AMS 572 Class Notes
November 19, 2010
Chapter 12 Analysis of Variance (ANOVA)
One-way ANOVA (fixed factors)
H 0 : 1  2 
 a
* Goal: compare the means from a (a≥2) different populations.
* It is an extension of the pooled variance t-test.
* Assumptions:
(i) Equal (unknown) population variances  12   22 
  a2   2
(ii) Normal populations
(iii) Independent samples
H a : these i ’s are not all equal.
Assumptions: a population, N ( i ,  2 ) , i=1,2,…,a.  2 is unknown.
Samples: a independent samples.
Data:
population1
populationi

sample1

samplei

sample a
Y11
Y
 12
n1 

Y1n
 1
Yi1

Yi 2
ni 


Yini
Ya1

Ya 2
na 

Yan
 a
Balanced design: ni  n
Unbalanced design: otherwise
Derivation of the test
(1) PQ, can be derived
population a
(2) * Union-intersection method. Best method for this type of test as in other regression
analysis related tests. Please see AMS 570/571 text book, and also the book by G.A.F.
Seber: Linear Regression Model, published by John Wiley for details.
(3) LRT (Likelihood Ratio test)
Test Statistic:
ni
a
F0 
MSA

MSE
 ( y
i
i 1 j 1
a ni
 ( y
ij
i 1 j 1
 y ) 2 /  a  1
 yi ) /  N  a 
H0
~ Fa 1, N  a
2
a
Total sample size N   ni .
i 1
Sample mean: Y1
Yi
Ya
iid
Yij ~ N (i ,  2 ),
grand mean Y
i  1,
j  1,
,a
Balanced design: ni  n
iid
Yij ~ N ( ,  2 )
Y1 ~ N (  ,
a
 (Y
i 1
i
n
 Y )2
2 /n
a
n
 (Y
ij
i 1 j 1

a
 (Y
i 1
2
i
),
n
)
)
~  a2( n1)   N2 a
 Y )2
H0
~  a21
iid
2
n
 Yi )2
Theorem Let X i ~ N (  ,  2 ), i  1,
(1) X ~ N (  ,
2
~  a21
2
2 /n
Yi ~ N (  ,
,n
, ni
n
(X
(2)
i 1
i

 X )2

2
(n  1) S 2

2
~  n21
(3) X and S 2 are independent.
Definition F 
W / k1
~ Fk1 ,k2 , where W ~ k21 ,V ~ k22 and they are independent.
V / k2
a
E ( MSA)   2 
 n (
i 1
i
i
  )2
a 1
, E ( MSE )   2 .

When H 0 is true: F0 ~1 , ( i   )
H a is true: F0  1 .
Intuitively, we reject H 0 in favor of H a if F0  C , where C is determined by the
significance level as usual:
  P(reject H 0 | H 0 )  P( F0  C | H 0 ) .
When a=2, H 0 : 1  2 H a : 1  2
T0 
y1  y2 H 0
~ Tn1  n2  2
1 1
S

n1 n2
Note: If T ~ tk , T 2 ~ F1,k . (One can prove this easily using the definitions of the t- and Fdistributions)
If we reject the ANOVA hypothesis, then we should do the pairwise comparisons.
 a  a(a  1)
 
2
2
H 01 : 1  2 H 02 : 1  3
H 0,a ( a 1) / 2 : a 1  a
H a1 : 1  2 H a 2 : 1  3
H a ,a ( a 1) / 2 : a 1  a
The multiple comparison problem
FWE: (Family wise error rate)
=P(reject at least 1 true null hypothesis)
Tukey’s Studentized Range test (* It is the preferred method to ensure the FWE)
H 0ij : i   j
H aij : i   j
At FWE  , reject H 0ij if
yi  y j
| tij |
1 1

