lecture12.ppt

More than two groups: ANOVA
and Chi-square

First, recent news…
 RESEARCHERS FOUND A NINE-
FOLD INCREASE IN THE RISK OF
DEVELOPING PARKINSON'S IN
INDIVIDUALS EXPOSED IN THE
WORKPLACE TO CERTAIN
SOLVENTS…

The data…
Table 3. Solvent Exposure Frequencies and Adjusted Pairwise
Odds Ratios in PD–Discordant Twins, n = 99 Pairsa

Which statistical test?
Outcome
Variable
Are the observations correlated? Alternative to the chi-
square test if sparse
cells:
independent correlated
Binary or
categorical
(e.g.
fracture,
yes/no)
Chi-square test:
compares proportions between
two or more groups
Relative risks: odds ratios
or risk ratios
Logistic regression:
multivariate technique used
when outcome is binary; gives
multivariate-adjusted odds
ratios
McNemar’s chi-square test:
compares binary outcome between
correlated groups (e.g., before and
after)
Conditional logistic
regression: multivariate
regression technique for a binary
outcome when groups are
correlated (e.g., matched data)
GEE modeling: multivariate
correlated (e.g., repeated measures)
Fisher’s exact test: compares
proportions between independent
groups when there are sparse data
(some cells <5).
McNemar’s exact test:
correlated groups when there are
sparse data (some cells <5).

Comparing more than two
groups…

Continuous outcome (means)
Outcome
Variable
Are the observations independent or correlated?
Alternatives if the normality
assumption is violated (and
small sample size):
Continuous
(e.g. pain
scale,
cognitive
function)
Ttest: compares means
between two independent
groups
ANOVA: compares means
between more than two
independent groups
Pearson’s correlation
coefficient (linear
correlation): shows linear
correlation between two
continuous variables
Linear regression:
multivariate regression technique
used when the outcome is
continuous; gives slopes
Paired ttest: compares means
between two related groups (e.g.,
the same subjects before and
after)
Repeated-measures
ANOVA: compares changes
over time in the means of two or
more groups (repeated
measurements)
Mixed models/GEE
modeling: multivariate
regression techniques to compare
changes over time between two
or more groups; gives rate of
change over time
Non-parametric statistics
Wilcoxon sign-rank test:
non-parametric alternative to the
paired ttest
Wilcoxon sum-rank test
(=Mann-Whitney U test): non-
parametric alternative to the ttest
Kruskal-Wallis test: non-
parametric alternative to ANOVA
Spearman rank correlation
coefficient: non-parametric
alternative to Pearson’s correlation
coefficient

ANOVA example
S1a, n=28 S2b, n=25 S3c, n=21 P-valued
Calcium (mg) Mean 117.8 158.7 206.5 0.000
SDe 62.4 70.5 86.2
Iron (mg) Mean 2.0 2.0 2.0 0.854
SD 0.6 0.6 0.6
Folate (μg) Mean 26.6 38.7 42.6 0.000
SD 13.1 14.5 15.1
Zinc (mg) Mean 1.9 1.5 1.3 0.055
SD 1.0 1.2 0.4
a School 1 (most deprived; 40% subsidized lunches).
b School 2 (medium deprived; <10% subsidized).
c School 3 (least deprived; no subsidization, private school).
d ANOVA; significant differences are highlighted in bold (P<0.05).
Mean micronutrient intake from the school lunch by school
FROM: Gould R, Russell J,
Barker ME. School lunch menus
and 11 to 12 year old children's
food choice in three secondary
schools in England-are the
nutritional standards being met?
Appetite. 2006 Jan;46(1):86-92.

ANOVA
(ANalysis Of VAriance)
 Idea: For two or more groups, test
difference between means, for
quantitative normally distributed
variables.
 Just an extension of the t-test (an
ANOVA with only two groups is
mathematically equivalent to a t-test).

