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If there are only 2 groups to be compared, the two-sided two sample t-test is equivalent to the ANOVA F-test? How can I prove this intuitively, and mathematically?

Thanks!

Thanks!

1 Solution

ozo is right, but the author is asking for an intuitive explanation or basic pointers toward proof.

SS = sum of squared deviations, e.g., SUM(X - XBar)^2.

SS of random deviations is distributed Chi-square (scaled by (df)).

an F distribution is a ratio of scaled Chi-squares (by definition, see any math stat book)

Now take t^2

Numerator of t = [(M1 - M2) - (Xbar1 - Xbar2)]

Denominator of t = s = standard error of the difference.

Square the numerator? You get a Sum of Squared deviations (actually, a "sum" of one deviation)

Square the denominator? You get s^2 the variance which is a Sum of Squared deviations over (df).

You just found that t^2 is a ratio of chi-squares, i.e., because num and denom are sum of squares of random deviations.

which makes it exactly equal to an F statistic with (1 and df denominator) degrees of freedom.

This should be enough to get you to construct a proof.

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