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 specific differences among them. Although it may seem overwhelming due to the number of comparisons, the process is straightforward and involves repetitive steps similar to t-tests.

Before performing the Tukey test, it is essential to confirm that the ANOVA test has rejected the null hypothesis. For our example, we have study time data for grades 10, 11, and 12, each with a sample size of 10. After confirming the rejection of the null hypothesis, we set the significance level (alpha) to 0.05 and refer to the studentized range distribution (q table) to find the critical value. The degrees of freedom for the test is calculated as the total number of observations minus the number of groups, which in this case is 30 - 3 = 27. Using the q table, we find a critical value of 3.05.

In the Tukey test, each pair of means is tested against this critical value. The q statistic for each pair 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 being compared, \( MS_{error} \) is the mean squares due to error from the ANOVA output, and \( n \) is the sample size of each group.

For the first pair (grades 10 and 11), the null hypothesis states that their means are equal. After calculating the q statistic, if it is less than the critical value, we fail to reject the null hypothesis, indicating no significant difference in study time between these grades. The same process is repeated for the next pairs (grades 11 and 12, and grades 10 and 12). In the final comparison, if the q statistic exceeds the critical value, we reject the null hypothesis, concluding that there is a significant difference in study time between those grades.

Through this systematic approach, the Tukey test provides clarity on which means differ, enhancing our understanding of the data and allowing for informed conclusions.

In this analysis, we explore the effectiveness of three diet plans (A, B, and C) by employing a Tukey test following an ANOVA test. The initial ANOVA test indicated that at least one diet plan resulted in a different average weight loss, prompting the need for further investigation into which specific pairs of means differ.

The Tukey test begins with the critical step of confirming the rejection of the null hypothesis from the ANOVA test. With this established, we proceed to determine the critical value using the Studentized Range distribution table at an alpha level of 0.05. For three groups and 12 degrees of freedom (calculated as the total number of observations minus the number of groups), the critical value is found to be 3.773.

Next, we calculate the Q statistic for each pair of diet plans. The formula for the Q statistic is given by:

Q = \frac{|\bar{x}_1 - \bar{x}_2|}{\sqrt{\frac{MSE}{n}}}

where:

• Q is the test statistic.
• \bar{x}_1 and \bar{x}_2 are the means of the two groups being compared.
• MSE is the mean square error from the ANOVA output.
• n is the sample size for each group.

For our analysis, the means for the diet plans are:

• Plan A: 8
• Plan B: 4.8
• Plan C: 11

The MSE from the ANOVA output is 1.9, and the sample size for each group is 5.

We begin by comparing Plans A and B:

Null Hypothesis: \( \mu_A = \mu_B \)

Alternative Hypothesis: \( \mu_A \neq \mu_B \)

Calculating the Q statistic:

Q = \frac{8 - 4.8}{\sqrt{\frac{1.9}{5}}} = 5.19

Since 5.19 is greater than the critical value of 3.773, we reject the null hypothesis, concluding that the average weight loss for Plans A and B is significantly different.

Next, we compare Plans B and C:

Null Hypothesis: \( \mu_B = \mu_C \)

Alternative Hypothesis: \( \mu_B \neq \mu_C \)

Calculating the Q statistic:

Q = \frac{11 - 4.8}{\sqrt{\frac{1.9}{5}}} = 10.06

Again, since 10.06 exceeds the critical value of 3.773, we reject the null hypothesis, indicating a significant difference in average weight loss between Plans B and C.

Finally, we compare Plans A and C:

Null Hypothesis: \( \mu_A = \mu_C \)

Alternative Hypothesis: \( \mu_A \neq \mu_C \)

Calculating the Q statistic:

Q = \frac{11 - 8}{\sqrt{\frac{1.9}{5}}} = 4.87

Since 4.87 is also greater than 3.773, we reject the null hypothesis, concluding that the average weight loss for Plans A and C is significantly different as well.

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