渥太華大學(xué)統(tǒng)計(jì)英文課件Lecture 8

1、Universit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.1lWhat are multiple comparisons?lThe problem of experiment-wise a a errorlWhen do we do multiple comparisons?lStatistical tests which control a aelEstimating treatment effectsUniversit dOttawa / University of Ottawa1999Bio

2、4118 Applied BiostatisticsL8.2lPair-wise comparisons of different treatmentslThese comparisons may involve group means, medians, variances, etc.lfor means, done after ANOVA lIn all cases, H0 is that the groups in question do not differ.Yieldm mCm mNm mN+PControlExperimental (N)Experimental (N+P)m mc

3、:m mNm mN:m mN+Pm mC: m mN+PFrequencyUniversit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.3lplanned (a priori): independent of ANOVA results; theory predicts which treatments should be different.lunplanned (a posteriori): depend on ANOVA results; unclear which treatments shou

4、ld be different.lTest of significance are very different between the two!YYX1X2X3X4X5X1X2X3X4X5PlannedunplannedUniversit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.4lComparisons of interest are determined by experimenter beforehand based on theory and do not depend on ANOVA r

5、esults.lPrediction from theory: catecholamine levels increase above basal levels only after threshold PAO2 = 30 torr is reached.lSo, compare only treatments above and below 30 torr (NT = 12).Predicted thresholdPAO2 (torr)1020304050Catecholamine0.00.10.20.30.40.50.60.7Universit dOttawa / University o

6、f Ottawa1999Bio 4118 Applied BiostatisticsL8.5lComparisons are determined by ANOVA results.lPrediction from theory: catecholamine levels increase with increasing PAO2 .lSo, comparisons between any pairs of treatments may be warranted (NT = 21).Predicted relationshipPAO2 (torr)1020304050Catecholamine

7、0.00.10.20.30.40.50.60.7Universit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.6lFor k comparisons, the probability of accepting H0 (no difference) is (1 - a a)k.lFor 4 treatments, (1 - a a)k = (0.95)6 = .735, so experiment-wise a a (a ae) = 0.265.lThus we would expect to rejec

8、t H0 for at least one paired comparison about 27% of the time, even if all four treatments are identical.Nominal a a = .05Number of treatments0246810Experiment-wise a a (a ae) 0.00.20.40.60.81.0Universit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.7lTo maintain a ae at nominal

9、 a a, we need to adjust a a for each comparison by the total number of comparisons.lIn this manner, a ae becomes independent of the number of treatments and/or comparisons.Number of treatments0246810Experiment-wise a a (a ae) 0.00.20.40.60.81.0Nominal a a = .05Universit dOttawa / University of Ottaw

10、a1999Bio 4118 Applied BiostatisticsL8.8lUse modified test-statistic S for pair-wise comparisons whose distribution depends on the total number of comparisons NT such that p(S) increases with NT.05101520Value of test statistic (S)00.20.3Probability (p)S, NT = 1 S, , NT = 2S, NT = 3Universit dOttawa /

11、 University of Ottawa1999Bio 4118 Applied BiostatisticsL8.9lUse only after H0 is rejected on the basis of an ANOVA because.l. ANOVA is more robust and reliable than multiple comparisons.lSo, if H0 is accepted in original ANOVA, do not proceed to do multiple comparisons.lNote, however, that there is

12、no universally agreed-upon method for doing multiple comparisons, and.lresults may differ depending on which method you use.lSo, proceed with caution!Universit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.10MethodTeststatistica a for individualcomparisonsRemarksBonferronita a a

13、 a kconservativeSidakta a a a kconservativeUncorrectedtnominal a aliberalUniversit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.11UncorrectedBonferonniSidakkpa aH0a aH0a aH01.04.05reject.05reject.05reject2.04.05reject.025accept.0253accept3.04.05reject.0167accept.0167acceptp is

14、probability associated with t-test of difference between1 pair of means; k is total number of comparisons.Universit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.12lPrediction: dam construction resulted in loss of large sturgeonlTest: compare sturgeon size before and after dam c

15、onstructionlH0: mean size is the same for all years1954 195819651966YEAR35.038.842.646.450.254.0FKLNGTHDam constructionUniversit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.13 Analysis of VarianceSource Sum-of-Squares df Mean-Square F-ratio PYEAR 485.264 3 161.755 5 .957 0.001

16、Error 3095.295 114 27.152Conclusion: reject H0Universit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.14U sin g m odel M S E of 27.152 w ith 114 df.M atrix of pairw ise m ean differen ces: 1 2 3 4 1 0.0 2 0.187 0.0 3 -5.508 -5.695 0.0 4 -3.313 -3.500 2.195 0.0B on ferron i A dju

17、stm en t.M atrix of pairw ise com parison prob ab ilities: 1 2 3 4 1 1.000 2 1.000 1.000 3 0.011 0.013 1.000 4 0.033 0.047 1.000 1.0001954 195819651966YEAR35.038.842.646.450.254.0FKLNGTHDam constructionUniversit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.15MethodTeststatistic

18、 Test statisticdistributionRemarksScheffsSFpowerful, consistentwith ANOVATukeysqspecialdistributionOKGT2qspecialdistributionOKSNKqspecialdistributionpowerful, but muchtoo liberalDuncansqspecialdistributionpowerful, but tooliberalDunnettsqspecialdistributionCompare treatmentswith controlUniversit dOt

19、tawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.16lTukeys, GT2 lTry several methods, and if you get similar results, youre on safe ground.lIf you get differences, they are due to:nhow conservative/liberal the test isnhow powerful it islIf comparisons using Bonferroni are still signif

20、icant, youre O.K.lIf comparisons using Sidak are still non-significant, youre also O.K.Universit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.17lConcern is not just with whether treatments differ, but with how much they differ.lFor example, what effect does a change in water te

21、mperature of 4C have on trout growth rate?lSince each treatment mean has a certain precision, so too will the estimate of the effect.Water temperature (C)162024280.000.040.080.120.160.20Growth rate l l (cm/day)Estimated effectRange of possibleeffectsUniversit dOttawa / University of Ottawa1999Bio 4118 Applied BiostatisticsL8.18lUse MSerror from ANOVA table as an estimate of the variance of each treatment to calculate confidence intervals for treatment means.ANOVA TableEffectSSdfMSFpYea