Repeated measures ANOVA in SPSS Cross tabulations Survival analysis

Repeated measures ANOVA in SPSS 2

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? 3

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? ??? 4

Repeated measures ANOVA Time is a within subject factor (4 levels) Anxiety is a between subjects factor (2 levels) 5

Repeated measures ANOVA Plotting the data 6

Repeated measures ANOVA – between subjects effects 7

Repeated measures ANOVA – within subjects effects Is there a time effect? Univariate or Multivariate? Univariate assumes sphericity of the covariance matrix: 8 Residual SSCP Matrix Trial1Trial2Trail3Trial4 Trial14,5672,4171,3831,017 Trial22,4176,2835,1335,167 Trail31,3835,1336,4176,083 Trial41,0175,1676,0837,617 Based on Type III Sum of Squares

Repeated measures ANOVA – within subjects effects Univariate Mauchly's Test of Sphericity b Measure:MEASURE_1 Within Subjects Effect Mauchly's W Approx. Chi- SquaredfSig. dimension1 time,3379,4965,093 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 Trial1Trial2Trail3Trial4 Trial14,5672,4171,3831,017 Trial22,4176,2835,1335,167 Trail31,3835,1336,4176,083 Trial41,0175,1676,0837,617 Based on Type III Sum of Squares

Repeated measures ANOVA – within subjects effects Univariate 10 Greenhouse-Geisser is newer than Huynh-Feldt methods Lower-bound is the most konservative Alternatively use Multible ANOVA

Repeated measures ANOVA – within subjects effects Multivariate 11 The covariance matrices must be equal for all levels of the between groups Tested by Box’s test

Repeated measures ANOVA – within subjects effects Multivariate 12 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 effects Which method to choose? 13

Repeated measures ANOVA – within subjects effects Pairwise comparisons 14 Does all these comparisons make sense?

Repeated measures ANOVA – within subjects effects Contrasts? 15

Cross tabulations 16

Delivery and housing tenure Housing tenurePretermTerm Owner-occupier50849 Council tentant29229 Private tentant11164 Lives with parents666 Other336

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

Delivery and housing tenure Housing tenurePretermTermTotal Owner-occupier Council tentant Private tentant Lives with parents66672 Other33639 Total

Delivery and housing tenure Expected number without any association between delivery and housing tenure Housing tenurePreTermTotal Owner-occupier899 Council tenant258 Private tenant175 Lives with parents72 Other39 Total

Delivery and housing tenure If 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 Housing tenurePreTermTotal Owner-occupier899 Council tenant258 Private tenant175 Lives with parents72 Other39 Total

Delivery and housing tenure If the null-hypothesis is true 899/1443 = 62.3% are house owners. 62.3% of the ‘Term’s should be house owners: 1344*899/1443 = Housing tenurePreTermTotal Owner-occupier Council tenant258 Private tenant175 Lives with parents72 Other39 Total

Delivery and housing tenure If the null-hypothesis is true 258/1443 = 17.9% are council tenant. 17.9% of the ‘preterm’s should be council tenant: 99*258/1443 = 17.7 Housing tenurePreTermTotal Owner-occupier Council tenant258 Private tenant175 Lives with parents72 Other39 Total

Delivery and housing tenure If the null-hypothesis is true In general Housing tenurePreTermTotal Owner-occupier Council tenant Private tenant Lives with parents Other Total row total * column total grand total

Delivery and housing tenure If the null-hypothesis is true Housing tenurePreTermTotal Owner-occupier50(61.7)849(837.3)899 Council tenant29(17.7)229(240.3)258 Private tenant11(12.0)164(163.0)175 Lives with parents6(4.9)66(67.1)72 Other3(2.7)36(36.3)39 Total

Delivery and housing tenure test for association If the numbers are large this will be chi-square distributed. The degree of freedom is (r-1)(c-1) = 4 From Table 13.3 there is a 1 - 5% probability that delivery and housing tenure is not associated

Chi Squared Table

Delivery and housing tenure If the null-hypothesis is true It is difficult to say anything about the nature of the association. Housing tenurePreTermTotal Owner-occupier50(61.7)849(837.3)899 Council tenant29(17.7)229(240.3)258 Private tenant11(12.0)164(163.0)175 Lives with parents6(4.9)66(67.1)72 Other3(2.7)36(36.3)39 Total

Chi-squared test for small samples Expected valued > 80% >5 All >1 StreptomycinControlTotal Improvement13 (8.4)5 (9.6)18 Deterioration2 (4.2)7 (4.8)9 Death0 (2.3)5 (2.7)5 Total151732

Chi-squared test for small samples Expected valued > 80% >5 All >1 StreptomycinControlTotal Improvement13 (8.4)5 (9.6)18 Deterioration and death 2 (6.6)12 (7.4)14 Total151732

Fisher’s exact test An example SDT A314 B SDT A404 B SDT A134 B SDT A224 B

Fisher’s exact test Survivors: a, b, c, d, e Deaths: f, g, h Table 1 can be made in 5 ways Table 2: 30 Table 3: 30 Table 4: 5 70 ways in total The properties of finding table 2 or a more extreme is: SDT A314 B SDT A404 B SDT A134 B SDT A224 B

Fisher’s exact test SDT A f 11 f 12 r1r1 B f 21 f 22 r2r2 c1c1 c2c2 n SDT A314 B SDT A f 11 f 12 r1r1 B f 21 f 22 r2r2 c1c1 c2c2 n SDT A404 B

Odds and odds ratios Odds, p is the probability of an event Log odds / logit

Odds The probability of coughs in kids with history of bronchitis. p = 26/273 = o = 26/247 = The probability of coughs in kids with history without bronchitis. p = 44/1046 = o = 44/1002 = BronchitisNo bronchitisTotal Cough26 (a)44 (b)70 No Cough247 (c)1002 (d)1249 Total

Odds ratio The odds ratio; the ratio of odds for experiencing coughs in kids with and kids without a history of bronchitis. BronchitisNo bronchitisTotal Cough26; (a)44; (b)70 No Cough247; 9.50 (c)1002; 22.8 (d)1249 Total

Is the odds ratio different form 1? BronchitisNo bronchitisTotal Cough26 (a)44 (b)70 No Cough247 (c)1002 (d)1249 Total 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 1.38 Returning to the odds ratio itself: e to e = 1.45 to 3.97 The interval does not contain 1, indicating a statistically significant difference BronchitisNo bronchitisTotal Cough26 (a)44 (b)70 No Cough247 (c)1002 (d)1249 Total

McNemar’s test Colds at 14 Total YesNo Colds at 12 Yes No Total

Survival analysis 40

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

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

Comparing Two Survival Curves The logrank statistics Aka Mantel-logrank statistics Aka Cox-Mantel-logrank statistics

Comparing Two Survival Curves Five steps to the logrank statistics table 1.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 total 4.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

Comparing Two Survival Curves Smells like Chi-Square statistics

Comparing Two Survival Curves