Testing Multiple Means and the Analysis of Variance (

1
Testing Multiple Means and the Analysis of
Variance (8.1, 8.2, 8.6)

• Situations where comparing more than two means is
  important.
• The approach to testing equality of more than two
  means.
• Introduction to the analysis of variance table,
  its construction and use.

2
Study Designs and Analysis Approaches

1. Simple Random Sample from a population with known
   s - continuous response.
2. Simple Random Sample from a population with
   unknown s - continuous response.
3. Simple RandomSamples from 2 popns with known s.
4. Simple Random Samples from 2 popns with unknown s.

• One sample z-test.
• One sample t-test.
• Two sample z-test.
• Two sample t-test.

3
Sampling Study with tgt2 Populations
One sample is drawn independently and randomly
from each of t gt 2 populations.
Objective to compare the means of the t
populations for statistically significant
differences in responses.
Initially we will assume all populations have
common variance, later, we will test to see if
this is indeed true. (Homogeneity of variance
tests).
4
Sampling Study
Vegetarians
Meat Potato Eaters
Health Eaters
Random Sample
Random Sample
Random Sample
Cholesterol Levels
5
Experimental Studywith tgt2 treatments
Experimental Units samples of size n1, n2, ,
nt, are independently and randomly drawn from
each of the t populations.
Separate treatments are applied to each sample.
A treatment is something done to the experimental
units which would be expected to change the
distribution (usually only the mean) of the
response(s). Note if all samples are drawn
from the same population before application of
treatments, the homogeneity of variances
assumption might be plausible.
6
Experimental Study
Male College Undergraduate Students
Veg. Diet
Health Diet
Random Sampling
M P Diet
Set of Experimental Units
Set of Experimental Units
Set of Experimental Units
Responses
7
Hypothesis
Let mi be the true mean of treatment group i (or
population i ).
Hence we are interested in whether all the groups
(populations) have exactly the same true means.
The alternative is that some of the groups
(populations) differ from the others in their
means. Let ?0 be the hypothesized common mean
under H0.
8
A Simple Model
Let yij be the response for experimental unit j
in group i, i1,2, ..., t, j1,2, ..., ni. The
model is
E(yij) ?i we expect the group mean to be mi.
eij is the residual or deviation from the group
mean.
Each population has normally distributed
responses around their own means, but the
variances are the same across all populations.
Assuming yij N(mi, s2), then eij N(0, s2)
If H0 holds, yij m0 eij , that is, all groups
have the same mean and variance.
9
A Naïve Testing Approach
Test each possible pair of groups by performing
all pair-wise t-tests.

• Assume each test is performed at the a0.05
  level.
• For each test, the probability of not rejecting
  Ho when Ho is true is 0.95 (1-a).
• The probability of not rejecting Ho when Ho is
  true for all three tests is (0.95)3 0.857.
• Thus the true significance level (type I error)
  for the overall test of no difference in the
  means will be 1-0.857 0.143, NOT the a0.05
  level we thought it would be.

Also, in each individual t-test, only part of the
information available to estimate the underlying
variance is actually used. This is inefficient -
WE CAN DO MUCH BETTER!
10
Testing Approaches - Analysis of Variance
The term analysis of variance comes from the
fact that this approach compares the variability
observed between sample means to a (pooled)
estimate of the variability among observations
within each group (treatment).
11
Extreme Situations
12
Pooled Variance
From two-sample t-test with assumed equal
variance, s2, we produced a pooled
(within-group) sample variance estimate.
13
Variance Between Group Means
Consider the variance between the group means
computed as
If we assume each group is of the same size, say
n, then under H0, s2 is an estimate of s2/n (the
variance of the sampling distribution of the
sample mean). Hence, n times s2 is an estimate
of s2. When the sample sizes are unequal, the
estimate is given by
where
14
F-test
Now we have two estimates of s2, within and
between means. An F-test can be used to
determine if the two statistics are equal. Note
that if the groups truly have different means,
sB2 will be greater than sw2. Hence the
F-statistic is written as
If H0 holds, the computed F-statistics should be
close to 1. If HA holds, the computed F-statistic
should be much greater than 1. We use the
appropriate critical value from the F - table to
help make this decision.
Hence, the F-test is really a test of equality of
means under the assumption of normal populations
and homogeneous variances.
15
Partition of Sums of Squares
SSB
SSW