One-Way Analysis of Variance
 Assumptions, same as ttest
 Normally distributed outcome
 Equal variances between the groups
 Groups are independent

Hypotheses of One-Way
ANOVA



 3
2
1
0 μ
μ
μ
:
H
same
the
are
means
population
the
of
all
Not
:
1
H

ANOVA
 It’s like this: If I have three groups to
compare:
 I could do three pair-wise ttests, but this
would increase my type I error
 So, instead I want to look at the pairwise
differences “all at once.”
 To do this, I can recognize that variance is
a statistic that let’s me look at more than
one difference at a time…

The “F-test”
groups
within
y
Variabilit
groups
between
y
Variabilit
F 
Is the difference in the means of the groups more
than background noise (=variability within groups)?
Recall, we have already used an “F-test” to check for equality of variances If F>>1 (indicating
unequal variances), use unpooled variance in a t-test.
Summarizes the mean differences
between all groups at once.
Analogous to pooled variance from a ttest.

The F-distribution
 The F-distribution is a continuous probability distribution that
depends on two parameters n and m (numerator and denominator
degrees of freedom, respectively):
http://www.econtools.com/jevons/java/Graphics2D/FDist.html

The F-distribution
 A ratio of variances follows an F-distribution:
2
2
2
2
0
:
:
within
between
a
within
between
H
H






The F-test tests the hypothesis that two variances
are equal.
F will be close to 1 if sample variances are equal.
m
n
within
between
F ,
2
2
~



How to calculate ANOVA’s by
hand…
Treatment 1 Treatment 2 Treatment 3 Treatment 4
y11 y21 y31 y41
y12 y22 y32 y42
y13 y23 y33 y43
y14 y24 y34 y44
y15 y25 y35 y45
y16 y26 y36 y46
y17 y27 y37 y47
y18 y28 y38 y48
y19 y29 y39 y49
y110 y210 y310 y410
n=10 obs./group
k=4 groups
The group means
10
10
1
1
1


 
j
j
y
y
10
10
1
2
2


 
j
j
y
y
10
10
1
3
3


 
j
j
y
y 10
10
1
4
4


 
j
j
y
y
The (within)
group variances
1
10
)
(
10
1
2
1
1





j
j y
y
1
10
)
(
10
1
2
2
2





j
j y
y
1
10
)
(
10
1
2
3
3





j
j y
y
1
10
)
(
10
1
2
4
4





j
j y
y

Sum of Squares Within (SSW),
or Sum of Squares Error (SSE)
The (within) group
variances
1
10
)
(
10
1
2
1
1





j
j y
y
1
10
)
(
10
1
2
2
2





j
j y
y
1
10
)
(
10
1
2
3
3





j
j y
y
1
10
)
(
10
1
2
4
4





j
j y
y

 



4
1
10
1
2
)
(
i j
i
ij y
y
+




10
1
2
1
1 )
(
j
j y
y 



10
1
2
2
2 )
(
j
j y
y 



10
3
2
3
3 )
(
j
j y
y 



10
1
2
4
4 )
(
j
j y
y
+
+
Sum of Squares Within (SSW)
(or SSE, for chance error)

Sum of Squares Between (SSB), or
Sum of Squares Regression (SSR)
Sum of Squares Between
(SSB). Variability of the
group means compared to
the grand mean (the
variability due to the
treatment).
Overall mean of
all 40
observations
(“grand mean”)
40
4
1
10
1

 

 
i j
ij
y
y
2
4
1
)
(
10 



 
i
i y
y
x

Total Sum of Squares (SST)
Total sum of squares(TSS).
Squared difference of every
observation from the overall
mean. (numerator of
variance of Y!)

 



4
1
10
1
2
)
(
i j
ij y
y

Partitioning of Variance

 


4
1
10
1
2
)
(
i j
i
ij y
y 



 
4
1
2
)
(
i
i y
y 
 



4
1
10
1
2
)
(
i j
ij y
y
=
+
SSW + SSB = TSS
x
10

ANOVA Table
Between
(k groups)
k-1 SSB
(sum of squared
deviations of
group means from
grand mean)
SSB/k-1 Go to
Fk-1,nk-k
chart
Total
variation
nk-1 TSS
(sum of squared deviations of
observations from grand mean)
Source of
variation d.f.
Sum of
squares
Mean Sum
of Squares
F-statistic p-value
Within
(n individuals per
group)
nk-k SSW
(sum of squared
deviations of
observations from
their group mean)
s2=SSW/nk-k
k
nk
SSW
k
SSB