ni n j
s*

qa , N a ,
2
Finally, the name ANOVA came from the partitioning of the variations:
a
ni
 ( y
i 1 j 1
ij
a
ni
a
ni
 y )   ( yij  yi )   ( yi  y ) 2
SST
2
2
i 1 j 1
i 1 j 1
SSE
For more details, please refer to the text book.
Attachment: Handout.
SSA
Ronald Fisher (1890-1962)
Sir Ronald Aylmer Fisher was a British statistician, evolutionary biologist, and geneticist.
He has been described as: “a genius who almost single-handedly created the foundations
for modern statistical science” and “the greatest of Darwin's successors”.
Fisher was born in East Finchley in London, to George and Katie Fisher. Although Fisher
had very poor eyesight, he was a precocious student, winning the Neeld Medal (a
competitive essay in Mathematics) at Harrow School at the age of 16. Because of his
poor eyesight, he was tutored in mathematics without the aid of paper and pen, which
developed his ability to visualize problems in geometrical terms, as opposed to using
algebraic manipulations. He was legendary in being able to produce mathematical results
without setting down the intermediate steps. In 1909 he won a scholarship to Gonville
and Caius College, Cambridge, and graduated with a degree in mathematics in 1913.
During his work as a statistician at the Rothamsted Agricultural Experiment Station, UK,
Fisher pioneered the principles of the design of experiments and elaborated his studies of
"analysis of variance". In addition to "analysis of variance", Fisher invented the technique
of maximum likelihood and originated the concepts of sufficiency, ancillarity, Fisher's
linear discriminator and Fisher information. The contributions Fisher made also included
the development of methods suitable for small samples, like those of Gosset, and the
discovery of the precise distributions of many sample statistics. Fisher published a
number of important texts including Statistical Methods for Research Workers (1925),
The design of experiments (1935) and Statistical tables (1947). Fisher's important
contributions to both genetics and statistics are emphasized by the remark of L.J. Savage,
"I occasionally meet geneticists who ask me whether it is true that the great geneticist
R.A. Fisher was also an important statistician" (Annals of Statistics, 1976).
John Tukey (1915-2000)
John Tukey, 85, Statistician; Coined the Word 'Software'
John Wilder Tukey, one of the most influential statisticians of the last 50 years and a
wide-ranging thinker credited with inventing the word ''software,'' died on Wednesday in
New Brunswick, N.J. He was 85.
The cause was a heart attack after a short illness, said Phyllis Anscombe, his sister-in-law.
Mr. Tukey developed important theories about how to analyze data and compute series of
numbers quickly. He spent decades as both a professor at Princeton University and a
researcher at AT&T's Bell Laboratories, and his ideas continue to be a part of both
doctoral statistics courses and high school math classes. In 1973, President Richard M.
Nixon awarded him the National Medal of Science.
But Mr. Tukey frequently ventured outside of the academy as well, working as a
consultant to the government and corporations and taking part in social debates.
In the 1950's, he criticized Alfred C. Kinsey's research on sexual behavior. In the 1970's,
he was chairman of a research committee that warned that aerosol spray cans damaged
the ozone layer. More recently, he recommended that the 1990 Census be adjusted by
using statistical formulas in order to count poor urban residents whom he believed it had
missed.
''The best thing about being a statistician,'' Mr. Tukey once told a colleague, ''is that you
get to play in everyone's backyard.''
An intense man who liked to argue and was fond of helping other researchers, Mr. Tukey
was also an amateur linguist who made significant contributions to the language of
modern times. In a 1958 article in American Mathematical Monthly, he became the first
