Multiple Comparisons: Tukey Test: Videos & Practice Problems

When conducting an ANOVA test to compare three means, rejecting the null hypothesis indicates that at least one mean is different, but it does not specify which one. To address this uncertainty, post hoc tests, such as the Tukey test, are employed. The Tukey test systematically compares pairs of means to identify which specific means differ from one another.

To perform a Tukey test, it is essential first to confirm that the ANOVA test has rejected the null hypothesis. This ensures that there is at least one significant difference among the means. For our example, we will analyze study time data for grades 10, 11, and 12, where each group has a sample size of 10. We set the significance level (alpha) at 0.05 and utilize the studentized range distribution, or q table, to find the critical value necessary for our comparisons.

The degrees of freedom for the Tukey test is calculated as the total number of observations minus the number of groups. In this case, with 30 total observations (10 from each of the three grades), the degrees of freedom is 30 - 3 = 27. Referring to the q table for alpha = 0.05, we find a critical value of 3.05.

In each pairwise comparison, we calculate a q statistic, which we then compare to the critical value. If the q statistic exceeds the critical value, we reject the null hypothesis for that pair, indicating a significant difference between the means. Conversely, if the q statistic is less than the critical value, we fail to reject the null hypothesis, suggesting no significant difference.

For the first pair, comparing grades 10 and 11, the null hypothesis states that their means are equal. The test statistic is calculated using the formula:

$$ q = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{\frac{MS_{error}}{n}}} $$

where $\bar{X}_1$ and $\bar{X}_2$ are the means of the two groups, $MS_{error}$ is the mean squares due to error from the ANOVA output, and $n$ is the sample size. After calculating, we find a q statistic of 1.949, which is less than the critical value of 3.05, leading us to fail to reject the null hypothesis. Thus, the average study times for grades 10 and 11 are not significantly different.

Next, we compare grades 11 and 12. Again, we set up the null hypothesis and calculate the q statistic, resulting in a value of 2.549, which is still less than 3.05. Therefore, we fail to reject the null hypothesis, indicating no significant difference between grades 11 and 12.

Finally, we compare grades 10 and 12. The q statistic for this pair is calculated to be 4.498, which exceeds the critical value of 3.05. This leads us to reject the null hypothesis, concluding that there is a significant difference in average study time between grades 10 and 12.

In summary, the Tukey test provides a structured approach to identify specific differences between group means following an ANOVA test. By systematically comparing pairs and utilizing critical values, we can draw meaningful conclusions about the data.

In this analysis, we explore the effectiveness of three diet plans (A, B, and C) by conducting a Tukey test following an ANOVA test, which previously indicated that at least one diet plan leads to a different average weight loss. The Tukey test is used to identify which specific pairs of means differ significantly, with a significance level (alpha) set at 0.05.

Before performing the Tukey test, it is essential to confirm that the null hypothesis from the ANOVA test has been rejected. This step is crucial because if the null hypothesis is not rejected, the Tukey test would not be meaningful. In our case, we proceed with the Tukey test since the ANOVA results were significant.

To begin, we need to determine the critical value for the Tukey test using the Studentized Range Distribution table. With three groups and a total of 15 observations, the degrees of freedom is calculated as 15 - 3 = 12. Referring to the table for alpha = 0.05, we find the critical value to be 3.773.

Next, we gather necessary data, including the Mean Squares Error (MSE) from the ANOVA output, which is 1.9. This value will be used in our calculations for the Q statistic. The means for each diet plan are as follows: Plan A has a mean of 8, Plan B has a mean of 4.8, and Plan C has a mean of 11. Each group consists of 5 samples.

We will now conduct pairwise comparisons:

1. **Comparison of Plans A and B**: The null hypothesis states that the average weight loss for both plans is equal (μ_A = μ_B), while the alternative hypothesis states they are not equal (μ_A ≠ μ_B). The Q statistic is calculated as follows:

Q = \(\frac{\bar{x}_A - \bar{x}_B}{\sqrt{\frac{MSE}{n}}}\) = \(\frac{8 - 4.8}{\sqrt{\frac{1.9}{5}}}\)

Calculating this gives Q = 5.19, which exceeds the critical value of 3.773. Therefore, we reject the null hypothesis, concluding that the average weight loss for Plans A and B is significantly different.

2. **Comparison of Plans B and C**: The null hypothesis is μ_B = μ_C, and the alternative is μ_B ≠ μ_C. The Q statistic is calculated as:

Q = \(\frac{\bar{x}_C - \bar{x}_B}{\sqrt{\frac{MSE}{n}}}\) = \(\frac{11 - 4.8}{\sqrt{\frac{1.9}{5}}}\)

This results in Q = 10.06, which is also greater than 3.773. Thus, we reject the null hypothesis, indicating a significant difference in average weight loss between Plans B and C.

3. **Comparison of Plans A and C**: The null hypothesis states μ_A = μ_C, while the alternative states μ_A ≠ μ_C. The Q statistic is calculated as:

Q = \(\frac{\bar{x}_C - \bar{x}_A}{\sqrt{\frac{MSE}{n}}}\) = \(\frac{11 - 8}{\sqrt{\frac{1.9}{5}}}\)

This yields Q = 4.87, which again exceeds the critical value of 3.773. Therefore, we reject the null hypothesis, concluding that the average weight loss for Plans A and C is significantly different.

In summary, the Tukey test results indicate that all pairs of diet plans (A vs. B, B vs. C, and A vs. C) show significant differences in average weight loss, confirming the effectiveness of the different diet plans.