Some Chapter 2 Notes on Completely Randomized Designs
Want to compare the weight gains of hogs on 3 diets (A,B,C)
12 hogs available for study
Can use table of random digits on p. 624 of Kuehl. Arbitrarily start at Line 6, Column 6.
ID Number
1
2
3
4
5
6
7
8
9
10
11
12
Random Number
29
69
24
14
90
78
59
62
39
67
53
73
RANDOMIZE
−→ −→−→−→
ID Number
4
3
1
9
11
7
8
10
2
12
6
5
Diet
A
A
A
A
B
B
B
B
C
C
C
C
Weight Gain
28
33
30
37
23
28
25
24
31
28
33
28
Response Label
y11
y12
y13
y14
y21
y22
y23
y24
y31
y32
y33
y34
Assumptions All observations are independent and
y11 , y12 , y13 , y14 ∼ N (µ1 , σ 2 )
y21 , y22 , y23 , y24 ∼ N (µ2 , σ 2 )
y31 , y32 , y33 , y34 ∼ N (µ3 , σ 2 )
yij
= µi + eij
i = 1, . . . , t (t = 3 in this case)
j = 1, . . . , ri (r1 = r2 = r3 = 4 in this case)
eij
∼
independent N (0, σ 2 )
eij are experimental errors. If there was no experimental error, the weight gains of all hogs given the same treatment
would be exactly the same.
If µ1 = 33, what is e13 ?
How about e14 ?
We want to test
H0 : µ1 = µ2 = µ3 versus HA : µi 6= µk for some i, k.
If H0 is true, there exists µ such that
yij = µ + eij i = 1, . . . , t j = 1, . . . , ri .
This is known as the reduced model.
The full model is
yij + µi + eij i = 1, . . . , t j = 1, . . . , ri .
Under the reduced model, all diets are equivalent. Diets don’t affect weight gain. Under the full model, diets
matter. Which model seems to fit data the best?
(yij = µi + eij i = 1, . . . , t j = 1, . . . , ri )
For the Full Model:
We want estimates of the model parameters µ1 , µ2 , µ3 , and σ 2 .
Choose estimates of µ1 , µ2 , and µ3 that will make residuals as close to zero as possible.
experimental error eij = yij − µi
residual êij = yij − µ̂i
i
i
Choose µ̂i so that SSE = ti=1 rj=1
ê2ij = ti=1 rj=1
(yij − µ̂i )2 is as small as possible.
Such estimates are called least squares estimates.
P
P
P
P
Can use calculus or some slightly tricky algebra to show that
Pi
the least squares estimate of µi is µ̂i = ȳi· = rj=1
yij /ri .
What is the least squares estimates of µ2 in the weight gain experiment?
Note that
SSE =
ri
t X
X
2
(yij − µ̂i ) =
i=1 j=1
ri
t X
X
2
(yij − ȳi· ) =
i=1 j=1
t
X
r
(ri −
1)s2i
where
s2i
i=1
i
1 X
(yij − ȳi· )2
=
ri − 1 j=1
s2i is the sample variance in the ith group.
The error variance σ 2 is estimated by
M SE =
(N =
SSE
(r1 − 1)s21 + · · · + (rt − 1)s2t
=
= weighted average of within-group sample variances.
N −t
(r1 − 1) + · · · + (rt − 1)
Pt
i=1 ri
=total sample size)
(yij = µ + eij i = 1, . . . , t j = 1, . . . , ri )
For the Reduced Model:
êij = yij − µ̂
We want to estimate µ.
Least squares estimate of µ is µ̂ = ȳ·· =
SSER =
Pt
i=1
Pri
j=1 (yij
− µ̂)2 =
Pt
SSER =
Pt
i=1
i=1
Pt
i=1
Pri
j=1 (yij
− µ̂)2
Pri
j=1 yij /N .
Pri
j=1 (yij
− ȳ·· )2 = SSTotal
SSER ≥ SSEF
SSER − SSEF = SST = SSTreatment =sum of squares treatment
(SST is reduction in sum of squares error due to including treatment in the model.)
SSTotal
Pt
i=1
Pri
j=1 (yij
− ȳ·· )2
=
SSTreatment
+
SSError
=
Pt
+
Pt
i=1 ri (ȳi·
− ȳ·· )2
i=1
Pri
j=1 (yij
− ȳi· )2
Source
DF
Sum of Squares
Mean Square
Fo
Treatment
t−1
Pt
SST /(t − 1)
MST/MSE
Error
N −t
Pt
Pri
− ȳi· )2
Total
N −1
Pt
Pri
− ȳ·· )2
i=1 ri (ȳi·
i=1
i=1
− ȳ·· )2
j=1 (yij
j=1 (yij
SSE/(N − t)
Fo ∼ F with t − 1 and N − t degrees of freedom when H0 : µ1 = · · · = µt is true.