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I am looking at two-factor experiments with fixed factors. I've seen that for MSE, the expected value is $\sigma^2$. For E(MSA), E(MSB), and E(MSAB) I think these will be linear combinations of $\sigma^2$ such as $\sigma^2$+ something. How can I go about finding these?

Research: I found for one-way ANOVA that E(MSE) can be obtained by dividing SSE by its' degrees of freedom and denoting it as a chi-square random variable. Can this be done in two-way ANOVA and if so, would it work for the factors MSA, MSB, and MSE?

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It is similar for a two-factor ANOVA, but maybe not quite as simple as you suggest.

If fixed factor A has $a$ levels, fixed factor B has $b$ levels, and there are $n$ replications in each of the $ab$ cells, then let the model be $$Y_{ijk} = \mu + \alpha_i + \beta_j + \gamma_{ij} + e_{ijk},$$ where $\sum_i \alpha_i = \sum_j \beta_j = \sum_i \gamma_{ij} = \sum_j \gamma_{ij} = 0,$ and $e_{ijk} \stackrel{iid}{\sim} \mathsf{NORM}(0, \sigma).$ Then define $\theta_\alpha = \frac{1}{a-1}\sum_i \alpha_i^2$, $\theta_\beta = \frac{1}{b-1}\sum_i \beta_i^2$, and $\theta_\gamma = \frac{1}{(a-1)(b-1)}\sum_i \sum_j \gamma_{ij}^2.$ With this notation

EMS(A) = $\sigma^2 + bn\theta_\alpha,\,$ EMS(B) = $\sigma^2 + an\theta_\beta,\,$ EMS(Interaction) = $\sigma^2 + n\theta_\gamma,\,$ and EMS(Error) = $\sigma^2.$

The important consequence of these EMS's is that the three F-ratios for testing each of the fixed factors and their interaction all have MS(Error) in the denominator.

The situation for models with a mixture of fixed and random effects is more complicated, and even controversial. (The controversy results from different kinds of restrictions on effect parameters, such as $\sum_i \alpha_i = 0,$ but trickier for random effects.)

Many statistical software packages print out EMS tables, and (even if not explicitly printed) use them to form appropriate F-ratios for various F-tests. The algebra for deriving EMS's can get a bit messy for more complex ANOVA models, and there are algorithms for finding EMS's; one of these is called the 'Bennett-Franklin' algorithm.

If you are taking a course in ANOVA designs, you should look for the definitions and derivations of EMS's in your text.

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  • $\begingroup$ How would you derive something like that then? For example, MSA=$\frac{nb\sum_{i=1}^n\alpha_i^2}{a-1}$ What is the process of taking its' expected value? The formulas look the same except for the addition of $\sigma^2$. $\endgroup$ – user408688 Apr 22 '17 at 16:59
  • $\begingroup$ Write out the formula for MSA, perform the square, find the expectation of each term (some of which are 0 because of independence), simplify. It's a bit messy; more work than I'm going to attempt here. Maybe you can find this process done in your text for one such EMS. K.A.Brownlee's Wiley book from the 1950s may be in your Univ library; it derives many EMSs, maybe this one. // In your question, you say 'chi-sq', but unless you assume effects are 0 (counterproductive), all but EMS(Error) are noncentral chi-sq. $\endgroup$ – BruceET Apr 23 '17 at 1:41

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