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STAT 211
Handout 10 (Chapter 10):The Analysis of Variance
When two or more populations or treatments are being compared, the characteristic that distinguishes the
populations or treatments from one another is called the factor under investigation.
Single Factor ANOVA :
Model: X ij   i   ij where i=1,...,I (number of treatments), j=1,...,J (number of observations in each
treatment).
X ij : observations
 i : ith treatment mean.
 ij : errors which are normally distributed with mean, 0 and the constant variance,  2 .
Alternative model:
X ij     i   ij where i=1,...,I (number of treatments), j=1,...,J (number of observations in each treatment).
X ij : observations
 :overall mean
 i   i   : ith treatment effect.
 ij : errors which are normally distributed with mean, 0 and the constant variance,  2 .
Assumptions:
(ii) X ij 's are independent (  ij 's are independent)
(ii)  ij 's are normally distributed with mean, 0 and the constant variance,  2 .
(iii) X ij 's are normally distributed with mean,  i and the constant variance,  2 .
Hypothesis: H 0 : 1  ...   I 1   I versus H a : at least one  i   j for i  j where  i .and  j 's are
treatment means. Or H 0 :  i  0 for all i versus H a :  i  0 for at least one i where  i is the ith treatment
effect.
Analysis of Variance Table:
Source
df
SS
MS
F
Prob > F
Treatments I-1
SSTr
MSTr = SStr / (I-1)
MSTr / MSE P-value
Error
I(J-1) SSE
MSE = SSE / [I(J-1)]
Total
IJ-1
SSTotal
where df is the degrees of freedom, SS is the sum of squares, MS is the mean square.
Reject H0 if the P-value or if the test statistics F > F;I-1,I(J-1).
If you reject the null hypothesis, you need to use multiple comparison test such as Tukey-Kramer, page 414 to
see which means are different.
_
_
100(1-)% simultaneous confidence interval for i-j : ( x i  x j )  Q , I ; I ( J 1)
MSE
.
J
or write the sample means in increasing order and look at their pairwise differences. Reject H 0 :  i   j if
_
_
x i  x j  Q , I ;I ( J 1)
MSE
in the multiple comparison test.
J
Q , I ; I ( J 1) is the critical value in studentized range distribution (Table A10).
Example 1: The data on Calcium content of wheat is observed. Four different storage times are considered.
Storage Period
Observations
0 months
58.75 57.94 58.91 56.85
1 month
58.87 56.43 56.51 57.67
2 months
59.13 60.38 58.01 59.95
4 months
62.32 58.76 60.03 59.36
Is there sufficient evidence to conclude that the
storage times? Use =0.05.
55.21 57.30
59.75 58.48
59.51 60.34
59.61 61.95
mean calcium content is not the same for the four different
You are testing H 0 : 1   2   3   4 versus H a : at least one  i   j for i  j , i,j=1,2,3,4.
Assumptions:
(i) Each month's distribution is normal.
(ii) Each month's distribution has identical standard deviations.
(iii) The observations selected for each month are independent from one another.
(iv) The samples selected for each month are independent from one another.
Analysis of Variance
Source
DF
SS
Factor
3
32.1381669
Error
20
32.9010529
Total
23
65.0392198
Level
Month0
Month1
Month2
Month4
Total
N
6
6
6
6
24
Mean
57.493333
57.951666
59.553333
60.338333
58.834166
MS
10.7127223
1.64505264
F
6.51
P
0.0030
StDev
1.3748118
1.3288857
0.89694374
1.4559044
1.681604
Bartlett's test for equal variances(normal distribution)
Test Statistic: 1.1633
P-Value
: 0.762
Tukey-Kramer multiple comparison test gives 95% simultaneous confidence intervals for
1   2 : (-2.5319, 1.6152)
1   3 : (-4.1335, 0.0135)
1   4 : (-4.9185, -0.7715)
 2   3 : (-3.6752, 0.4719)
 2   4 : (-4.4602, -0.3131)
 3   4 : (-2.8585, 1.2885)
Example 2: An engineer conducted a study of the factors influencing the lengths of steel bars. The lengths of
twelve bars were taken from a screw machine, 4 being subjected to W heat treatment, 4 to L heat treatment, and
4 to D heat treatment. The lengths (less 438) were as follows:
Heat Treatment
W L
D
6
4
7
7
6
9
1
-1 10
6
4
6
Boxplots of W - D
(means are indicated by solid circles)
10
5
D
L
W
0
Analysis of Variance
Source
DF
SS
Factor
2
46.17
Error
9
58.75
Total
11
104.92
MS
23.08
6.53
Level
W
L
D
StDev
2.708
2.986
1.826
N
4
4
4
Mean
5.000
3.250
8.000
F
3.54
Bartlett's Test (normal distribution)
Test Statistic: 0.637
P-Value
: 0.727
Levene's Test (any continuous distribution)
Test Statistic: 0.022
P-Value
: 0.979
We will answer the question using the output above.
