[zetablue1]

I recently completed a simulation and I got this problem wrong. I don't understand what formula to use for this question. Insight anyone?

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, so that each version was given to students. After the exam, the professor computed the following information about the scores (the exam was worth points):
Group Size mean Variance
Version A 72 150.5 454.2
Version B 72 152.9 490.4
Version C 72 147.5 395.1
Version D 72 158.0 391.8
The professor is willing to assume that the populations of scores from which the above samples were drawn are approximately normally distributed and that each has the same mean and the same variance.

Give an estimate of this common population variance based on the variance of the sample means given.

What is the value of the mean square for treatments (the between groups mean square) that would be reported in the ANOVA test?

I think this is the same question asked differently but again I just had no clue what formula to use to solve.

[Unknown008]

Variance is given by the mean square minus the square mean.

or