Share
Explore BrainMass

# ANOVA, Mean Square and Common Pop Variance

In an effort to counteract student cheating, the professor of a large class created four versions of a midterm exam, distributing the four versions among the students in the class. After the exam, students from the class got together and petitioned to nullify the results on the grounds that the four versions were not equal in difficulty.

To investigate the students' assertion, the professor examined the means and variances of the scores on the different versions of the exam, obtaining the following information (the exam had possible points).

Sample Sample Sample
Size Mean Variance
Version A 84 151.8 347.1
Version B 84 152.3 333.7
Version C 84 157.9 423.6
Version D 84 155.8 465.5

Taking the scores for each version of the exam as a sample of scores for that version, the professor performed a one-way, independent-samples ANOVA test of the equality of the population mean scores for the four versions.

What is the value of the mean square for error (the "within groups" mean square) that would be reported in the ANOVA test?
What is the value of the mean square for treatments (the "between groups" mean square) that would be reported in the ANOVA test?
Carry your intermediate computations to at least three decimal places, and round your responses to at least one decimal place.

#### Solution Summary

The solution performs ANOVA for anti-cheating measures in a large class.

\$2.19