6. Post hoc tests with repeated-measures ANOVA Suppose that you conduct a repeated-measures ANOVA with more than two treatment conditions, and the null hypothesis is rejected. To determine exactly which mean differences are significant and which are not, you conduct post hoc tests. Which of the following statements about the use of Tukey's HSD test is correct? Tukey's HSD test can be conducted using HSD = q?(MSwithin treatments/n) and using dfwithin treatments to find the value of q. Tukey's HSD test can be conducted using HSD = q?(MSerror/n) and using dferror to find the value of q. Tukey's HSD test can't be conducted.
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Suppose that you have performed a one-factor ANOVA on a data set and obtained a p-value that is less than the significance level (i.e., rejected the null hypothesis of equal means across groups). Select which of the following statements are true: All of the group means are different from each other. The variability of the data between groups must be larger than the variability of the data within groups. The F statistic for the ANOVA test must be less than 1. A follow-up pairwise analysis (e.g., the Tukey HSD test) will identify at least one pair of group means that are significantly different.
Madhur L.
3. Multiple comparison tests often must be run after an ANOVA because: a. ANOVA doesn't tell us anything about differences among means. b. ANOVA greatly increases the potential to make a Type I Error. c. a significant ANOVA tells us that there is at least one difference among means, but not specifically which groups differ. d. ANOVA cannot handle multiple group means, but multiple comparison tests can. e. This is all a lie; multiple comparison tests must be run after a multiple regression, not an ANOVA. 4. The most realistic way to decrease the chance of both Type I and Type II Errors simultaneously is to: a. use a larger value for α. b. use a smaller value for α. c. use a non-parametric analysis. d. increase sample size. e. It can't be done. Only one or the other can be reduced at a time.
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If the null hypothesis is rejected in a one-way ANOVA test of three or more means, then a Scheffé Test can be performed to find which means have a significant difference. In a Scheffé Test, the means are compared two at a time. For instance, with three means you would have the following comparisons: $\bar{x}_{1}$ versus $\bar{x}_{2}, \bar{x}_{1}$ versus $\bar{x}_{3},$ and $\bar{x}_{2}$ versus $\bar{x}_{3} .$ For each comparison, calculate $$\frac{\left(\bar{x}_{a}-\bar{x}_{b}\right)^{2}}{\frac{S S_{W}}{\Sigma\left(n_{i}-1\right)}\left(\frac{1}{n_{a}}+\frac{1}{n_{b}}\right)}$$ where $\bar{x}_{a}$ and $\bar{x}_{b}$ are the means being compared and $n_{a}$ and $n_{b}$ are the corresponding sample sizes. Calculate the critical value by multiplying the critical value of the one-way ANOVA test by $k-1 .$ Then compare the value that is calculated using the formula above with the critical value. The means have a significant difference when the value calculated using the formula above is greater than the critical value. Refer to the data in Exercise $5 .$ At $\alpha=0.05,$ perform a Scheffé Test to determine which means have a significant difference.
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Analysis of Variance
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