AMS 572 Class Notes
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
population a

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
(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.
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(3) LRT (Likelihood Ratio test)
Test Statistic:
ni
a
MSA
F0 

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
grand mean Y
Ya
iid
Yij ~ N (i ,  2 ),
i  1,
j  1,
,a
, ni
Balanced design: ni  n
iid
Yij ~ N ( ,  2 )
Y1 ~ N (  ,
a
 (Y
i 1
i
2
 Y )2
 /n
n
 (Y
ij
i 1 j 1

a
 (Y
i 1
i
n
)
~  a21
2
a
Yi ~ N (  ,
),
n
2
 Yi )2
~  a2( n1)   N2 a
2
 Y )2
H0
 /n
2
~  a21
iid
Theorem Let X i ~ N (  ,  2 ), i  1,
(1) X ~ N (  ,
n
(X
(2)
i 1
i
2
n
)
 X )2
2
,n

(n  1) S 2
2
~  n21
(3) X and S 2 are independent.
2
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.
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 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
Familywise Error (FWE):
Family wise error rate = αFWE = P(reject at least 1 true null hypothesis)
As more tests are conducted, the likelihood that one or more are significant just due to
chance (Type I error) increases. One can estimate the familywise error rate with the
following formula:
 FWE  1  (1   EC )c
where αFWE is the familywise error rate, αEC is the alpha rate for an individual test (often
set at .05), and c is the number of comparisons.
Bonferroni Correction:
The Bonferroni correction method simply calculates a new pairwise alpha to keep
the familywise alpha value at .05 (or another specified value). The formula for doing this
is as follows:
B 
 FWE
c
where αB is the new alpha based on the Bonferroni test that should be used to evaluate
each comparison or significance test, αFWE is the familywise error rate as computed in
the first formula, and c is the number of comparisons (statistical tests).
Until the advent of modern multiple comparison methods, the Bonferroni correction was
perhaps the most commonly used post hoc test, because it is highly flexible, very simple
to compute, and can be used with any type of statistical test (e.g., correlations)—not just
post hoc tests with ANOVA. The traditional Bonferroni, however, tends to lack power.
The loss of power occurs for several reasons: (1) the familywise error calculation
depends on the assumption that, for all tests, the null hypothesis is true. This is unlikely
to be the case, especially after a significant omnibus test; (2) all tests are assumed to be
orthogonal (i.e., independent or nonoverlapping) when calculating the familywise error
test, and this is usually not the case when all pairwise comparisons are made; (3) the test
does not take into account whether the findings are consistent with theory and past
research. If consistent with previous findings and theory, an individual result should be
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less likely to be a Type I error; and (4) Type II error rates are too high for individual tests.
In other words, the Bonferroni overcorrects for Type I error.
Tukey’s Studentized Range test: (* It is the preferred method to ensure the FEW when
all pairwise comparisons are of interest – the resulting FWE is exact when the sample
sizes are equal, and smaller (more conservative) when the sample sizes are unequal.)
H 0ij : i   j
H aij : i   j
At FWE  , reject H 0ij if
| tij |
yi  y j
s*
1 1

ni n j

qa , N a ,
2
where s 2 = MSE.
Fisher’s LSD (Least Significant Difference) test: (It does not control for the familywise
error rate!)
One could just use multiple t-tests to make each comparison desired, but one runs the risk
of greatly inflating the familywise error rate when doing so as discussed earlier. If you
Yi Y j
have equal sample sizes and homogeneity of variance, you can use t 0 
,
2  MSE
n
which pools the error variance across all a groups, giving you N - a degrees of freedom.
Y i Y j
If you have homogeneity of variance but unequal n’s use: t 0 
. MSE
1

1
MSE   
n n 
j 
 i
is the error mean square from the ANOVA.
Finally, the name ANOVA came from the partitioning of the variations:
a
ni
a
ni
a
ni
 ( yij  y )2   ( yij  yi )2   ( yi  y )2
i 1 j 1
i 1 j 1
SST
i 1 j 1
SSE
SSA
For more details, please refer to our text book.
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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).
Source: http://en.wikipedia.org/wiki/Ronald_Fisher
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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
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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.
8
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
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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
10
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 ;
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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
12
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.
Example 3. The effect of caffeine levels on performing a simple finger tapping task was
investigated in a double blind study. Thirty male college students were trained in finger
tapping and randomly assigned to receive three different doses of caffeine (0, 100, or 200
mg) with 10 students per dose group. Two hours following the caffeine treatment,
students were asked to finger tap and the numbers of taps per minute were counted. The
data are tabulated below.
Caffeine Dose
0 mg
242 245 244
100 mg
248 246 245
200 mg
246 248 250
Finger Taps per Minute
248 247 248 242 244 246 242
247 248 250 247 246 243 244
252 248 250 246 248 245 250
(a) Construct an ANOVA table and test if there are significant differences in finger
tapping between the groups at α =.05.
(b) Compare the finger tapping speed between the 0 mg and the 200 mg groups at α =.05.
List assumptions necessary – and, please perform tests for the assumptions that you
can test in an exam setting.
(c) Please write up the entire SAS program necessary to answer question raised in (a),
including the data step.
(d) Please write up the entire SAS program necessary to answer question raised in (b),
including the data step, and the tests for all assumptions necessary.
Answer:
(a) This is Problem 12.2(b) in our text book, one-way ANOVA. We are testing whether
the mean tapping speed in the three groups are equal or not. That is:
H 0 : 1  2  3 versus H a : The above is not true
13
Since F0 = 6.181 > F2,27,0.05 = 3.35, there do appear to be significant differences in
the number of finger taps for different doses of caffeine.
(b) This is inference on two population means, independent samples. The first
assumption is that both populations are normal. The second is the equal variance
assumption which we can test in the exam setting as the follows.
Group 1 (dose 0 mg): X 1  244.8 , s12  5.73 , n1  10
Group 2 (dose 200 mg): X 2  248.3 , s22  4.9 , n2  10
Under the normality assumption, we first test if the two population variances are equal. That is,
H 0 :  12   22 versus H a :  12   22 . The test statistic is
F0 
s12 5.73

