1 clinical research training program 2021 anova and ancova fall 2004

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1

Clinical Research Training Program 2021

ANOVA and ANCOVAANOVA and ANCOVA

Fall 2004

www.edc.gsph.pitt.edu/faculty/dodge/clres2021.html

2

ANOVA vs. REGRESSION

• ANOVA can be regarded as a special type of linear regressions.

• By using dummy coding, we can get coefficients which indicate difference in means in various groups.

3

REGRESSIONREGRESSION METHODSMETHODS

Regression

ANOVA

ANCOVA/Regression

4

Analysis of VarianceAnalysis of Variance

The null and alternative hypotheses are

H0:

H1: for some

kμμμ 21

ji μμ kji ,1,,

, where i represents the mean of population i

Hypotheses - Whether all group means are equal versus at least two group means are different

5

Assumptions - k independent random samples from k

normal populations with distributions N(

1, 2), …, N(

k, 2), respectively.

Outcome variable should be continuous. All the populations have the same unknown

variance 2 (homogeneous variance).

Analysis of VarianceAnalysis of Variance

6

1x 2x• •

• •1x 2x

• • • • • • •• • • • • • •

• • • • • • •• • • • • • •

• • • • • • •• • • • • • •

••••••••••••••

7

Analysis of VarianceAnalysis of Variance

The main idea for comparing means: what matters is not how far apart the sample means are but how far apart they are relative to the variability of individual observations

ANOVA compares the variation due to specific sources within the variation among individuals who should be similar. In particular, ANOVA tests whether several populations have the same mean by comparing how far apart the sample means are with how much variation there is within the sample

8

Analysis of VarianceAnalysis of Variance

Between Sum of Squares (SSB)

Within Sum of Squares (SSW)

Total Sum of Squares (SST)

SST = SSB + SSW

k

i

n

j

i

i

xx

1 1

2)(

k

i

n

j

ij

i

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1 1

2)(

k

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j

iij

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1 1

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9

Analysis of VarianceAnalysis of Variance

Hypotheses: H0:

H1: for some

Test statistic:

kμμμ 21

ji μμ kji ,1,,

knkF 1,~ )(

)(

)1(

)(

1 1

2

1 1

2

kn

xx

k

xx

k

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n

jiij

k

i

n

ji

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i

MSW

MSB

samples within sindividual amongVariation

means sample theamongVariation F

10

Analysis of VarianceAnalysis of Variance

Between Mean Squares (MSB)

Within Mean Squares (MSW)

kn

xxk

i

n

jiij

i

1 1

2)(

1

)(1 1

2

k

xxk

i

n

ji

i

knnnn 21

11

Analysis of VarianceAnalysis of VarianceSummary Table

Sourceof

variation

Sum of Squares(SS)

df MeanSquares

(MS)

FStatistic

P-value

Between

k

i

n

ji

i

xx1 1

2)(SSB 1k 1

SSBMSB

k MSE

MSBF ) calculated Pr( FF

Within(Errors)

k

i

n

jiij

i

xx1 1

2)(SSEkn kn

SSE

MSE

Total

k

i

n

jij

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xx1 1

2)(SST1n

(ANOVA Table)

12

ANOVA vs. REGRESSION

SSR: Between Sum of Squares (SSB)

SSE: Within Sum of Squares (SSW)

SST: Total Sum of Squares (SST)

SST = SSB + SSW

k

i

n

jiij

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xx1 1

2)(

k

i

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ji

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xx1 1

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k

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13

ANOVA vs. REGRESSION

• By using a reference coding, we can get similar results to ANOVA with additional information.

• Additional information Difference in means between the

reference group and other groups Difference in means between each group

and overall mean (when each group has equal N).

14

REFERENCE CODING

• Each variable takes on only values of 1 and 0.

• 3 groups: groups 1, 2, and 3

need 2 dummy variables

X2 1 if group 2

0 Otherwise

X3 1 if group 3

0 Otherwise

15

REFERENCE CODING

For Group 1: X2=0, X3=0:

For Group 2: X2=1, X3=0:

For Group 3: X2=0, X3=1:

ˆˆˆ Y

33 ˆˆˆˆ Y22 ˆˆˆˆ Y

3322 ˆˆˆˆ XXY

16

REFERENCE CODING

33

22

1

ˆˆˆ

ˆˆˆ

ˆˆ

1333

1222

1

ˆˆˆˆˆ

ˆˆˆˆˆ

ˆˆ

17

REFERENCE CODING

• Intercept μ is the mean of group 1 (mean of reference group).