person to define the programs on which electronic calculators ran, said Fred R. Shapiro, a
librarian at Yale Law School who is editing a dictionary of quotations with information
on the origin of terms. Three decades before the founding of Microsoft, Mr. Tukey saw
that ''software,'' as he called it, was gaining prominence. ''Today,'' he wrote at the time, it
is ''at least as important'' as the '' 'hardware' of tubes, transistors, wires, tapes and the like.''
Twelve years earlier, while working at Bell Laboratories, he had coined the term ''bit,'' an
abbreviation of ''binary digit'' that described the 1's and 0's that are the basis of computer
programs.
Both words caught on, to the chagrin of some computer scientists who saw Mr. Tukey as
an outsider. ''Not everyone was happy that he was naming things in their field,'' said
Steven M. Schultz, a spokesman for Princeton.
Mr. Tukey had no immediate survivors. His wife of 48 years, Elizabeth Rapp Tukey, an
antiques appraiser and preservation activist, died in 1998.
Mr. Tukey was born in 1915 in New Bedford, a fishing town on the southern coast of
Massachusetts, and was the only child of Ralph H. Tukey and Adah Tasker Tukey. His
mother was the valedictorian of the class of 1898 at Bates College in Lewiston, Me., and
her closest competition was her eventual husband, who became the salutatorian.
Classmates referred to them as the couple most likely to give birth to a genius, said Marc
G. Glass, a Bates spokesman.
The elder Mr. Tukey became a Latin teacher at New Bedford's high school, but, because
of a rule barring spouses from teaching at the school, Mrs. Tukey was a private tutor, Mrs.
Anscombe said. Mrs. Tukey's main pupil became her son, who attended regular classes
only for special subjects like French. ''They were afraid that if he went to school, he'd get
lazy,'' said Howard Wainer, a friend and former student of John Tukey's.
In 1936, Mr. Tukey graduated from nearby Brown University with a bachelor's degree in
chemistry, and in the next three years earned three graduate degrees, one in chemistry at
Brown and two in mathematics at Princeton, where he would spend the rest of his career.
At the age of 35, he became a full professor, and in 1965 he became the founding
chairman of Princeton's statistics department.
Mr. Tukey worked for the United States government during World War II. Friends said
he did not discuss the details of his projects, but Mrs. Anscombe said he helped design
the U-2 spy plane.
In later years, much of his important work came in a field that statisticians call robust
analysis, which allows researchers to devise credible conclusions even when the data
with which they are working are flawed. In 1970, Mr. Tukey published ''Exploratory Data
Analysis,'' which gave mathematicians new ways to analyze and present data clearly.
One of those tools, the stem-and-leaf display, continues to be part of many high school
curriculums. Using it, students arrange a series of data points in a series of simple rows
and columns and can then make judgments about what techniques, like calculating the
average or median, would allow them to analyze the information intelligently.
That display was typical of Mr. Tukey's belief that mathematicians, professional or
amateur, should often start with their data and then look for a theorem, rather than vice
versa, said Mr. Wainer, who is now the principal research scientist at the Educational
Testing Service.
''He legitimized that, because he wasn't doing it because he wasn't good at math,'' Mr.
Wainer said. ''He was doing it because it was the right thing to do.''
Along with another scientist, James Cooley, Mr. Tukey also developed the Fast Fourier
Transform, an algorithm with wide application to the physical sciences. It helps
astronomers, for example, determine the spectrum of light coming from a star more
quickly than previously possible.
As his career progressed, he also became a hub for other scientists. He was part of a