1
TSS=SSB + SSW

ANOVA=t-test
Between
(2 groups)
1 SSB
(squared
difference
in means
multiplied
by n)
Squared
difference
in means
times n
Go to
F1, 2n-2
Chart
notice
values are
just (t 2n-
2)2
Total
variation
2n-1 TSS
Source of
variation d.f.
Sum of
squares
Mean
Sum of
Squares F-statistic p-value
Within 2n-2 SSW
equivalent to
numerator of
pooled
variance
Pooled
variance
2
2
2
2
2
2
2
2
)
(
)
)
(
(
)
(






n
p
p
p
t
n
s
n
s
Y
X
s
Y
X
n
2
2
2
2
2
2
2
2
1
2
1
2
1
2
1
)
(
)
*
2
(
)
2
*
2
)
2
(
)
2
(
2
*
2
)
2
(
)
2
((
)
2
2
(
)
2
2
(
))
2
(
(
))
2
(
(
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
n
i
n
n
n
i
n
n
n
n
i
n
n
n
n
i
Y
X
n
Y
Y
X
X
n
Y
X
X
Y
Y
X
Y
X
n
X
Y
n
Y
X
n
Y
X
Y
n
Y
X
X
n
SSB





























Example
60 inches 50 48 47
67 52 49 67
42 43 50 54
67 67 55 67
56 67 56 68
62 59 61 65
64 67 61 65
59 64 60 56
72 63 59 60
71 65 64 65

Example
60 inches 50 48 47
67 52 49 67
42 43 50 54
67 67 55 67
56 67 56 68
62 59 61 65
64 67 61 65
59 64 60 56
72 63 59 60
71 65 64 65
Step 1) calculate the sum
of squares between groups:
Mean for group 1 = 62.0
Grand mean= 59.85
SSB = [(62-59.85)2 + (59.7-59.85)2 + (56.3-59.85)2 + (61.4-59.85)2 ] xn per
group= 19.65x10 = 196.5

Example
60 inches 50 48 47
67 52 49 67
42 43 50 54
67 67 55 67
56 67 56 68
62 59 61 65
64 67 61 65
59 64 60 56
72 63 59 60
71 65 64 65
Step 2) calculate the sum
of squares within groups:
(60-62) 2+(67-62) 2+ (42-62)
2+ (67-62) 2+ (56-62) 2+ (62-
62) 2+ (64-62) 2+ (59-62) 2+
(72-62) 2+ (71-62) 2+ (50-
59.7) 2+ (52-59.7) 2+ (43-
59.7) 2+67-59.7) 2+ (67-
59.7) 2+ (69-59.7)
2…+….(sum of 40 squared
deviations) = 2060.6

Step 3) Fill in the ANOVA table
3 196.5 65.5 1.14 .344
36 2060.6 57.2
Source of variation d.f. Sum of squares Mean Sum of
Squares
F-statistic p-value
Between
Within
Total 39 2257.1

Step 3) Fill in the ANOVA table
3 196.5 65.5 1.14 .344
36 2060.6 57.2
Squares
F-statistic p-value
Between
Within
Total 39 2257.1
INTERPRETATION of ANOVA:
How much of the variance in height is explained by treatment group?
R2=“Coefficient of Determination” = SSB/TSS = 196.5/2275.1=9%

Coefficient of Determination
SST
SSB
SSE
SSB
SSB
R 


2
The amount of variation in the outcome variable (dependent
variable) that is explained by the predictor (independent variable).

Beyond one-way ANOVA
Often, you may want to test more than 1
treatment. ANOVA can accommodate
more than 1 treatment or factor, so long
as they are independent. Again, the
variation partitions beautifully!
TSS = SSB1 + SSB2 + SSW