 1.17 , F9,9,0.05,U  3.18 .
s22 4.9
Since F0 < 3.18, we cannot reject H0 . Therefore it is reasonable to assume that  12   22 .
Next we perform the pooled-variance t-test with hypotheses H 0 : 1   2  0 versus
H a : 1  2  0
t0 
X 1  X 2  0  244.8  248.3  0

 3.39
1 1
1 1
sp

5.315

n n2
10 10
Since t0  3.39 is smaller than t18,0.025  2.10092 , we reject H0 and claim that the
finger tapping speed are significantly different between the two groups at the significance
level of 0.05.
(c)
data finger;
input group taps @@;
datalines;
0 242 0 245 0 244 0 248 0 247 0 248 0 242 0 244 0 246
0 242
1 248 1 246 1 245 1 247 1 248 1 250 1 247 1 246 1 243
1 244
2 246 2 248 2 250 2 252 2 248 2 250 2 246 2 248 2 245
2 250
;
run;
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proc anova data = finger;
class group;
model taps = group;
means group/tukey;
run;
/*the means step is not necessary for the given
problem.*/
(d)
data finger2;
set finger;
where group ne 1;
run;
proc univariate data = finger2 normal;
class group;
var taps;
run;
proc ttest data = finger2;
class group;
var taps;
run;
proc npar1way data = finger2;
class group;
var taps;
run;
/* the data step from part (d) follows immediately after that from part (c).*/
/* alternatively, one can save the data finger as a permanent sas data, and then you
can use that later*/
Example 4. A taste test was conducted by general Foods to determine the palatability
resulting from using a high or low amount of liquid and making the texture fine or
coarse. The palatability was recorded from -3 (terrible) to =3 (excellent). Sixteen
consumer groups (50 people per group) were given one of the four liquid-texture
combinations to score, four groups per combination. The response for each group is the
total of their individual palatability scores, which is tabled below.
Low liquid,
Low liquid,
High liquid,
High liquid,
Coarse texture
Fine texture
Coarse texture
Fine texture
35
104
24
65
39
129
21
94
77
97
39
86
16
84
60
64
(a) Please perform suitable statistical tests at the significance level of α =.05 to examine
15
whether the palatability results are significantly influenced by [1] a high or low
amount of liquid, or [2] by fine or coarse texture.
(b) Please discuss which factor appears to influence the palatability more: liquid amount
or food texture?
Answers:
This problem can be done in at least 3 approaches:
(1) Independent samples t-tests;
(2) One-way ANOVA with pairwise comparisons;
(3) Two-way ANOVA with interaction.
The best solution is approach 3, two-way ANOVA; however, since it is not required, we
expect you to get the problem done using either approach (1) t-tests on the combined
groups, or approach (2) one-way ANOVA with pairwise comparisons – since the table
for Tukey’s studentized range test is not provided, you can simply replace it with the ttests on individual groups.
In below, I have provided the SAS programs (excluding tests on some assumptions such
as normality etc for simplicity) and the corresponding output for your reference. However,
in the exam situation, you are expected to write up the complete formula and solutions as
usual.
DATA taste;
INPUT liquid $ texture $ comb $ palat @@;
DATALINES;
L C LC 35 L C LC 39 L C LC 77 L C LC 16 L F LF 104 L F LF
129 L F LF 97 L F LF 84
H C HC 24 H C HC 21 H C HC 39 H C HC 60 H F HF 65 H F HF
94 H F HF 86 H F HF 64
;
RUN;
/*t-tests*/
proc ttest data=taste;
class liquid;
var palat;
run;
proc ttest data=taste;
class texture;
var palat;
run;
/*one-way ANOVA*/
PROC ANOVA DATA=taste;
CLASS comb;
MODEL palat = comb;
16
MEANS comb / TUKEY;
RUN;
/*two-way ANOVA*/
PROC ANOVA DATA=taste;
CLASS liquid texture;
MODEL palat = liquid|texture;
RUN;
17
Homework 6 & Solutions:
Chapter 12: 1, 2, 5, 8, 10, 33 (*Dear students, Please do the homework by yourself first
before looking at the solutions!)
Solutions to the above homework problems:
12.1
(a) Sugar :
(2.138) 2  (1.985) 2  (1.865) 2
s 
 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.
2
(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 :
18
Source
Shelves
Error
Total
Analysis of Variance
SS
d.f.
MS
275.03
2
137.5
227.80
57
3.996
502.83
59
F
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.
19
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 F0  F2,.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
20
Comparison
yi  y j
1 vs. 2
1 vs. 3
2 vs. 3
0.73
0.49
1.22
Significant?
Bonferroni
Tukey
No
No
No
No
Yes
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.
21
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
2
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
2
F0 
 2
 t0
MSE s (1 n1  1 n2 )
(c)
F0  F1, ,   F   f1, ,
 t  t , 2
22