Repeated measures ANOVA in SPSS Cross tabulations Survival analysis

Repeated measures ANOVA in SPSSCross tabulationsSurvival analysis

Repeated measures ANOVA in SPSS

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The data from Chen 2004

• The number of errors are recorded for the same test repeated 4 times

• The degree of anxiety was different between the subjects.

• Stuff we would like to know:• Is there a difference in the

performance between the anxiety groups?

• Is there a reduction in errors over time – a learning effect?

• Does one of the groups learn faster than the other?

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What to do? A mediocre suggestion and a really bad one

• Is there a difference in the performance between the anxiety groups?• We could calculate the mean

(or sum, time-to-max, min, max, etc.) for each subject and do a t-test between the anxiety groups

• Is there a reduction in errors over time – a learning effect?• We could make 6 paired t-tests

to test the difference between • Does one of the groups learn

faster than the other?• ???

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Repeated measures ANOVA

• Time is a within subject factor (4 levels)

• Anxiety is a between subjects factor (2 levels)

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Repeated measures ANOVA

• Plotting the data

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Repeated measures ANOVA – between subjects effects

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Repeated measures ANOVA – within subjects effects

• Is there a time effect?• Univariate or Multivariate?• Univariate assumes

sphericity of the covariance matrix:

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Residual SSCP Matrix Trial1 Trial2 Trail3 Trial4

Trial1 4,567 2,417 1,383 1,017Trial2 2,417 6,283 5,133 5,167Trail3 1,383 5,133 6,417 6,083Trial4 1,017 5,167 6,083 7,617Based on Type III Sum of Squares

Repeated measures ANOVA – within subjects effectsUnivariate

Mauchly's Test of Sphericityb

Measure:MEASURE_1Within Subjects Effect Mauchly'

s W

Approx. Chi-

Square df Sig.dimensio

n1time ,337 9,496 5 ,093

Tests the null hypothesis that the error covariance matrix of the orthonormalized transformed dependent variables is proportional to an identity matrix. b. Design: Intercept + Anxiety Within Subjects Design: time 9

Residual SSCP Matrix Trial1 Trial2 Trail3 Trial4

Trial1 4,567 2,417 1,383 1,017Trial2 2,417 6,283 5,133 5,167Trail3 1,383 5,133 6,417 6,083Trial4 1,017 5,167 6,083 7,617Based on Type III Sum of Squares

Repeated measures ANOVA – within subjects effectsUnivariate

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• Greenhouse-Geisser is newer than Huynh-Feldt methods• Lower-bound is the most konservative• Alternatively use Multible ANOVA

Repeated measures ANOVA – within subjects effectsMultivariate

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• The covariance matrices must be equal for all levels of the between groups

• Tested by Box’s test

Repeated measures ANOVA – within subjects effectsMultivariate

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• If the 4 methods disagree use Wilk’s Lambda• Multivariate and Univariate renders the same conclusion in this case• Which method to choose?

Repeated measures ANOVA – within subjects effectsWhich method to choose?

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Repeated measures ANOVA – within subjects effectsPairwise comparisons

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• Does all these comparisons make sense?

Repeated measures ANOVA – within subjects effectsContrasts?

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Cross tabulations

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Delivery and housing tenure

Housing tenure Preterm Term

Owner-occupier 50 849

Council tentant 29 229

Private tentant 11 164

Lives with parents 6 66

Other 3 36

Cross-tabulations

Tables of countable entities or frequencies Made to analyze the association, relationship, or connection between

two variables This association is difficult to describe statistically Null- Hypothesis: “There is no association between the two variables”

can be tested Analysis of cross-tabulations with larges samples


Housing tenure Preterm Term Total

Owner-occupier 50 849 899

Council tentant 29 229 258

Private tentant 11 164 175

Lives with parents 6 66 72

Other 3 36 39

Total 99 1344 1443


Expected number without any association between delivery and housing tenure

Housing tenure Pre Term Total

Owner-occupier 899

Council tenant 258

Private tenant 175

Lives with parents 72

Other 39

Total 99 1344 1443

Delivery and housing tenureIf the null-hypothesis is true

899/1443 = 62.3% are house owners.62.3% of the Pre-terms should be house owners: 99*899/1443 = 61.7


