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Unformatted text preview: Lecture L: Oneway Analysis of Variance (ANOVA)
Text Sections 9.1, 9.2, and 9.3
Testing for difference in the means among several (>2) groups
Example A (Cereal)
Acme Food Company wished to test five different package designs for a new breakfast cereal. Fifty stores, with
approximately equal sales volumes, were selected as the experimental units. Each store was randomly assigned one
of the package designs, with each package design assigned to ten stores. Sales, in numbers of cases, were observed
for the study period. x
s
s2 Design A
73
83
78
86
78
78
79
76
72
72
77.5
4.57651
20.94444 Design B
58
82
70
76
81
104
67
79
67
69
75.3
12.5614
157.7889 Design C
82
67
94
83
80
71
104
87
73
77
81.8
11.1036
123.2889 Design D
94
85
93
79
69
81
103
114
77
87
88.2
13.2648
175.9556 Design E
104
81
88
94
101
81
80
91
90
86
89.6
8.26236
68.26667 Combined 82.48
11.5339
133.0302 120
110 100
90
80
70
60
Design A Test: H0: μ1 = μ2 = . . . = μk Design B vs. Design C Design D Design E Ha: not all the means are equal. Explanatory variable (a.k.a. Independent variable, Predictor Variable): a categorical variable.
In ANOVA, also called treatment or factor.
Each treatment/factor has various levels or treatment levels
Response variable (a.k.a. Dependent variable): a quantitative variable. Knapp Stat 350
Spring 2009 Lecture L: 1way ANOVA
Page 1 of 1 ⎛ between group variation ⎞
Test statistic = ⎜
⎟
⎝ within group variation ⎠
100 140
120 80 100
60
80
60 40 40
20
20
0 A B C D 0 1 2 3 4 Oneway Analysis of Variance (ANOVA) a.k.a. SingleFactor ANOVA
Design: completely randomized design
Notation
Assume we are testing for differences among k treatment levels
xi,j = the jth observation in the ith treatment level.
n1, n2, . . . , nk are the sample sizes in groups 1, 2, . . ., k respectively
n = n1 + n2 + . . . + nk is the total sample size
x1 , x2 ,… , xk are the sample means in groups 1, 2, . . . , k respectively.
2
s12 , s2 ,… , sk2 are the sample variances in groups 1, 2, . . . , k respectively x is the overall average (the average of all n observations) Knapp Stat 350
Spring 2009 Lecture L: 1way ANOVA
Page 2 of 2 Assumptions
2
(1) The variances are the same across the k populations: σ 12 = σ 2 =
(2) Each of the k populations follow a normal distribution Treatment Sum of Squares: SSTr ) ( ( 2 ) 2 SSTr = n1 x1 − x + n2 x2 − x + ( + nk xk − x ) = σ k2 2 Error Sum of Squares: SSE
n1 SSE = ∑ ( x1 j − x1 ) n2 nk + ∑ ( x2 j − x2 ) + j =1 + ∑ ( xkj − xk ) 2
+ ( n2 − 1) s2 2 2
+ ( nk − 1) sk 2 j =1 = ( n1 − 1) s12 + 2 j =1 Total Sum of Squares: SST
SST = SSTr + SSE
k ni ( SST = ∑∑ xij − x
i =1 j =1 ) 2 = ( n − 1) s 2 Degrees of Freedom
• Total degrees of freedom = n 1 (associated with SST)
• Treatment degrees of freedom = k – 1 (associated with SSTr)
• Error degrees of freedom = n – k (associated with SSE) Mean square for treatments (betweengroups): MSTr
SSTr
MSTr =
k −1
Mean square error (withingroups): MSE
SSE
MSE =
n−k
Test Statistic: F
MSTr SSTr ( k − 1)
F=
=
MSE SSE ( n − k )
Under the null hypothesis, this statistic has an Fdistribution with
k – 1 numerator degrees of freedom
n – k denominator degrees of freedom Knapp Stat 350
Spring 2009 Lecture L: 1way ANOVA
Page 3 of 3 Example Data Set B: A simple data set for illustration
A
73
83
78 mean
variance 78
25 B
58
82
70
76
81
73.4
96.8 C
61
105
59
103 Overall 82
77.41667
646.6667 231.1742 SSTr SSE SST Knapp Stat 350
