Unformatted text preview: 114 Solutions Manual for Statistical Inference where N = ni . Therefore, the test is to reject H0 if ai Yi  2 Sp a2 /ni i > tN k, . 2 b. Similarly for H0 : ai i vs. H1 : ai i > , we reject H0 if ai Yi  > tN k, . 2 Sp a2 /ni i i 11.10 a. Let H0 , i = 1, . . . , 4 denote the null hypothesis using contrast ai , of the form i H0 : j 1 If H0 is rejected, it indicates that the average of 2 , 3 , 4 , and 5 is bigger than 1 which i is the control mean. If all H0 's are rejected, it indicates that 5 > i for i = 1, 2, 3, 4. To see 4 5 this, suppose H0 and H0 are rejected. This means 5 > 5 +4 > 3 ; the first inequality is 2 5 4 implied by the rejection of H0 and the second inequality is the rejection of H0 . A similar argument implies 5 > 2 and 5 > 1 . But, for example, it does not mean that 4 > 3 or 3 > 2 . It also indicates that aij j 0. 1 (5 + 4 ) > 3 , 2 1 (5 + 4 + 3 ) > 2 , 3 1 (5 + 4 + 3 + 2 ) > 1 . 4 b. In part a) all of the contrasts are orthogonal. For example, 0 0 5 1 1 1 1 =  1 + 1 + 1 = 0, a2i a3i = 0, 1,  ,  ,  1 3 6 6 3 3 3 i=1
2 1 2 and this holds for all pairs of contrasts. Now, from Lemma 5.4.2, Cov
i aji Yi ,
i aj i Yi = 2 n aji aj i ,
i which is zero because the contrasts are orthogonal. Note that the equal number of observations per treatment is important, since if ni = ni for some i, i , then
k k Cov
i=1 aji Yi ,
i=1 aj i Yi 2 aji aj = aji aj i = 2 ni ni i=1 i=1 k k i = 0. c. This is not a set of orthogonal contrasts because, for example, a1 a2 = 1. However, each contrast can be interpreted meaningfully in the context of the experiment. For example, a1 tests the effect of potassium alone, while a5 looks at the effect of adding zinc to potassium. 11.11 This is a direct consequence of Lemma 5.3.3. 11.12 a. This is a special case of (11.2.6) and (11.2.7). ...
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This note was uploaded on 02/03/2012 for the course STA 1014 taught by Professor Dr.hackney during the Spring '12 term at UNF.
 Spring '12
 Dr.Hackney
 Statistics

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