ANOVA example
S1a, n=25 S2b, n=25 S3c, n=25 P-valued
Calcium (mg) Mean 117.8 158.7 206.5 0.000
SDe 62.4 70.5 86.2
Iron (mg) Mean 2.0 2.0 2.0 0.854
SD 0.6 0.6 0.6
Folate (μg) Mean 26.6 38.7 42.6 0.000
SD 13.1 14.5 15.1
Zinc (mg)
Mean 1.9 1.5 1.3 0.055
SD 1.0 1.2 0.4
a School 1 (most deprived; 40% subsidized lunches).
b School 2 (medium deprived; <10% subsidized).
c School 3 (least deprived; no subsidization, private school).
d ANOVA; significant differences are highlighted in bold (P<0.05).
Table 6. Mean micronutrient intake from the school lunch by school
FROM: Gould R, Russell J,
Barker ME. School lunch menus
and 11 to 12 year old children's
food choice in three secondary
schools in England-are the
nutritional standards being met?
Appetite. 2006 Jan;46(1):86-92.

Answer
Step 1) calculate the sum of squares between groups:
Mean for School 1 = 117.8
Grand mean: 161
SSB = [(117.8-161)2 + (158.7-161)2 + (206.5-161)2] x25 per
group= 98,113

Answer
Step 2) calculate the sum of squares within groups:
S.D. for S1 = 62.4
S.D. for S2 = 70.5
S.D. for S3 = 86.2
Therefore, sum of squares within is:
(24)[ 62.42 + 70.5 2+ 86.22]=391,066

Answer
Step 3) Fill in your ANOVA table
Squares
F-statistic p-value
Between 2 98,113 49056 9 <.05
Within 72 391,066 5431
Total 74 489,179
**R2=98113/489179=20%
School explains 20% of the variance in lunchtime calcium
intake in these kids.

ANOVA summary
 A statistically significant ANOVA (F-test)
only tells you that at least two of the
groups differ, but not which ones differ.
 Determining which groups differ (when
it’s unclear) requires more sophisticated
analyses to correct for the problem of
multiple comparisons…

Question: Why not just do 3
pairwise ttests?
 Answer: because, at an error rate of 5% each test,
this means you have an overall chance of up to 1-
(.95)3= 14% of making a type-I error (if all 3
comparisons were independent)
 If you wanted to compare 6 groups, you’d have to
do 6C2 = 15 pairwise ttests; which would give you
a high chance of finding something significant just
by chance (if all tests were independent with a
type-I error rate of 5% each); probability of at
least one type-I error = 1-(.95)15=54%.

Correction for multiple comparisons
How to correct for multiple comparisons post-
hoc…
• Bonferroni correction (adjusts p by most
conservative amount; assuming all tests
independent, divide p by the number of tests)
• Tukey (adjusts p)
• Scheffe (adjusts p)
• Holm/Hochberg (gives p-cutoff beyond which
not significant)

Procedures for Post Hoc
Comparisons
If your ANOVA test identifies a difference
between group means, then you must identify
which of your k groups differ.
If you did not specify the comparisons of interest
(“contrasts”) ahead of time, then you have to pay a
price for making all kCr pairwise comparisons to
keep overall type-I error rate to α.
Alternately, run a limited number of planned comparisons
(making only those comparisons that are most important to your
research question). (Limits the number of tests you make).

1. Bonferroni
Obtained P-value Original Alpha # tests New Alpha Significant?
.001 .05 5 .010 Yes
.011 .05 4 .013 Yes
.019 .05 3 .017 No
.032 .05 2 .025 No
.048 .05 1 .050 Yes
For example, to make a Bonferroni correction, divide your desired alpha cut-off
level (usually .05) by the number of comparisons you are making. Assumes
complete independence between comparisons, which is way too conservative.

2/3. Tukey and Sheffé
 Both methods increase your p-values to
account for the fact that you’ve done multiple
comparisons, but are less conservative than
Bonferroni (let computer calculate for you!).
 SAS options in PROC GLM:
 adjust=tukey
 adjust=scheffe

4/5. Holm and Hochberg
 Arrange all the resulting p-values (from
the T=kCr pairwise comparisons) in
order from smallest (most significant) to
largest: p1 to pT

Holm
1. Start with p1, and compare to Bonferroni p (=α/T).
2. If p1< α/T, then p1 is significant and continue to step 2.
If not, then we have no significant p-values and stop here.
3. If p2< α/(T-1), then p2 is significant and continue to step.
If not, then p2 thru pT are not significant and stop here.
4. If p3< α/(T-2), then p3 is significant and continue to step
If not, then p3 thru pT are not significant and stop here.
Repeat the pattern…

