10.6 - Example: Exam Grade by Professor

10.6 - Example: Exam Grade by Professor


Three professors were each teaching one section of a course. They all gave the same final exam and they want to know if there are any differences between their sections’ mean scores.

Step 1: Write Hypotheses


\(H_a: Not\;all\;\mu\;are\;equal\)

Instructor N Mean StDev 95% CI
Dr. Al 60 68.367 17.719 (63.977, 72.756)
Dr. Oh 87 71.448 16.702 (67.803, 75.094)
Dr. Pa 98 67.939 17.465 (64.504, 71.373)

Pooled StDev = 17.2609

The standard deviations for all three classes are all similar.

Step 2: Compute Test Statistic
  1. Open the Minitab file: ANOVA_Exam_Profs.mpx
  2. Using Minitab: Stat > ANOVA > One-Way

The result is the following ANOVA source table:

Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Instructor 2 635.3 317.7 1.07 0.346
Error 242 72101.1 297.9    
Total 244 72736.4      

F (2, 242) = 1.07

Step 3: Find the p-Value

From our ANOVA source table, p = .346

Step 4: Make a Decision

Because \(p > \alpha\), we fail to reject the null hypothesis.

Step 5: State Conclusion

There is not enough evidence to conclude that the mean scores from the three different professors’ sections are different.

Tukey Pairwise Comparisons

There is some debate as to whether pairwise comparisons are appropriate when the overall one-way ANOVA is not statistically significant. Some argue that if the overall ANOVA is not significant then pairwise comparisons are not necessary. Others argue that if the pairwise comparisons were planned before the ANOVA was conducted (i.e., "a priori") then they are appropriate.

The results of our Tukey pairwise comparisons were as follows:

Grouping Information Using the Tukey Method and 95% Confidence
Instructor N Mean Grouping
Dr. Oh 87 71.448 A
Dr. Al 60 68.367 A
Dr. Pa 98 67.939 A

Means that do not share a letter are significantly different.

Tukey Simultaneous Tests for Differences of Means
Difference of Levels Difference of Means SE of Difference 95% CI T-Value Adjusted P-Value
Dr. Oh-Dr. Al 3.08 2.90 (-3.70, 9.86) 1.06 0.537
Dr. Pa-Dr. Al -0.43 2.83 (-7.05, 6.19) -0.15 0.987
Dr. Pa-Dr. Oh -3.51 2.54 (-9.46, 2.44) -1.38 0.351

Individual confidence level = 97.99%

Looking at the first table, all three instructors are in group A. Means that share a letter are not significantly different from one another (i.e., they are in the same group). Because all three instructors share the letter A, there are no significantly different pairs of instructors.

We could also look at the second table which gives us the t-test statistic and adjusted p-value for each possible pairwise comparison. This p-value is adjusted to take into account that multiple tests are being conducted. You can compare these p-values to the standard alpha level of .05.  All p-values are greater than .05, therefore no pairs are significantly different from one another. 

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