group of Princeton professors that gathered regularly and included Lyman Spitzer Jr.,
who inspired the Hubble Space Telescope. Mr. Tukey also persuaded a group of the
nation's top statisticians to spend a year at Princeton in the early 1970's working together
on robust analysis problems, said David C. Hoaglin, a former student of Mr. Tukey.
Mr. Tukey was a consultant to the Educational Testing Service, the Xerox Corporation
and Merck & Company. From 1960 to 1980, he helped design the polls that the NBC
television network used to predict and analyze elections.
His first brush with publicity came in 1950, when the National Research Council
appointed him to a committee to evaluate the Kinsey Report, which shocked many
Americans by describing the country's sexual habits as far more diverse than had been
thought. From their first meeting, when Mr. Kinsey told Mr. Tukey to stop singing a
Gilbert and Sullivan tune aloud while working, the two men clashed, according to ''Alfred
C. Kinsey,'' a biography by James H. Jones.
In a series of meetings over two years, Mr. Kinsey vigorously defended his work, which
Mr. Tukey believed was seriously flawed, relying on a sample of people who knew each
other. Mr. Tukey said a random selection of three people would have been better than a
group of 300 chosen by Mr. Kinsey.
By DAVID LEONHARDT, July 28, 2000 © The New York Times Company
Example 1. A deer (definitely not reindeer) hunter prefers to practice with several
different rifles before deciding which one to use for hunting. The hunter has chosen five
particular rifles to practice with this season. In one test to see which rifles could shoot the
farthest and still have sufficient knock-down power, each rifle was fired six times and the
distance the bullet traveled recorded. A summary of the sample data is listed below,
where the distances are recorded in yards.
Rifle
1
2
3
4
5
Mean
341.7
412.5
365.8
505.0
430.0
Std. Dev.
40.8
23.6
62.2
28.3
38.1
(a) Are these rifles equally good? Test at =0.05.
Answer: This is one-way ANOVA with 5 “samples” (a=5), and 6 observations per
sample ( ni  n  6 ), and thus the total sample size is N=30. The grand mean is
341.7  412.5   430.0
y
 411
5
We are testing H 0 : 1  2   5 versus H a : at least one of these equalities is not
true. The test statistic is
MSA
F0 
MSE
where
a
MSA 
 n ( y  y)
i 1
i
i
a 1
2
6
 [(341.7  411) 2 
4
 (430.0  411) 2 ]  24067.17
and
a
MSE  s 2 
 (n  1)s
i 1
a
2
i
i
 (n  1)
i 1
1
 [40.82 
5
 38.12 ]  1668.588
i
Therefore
F0 
24067.17
 14.42
1668.588
Since F0  14.42  F4,25,0.05  2.67 , we reject the ANOVA hypothesis H 0 and claim that
the five rifles are not equally good.
(b) Use Tukey’s procedure with =0.05 to make pairwise comparisons among the five
population means.
Answer: Now we will do the pairwise comparison using Tukey’s method. The Tukey
method will reject any pairwise null hypothesis H 0ij : i   j at FWE= if
yi  y j
 qa , N a ,
s/ n
In our case, a=5, n=6, s  1668.588  40.85 , N-a=25, =0.05, and
qa , N  a ,  q5,30,0.05  4.17 . Therefore, we would reject H 0ij if
yi  y j  s * qa , N  a , / n  69.54
The conclusion is that at the familywise error rate of 0.05, we declare that the following
rifle pairs are significantly different: 4/1, 4/2, 4/3, 4/5, 5/1, 2/1.
Example 2. Fifteen subjects were randomly assigned to three treatment groups X, Y and
Z (with 5 subjects per treatment). Each of the three groups has received a different
method of speed-reading instruction. A reading test is given, and the number of words per
minute is recorded for each subject. The following data are collected:
X
700
850
820
640
920
Y
480
460
500
570
580
Z
500
550
480
600
610
Please write a SAS program to answer the following questions.
(a) Are these treatments equally effective? Test at α = 0.05.
(b) If these treatments are not equally good, please use Tukey’s procedure with α =
0.05 to make pairwise comparisons.