Hochberg
1. Start with largest (least significant) p-value, pT,
and compare to α. If it’s significant, so are all
the remaining p-values and stop here. If it’s not
significant then go to step 2.
2. If pT-1< α/(T-1), then pT-1 is significant, as are all
remaining smaller p-vales and stop here. If not,
then pT-1 is not significant and go to step 3.
Repeat the pattern…
Note: Holm and Hochberg should give you the same results. Use
Holm if you anticipate few significant comparisons; use Hochberg if
you anticipate many significant comparisons.

Practice Problem
A large randomized trial compared an experimental drug and 9 other standard
drugs for treating motion sickness. An ANOVA test revealed significant
differences between the groups. The investigators wanted to know if the
experimental drug (“drug 1”) beat any of the standard drugs in reducing total
minutes of nausea, and, if so, which ones. The p-values from the pairwise
ttests (comparing drug 1 with drugs 2-10) are below.
a. Which differences would be considered statistically significant using a
Bonferroni correction? A Holm correction? A Hochberg correction?
Drug 1 vs. drug
…
2 3 4 5 6 7 8 9 10
p-value .05 .3 .25 .04 .001 .006 .08 .002 .01

Answer
Bonferroni makes new α value = α/9 = .05/9 =.0056; therefore, using Bonferroni, the
new drug is only significantly different than standard drugs 6 and 9.
Arrange p-values:
6 9 7 10 5 2 8 4 3
.001 .002 .006 .01 .04 .05 .08 .25 .3
Holm: .001<.0056; .002<.05/8=.00625; .006<.05/7=.007; .01>.05/6=.0083; therefore,
new drug only significantly different than standard drugs 6, 9, and 7.
Hochberg: .3>.05; .25>.05/2; .08>.05/3; .05>.05/4; .04>.05/5; .01>.05/6; .006<.05/7;
therefore, drugs 7, 9, and 6 are significantly different.

Practice problem
 b. Your patient is taking one of the standard drugs that was
shown to be statistically less effective in minimizing
motion sickness (i.e., significant p-value for the
comparison with the experimental drug). Assuming that
none of these drugs have side effects but that the
experimental drug is slightly more costly than your
patient’s current drug-of-choice, what (if any) other
information would you want to know before you start
recommending that patients switch to the new drug?

Answer
 The magnitude of the reduction in minutes of nausea.
 If large enough sample size, a 1-minute difference could be
statistically significant, but it’s obviously not clinically
meaningful and you probably wouldn’t recommend a
switch.

Non-parametric ANOVA
Kruskal-Wallis one-way ANOVA
(just an extension of the Wilcoxon Sum-Rank (Mann
Whitney U) test for 2 groups; based on ranks)
Proc NPAR1WAY in SAS

Binary or categorical outcomes
(proportions)
Outcome
Variable
cells:
Binary or
categorical
(e.g.
fracture,
yes/no)
Chi-square test:
two or more groups
or risk ratios
ratios
after)
(some cells <5).
sparse data (some cells <5).

Chi-square test
for comparing proportions
(of a categorical variable)
between >2 groups
I. Chi-Square Test of Independence
When both your predictor and outcome variables are categorical, they may be cross-
classified in a contingency table and compared using a chi-square test of
independence.
A contingency table with R rows and C columns is an R x C contingency table.

Example
 Asch, S.E. (1955). Opinions and social
pressure. Scientific American, 193, 31-
35.

The Experiment
 A Subject volunteers to participate in a
“visual perception study.”
 Everyone else in the room is actually a
conspirator in the study (unbeknownst
to the Subject).
 The “experimenter” reveals a pair of
cards…

The Task Cards
Standard line Comparison lines
A, B, and C

The Experiment
 Everyone goes around the room and says
which comparison line (A, B, or C) is correct;
the true Subject always answers last – after
hearing all the others’ answers.
 The first few times, the 7 “conspirators” give
the correct answer.
 Then, they start purposely giving the
(obviously) wrong answer.
 75% of Subjects tested went along with the
group’s consensus at least once.