P
0.074
Normal Probability Plot for x
ML Estimates - 95% CI
99
ML Estimates
95
Mean
5.41667
StDev
2.95687
90
Goodness of Fit
Percent
80
AD*
70
60
50
40
30
1.319
20
10
5
1
-5
0
5
10
15
Data
Normal Probability Plot for W...D
ML Estimates - 95% CI
W
99
L
D
95
Goodness of Fit
90
AD*
Percent
80
2.953
2.802
2.619
70
60
50
40
30
20
10
5
1
-5
0
5
Data
10
15
Example 3: A tire manufacturer wants to test whether the mean diameters of tires produced at its three plants
(New York, Illinois, and California) are equal. Last month, he took a random sample of tires at each plant, and
their diameters (in inches) were as follows:
New York
Illinois
California
24.2 24.2 24.1 24.4 24.2 24.3 24.4 24.3 24.2
24.1 24.2 24.4 24.3 24.1 24.4 24.5 24.4 24.2
24.3 24.3 24.4 24.3 24.4 24.4 24.4 24.5 24.4
Boxplots of NY - CA
(means are indicated by solid circles)
24.5
24.4
24.3
24.2
Analysis of Variance
Source
DF
SS
Factor
2
0.0674
Error
24
0.2911
Total
26
0.3585
Level
NY
IL
CA
N
9
9
9
Mean
24.244
24.311
24.367
CA
IL
NY
24.1
MS
0.0337
0.0121
F
2.78
P
0.082
StDev
0.113
0.105
0.112
Bartlett's Test (normal distribution)
Test Statistic: 0.042
P-Value
: 0.979
Levene's Test (any continuous distribution)
Test Statistic: 0.063
P-Value
: 0.939
Confidence Interval for  
 ci  i :
i
_
 ci x i  t / 2;I ( J 1)
i
MSE  ci2
J
with equal sample sizes will be
discussed in class.
If we go back to example 2, I do not approve L treatment. I really like to test see the differences between the
average of L and the combined average of D and W.
  L 
_
_
_
^
1
W   D  then   x L  1  xW  x D   3.250  1 (5  8)  3.25
2
2
2

MSE  ci2
_
 ci x i  t / 2;I ( J 1)
J
i
6.53(12  (0.5) 2  (0.5) 2 )
=(-6.79 , 0.29) is the 95%
4
 3.25  t 0.025;9
C.I. for  where t 0.025;9 =2.262.
For the case of unequal sample sizes, let n 
I
J
i 1
i
and j=1,…,Ji . Then the difference in the analysis of
variance table and multiple comparison test is as follows.
Source
Treatments
Error
Total
df
I-1
n-I
n-1
SS
MS
F
Prob > F
SSTr
MSTr = SStr / (I-1) MSTr / MSE P-value
SSE
MSE = SSE / (n-I)
SSTotal
where df is the degrees of freedom, SS is the sum of squares, MS is the mean square.
Reject H0 if the P-value or if the test statistics F > F;I-1,n-I.
If you reject the null hypothesis, you need to use multiple comparison test such as Tukey-Kramer to see which
means are different.
_
_
100(1-)% simultaneous confidence interval for i-j : ( x i  x j )  Q , I ;n  I
MSE  1
1 
  .
2  J i J j 
or write the sample means in increasing order and look at their pairwise differences.
_
_
Reject H 0 :  i   j if x i  x j  Q , I ;n  I
MSE  1
1 
  for the Tukey-Kramer.
2  J i J j 
Q , I ;n  I is the critical value in studentized range distribution (Table A10).
Example 4 (Exercise 10.22): The data is about the yield of tomatoes for four different levels of salinity. Using
=0.05, test for any differences in true average yield due to the different salinity levels.
Analysis of Variance
Source
DF
SS
Factor
3
456.50
Error
14
124.50
Total
17
581.00
Level
level
level
level
level
1.6
3.8
6.0
10.2
N
5
4
4
5
Pooled StDev =
Mean
58.280
55.400
50.850
45.500
2.982
MS
152.17
8.89
StDev
3.602
2.665
2.426
2.901
F
17.11
P
0.000
Individual 95% CIs For Mean
Based on Pooled StDev
---------+---------+---------+------(----*----)
(----*-----)
(-----*----)
(----*----)
---------+---------+---------+------48.0
54.0
60.0
Bartlett's Test (normal distribution)
Test Statistic: 0.558
P-Value
: 0.906
Levene's Test (any continuous distribution)
Test Statistic: 0.130
P-Value
: 0.940
Tukey's pairwise comparisons
Family error rate = 0.0500
Individual error rate = 0.0115
Critical value = 4.11
Intervals for (column level mean) - (row level mean)
1.6
3.8
3.8
-2.934
8.694
6.0
1.616
13.244
-1.578
10.678
10.2
7.299
18.261
4.086
15.714
6.0
-0.464
11.164
Differences between fixed effect and random effect models will be discussed in class.