TSS
Total Sums of Squares
Sums of Squares Between Means
Sums of Squares Within Groups


TSS measures variability about the overall mean
16
The AOV (Analysis of Variance) Table
The computations needed to perform the F-test for
equality of variances are organized into a table.
17
Example-Excel
average(b6b10) var(b6b10) sqrt(b13) count(b6
b10) (B15-1)B13
(sum(B15D15)-1)var(B6D10) sum(b16d16) b18-b
19
18
Excel Analysis Tool Pac
19
Example SAS
proc anova
class popn
model resp popn
title 'Table 13.1 in Ott -
Analysis of Variance' run


Table 13.1 in Ott - Analysis of Variance
31

Analysis of Variance Procedure


Dependent Variable RESP

Sum of Mean
Source DF
Squares Square F Value Pr gt F

Model
2 2.03333333 1.01666667 5545.45
0.0001
Error
12 0.00220000 0.00018333


Corrected Total 14 2.03553333


R-Square
C.V. Root MSE RESP Mean


0.998919 0.247684 0.013540
5.466667


Source
DF Anova SS Mean Square F Value
Pr gt F
POPN
2 2.03333333 1.01666667
5545.45 0.0001


20
GLM in SAS


General Linear Models Procedure


Dependent Variable RESP

Sum of Mean
Source DF
Squares Square F Value Pr gt F

Model
2 2.03333333 1.01666667 5545.45
0.0001
Error
12 0.00220000 0.00018333


Corrected Total 14 2.03553333


R-Square
C.V. Root MSE RESP Mean


0.998919 0.247684 0.013540
5.466667


Source
DF Type I SS Mean Square F Value
Pr gt F
POPN
2 2.03333333 1.01666667
5545.45 0.0001

Source DF Type III SS
Mean Square F Value Pr gt F

POPN 2
2.03333333 1.01666667 5545.45 0.0001




T for H0 Pr gt T Std Error of
Parameter Estimate
Parameter0 Estimate

INTERCEPT
5.000000000 B 825.72 0.0001
0.00605530 POPN 1
0.900000000 B 105.10 0.0001
0.00856349 2
0.500000000 B 58.39 0.0001
0.00856349 3
0.000000000 B . . .

NOTE The
X'X matrix has been found to be singular and a
generalized inverse was used to solve
the normal equations. Estimates followed by the
letter 'B' are biased, and are not
unique estimators of the parameters.
proc glm
class popn
model resp popn / solution
title 'Table 13.1 in Ott
run
21
Minitab Example
STAT gt ANOVA gt OneWay (Unstacked)
One-way Analysis of Variance Analysis of
Variance Source DF SS MS
F P Factor 2 2.033333 1.016667
5545.45 0.000 Error 12 0.002200
0.000183 Total 14 2.035533
Individual 95 CIs For Mean
Based on Pooled
StDev Level N Mean StDev
--------------------------------- EG1
5 5.9000 0.0158
( EG2 5 5.5000 0.0071
) EG3 5 5.0000 0.0158
(
--------------------------------- Pooled
StDev 0.0135 5.10 5.40
5.70 6.00
22
R Example
Function lm( )

• gt hwages lt- c(5.90,5.92,5.91,5.89,5.88,5.51,5.50,5
  .50,5.49,5.50,5.01,
  5.00,4.99,4.98,5.02)
• gt egroup lt- factor(c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3
  ))
• gt wages.fit lt- lm(hwagesegroup)
• gt anova(wages.fit)