Further Results
 In a further experiment, group size
(number of conspirators) was altered
from 2-10.
 Does the group size alter the proportion
of subjects who conform?

The Chi-Square test
Conformed?
Number of group members?
2 4 6 8 10
Yes 20 50 75 60 30
No 80 50 25 40 70
Apparently, conformity less likely when less or more group
members…

 20 + 50 + 75 + 60 + 30 = 235
conformed
 out of 500 experiments.
 Overall likelihood of conforming =
235/500 = .47

Calculating the expected, in
general
 Null hypothesis: variables are
independent
 Recall that under independence:
P(A)*P(B)=P(A&B)
 Therefore, calculate the marginal
probability of B and the marginal
probability of A. Multiply P(A)*P(B)*N to
get the expected cell count.

Expected frequencies if no
association between group
size and conformity…
Conformed?
Number of group members?
2 4 6 8 10
Yes 47 47 47 47 47
No 53 53 53 53 53

 Do observed and expected differ more
than expected due to chance?

Chi-Square test


expected
expected)
-
(observed 2
2

Degrees of freedom = (rows-1)*(columns-1)=(2-1)*(5-1)=4
85
53
)
53
70
(
53
)
53
40
(
53
)
53
25
(
53
)
53
50
(
53
)
53
80
(
47
)
47
30
(
47
)
47
60
(
47
)
47
75
(
47
)
47
50
(
47
)
47
20
(
2
2
2
2
2
2
2
2
2
2
2
4























The Chi-Square distribution:
is sum of squared normal deviates
The expected
value and
variance of a chi-
square:
E(x)=df
Var(x)=2(df)
)
Normal(0,1
~
Z
where
;
1
2
2



df
i
Z
df


Chi-Square test


expected
expected)
-
(observed 2
2

Degrees of freedom = (rows-1)*(columns-1)=(2-1)*(5-1)=4
Rule of thumb: if the chi-square statistic is much greater than it’s degrees of freedom,
indicates statistical significance. Here 85>>4.
85
53
)
53
70
(
53
)
53
40
(
53
)
53
25
(
53
)
53
50
(
53
)
53
80
(
47
)
47
30
(
47
)
47
60
(
47
)
47
75
(
47
)
47
50
(
47
)
47
20
(
2
2
2
2
2
2
2
2
2
2
2
4























22
.
1
0156
.
019
.
91
)
982
)(.
018
(.
352
)
982
)(.
018
(.
)
033
.
014
(.
018
.
453
8
;
)
1
)(
(
)
1
)(
(
0
)
ˆ
ˆ
(
033
.
91
3
;
014
.
352
5
2
1
2
1
/
/



















Z
p
n
p
p
n
p
p
p
p
Z
p
p nophone
tumor
cellphone
tumor
Brain tumor No brain tumor
Own a cell
phone
5 347 352
Don’t own a
cell phone
3 88 91
8 435 453
Chi-square example: recall data…

Same data, but use Chi-square test
48
.
1
22
.
1
:
note
48
.
1
7
.
345
345.7)
-
(347
3
.
89
88)
-
(89.3
7
.
1
1.7)
-
(3
3
.
6
6.3)
-
(8
df
1
1
1
1
1
d
cell
in
89.3
b;
cell
in
345.7
c;
cell
in
1.7
6.3;
453
*
.014
a
cell
in
Expected
014
.
777
.
*
018
.
777
.
453
352
;
018
.
453
8
2
2
2
2
2
2
1
2

















Z
NS
*
)
)*(C-
(R-
xp
p
p
p
cellphone
tumor
cellphone
tumor

Brain tumor No brain tumor
Own 5 347 352
Don’t own 3 88 91
8 435 453
Expected value in
cell c= 1.7, so
technically should
use a Fisher’s exact
here! Next term…

Caveat
**When the sample size is very small in
any cell (expected value<5), Fisher’s
exact test is used as an alternative to
the chi-square test.

Binary or categorical outcomes
(proportions)
Outcome
Variable
cells:
Binary or
categorical
(e.g.
fracture,
yes/no)
Chi-square test:
two or more groups
or risk ratios
ratios
after)
(np <5).
sparse data (np <5).

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