all group means are the same. 2. Unlike t-tests, an ANOVA may be used to test for differences among more than 2 groups. True, an analysis of variance (ANOVA) is most often used to determine if there are differences among 3 or more group means. However, when there are more than 2 groups, an ANOVA does not provide information regarding where the differences lie. 3. It is valid to do the Tukey HSD test without first finding a significant effect with an ANOVA. True, the Tukey HSD controls the Type I error rate and is valid without first running an ANOVA. Question 2. In a two-factor ANOVA in which one variable has 4 levels and the other has 2, what is the degrees of freedom for the interaction? The df is (4-1) x (2-1) = 3. Question 3. To compare a control group with the average of the other three groups, you would use Tukey s test An interaction test A specific comparison A specific comparison Question 4. To compare each mean with each other mean, you would use Tukey s test An interaction test A specific comparison ANOVA Tukey s test

2 ONE-WAY ANOVA: THE SMILES AND LENIENCY STUDY 2 2 One-way ANOVA: the smiles and leniency study A case study prepared by David Lane ( leniency.html) The means and variances of the four groups in the "Smiles and Leniency" case study are shown in Table 1. Note that there are 34 subjects in each of the four conditions (False, Felt, Miserable, and Neutral).

3 ONE-WAY ANOVA: THE SMILES AND LENIENCY STUDY 3 Table 1: Means and Variances from the "Smiles and Leniency" Study. Condition Mean Variance False Felt Miserable Neutral State the null and alternative hypotheses in this one-way ANOVA test. H 0 : On average, there is no difference in leniency measure between different types of smiles. H 1 : at least one type of smile gives a different leniency measure, on average, than the other types of smile. 2. Fill the grey cells in the ANOVA summary table 2 below. Table 2: ANOVA Summary Table. Source of Degrees of Sum of Squares Mean Sum of Squares F statistic p-value Variation Freedom Condition 4 1 = = = Error = = Total Conclude on the ANOVA test. The p-value associated to the one way ANOVA test is less than our significance level of 5%. Therefore we reject H 0 and can conclude that the type of smiles has an effect on leniency. Note that a post-hoc test would be needed to assess which type of smile exactly increases leniency.

4 TWO-WAY ANOVA: TOOTH GROWTH IN GUINEA PIGS 4 3 Two-way ANOVA: tooth growth in Guinea Pigs We examine data on the effect of vitamin C on tooth growth in 60 guinea pigs. The data contain the length of the tooth, which is the response variable, the way the supplement was administered (supp is either OJ = Orange Juice or VC = Vitamin C, i.e. ascorbic acid) and the dose (dose is either low, medium or high). pig ID length supp dose pig ID length supp dose OJ low VC low OJ low VC low OJ low VC low OJ low VC low OJ low VC low OJ low VC low OJ low VC low OJ low VC low OJ low VC low OJ low VC low OJ med VC med OJ med VC med OJ med VC med OJ med VC med OJ med VC med OJ med VC med OJ med VC med OJ med VC med OJ med VC med OJ med VC med OJ high VC high OJ high VC high OJ high VC high OJ high VC high OJ high VC high OJ high VC high OJ high VC high OJ high VC high OJ high VC high OJ high VC high 1. Explain why we should use a two-way rather than a one-way ANOVA for this experiment. What are the different treatment factors considered, and the number of levels per factor? The response variable is the tooth length. We want to test the effect of two factors: the type of supplement (2 levels) and the type of dose (3 levels). 2. We obtain the following ANOVA table in R. State the null and alternative hypotheses of each of the tests performed in the two-way ANOVA. Df Sum Sq Mean Sq F value Pr(>F) supp *** dose < 2e-16 *** supp:dose * Residuals Signif. codes: 0 *** ** 0.01 * First row: H 0 : There is no significant effect of the type of supplement in the mean tooth length of guinea pigs vs. H 1 : there is a significant effect of the type of supplement in the mean tooth length of guinea pigs.

5 TWO-WAY ANOVA: TOOTH GROWTH IN GUINEA PIGS 5 Second row: H 0 : There is no significant effect of the type of dose in the mean tooth length of guinea pigs vs. H 1 : there is a significant effect of the type of dose in the mean tooth length of guinea pigs. Third row: H 0 : There is no significant interaction between dose and supplement on the mean tooth length of guinea pigs vs. H 1 : there is a significant interaction between dose and supplement on the mean tooth length of guinea pigs. 3. Conclude on these tests. The p-values associated to these tests are all significant (< 0.05), therefore we can conclude that there is a significant effect of supplement, dose, as well as a significant interaction between dose and supplement on the means of the tooth length. Following the results obtained, we perform Tukey s HSD test for post-hoc analyses. Tukey s HSD test for the Supplement main effect: diff p adj Tukey s HSD test for the Dose main effect: diff p adj med-low e-10 high-low e-13 high-med e-06 Tukey s HSD test for the interaction Supplement * Dose: VC:low-OJ:low OJ:med-OJ:low VC:med-OJ:low OJ:high-OJ:low VC:high-OJ:low OJ:med-VC:low VC:med-VC:low OJ:high-VC:low VC:high-VC:low VC:med-OJ:med OJ:high-OJ:med VC:high-OJ:med OJ:high-VC:med VC:high-VC:med VC:high-OJ:high diff p adj e e e e e e e e e e e e e e e For the Dose pairwise comparisons explain the hypotheses that are tested and interpret the results obtained. In the post-hoc test we test whether there is no difference in the means of the tooth length between 1/ medium and low doses, 2/ high and low doses and 3/ medium and high doses (null

6 TWO-WAY ANOVA: TOOTH GROWTH IN GUINEA PIGS 6 hypotheses) vs. the alternative hypotheses: there is a difference in the means for each of these types of doses. All the adjusted p-values are significant, indicating that there is a significant difference in the means of the tooth lengths between all types of doses. 5. Explain why the p-values are adjusted in the Tukey s HSD test. The Tukey s HSD test tests if the difference between every possible pairs of means is equal to zero (null hypothesis). When we have K groups it corresponds to K (K 1)/2 number of comparisons for the same response variable (here tooth growth). Therefore, we need to ajust for multiple testing as we are performing more than one test. 6. For the interaction Supp * Dose, discuss one (1) non-significant result of your choice using the Tukeys HSD test output and the profile plot below (i.e. what would you tell and show to the researcher who performed the experiment regarding that particular test?). From the Tukeys HSD test we can see that there is no significance difference in the means between the groups as the associated adjusted pvalues are > 0.05 for the following cases: (a) OJ:high-OJ:med (b) VC:high-OJ:med (c) VC:high-OJ:high For the case 1 OJ:high-OJ:med we can visualize a very small difference in the tooth lengths between the high vs. medium doses for the guinea pigs who had the supplement OJ. This difference is estimated to 3.36 according to the R output but is not declared significant. Interaction Plot Tooth Length dose high med low OJ VC supp