Spring 2009 Lecture L: 1way ANOVA
Page 4 of 4 Example B. (continued)
A
73
83
78 mean
variance 78
25 B
58
82
70
76
81
73.4
96.8 C
61
105
59
103 Overall 82
77.41667
646.6667 231.1742 MSTr MSE TestStatistic F ANOVA Table – summarizes results of an ANOVA
Source of variation
Treatments
Error
Total Knapp Stat 350
Spring 2009 df
k–1
n–k
n–1 SS
MS
F
SSTr MSTr MSTr/MSE
SSE MSE
SST Lecture L: 1way ANOVA
Page 5 of 5 SAS code for Example A (Cereal)
data cereal;
input design $ cases;
cards;
A
73
A
83
A
78
A
86
A
78
A
78
A
79
A
76
A
72
A
72
B
58
B
82
B
70
B
76
B
81
B
104
B
67
B
79
B
67
B
69
C
82
C
67
C
94
C
83
C
80
C
71
C
104
C
87
C
73
C
77
D
94
D
85
D
93
D
79
D
69
D
81
D
103
D
114
D
77
D
87
E
104
E
81
E
88
E
94
E
101
E
81
E
80
E
91
E
90
E
86
;
proc anova data=cereal;
class design;
model cases = design;
means design; run;
Knapp Stat 350
Spring 2009 Lecture L: 1way ANOVA
Page 6 of 6 SAS Output For Example A (Cereal)
The SAS System 1 The ANOVA Procedure
Class Level Information
Class Levels design 5 Values
ABCDE Number of Observations Read
Number of Observations Used 50
50 The SAS System 2 The ANOVA Procedure
Dependent Variable: cases DF Sum of
Squares Mean Square F Value Pr > F Model 4 1602.280000 400.570000 3.67 0.0114 Error 45 4916.200000 109.248889 Corrected Total 49 6518.480000 Source RSquare Coeff Var Root MSE cases Mean 0.245806 12.67243 10.45222 82.48000 Source DF Anova SS Mean Square F Value Pr > F design 4 1602.280000 400.570000 3.67 0.0114 The SAS System 3 The ANOVA Procedure
Level of
design
A
B
C
D
E Knapp Stat 350
Spring 2009 N
10
10
10
10
10 casesMean
Std Dev
77.5000000
75.3000000
81.8000000
88.2000000
89.6000000 4.5765101
12.5614047
11.1035530
13.2648240
8.2623645 Lecture L: 1way ANOVA
Page 7 of 7 Main Effects Plots – plot of the mean for each treatment group
Example A Data Set: Main Effects Plot
Data Means
90.0
87.5 Mean 85.0
82.5
80.0
77.5
75.0
1 2 3
group 4 5 Interval Plots  plot of the mean for each treatment group with bars representing a confidence
interval for the mean
Example A Data Set: Interval Plot
95% CI for the Mean
100
95 Combined 90
85
80
75
70
65
1 Knapp Stat 350
Spring 2009 2 3
group 4 5 Lecture L: 1way ANOVA
Page 8 of 8 Multiple Comparisons Procedures
(1) Tukey Test
(2) Dunnett's Method – Multiple Comparisons to a Control ** only do these if the ANOVA test was significant
Multiple Comparison Procedure: Tukey Test
Familywise error rate – for a significance level α, the Tukey procedure assures that the
probability of making a Type I error is at most α for the entire collection of pairwise hypothesis
tests. How the Tukey test works
• The test determines how far apart a pair of treatment means would have to be to be
"significantly different", based on the significance level α, the MSE from the ANOVA
test, and the sample size.
• Let T = the minimum difference between treatment means that we will call "significant".
This is the "threshold value". If a pair of treatment means differ by more than T, we will
conclude that those groups have significantly different means.
MSE
• T = qα
ni
o qα can be found in Appendix Table IX, based on
k – the number of treatment groups
Error df = n – k for a oneway ANOVA
o ni is the number of observations in each group • For each pair of treatments see if xi − x j > T .
If so, declare that Treatment i and Treatment j have significantly different means. Conveying results of Tukey Test
A common approach is to lineup the sample means in increasing (or decreasing) order.