Owner-occupier 899

Council tenant 258

Private tenant 175


Other 39

Total 99 1344 1443


899/1443 = 62.3% are house owners.62.3% of the ‘Term’s should be house owners: 1344*899/1443 =

837.3


Owner-occupier 61.7 899

Council tenant 258

Private tenant 175


Other 39

Total 99 1344 1443


258/1443 = 17.9% are council tenant.17.9% of the ‘preterm’s should be council tenant: 99*258/1443 =

17.7


Owner-occupier 61.7 837.3 899

Council tenant 258

Private tenant 175


Other 39

Total 99 1344 1443


In general


Owner-occupier 61.7 837.3 899

Council tenant 17.7 240.3 258

Private tenant 12.0 163.0 175

Lives with parents 4.9 67.1 72

Other 2.7 36.3 39

Total 99 1344 1443

row total * column total

grand total



Owner-occupier 50(61.7) 849(837.3) 899

Council tenant 29(17.7) 229(240.3) 258

Private tenant 11(12.0) 164(163.0) 175

Lives with parents 6(4.9) 66(67.1) 72

Other 3(2.7) 36(36.3) 39

Total 99 1344 1443

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Delivery and housing tenuretest for association

If the numbers are large this will be chi-square distributed.The degree of freedom is (r-1)(c-1) = 4From Table 13.3 there is a 1 - 5% probability that delivery and

housing tenure is not associated

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Chi Squared Table


It is difficult to say anything about the nature of the association.


Owner-occupier 50(61.7) 849(837.3) 899

Council tenant 29(17.7) 229(240.3) 258

Private tenant 11(12.0) 164(163.0) 175

Lives with parents 6(4.9) 66(67.1) 72

Other 3(2.7) 36(36.3) 39

Total 99 1344 1443

Chi-squared test for small samples

Expected valued• > 80% >5• All >1

Streptomycin Control Total

Improvement 13 (8.4) 5 (9.6) 18

Deterioration 2 (4.2) 7 (4.8) 9

Death 0 (2.3) 5 (2.7) 5

Total 15 17 32

Chi-squared test for small samples

Expected valued• > 80% >5• All >1

Streptomycin Control Total

Improvement 13 (8.4) 5 (9.6) 18

Deterioration and death

2 (6.6) 12 (7.4) 14

Total 15 17 32

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Fisher’s exact testAn example

S D T

A 3 1 4

B 2 2 4

5 3 8

S D T

A 4 0 4

B 1 3 4

5 3 8

S D T

A 1 3 4

B 4 0 4

5 3 8

S D T

A 2 2 4

B 3 1 4

5 3 8

Fisher’s exact test

Survivors: • a, b, c, d, e

Deaths: • f, g, h

Table 1 can be made in 5 waysTable 2: 30Table 3: 30Table 4: 570 ways in totalThe properties of finding table 2

or a more extreme is:

S D T

A 3 1 4

B 2 2 4

5 3 8

S D T

A 4 0 4

B 1 3 4

5 3 8

S D T

A 1 3 4

B 4 0 4

5 3 8

S D T

A 2 2 4

B 3 1 4

5 3 8

5 30 1

70 70 2

Fisher’s exact test

S D T

A f11 f12 r1

B f21 f22 r2

c1 c2 n

S D T

A 3 1 4

B 2 2 4

5 3 8

1 2 1 2

11 12 21 22

! ! ! !

! ! ! ! !

4!4!5!3!0.4286

8!3!1!2!2!

r r c cp

n f f f f

S D T

A f11 f12 r1

B f21 f22 r2

c1 c2 n

S D T

A 4 0 4

B 1 3 4

5 3 8

1 2 1 2

11 12 21 22

! ! ! !