• α2 indicates difference in mean between group 1 (reference group) and group2

• α3 indicates difference in mean between group 1 (reference group) and group3

18

• . list y x x2 x3 x4 x5

• y x x2 x3 x4 x5 • 1. 5 1 0 0 0 0 • 2. 8 1 0 0 0 0 • 3. 7 1 0 0 0 0 • 4. 7 1 0 0 0 0 • 5. 10 1 0 0 0 0 • 6. 8 1 0 0 0 0 • 7. 4 2 1 0 0 0 • 8. 6 2 1 0 0 0 • 9. 6 2 1 0 0 0 • 10. 3 2 1 0 0 0 • 11. 5 2 1 0 0 0 • 12. 6 2 1 0 0 0 • 13. 6 3 0 1 0 0 • 14. 4 3 0 1 0 0 • 15. 4 3 0 1 0 0 • 16. 5 3 0 1 0 0 • 17. 4 3 0 1 0 0 • 18. 3 3 0 1 0 0 • 19. 7 4 0 0 1 0 • 20. 4 4 0 0 1 0 • 21. 6 4 0 0 1 0 • 22. 6 4 0 0 1 0 • 23. 3 4 0 0 1 0 • 24. 5 4 0 0 1 0 • 25. 9 5 0 0 0 1 • 26. 3 5 0 0 0 1 • 27. 5 5 0 0 0 1 • 28. 7 5 0 0 0 1 • 29. 7 5 0 0 0 1 • 30. 6 5 0 0 0 1

19

• sort x

• . by x:summarize y

• _______________________________________________________________________________• -> x = 1

• Variable | Obs Mean Std. Dev. Min Max• -------------+-----------------------------------------------------• y | 6 7.5 1.643168 5 10

• _______________________________________________________________________________• -> x = 2

• Variable | Obs Mean Std. Dev. Min Max• -------------+-----------------------------------------------------• y | 6 5 1.264911 3 6

• _______________________________________________________________________________• -> x = 3

• Variable | Obs Mean Std. Dev. Min Max• -------------+-----------------------------------------------------• y | 6 4.333333 1.032796 3 6

• _______________________________________________________________________________• -> x = 4

• Variable | Obs Mean Std. Dev. Min Max• -------------+-----------------------------------------------------• y | 6 5.166667 1.47196 3 7

• _______________________________________________________________________________• -> x = 5

• Variable | Obs Mean Std. Dev. Min Max• -------------+-----------------------------------------------------• y | 6 6.166667 2.041241 3 9

20

• regress y x2 x3 x4 x5

• Source | SS df MS Number of obs = 30• -------------+------------------------------ F( 4, 25) = 3.90• Model | 36.4666667 4 9.11666667 Prob > F = 0.0136• Residual | 58.50 25 2.34 R-squared = 0.3840• -------------+------------------------------ Adj R-squared = 0.2854• Total | 94.9666667 29 3.27471264 Root MSE = 1.5297

• ------------------------------------------------------------------------------• y | Coef. Std. Err. t P>|t| [95% Conf. Interval]• -------------+----------------------------------------------------------------• x2 | -2.5 .8831761 -2.83 0.009 -4.318935 -.6810648• x3 | -3.166667 .8831761 -3.59 0.001 -4.985602 -1.347731• x4 | -2.333333 .8831761 -2.64 0.014 -4.152269 -.5143981• x5 | -1.333333 .8831761 -1.51 0.144 -3.152269 .4856019• _cons | 7.5 .6244998 12.01 0.000 6.213819 8.786181• ------------------------------------------------------------------------------

MeanGroup1=7.5Group2=5 α2=5- 7.5= -2.5Group3=4.3 α3=4.3-7.5= -3.2Group4=5.2 α4=5.2-7.5= -2.3Group5=6.2 α5=6.2-7.5= -1.3

21

ANOVA vs. REGRESSION

SSR: Between Sum of Squares (SSB)

SSE: Within Sum of Squares (SSW)

SST: Total Sum of Squares (SST)

SST = SSB + SSW

k

i

n

jiij

i

xx1 1

2)(

k

i

n

ji

i

xx1 1

2)(

k

i

n

jij

i

xx1 1

2)(

22

ANOVA vs. REGRESSION

• anova y x

• Number of obs = 30 R-squared = 0.3840• Root MSE = 1.52971 Adj R-squared = 0.2854

• Source | Partial SS df MS F Prob > F• -----------+----------------------------------------------------• Model | 36.4666667 4 9.11666667 3.90 0.0136• |• x | 36.4666667 4 9.11666667 3.90 0.0136• |• Residual | 58.50 25 2.34 • -----------+----------------------------------------------------• Total | 94.9666667 29 3.27471264

23

EFFECT CODING

• 3 groups: groups 1, 2, and 3

need 2 dummy variables

X2 1 if group 2

0 if group 3

-1 if group 1

X3 1 if group 3

0 if group 2

-1 if group 1

24

EFFECT CODING

For Group 1: X2=-1, X3=-1:

For Group 2: X2=1, X3=0:

For Group 3: X2=0, X3=1:

32 ˆˆˆˆˆ Y

33 ˆˆˆˆ Y22 ˆˆˆˆ Y

3322 ˆˆˆˆ XXY

25

EFFECT CODING

3

ˆˆˆˆˆ

3

ˆˆˆˆˆ

3

ˆˆˆˆ

32133

32122

321

33

22

32

ˆˆˆ

ˆˆˆ

ˆˆˆˆ

26

EFFECT CODING

• Intercept μ is the unweighted average of the K group means. In the example here, K=3. If all groups have equal sample size, this is a grand mean.