Then draw bars across groups with nonsignificantly different means
Another option is to use letters – treatments shown with the same letter are not significantly
different Knapp Stat 350
Spring 2009 Lecture L: 1way ANOVA
Page 9 of 9 Example A – Tukey Test
Finding qα: T= Compare all pairs of means
x: Design B
75.3 Design A
77.5 Design C
81.8 Design D
88.2 Design E
89.6 Difference in Means
Design B
B
A
C
D
E A
2.2 C
6.5
4.3 D
12.9
10.7
6.4 E
14.3
12.1
7.8
1.4 This is how results should be displayed:
Design B
75.3 Design A
77.5 Design C
81.8 Design D
88.2 Design E
89.6 SAS Code for Tukey Test on Example A
proc anova data=cereal;
class design;
model cases = design;
means design/tukey alpha=0.05;
run; Note: if your design is unbalanced, to get the same output, you have to use the code:
means design/lines tukey; to get equivalent output Knapp Stat 350
Spring 2009 Lecture L: 1way ANOVA
Page 10 of 10 SAS Output for Tukey Test on Example A
The ANOVA Procedure
Tukey's Studentized Range (HSD) Test for cases
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
45
Error Mean Square
109.2489
Critical Value of Studentized Range 4.01842
Minimum Significant Difference
13.282 Means with the same letter are not significantly different. Tukey Grouping Mean N A
A
A
A
A
A
A 89.600 10 E 88.200 10 D 81.800 10 C 77.500 10 A 75.300 10 B B
B
B
B
B
B
B design Example A data set – bar chart displaying means of each group
100
95
90
85
80
75
70
65
60
55
50 AB 4 5 AB
A AB 2 Response B 1 3
Group Letters represent results of Tukey Test
error bars are the standard error of the means Knapp Stat 350
Spring 2009 Lecture L: 1way ANOVA
Page 11 of 11 Multiple Comparison Procedure: Dunnett's Method
Many scientific studies include a "control" If you don't care whether the various treatment groups differ from each other, only whether they
differ from the control, Dunnett's test is appropriate.
Note that for k treatment groups, there are k(k – 1)/2 pairs of groups to test, but if we only care
about testing each group to the control there are only k – 1 tests. How Dunnett's Method works
As with the Tukey test, we will calculate a threshold (T). Groups with means more than T from
the control group mean will be declared significantly different from the control group. ⎛1 1⎞
T = tα ,( k −1, n − k ) MSE ⎜ + ⎟
⎝ ni nc ⎠
MSE is the Mean Squared Error from the ANOVA
nc = the number of observations in the control group
ni = the number of observations in the ith group
tα ,( k −1, n − k ) is the critical value from Appendix X. Dunnett's t, based on k – 1 and n – k degrees of freedom
Example A – Dunnett's Method
for the sake of illustration, pretend that Design A was the control group.
tα ,( k −1, n − k ) = T= x: Knapp Stat 350
Spring 2009 Design A
77.5 Design B
75.3 Design C
81.8 Design D
88.2 Design E
89.6 Lecture L: 1way ANOVA
Page 12 of 12 SAS Code for Dunnett's Method on Example A
proc anova data=cereal;
class design;
model cases = design;
means design/dunnett('A') alpha = 0.05;
run; SAS Output for Dunnett's Method on Example A
Dunnett's t Tests for cases
NOTE: This test controls the Type I experimentwise error for comparisons of all treatments
against a control.
Alpha
0.05
Error Degrees of Freedom
45
Error Mean Square
109.2489
Critical Value of Dunnett's t
2.53129
Minimum Significant Difference
11.832
Comparisons significant at the 0.05 level are indicated by ***. design
Comparison
E
D
C
B  A
A
A
A Difference
Between
Means
12.100
10.700
4.300
2.200 Simultaneous 95%
Confidence Limits
0.268
1.132
7.532
14.032 23.932
22.532
16.132
9.632 *** You may wish to conduct 1tailed tests.
For example, you may wish to test whether the other designs have significantly higher sales than
the current design (not just significantly different). Knapp Stat 350
Spring 2009 Lecture L: 1way ANOVA
Page 13 of 13 SAS code to test if the other treatments are significantly greater than control
proc anova data=cereal;
class design;
model cases = design;
means design/dunnettu('A') alpha = 0.05;
run;
Dunnett's Onetailed t Tests for cases
NOTE: This test controls the Type I experimentwise error for comparisons of all treatments
against a control.
Alpha
0.05
Error Degrees of Freedom
45
Error Mean Square
109.2489
Critical Value of Dunnett's t
2.22241
Minimum Significant Difference
10.388
Comparisons significant at the 0.05 level are indicated by ***. design
Comparison
EA
DA
CA
BA Difference
Between
Means
12.100
10.700
4.300
2.200 Simultaneous 95%
Confidence Limits
1.712 Infinity
0.312 Infinity
6.088 Infinity
12.588 Infinity ***
*** SAS code to test if the other treatments are significantly less than control
proc anova data=cereal;
class design;
model cases = design;
means design/dunnettl('A');
run;
Dunnett's Onetailed t Tests for cases
NOTE: This test controls the Type I experimentwise error for comparisons of all treatments
against a control.