! ! ! ! !

4!4!5!3!0.0714

8!4!0!1!3!

r r c cp

n f f f f

Odds and odds ratios

Odds, p is the probability of an event

Log odds / logit

1

po

p

ln( ) ln1

po

p

Odds

The probability of coughs in kids with history of bronchitis.p = 26/273 = 0.095o = 26/247 = 0.105

The probability of coughs in kids with history without bronchitis.

p = 44/1046 = 0.042o = 44/1002 = 0.044

Bronchitis No bronchitis Total

Cough 26 (a) 44 (b) 70

No Cough 247 (c) 1002 (d) 1249

Total 273 1046 1319

1

po

p

Odds ratio

The odds ratio; the ratio of odds for experiencing coughs in kids with and kids without a history of bronchitis.


Cough 26; 0.105 (a) 44; 0.0439 (b) 70

No Cough 247; 9.50 (c) 1002; 22.8 (d) 1249

Total 273 1046 1319

ac

bd

ador

bc

ab

cd

ador

bc

26247

441002

26*10022.40

247*44or

Is the odds ratio different form 1?


Cough 26 (a) 44 (b) 70

No Cough 247 (c) 1002 (d) 1249

Total 273 1046 1319

1 1 1 1 1 1 1 126 44 247 1002SE ln 0.257a b c dor

0.874 1.96 0.257 _ to_0.874 1.96 0.257 0.37 _ _1.38to

ln( ) ln(2.40) 0.874or We could take ln to the odds ratio. Is ln(or) different from

zero?

95% confidence (assumuing normailty)

Confidence interval of the Odds ratio

ln (or) ± 1.96*SE(ln(or)) = 0.37 to 1.38Returning to the odds ratio itself:e0.370 to e1.379 = 1.45 to 3.97The interval does not contain 1, indicating a statistically significant

difference


Cough 26 (a) 44 (b) 70

No Cough 247 (c) 1002 (d) 1249

Total 273 1046 1319

McNemar’s test

Colds at 14

TotalYes No

Colds at 12

Yes 212 144 356

No 256 707 963

Total 468 851 1319

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Survival analysis

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First example of the day

Problem

Do patients survive longer after treatment 1 than after treatment 2?Possible solutions:

• ANOVA on mean survival time?• ANOVA on median survival time?

Progressively censored observations

Current life table• Completed dataset

Cohort life table• Analysis “on the fly”

Actuarial / life table anelysis

Treatment for lung cancer

Actuarial / life table anelysis

A sub-set of 13 patients undergoing the same treatment

Kaplan-Meier

Simple example with only 2 ”terminal-events”.

Confidence interval of the Kaplan-Meier method

Fx after 32 months

( ) ii i

i i i

dSE S S

n n d

1

( ) 0.9 0.094910 10 1iSE S

Confidence interval of the Kaplan-Meier method

• Survival plot for all data on treatment 1

• Are there differences between the treatments?

Comparing Two Survival Curves

• One could use the confidence intervals…

• But what if the confidence intervals are not overlapping only at some points?

• Logrank-stats


The logrank statistics Aka Mantel-logrank statisticsAka Cox-Mantel-logrank statistics


Five steps to the logrank statistics table1. Divide the data into intervals (eg. 10 months)2. Count the number of patients at risk in the groups and in

total 3. Count the number of terminal events in the groups and in

total4. Calculate the expected numbers of terminal events

e.g. (31-40) 44 in grp1 and 46 in grp2, 4 terminal events. expected terminal events 4x(44/90) and 4x(46/90)

5. Calculate the total


Smells like Chi-Square statistics

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2

all_treatments

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2 2

2 23 17.07 12 17.934.02

17.07 17.93

1df

0.05p


1 1

2 2

23 17.07Hazard ratio 2.01

12 17.93

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Repeated measures ANOVA in SPSS Cross tabulations Survival analysis