Answer: This is one-way ANOVA with 3 samples and 5 observations per sample.
The SAS code is as follows:
DATA READING;
INPUT GROUP $ WORDS @@;
DATALINES;
X 700 X 850 X 820 X 640 X 920 Y 480 Y 460 Y 500
Y 570 Y 580 Z 500 Z 550 Z 480 Z 600 Z 610
;
RUN ;
PROC ANOVA DATA=READING ;
TITLE ‘Analysis of Reading Data’ ;
CLASS GROUP;
MODEL WORDS = GROUP;
MEANS GROUP / TUKEY;
RUN;
The SAS output is as follows:
Analysis of Reading Data
The ANOVA Procedure
Class Level Information
Class
Levels
GROUP
3
Values
X Y Z
Number of observations
15
The ANOVA Procedure
Dependent Variable: WORDS
Source
DF
Sum of
Squares
Mean Square
F Value
Pr > F
Model
2
215613.3333
107806.6667
16.78
0.0003
Error
12
77080.0000
6423.3333
Corrected Total
14
292693.3333
R-Square
Coeff Var
Root MSE
WORDS Mean
0.736653
12.98256
80.14570
617.3333
Source
DF
Anova SS
Mean Square
F Value
Pr > F
GROUP
2
215613.3333
107806.6667
16.78
0.0003
The ANOVA Procedure
Tukey's Studentized Range (HSD) Test for WORDS
NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type
II error rate than REGWQ.
Alpha
0.05
Error Degrees of Freedom
12
Error Mean Square
6423.333
Critical Value of Studentized Range 3.77278
Minimum Significant Difference
135.22
Means with the same letter are not significantly different.
Tukey Grouping
Mean
N
GROUP
A
786.00
5
X
B
B
B
548.00
5
Z
518.00
5
Y
Conclusion:
(a) The p-value of the ANOVA F-test is 0.0003, less than the significance level α =
0.05. Thus we conclude that the three reading methods are not equally good.
(b) Furthermore, the Tukey’s procedure with α = 0.05 shows that method X is
superior to methods Y and Z, while methods Y and Z are not significantly
different.
http://www.stat.sfu.ca/~sgchiu/Grace/S330/handouts/1-way/
Example 3. One-Way ANOVA on SAS -- Motor Oil Example. The SAS code:
/****************************************************
A sample SAS program to analyze the Motor Oil data
****************************************************/
title 'Motor Oil analysis';
options nodate noovp linesize=68;
/****************************************************
Virtually all SAS programs consist of a DATA step where
the raw data is read into a SAS file, and procedure (PROC)
step which perform various analyses
****************************************************/
data oil;
/* this will read the data into an internal SAS dataset called 'oil' */
input type $ visc;
/* the $ indicates 'type' is a character variable */
label type='oil type';
label visc='viscosity';
datalines;
CNVNTNL
CNVNTNL
CNVNTNL
CNVNTNL
SYNTHET
SYNTHET
SYNTHET
SYNTHET
HYBRID
HYBRID
HYBRID
;
/* data follows */
44
49
37
38
42
49
52
57
60
58
78
proc print
data=oil;
/* this will print out the raw data for checking */
title2 'raw data';
proc sort
by type;
data=oil;
proc means
data=oil maxdec=2 n mean std;
/* get simple summary statistics (sample size, sample mean and SD) with
max of 2 decimal places */
title2 'simple summary statistics';
by type; /* statistics computed for each oil type... */
var visc; /* ... on the variable 'visc' */
proc plot
data=oil;
/* request a plot of the raw data */
title2 'plot of the raw data';
plot visc*type;
proc anova
data=oil;
title2 'Analysis';
class type;
/* class statement indicates that 'oil type' is a factor */
model visc = type;
/* assumes 'oil type' influences 'viscosity' */
means type / tukey cldiff;
/* multiple comparison by Tukey's method -- get actual C.I.'s */
means type / tukey lines;
/* get pictorial display of comparisons */
proc glm
data=oil;
title2 'Proc glm Analysis';
/* same as 'proc anova' except
'glm' allows residual plots but gives more junk output */
class type;
model visc = type;
output
out=oilfit p=yhat r=resid;
/* store fitted values and fitted residuals
in dataset called 'oilfit' for later use */
proc univariate
data=oilfit plot normal;
var resid;
/* plot qq-plot of fitted residuals */
proc plot;
plot resid*type;
plot resid*yhat;
/* two residual plots to check
independence and constant variance */
run;
