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Suppose a test statistic $\frac{MSS(X)}{MSS(Y)}$, where $MSS$ denotes Mean Sum of Squares, is to be used for testing the significance of the factor $X$. Do we need the assumption $$\mathbb E[MSS(X)]=\mathbb E[MSS(Y)]$$ to be satisfied? Why?

As far i know if the assumption is true , then the test statistic follows F-distribution .

Is that the only case?Is their any detail explanation?

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  • $\begingroup$ unless X and Y are normally distributed, then their ratio will not be an F distribution. Also, if their MSSs are equal, then why are you doing the test? $\endgroup$ – user76844 Sep 10 '14 at 2:59
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I take it that by "mean sum of squares" you mean sum of squares divided by degrees of freedom. Sometimes those are called mean squares; I don't think I've heard "mean sum of squares" before.

The F-distribution is usually characterized as the distribution of $$ \frac{\chi^2_n/n}{\chi^2_m/m} $$ where the two chi-square random variables have respective degrees of freedom $n$ and $m$ and are independent and are indeed chi-square random variables.

If they actually are chi-square random variables with the specified number of degrees of freedom, and they are independent, then the quotient above has an F-distribution.

And if they actually are chi-square random variables with the specified number of degrees of freedom, then the expected values of the numerator and denominator are equal to $1$.

But it is certainly not true that every time you divide a random variable with expected value $1$ by another random variable with expected value $1$, then the quotient has an F-distribution.

Perhaps the most well known situation in which an F-distribution arises is in the following situation: $$ \overbrace{\sum_{i=1}^n \sum_{j=1}^{m_i} (Y_{ij} - \bar Y_{\bullet\bullet})^2}^{\text{total sum of squares}} = \overbrace{\sum_{i=1}^n m_i(\bar Y_{i\bullet} - \bar Y)^2}^{\text{between-group sum of squares}} + \overbrace{\sum_{i=1}^n \sum_{j=1}^{n_i} (Y_{ij} - \bar Y_{i\bullet})^2}^{\text{within-group sum of squares}}. $$ where $$ \bar Y_{\bullet\bullet} = \frac{\sum_{i=1}^n \sum_{j=1}^{m_i} Y_{ij}}{\sum_{i=1}^n(m_i-1)} \text{is the grand mean and } \bar Y_{i\bullet} = \frac{\sum_{j=1}^{n_i} Y_{ij} }{m_i} \text{ are the group means}. $$ With the usual assumptions on normality, independence, and homoskedasticity, and assuming the null hypothesis that there is no difference between groups in the population from which the sample was taken, the following statistic has an F-distribution: $$ \frac{\text{between-group sum of squares}/(n-1)}{\text{within-group sum of squares}/\sum_{i=1}^n(m_i-1)}. $$

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