Lec2 - These last two results can be demonstrated as follows SS A = n X i(¯ y i ¯ y · 2 since ¯ y i = μ a i ¯ e i and ¯ y · = μ ¯

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Unformatted text preview: These last two results can be demonstrated as follows: SS A = n X i (¯ y i ·- ¯ y ·· ) 2 since ¯ y i · = μ + a i + ¯ e i · and ¯ y ·· = μ + ¯ a · + ¯ e ·· , it follows that SS A = n X i ( a i + ¯ e i ·- ¯ a ·- ¯ e ·· ) 2 = n X i ( γ i- ¯ γ · ) 2 where γ i = a i + ¯ e i · , ¯ γ · = ¯ a · + ¯ e ·· . Notice that γ 1 , . . . , γ a iid ∼ N (0 , σ 2 a + σ 2 /n ). Therefore, γ 1 / p σ 2 a + σ 2 /n, . . . , γ a / p σ 2 a + σ 2 /n iid ∼ N (0 , 1). It follows that SS A n ( σ 2 a + σ 2 /n ) = X i ( γ i- ¯ γ · ) 2 ( σ 2 a + σ 2 /n ) = X i γ i p σ 2 a + σ 2 /n- ¯ γ · p σ 2 a + σ 2 /n ! 2 is a sum of a squared deviations from the mean in standard normal random variables (the γ i √ σ 2 a + σ 2 /n ’s). Therefore, SS A nσ 2 a + σ 2 ∼ χ 2 ( a- 1) . And since the expected value of a χ 2 (d . f . ) random variable is d . f . , we have E SS A nσ 2 a + σ 2 = a- 1 ⇒ E SS A a- 1 = nσ 2 a + σ 2 . 101 Under H : σ 2 a = 0, F = MS A MS E ∼ F ( a- 1 , N- a ) and we reject H if F > F α ( a- 1 , N- a ). ANOVA Table: Source of Sum of d.f. Mean E( MS ) F Variation Squares Squares Treatments SS A a- 1 MS A nσ 2 a + σ 2 MS A MS E Error SS E N- a MS E σ 2 Total SS T N- 1 • For an unbalanced design, replace n with ( N- ∑ i n 2 i /N ) / ( a- 1) in the above ANOVA table. 102 Estimation of Variance Components Since MS E is an unbiased estimators of its expected value σ 2 , we use ˆ σ 2 = MS E to estimate σ 2 . In addition, E MS A- MS E n = nσ 2 a + σ 2- σ 2 n = σ 2 a , so we use ˆ σ 2 a = MS A- MS E n to estimate σ 2 a . • The validity of this estimation procedure isn’t dependent on normal- ity assumptions (on a i s and e ij s). In addition, it can be shown that (under certain assumptions) the proposed estimators are optimal in a certain sense. • Occasionally, MS A < MS E . In such a case we will get ˆ σ 2 a < 0. Since a negative estimate of a variance component makes no sense, in this case ˆ σ 2 a is set equal to 0. 103 Confidence Intervals for Variance Components: Since SS E σ 2 ∼ χ 2 ( N- a ) it must be true that Pr χ 2 1- α/ 2 ( N- a ) ≤ SS E σ 2 ≤ χ 2 α/ 2 ( N- a ) = 1- α Inverting all three terms in the inequality just reverses the ≤ signs to ≥ ’s: Pr 1 χ 2 1- α/ 2 ( N- a ) ≥ σ 2 SS E ≥ 1 χ 2 α/ 2 ( N- a ) ! = 1- α ⇒ Pr SS E χ 2 1- α/ 2 ( N- a ) ≥ σ 2 ≥ SS E χ 2 α/ 2 ( N- a ) ! = 1- α Therefore, a 100(1- α )% CI for σ 2 is SS E χ 2 α/ 2 ( N- a ) , SS E χ 2 1- α/ 2 ( N- a ) ! . It turns out that it is a good bit more complicated to derive a confidence interval for σ 2 a . However, we can more easily find exact CIs for the intra- class correlation coefficient ρ = σ 2 a σ 2 a + σ 2 and for the ratio of the variance components: θ = σ 2 a σ 2 ....
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This note was uploaded on 11/13/2011 for the course STAT 8200 taught by Professor Staff during the Fall '08 term at University of Georgia Athens.

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Lec2 - These last two results can be demonstrated as follows SS A = n X i(¯ y i ¯ y · 2 since ¯ y i = μ a i ¯ e i and ¯ y · = μ ¯

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