Related homework problems:
Chapter 12: 1, 2, 5, 8, 10, 33* (#33 for AMS students only)
Solutions to the above homework problems:
12.1
(a) Sugar :
(2.138) 2  (1.985) 2  (1.865) 2
 3.996 ,
3
So that s  3.996  1.999 with 3 19  57 d.f. Using the critical value t 57, 0.025  2.000 ,
the 95% CI’s are :
1.999
Shelf 1 : 4.80  (2.000) 
 [3.906,5.694]
20
1.999
Shelf 2 : 9.85  (2.000) 
 [8.956,10.744]
20
1.999
Shelf 3 : 6.10  (2.000) 
 [5.206,6.994]
20
Fiber :
(1.166) 2  (1.162) 2  (1.277) 2
2
s 
 1.447 ,
3
So that s  1.447  1.203 with 3 19  57 d.f. Using the critical value t 57, 0.025  2.000 ,
the 95% CI’s are :
1.203
Shelf 1 : 1.68  (2.000) 
 [1.142,2.218]
20
1.203
Shelf 2 : 0.95  (2.000) 
 [0.412,1.488]
20
1.203
Shelf 3 : 2.17  (2.000) 
 [1.632,2.708]
20
Shelf 2 cereals are higher in sugar content than shelves 1 and 3, since the CI for shelf 2 is
above those of shelves 1 and 3. Similarly, the shelf 2 fiber content CI is below that of
shelf 3. So in general, shelf 2 cereals are higher in sugar and lower in fiber.
s2 
(b) Sugar :
SSE  57  .3.996  227.80 ,
SSA  n yi2  Ny 2
 20[( 4.80) 2  (9.85) 2  (6.10) 2 ]  60  (6.92) 2
 275.03
Then the ANOVA table is below :
Analysis of Variance
Source
SS
d.f.
MS
F
Shelves
Error
Total
275.03
227.80
502.83
2
57
59
137.5
3.996
34.41
Since F  f 2.57, 0.05  3.15 , there do appear to be significant differences among the
shelves in terms of sugar content.
Fiber :
SSE  57  .1.447  82.47 ,
SSA  n yi2  Ny 2
 20[(1.68) 2  (0.95) 2  (2.17) 2 ]  60  (1.60) 2
 15.08
Then the ANOVA table is below :
Analysis of Variance
Source
SS
d.f.
MS
F
Shelves
15.08
2
7.54
5.21
Error
82.47
57
1.447
Total
97.55
59
Since F  f 2.57, 0.05  3.15 , there do appear to be significant differences among the
shelves in terms of fiber content, as well as sugar content.
(c) The grocery store strategy is to place high sugar/low fiber cereals at the eye height of
school children where they can easily see them.
12.2
(a) The boxplot indicates that the number of finger taps increases with higher doses of
caffeine.
(b)
Analysis of Variance
Source
SS
d.f.
MS
F
Dose
61.400
2
30.700
6.181
Error
134.100
27
4.967
Total
195.500
29
Since F  f 2.27, 0.05  3.35 , there do appear to be significant differences in the numbers of
finger taps for different doses of caffeine.
(c) From the plot of the residuals against the predicted values, the constant variance
assumption appears satisfied. From the normal plot of the residuals, the residuals appear
to follow the normal distribution.
12.5
(a) The boxplot indicates that HBSC has the highest average hemoglobin level, followed
by HBS, and then HBSS.
(b)
Analysis of Variance
Source
SS
d.f.
MS
F
Disease type
99.889
2
49.945
49.999
Error
37.959
38
0.999
Total
137.848
40
Since F  f 2.38,0.05  3.23 , there do appear to be significant differences in the hemoglobin
levels between patients with different types of sickle cell disease.
(c) From the plot of the residuals against the predicted values, the constant variance
assumption appears satisfied. From the normal plot of the residuals, the residuals appear
to follow the normal distribution.
12.8
Sugar : The number of comparisons is
 a 3 
      3 ,
 2   2
Then
t 0.05  t 57, 0.0083  2.468 ,
57,
23
and the Bonferroni critical value is
2
2
t 57,0.0083  s 
 2.468  1.999
 1.56 .
n
20
For the Tukey method,
q3,57, 0.05  3.40 ,
And the Tukey critical value is
1
1
q3,57,0.05  s
 3.40  1.999
 1.52
n
20
Comparison
yi  y j
1 vs. 2
1 vs. 3
2 vs. 3
5.05
1.30
3.75
Significant?
Bonferroni
Tukey
Yes
Yes
No
No
Yes
Yes
Fiber : The Bonferroni critical value is
2
2.468  1.203
 0.939
20
And the Tukey critical value is
1
3.40  1.203
 0.915
20
Comparison
yi  y j
1 vs. 2
0.73
Significant?
Bonferroni
Tukey
No
No
1 vs. 3
2 vs. 3
0.49
1.22
No
Yes
No
Yes
12.10
The number of comparisons is
 a 3 
      3 ,
 2   2
Then
t 0.01  t 38, 0.0017  3.136 ,
38,
23
And, since s  0.999  1 , the form of the Bonferroni confidence interval is
yi  y j  (3.136)(1)
1
1

