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

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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.

Simple Random Sample from a population with known s - continuous response.

Simple Random Sample from a population with unknown s - continuous response.

Simple RandomSamples from 2 popns with known s.

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.

Study Designs and Analysis ApproachesSampling Study with t>2 Populations

One sample is drawn independently and randomly from each of t > 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).

Sampling Study

Vegetarians

Meat &

Potato

Eaters

Health

Eaters

Random

Sample

Random

Sample

Random

Sample

Cholesterol Levels

Experimental Studywith t>2 treatments

Experimental Units: samples of size n1, n2, …, nt, are independently and randomly drawn from each of the t populations.

Separate treatmentsare 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.

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

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 0be the hypothesized common mean under H0.

A Simple Model

Let yij be the response for experimental unit j in group i,

i=1,2, ..., t, j=1,2, ..., 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.

A Naïve Testing Approach

Test each possible pair of groups by performing all pair-wise t-tests.

- Assume each test is performed at the a=0.05 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 a=0.05 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!

Between sample means variance

Testing Approaches - Analysis of VarianceThe 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).

Within group variance is small compared to variability between means.

Clear separation of means.

Within group variance is large compared to variability between means.

Unclear separation of means.

Extreme SituationsExtend the concept of a pooled variance to t groups as follows:

If all the ni are equal to n then this reduces to an average variance.

Pooled VarianceFrom two-sample t-test with assumed equal variance, s2, we produced a pooled (within-group) sample variance estimate.

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

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.

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

Partition of the sums of squares

Partition of the degrees of freedom

Test Statistic

Variance Estimates

The AOV (Analysis of Variance) TableThe computations needed to perform the F-test for equality of variances are organized into a table.

Example-Excel

=average(b6:b10)

=var(b6:b10)

=sqrt(b13)

=count(b6:b10)

=(B15-1)*B13

=(sum(B15:D15)-1)*var(B6:D10)

=sum(b16:d16)

=b18-b19

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 > 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 > F

POPN 2 2.03333333 1.01666667 5545.45 0.0001

GLM in SAS

General Linear Models Procedure

Dependent Variable: RESP

Sum of Mean

Source DF Squares Square F Value Pr > 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 > F

POPN 2 2.03333333 1.01666667 5545.45 0.0001

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

POPN 2 2.03333333 1.01666667 5545.45 0.0001

T for H0: Pr > |T| Std Error of

Parameter Estimate Parameter=0 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;

Minitab Example

STAT > ANOVA > 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

R Example

> hwages <- 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)

> egroup <- factor(c(1,1,1,1,1,2,2,2,2,2,3,3,3,3,3))

> wages.fit <- lm(hwages~egroup)

> anova(wages.fit)

Df Sum Sq Mean Sq F value Pr(>F)

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

R Example

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(>|t|)

(Intercept) 5.900000 0.006055 974.35 < 2e-16 ***

egroup2 -0.400000 0.008563 -46.71 6.06e-15 ***

egroup3 -0.900000 0.008563 -105.10 < 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: < 2.2e-16

(Intercept) egroup2 egroup3

5.9 -0.4 -0.9

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.

.

Kruskal - Wallis Test

Extension of the rank-sum test for t=2 to the t>2 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 > 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.

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 ls=78 ps=49 nodate;

data OneWay;

input popn resp @@ ;

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;

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 > F

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ

Among 2 2.033333 1.016667 5545.455 <.0001

Within 12 0.002200 0.000183

Average scores were used for ties.

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

ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ

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 > Chi-Square 0.0018

Stat > Nonparametrics > 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)

kruskal.test( )

> 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

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