Quiz 1 2011, St711, Dickey
Here is a SAS printout for analyzing data from a completely randomized design with equal replication of
5 treatments. The model is the usual one Yij = µ + τi + eij for replicate j of treatment i with the usual
assumption that the eij are independent and eij~N(0,σ2).
Dependent Variable: Y
Source
Model
Error
Corrected Total
DF
Sum of
Squares
(___)
(___)
19
2120
300
(______)
Parameter
Estimate
Intercept
trt
1
trt
2
trt
3
trt
4
trt
5
100.0000000
-15.0000000
-30.0000000
-10.0000000
-5.0000000
0.0000000
Mean Square
Standard
Error
B
B
B
B
B
B
(__________)
3.16227765
3.16227765
3.16227765
3.16227765
.
(_________)
20
t Value
xxxxx
-4.74
-9.49
(________)
-1.58
.
F Value
(_______)
Pr > |t|
xxxxxx
0.0003
<.0001
0.0064
0.1347
.
NOTE: The X'X matrix has been found to be singular, and a generalized inverse was used to solve
the normal equations. Terms whose estimates are followed by the letter 'B' are not
uniquely estimable.
(a) (4 points) How many replicates r=_____ of each treatment did I have?
(b) (28 points) Fill in the seven blanks (_) in the printout above.
(c) (10 points) Compute the sample means for the 5 treatments
Trt 1_______ Trt 2_______ Trt 3________ Trt 4_______ Trt 5 ________
(d) (5 points) Compute, if possible, the t test ________ for comparing the treatment 4 mean to the
treatment 5 mean. If not possible, put “NP”.
(e) (7 points) Compute, if possible, the t test ________ for comparing the treatment 2 mean to the
treatment 3 mean. If not possible, put “NP”.
(f) (8 points) Estimate, each of these if possible. If any is nonestimable, put “NE.” Do not make any
arbitrary assumptions here.
τ1-τ2 _________
2τ1-τ2 _________ µ+τ2 _________
µ _________
(g) (5 points) In terms of µ, τ1 , τ2 ,τ3 ,τ4 and τ5, what is being estimated by 2Y1• − Y2• ___________
(that is, what is the expected value of this combination of sample means)
(h) (5 points) Compute the standard error _________ of 2Y1• − Y2• .
2. (10 points) In which of these analysis of variance situations will the F test have more power, assuming
the same error variance σ2 in both. Explain briefly what you did to come up with your choice.
Situation 1: True treatment effects τ1 = 10, τ2=20, τ3 = 12 with 5 replicates of each treatment (note
τ • = 14 )
Situation 2: True treatment effects τ1 = 10, τ2=18, τ3 = 11 with 7 replicates of each treatment (note
τ • = 13 )
3. (18 points) For the same data as in question 1, suppose the analyst who ran the SAS program was
mistaken and that the design was really a randomized complete block design. Suppose further that the
block sum of squares was 100. Fill in the numbers that would result in the part of the printout shown
here, if he/she added a CLASS variable block to the model statement in PROC GLM.
Source
Model
Error
Corrected Total
DF
(___)
(___)
(___)
Sum of
Squares
(_______)
(_______)
(_______)
****************** Answers ******************************
From Corrected total df there are 20 observations and with 5 treatments
equally replicated, there must have been 4 replicates of each.
Source
DF
Model
Error
Corrected Total
Sum of
Squares
Mean Square
(5-1=4)
2120
(2120/4=530)
(19-4=15)
300
20
19
(2120+300=2420)
Parameter
Estimate
Intercept
trt
1
trt
2
trt
3
trt
4
trt
5
100.0000000
-15.0000000
-30.0000000
-10.0000000
-5.0000000
0.0000000
Standard
Error
B
B
B
B
B
B
t Value
F Value
(530/20=26.5)
Pr > |t|
( 2.236*)
xxxxx
xxxxxx
3.16227765
-4.74
0.0003
3.16227765
-9.49
<.0001
3.16227765 (-10/3.16=-3.16)0.0064
3.16227765
-1.58
0.1347
.
.
.
* In reference cell coding, intercept is mean of last treatment. Mean of r observations has variance estimated as
MSE/r so we have 20/4=5. Standard error is square root of estimated variance.
(C) These means are estimates of m+τi which are estimable so we can use any solution to compute them. We get
100-15=85, 70, 90, 95, and 100+0=100.
(D) This t is on the printout, -1.58
(E) This t is not on the printout but because of balance, all comparisons of 2 means have the same standard error.
We have an estimate, -30, of τ2-τ5 and an estimate, -10, of τ3-τ5 so our estimate is -30+10=-20 and the standard
error is 3.16227765 so the t test statistic is -6.3246.
(F) τ1-τ2 __15_______
2τ1-τ2 __NE______ µ+τ2 ___70______
µ ____NE_____
(g&h) 2Y1• − Y2• estimates 2(µ+τ1)-(µ+τ2) = µ + 2τ1 – τ2 with estimated variance 4(20/4) + 20/4 (three
rules) which is 25, standard error is 5.
2. Find largest noncentrality. Noncentrality is
r ∑ (τ i −τ ) 2 / σ 2 = 5(16 + 36 + 4) / σ 2 = 5(56) / σ 2 = 280 / σ 2 or
r ∑ (τ i −τ ) 2 / σ 2 = 7(9 + 25 + 4) / σ 2 = 7(38) / σ 2 = 266 / σ 2
We would expect situation 1 to have the most power (though this also depends on the denominator
degrees of freedom for the noncentral F)
3. If the replicates are blocks then you have 4 blocks, 3 block df and 4+3=7 model df. The error df
becomes 19-7=12 (or 4x3=12). The model sum of squares increases by 100 and the error sum of squares
decreases by 100.