ni n j
For the Tukey method,
q3,38,0.01 4.39

 3.104 ,
2
2
And the form of the Tukey confidence interval is
1
1
yi  y j  (3.104)(1)

ni n j
For the LSD method, t 38,0.01 2  2.712 , and the form of the LSD confidence interval is
yi  y j  (2.712)(1)
1
1

ni n j
The 99% confidence intervals are summarized in the table below :
Bonferroni
Tukey
LSD
yi  y j
Comparison
Lower Upper Lower Upper Lower Upper
HBSC(3) vs. HBS(2)
1.670
0.392
2.948
0.406
2.934
0.564 2.776
HBS(2) vs. HBSS(1)
1.918
0.656
3.179
0.669
3.166
0.825 3.010
HBSC(3) vs. HBSS
3.588
2.462
4.713
2.475
4.700
2.614 4.561
(1)
Since all of these intervals are entirely above 0, all of the types of disease have
significantly different hemoglobin levels from one another. Note that the Bonferroni
method has the widest intervals, followed by the Tukey. The LSD intervals are narrowest
because there is no adjustment for multiplicity.
12.33
(a) Note that
y
n1 y1  n2 y 2
n1  n2
Therefore,
SSA  MSA  n1 ( y1  y ) 2  n2 ( y 2  y ) 2
2


n y  n2 y 2 
n1 y1  n2 y 2 
 n1  y1  1 1
  n2  y 2 

n1  n2 
n1  n2 


2
 n ( y  y2 ) 
 n1 ( y1  y 2 ) 
 n1  2 1
  n2 

 n1  n2 
 n1  n2 
n n 2 ( y  y 2 ) 2 n2 n12 ( y1  y 2 ) 2
 1 2 1

(n1  n2 ) 2
(n1  n2 ) 2
n1 n2 (n1  n2 )( y1  y 2 ) 2

(n1  n2 ) 2

n1 n2 ( y1  y 2 ) 2
(n1  n2 )

( y1  y 2 ) 2
(1 n1  1 n2 )
(b)
( y1  y 2 ) 2
MSA
F
 2
 t2
MSE s (1 n1  1 n2 )
(c)
F  f1, ,   F   f1, ,
 t  t , 2
2
2
***The following is a past exam problem with solutions using SAS.
Fifteen subjects were randomly assigned to three treatment groups X, Y and Z (with 5 subjects
per treatment). Each of the three groups has received a different method of speed-reading
instruction. A reading test is given, and the number of words per minute is recorded for each
subject. The following data are collected:
X
700
850
820
640
920
Y
480
460
500
570
580
Z
500
550
480
600
610
(c) Are these treatments equally effective? Test at α = 0.05.
(d) If these treatments are not equally good, please use Tukey’s procedure with α = 0.05 to
make pairwise comparisons.
(e) Please write a SAS program to answer the above two questions.
Solution : This is one-way ANOVA with 3 samples and 5 observations per sample.
(a)
The ANOVA Procedure
Dependent Variable: WORDS
Source
DF
Sum of
Squares
Mean Square
F Value
Pr > F
Model
2
215613.3333
107806.6667
16.78
0.0003
Error
12
77080.0000
6423.3333
Corrected Total
14
292693.3333
The p-value of the ANOVA F-test is 0.0003, less than the significance level α = 0.05. Thus
we conclude that the three reading methods are not equally good.
(b)
The ANOVA Procedure
Tukey's Studentized Range (HSD) Test for WORDS
NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type
II error rate than REGWQ.
Alpha
0.05
Error Degrees of Freedom
12
Error Mean Square
6423.333
Critical Value of Studentized Range 3.77278
Minimum Significant Difference
135.22
Means with the same letter are not significantly different.
Tukey Grouping
Mean
N
GROUP
A
786.00
5
X
B
B
B
548.00
5
Z
518.00
5
Y
Thus we found that method X is superior to methods Y and Z, while methods Y and Z are not
significantly different at the family-wise error rate of 0.05 using Tukey’s Studentized Range test.
(c) The SAS code is as follows:
DATA READING;
INPUT GROUP $ WORDS @@;
DATALINES;
X 700 X 850 X 820 X 640 X 920 Y 480 Y 460 Y 500
Y 570 Y 580 Z 500 Z 550 Z 480 Z 600 Z 610
;
RUN ;
PROC ANOVA DATA=READING ;
TITLE ‘Analysis of Reading Data’ ;
CLASS GROUP;
MODEL WORDS = GROUP;
MEANS GROUP / TUKEY;
RUN;