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4. It is known that for any data sample variance s2 with divisor (n - 1) is an unbiased estimator of the population variance σ2. Then prove that E(SSE) = (n-v)o2 in one way ANOVA

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Answer #1

here u = m

(ny-1)W σ2 follows a chi-square distribution withn-1 degrees of freedom. Another theorem we learned back in Stat 414 states that if we add up a bunch of independent chi-square random variables, then we get a chi-square m variable with the degrees of freedom added up, too. So lets add up the above quantity for all η data ports, that is forja l to ni and z l to m. Doing so we get i-1j-1 SSE σ2 Because we assume independence of the observations Xt, we are adding up independent chi-square random variables. (By the way, the assumption of independence is a perfectly fine assumption as long as we take a SSE sample when we collect the data.) Therefore, the theorem tells us thatfollows a chi-square random variable with: (n1-1)+ (n2 -1)++(nm-1)-n-m of freedom... as was to be proved. Now, what can we say about the mean square eror MSE? Well, one thing is. Theorem. The mean square error MSE is (always) an unbiased estimator of σ f. Recall that to show that MSZ is an unbiased estinator ofo we need to show that E M E σ Also recall that the expected value of a chi-square random variable is its degrees offreedom The results of the s theorem therefore suggests that: SSE t said, heres the crux of the proof σ2 SSE TTI first equality comes frorn the definition of MSE The second equality comes from multiplying 4SE by l in a special way. The third equality comes from taking the expected value of SSEơ al equality comes from simple algebra. And the fourth and e E(MSE)-σ. we have shown that, no matter what, MSE IS an unbiased estimator of σ always!

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