Quiz 3, St711, Fall 2009, Dickey 1. (12 pts.) I have a calculated F statistic 2.91 on my output and a p‐value, Pr>F that is 0.0413. Thus we reject our null hypothesis at the 5% level. Exactly what does that 0.0413 represent on the graph of F shown here? Use labels on the F graph shown here and a few words to explain. 2. (18 pts.) I have a repeated measures design with 3 time measurements per subject. Here are 3 categories of correlation matrices. Fill in the 4 missing entries in each correlation matrix in such a way that it illustrates the correlation matrix type for its label. Compound Symmetry Toeplitz AR(1) ___ 0.36 ⎞
___ 0.36 ⎞
___ 0.36 ⎞
⎛ 1
⎛ 1
⎛ 1
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⎜ ___
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___ ⎟ ⎜ ___
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___ ⎟ ⎜ 0.36 ___
⎜ 0.36 ___
⎜ 0.36 ___
1 ⎟⎠
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3. We have talked about differences between complete and incomplete designs. (A) (8 pts.) In a randomized complete block design, would you want to recover interblock information in testing for treatment effects? If not, why not? If so, how would you do it (using SAS)? (B) (5 pts.) In a randomized complete block design, will the LSMEANS treating blocks as fixed (GLM) and the LSMEANS treating blocks as random (MIXED) be different? (yes, no). If yes, explain why they are different. (no need to comment on standard errors). (C) (5 pts.) In a balanced incomplete block design, will the standard error for the difference between two treatment LSMEANS depend on whether the blocks are fixed or random? (yes, no). 4. I have factors A,B,C,D,E, and F, each at 2 levels and I run an unreplicated experiment using every possible combination of the levels of these 6 treatments. (A) (6 pts.) How many effects (i.e. points) _____ will there be in the Daniels half normal plot if I estimate all available main effects and interactions? (B) (24 pts.) My Daniels plot indicates that only A and D and their interaction have an effect so I fit a model using the model statement MODEL Y=A D A*D; in PROC GLM. List the error degrees of freedom ______ I will see on my printout. From the table of treatment means below, compute the A main effect ________ , its sum of squares ____ , and the sum of squares for AD interaction ______ . Treatment means A low A high D low 81 77 D high 79 67 (D) (7 pts.) Suppose I ran only half of the treatment combinations, namely the ones where A and D were at opposite levels (high A and low D or else low A and high D) and I know there are only main effects of factors A and D (no interactions). Do you foresee any problems when I fit a main effects model in factors A and D? (yes, no) Explain. (for example, I might be running MODEL Y=A D; in PROC GLM) 5. (15 pts.) Can I run a balanced incomplete block design with 10 treatments in 6 blocks of size 5? (yes, no). If so how many replicates ____ will I have of each treatment and how many times ____ will each pair of treatments occur together in a block? If it’s not possible, explain how you know. Answers 1. Under H0, this p‐value is the probability of getting an F at or exceeding the observed F 2.91. 2. Toeplitz – any A is OK. Stat majors – it must be positive definite Compound Symmetry Toeplitz AR(1) 0.36 0.36 ⎞
⎛ 1
⎛ 1
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⎜ 0.36
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0.36 ⎟ ⎜ A
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⎜ 0.36 0.36
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0.6 0.36 ⎞
A 0.36 ⎞
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A ⎟ ⎜ 0.6
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0.6 ⎟ 1
⎜ 0.36 0.6
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A
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3 A. In a RCB there is no interblock information to recover. Every comparison, like A versus B, is a mean of differences within blocks. No, you would not attempt to recover the (nonexistent) interblock information. B. Recall that the reason PROC MIXED has no MEANS statement is the means and lsmeans are the same for this kind of design (RCB). It is when the data are unbalanced, that the lsmeans and ordinary means differ from each other. It is in that case that the way the blocks enter the model (fixed vs. random) first came up. C. Yes. We spent some time talking about exactly how this difference comes about and had several examples where this was illustrated. 4 A. There are 64 data points from which we estimate 1 intercept and 63 effects so the Daniels plot of effects has 63 points. B. With 63 corrected total df and a model that has only A, D, and AD each with 1 df, we have 60 error df. This model will have sums of squares for A, D, and AD that we can compute from the given table of means. C. The A means are 80 and 72 so the A effect is ‐8. The overall mean is 76 so the A means deviate by 4 and ‐4 from this with each level of A being an average of 32 numbers. Our sum of squares is therefore SS(A) = 32(16+16)=1024. The interaction can be obtained from these totals: 16(81) 16(77) 16(79) 16(67) 1 ‐1 ‐1 1 Q = 16(81‐77‐79+67) = 16(‐8) = ‐128 Denominator = 16(1+1+1+1) = 16(4)=64 Q*Q/denominator = 4(64) = 256 = SS(AD). 5. You need r(k‐1) = λ(t‐1) which is 3(4) = λ(9) so no integer value of λ can be found. The number of times a pair of treatments occurs together in a block has to be an integer. I cannot run this. I know that r would have to be 3 for such a design because n=bk=30=tr=10r that is, 10r=30.