Df Sum Sq Mean Sq F value
Pr(gtF) factor(egroup) 2 2.03333 1.01667
5545.5 2.2e-16 Residuals 12
0.00220 0.00018
--- Signif. codes 0 ' 0.001 ' 0.01 '
0.05 .' 0.1 ' 1
23
R Example
gt summary(wages.fit)
Call lm(formula hwages egroup) Residuals
Min 1Q Median 3Q
Max -2.000e-02 -1.000e-02 -1.941e-18 1.000e-02
2.000e-02 Coefficients Estimate
Std. Error t value Pr(gtt) (Intercept)
5.900000 0.006055 974.35 lt 2e-16 egroup2
-0.400000 0.008563 -46.71 6.06e-15
egroup3 -0.900000 0.008563 -105.10 lt
2e-16 --- Signif. codes 0 '' 0.001 ''
0.01 '' 0.05 '.' 0.1 ' ' 1 Residual standard
error 0.01354 on 12 degrees of freedom Multiple
R-Squared 0.9989, Adjusted R-squared 0.9987
F-statistic 5545 on 2 and 12 DF, p-value lt
2.2e-16
(Intercept) egroup2 egroup3
5.9 -0.4
-0.9
24
A Nonparametric Alternative to the AOV Test The
Kruskal - Wallis Test (8.5)
What can we do if the normality assumption is
rejected in the one-way AOV test? We can use the
standard nonparametric alternative the
Kruskal-Wallis Test. This is an extension of the
Wilcoxon Rank Sum Test to more than two samples.
.
25
Kruskal - Wallis Test
Extension of the rank-sum test for t2 to the tgt2
case.
H0 The center of the t groups are identical. Ha
Not all centers are the same.
Test Statistic
Ti denotes the sum of the ranks for the
measurements in sample i after the combined
sample measurements have been ranked.
Reject if H gt c2(t-1),a
With large numbers of ties in the ranks of the
sample measurements use
where tj is the number of observations in the jth
group of tied ranks.
26

OBS POPN RESP
1 1 5.90
2 1 5.92 3
1 5.91 4
1 5.89 5 1
5.88 6 2
5.51 7 2
5.50 8 2
5.50 9 2
5.49 10 2
5.50 11 3
5.01 12 3
5.00 13 3
4.99 14 3
4.98 15 3
5.02
options ls78 ps49 nodate
data OneWay
input popn resp _at__at_
cards
1 5.90 1 5.92 1 5.91 1
5.89 1 5.88 2 5.51 2 5.50 2
5.50 2 5.49 2 5.50 3 5.01 3
5.00 3 4.99 3 4.98 3 5.02

run
proc print
run Proc npar1way class
popn var resp run

27
Analysis of Variance for Variable resp
Classified by Variable popn
popn N Mean
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
1 5
5.90 2 5
5.50 3 5
5.00 Source DF Sum of Squares
Mean Square F Value Pr gt F
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ Among 2 2.033333
1.016667 5545.455 lt.0001 Within
12 0.002200 0.000183
Average scores were used for ties.
28
The NPAR1WAY Procedure
Wilcoxon Scores (Rank Sums) for
Variable resp Classified by
Variable popn Sum of
Expected Std Dev Mean popn
N Scores Under H0 Under H0
Score ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ 1 5
65.0 40.0 8.135753 13.0
2 5 40.0 40.0
8.135753 8.0 3 5
15.0 40.0 8.135753 3.0
Average scores were used for
ties. Kruskal-Wallis
Test Chi-Square
12.5899 DF
2 Pr gt Chi-Square
0.0018
29
MINITAB
Stat gt Nonparametrics gt Kruskal-Wallis
Kruskal-Wallis Test RESP versus
POPN Kruskal-Wallis Test on RESP POPN
N Median Ave Rank Z 1 5
5.900 13.0 3.06 2 5
5.500 8.0 0.00 3 5
5.000 3.0 -3.06 Overall 15
8.0 H 12.50 DF 2 P 0.002 H
12.59 DF 2 P 0.002 (adjusted for ties)
30
R
kruskal.test( )
gt kruskal.test(hwages,factor(egroup))
Kruskal-Wallis rank sum test data hwages and
factor(egroup) Kruskal-Wallis chi-squared
12.5899, df 2, p-value 0.001846
31
SPSS