Alpha
0.05
Error Degrees of Freedom
45
Error Mean Square
109.2489
Critical Value of Dunnett's t
2.22241
Minimum Significant Difference
10.388
Comparisons significant at the 0.05 level are indicated by ***. design
Comparison
EA
DA
CA
BA Knapp Stat 350
Spring 2009 Difference
Between
Means
12.100
10.700
4.300
2.200 Simultaneous 95%
Confidence Limits
Infinity
22.488
Infinity
21.088
Infinity
14.688
Infinity
8.188 Lecture L: 1way ANOVA
Page 14 of 14 Fixed Effects versus Random Effects
Fixed factor / fixed effects models – if the k treatment levels we are using are the only ones of
interest in the experiment and conclusions will not be drawn beyond those k treatments
Random factor / random effects models – if the k treatment levels are only a sample from the
population of possible levels of interest. We are interested not in the difference in means
between the particular levels we happened to sample, but instead we want to estimate how much
variability in the sample is due to the treatment levels versus due to experimental error.
σ 2 = σ τ2 + σ ε2 Hypotheses: H0: σ τ2 = 0 vs. Ha: σ τ2 > 0
ˆ
σ ε2 = MSE
ˆ
σ τ2 = MSTr − MSE
ni Calculation of the F statistic is the same for random and fixed factors in a 1way ANOVA Example A – assume for the sake of illustration that the design is a random factor. Knapp Stat 350
Spring 2009 Lecture L: 1way ANOVA
Page 15 of 15 OneWay ANOVA vs. Pooled ttest Recall that ANOVA assumes that all groups have the same standard deviation (estimated by
MSE , Root MSE). A 1way ANOVA with only 2 groups is equivalent to a Pooled ttest.
Example: Cereal, comparing only Designs A and B.
data cereal2;
input design $ cases;
cards;
A
73
A
83
A
78
A
86
A
78
A
78
A
79
A
76
A
72
A
72
B
58
B
82
B
70
B
76
B
81
B
104
B
67
B
79
B
67
B
69
;
proc anova data=cereal2;
class design;
model cases = design;
run;
proc ttest data=cereal2;
class design;
var cases;
run; Knapp Stat 350
Spring 2009 Lecture L: 1way ANOVA
Page 16 of 16 SAS output: ANOVA
The ANOVA Procedure
Class Level Information
Class Levels design Values 2 AB Number of Observations Read
Number of Observations Used 20
20 The ANOVA Procedure
Dependent Variable: cases Source
Model
Error
Corrected Total Sum of
Squares
24.200000
1608.600000
1632.800000 DF
1
18
19
RSquare
0.014821 Source
design Coeff Var
12.37355
DF
1 Mean Square
24.200000
89.366667 Root MSE
9.453394 Anova SS
24.20000000 F Value
0.27 Pr > F
0.6091 cases Mean
76.40000 Mean Square
24.20000000 F Value
0.27 Pr > F
0.6091 SAS output: ttest
The TTEST Procedure
Statistics Variable design N cases
cases
cases A
B
Diff (12) 10
10 Lower CL
Mean Mean Upper CL
Mean Lower CL
Std Dev Std Dev Upper CL
Std Dev Std Err 74.226
66.314
6.682 77.5
75.3
2.2 80.774
84.286
11.082 3.1479
8.6402
7.1431 4.5765
12.561
9.4534 8.3549
22.932
13.98 1.4472
3.9723
4.2277 TTests
Variable
cases
cases Method
Pooled
Satterthwaite Variable
cases Knapp Stat 350
Spring 2009 Variances
Equal
Unequal DF
18
11.3 Equality of Variances
Method
Num DF
Den DF
Folded F
9
9 t Value
0.52
0.52 F Value
7.53 Pr > t
0.6091
0.6128 Pr > F
0.0060 Lecture L: 1way ANOVA
Page 17 of 17 ...
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This note was uploaded on 02/16/2010 for the course MA 350 taught by Professor Sellke during the Spring '10 term at Purdue UniversityWest Lafayette.
 Spring '10
 SELLKE
 Variance

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