• α2 indicates difference between mean of group 2 and unweighted average of K group mean.

• α3 indicates difference between mean of group 3 and unweighted average of K group mean.

27

• list y x x2 x3 x4 x5

• y x x2 x3 x4 x5 • 1. 5 1 -1 -1 -1 -1 • 2. 8 1 -1 -1 -1 -1 • 3. 7 1 -1 -1 -1 -1 • 4. 7 1 -1 -1 -1 -1 • 5. 10 1 -1 -1 -1 -1 • 6. 8 1 -1 -1 -1 -1 • 7. 4 2 1 0 0 0 • 8. 6 2 1 0 0 0 • 9. 6 2 1 0 0 0 • 10. 3 2 1 0 0 0 • 11. 5 2 1 0 0 0 • 12. 6 2 1 0 0 0 • 13. 6 3 0 1 0 0 • 14. 4 3 0 1 0 0 • 15. 4 3 0 1 0 0 • 16. 5 3 0 1 0 0 • 17. 4 3 0 1 0 0 • 18. 3 3 0 1 0 0 • 19. 7 4 0 0 1 0 • 20. 4 4 0 0 1 0 • 21. 6 4 0 0 1 0 • 22. 6 4 0 0 1 0 • 23. 3 4 0 0 1 0 • 24. 5 4 0 0 1 0 • 25. 9 5 0 0 0 1 • 26. 3 5 0 0 0 1 • 27. 5 5 0 0 0 1 • 28. 7 5 0 0 0 1 • 29. 7 5 0 0 0 1 • 30. 6 5 0 0 0 1

• .

28

• regress y x2 x3 x4 x5

• Source | SS df MS Number of obs = 30• -------------+------------------------------ F( 4, 25) = 3.90• Model | 36.4666667 4 9.11666667 Prob > F = 0.0136• Residual | 58.50 25 2.34 R-squared = 0.3840• -------------+------------------------------ Adj R-squared = 0.2854• Total | 94.9666667 29 3.27471264 Root MSE = 1.5297

• ------------------------------------------------------------------------------• y | Coef. Std. Err. t P>|t| [95% Conf. Interval]• -------------+----------------------------------------------------------------• x2 | -.6333333 .5585696 -1.13 0.268 -1.783729 .5170623• x3 | -1.3 .5585696 -2.33 0.028 -2.450396 -.1496044• x4 | -.4666667 .5585696 -0.84 0.411 -1.617062 .683729• x5 | .5333333 .5585696 0.95 0.349 -.6170623 1.683729• _cons | 5.633333 .2792848 20.17 0.000 5.058136 6.208531• ------------------------------------------------------------------------------

• MeanGroup1=7.5 intercept=(7.5+5+4.3+5.2+6.2)/5=5.6Group2=5 α2=5- 5.6= -0.6Group3=4.3 α3=4.3- 5.6= -1.3Group4=5.2 α4=5.2- 5.6= -0.4Group5=6.2 α5=6.2- 5.6= 0.6

29

Analysis of Covariance (ANACOVA)

Why we need to consider control variables?

• Need to produce accurate estimates of coefficients. interaction confounding increase precision

30

Analysis of Covariance (ANACOVA)

A Question to be answered by using ANACOVA

• If each control variables have the same distributions between group A and B, what would be the mean response value for group A and B?

31

Analysis of Covariance(ANACOVA)

• Outcome---continuous

• Covariates

– Nominal (study factors of interests)

– Control variables involve any level of measurements

32

Analysis of Covariance(ANACOVA)

• Most importantly….. This method is applicable only when

there is no interaction effect between the variable of interest with covariates.

Y=β0+ β1X+ β2Z+ β3XZ+Ε

See first whether H0: β3=0 is supported.

33

Analysis of Covariance(ANACOVA)

• Blood pressure data example (X=age, Z=sex)

14.46

89.140)14.46(96.078.96)(ˆ

40.154)14.46(96.029.110)(ˆ

96.078.96ˆ:)1(

96.029.110ˆ:)0(

51.1396.029.110ˆ

X

adjY

adjY

XYZFemale

XYZMale

ZXY

F

M

F

M

34

Analysis of Covariance(ANACOVA)

sex Unadjusted mean BP

Adjusted mean BP

Male 155.15 154.40

Female 139.86 140.89

Using the adjusted mean scores removes the influence of age on the comparison of mean blood pressures by considering what the mean BP in the two groups would be if both groups had the same mean age.

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