【正文】 are your degrees of freedom here? Answer: Not obvious! Case 1: ttest, unpooled variances It is plicated to figure out the degrees of freedom here! A good approximation is given as df ≈ harmonic mean (or SAS will tell you!): ?tmsnsYXTyxmn ~22???mn112? Case 2: pooled variance If you assume that the standard deviation of the characteristic (., IQ) is the same in both groups, you can pool all the data to estimate a mon standard deviation. This maximizes your degrees of freedom (and thus your power). 2)()()()1( a n d 1)()()1( a n d 1)(: v a r ia n c e sp o o lin g12122122122122122???????????????????????????????mnyyxxsyysmmyysxxsnnxxsmimininipmimiymimiyninixninix2)1()1( 222??????mnsmsns yxpDegrees of Freedom! Estimated standard error (using pooled variance estimate) msns ppyx22????2)()(:12122???????????mnyyxxsw h e r emimininipThe degrees of freedom are n+m2 Case 2: ttest, pooled variances 222~ ????? mnppmntmsnsYXT2)1()1( 222??????mnsmsns yxpAlternate calculation formula: ttest, pooled variance 2~ ????? mnpmn tmnnmsYXT)()()11( 2222mnmnsmnmmnnsnmsnsmsppppp ???????Pooled vs. unpooled variance Rule of Thumb: Use pooled unless you have a reason not to. Pooled gives you more degrees of freedom. Pooled has extra assumption: variances are equal between the two groups. SAS automatically tests this assumption for you (“Equality of Variances” test). If p.05, this suggests unequal variances, and better to use unpooled ttest. Example: twosample ttest ? In 1980, some researchers reported that “men have more mathematical ability than women” as evidenced by the 1979 SAT’s, where a sample of 30 random male adolescents had a mean score 177。 =SD (lnOR) = .44 3. Empirical evidence: = 20*40/60*10 =800/600 = ? lnOR = .288 4. Z = (.2880)/.44 = .65 pvalue = P(Z.65 or Z) = .26*2 5. Not enough evidence to reject the null hypothesis of no association 401601202001 ???401601202001 ???TWOSIDED TEST TWOSIDED TEST: it would be just as extreme if the sample lnOR were .65 standard deviations or more below the null mean Key measures of relative risk: 95% CIs OR and RR: ????????????????????????dcbadcbae xp*OR ,e xp*OR ???????? ?????????????? ??????cdccabaacdccabaa )/(1)/()/(1)/(e xp*RR ,e xp*RR For an odds ratio, 95% confidence limits: For a risk ratio, 95% confidence limits: Continuous oute (means) Oute Variable Are the observations independent or correlated? Alternatives if the normality assumption is violated (and small sample size): independent correlated Continuous (. pain scale, cognitive function) Ttest: pares means between two independent groups ANOVA: pares means between more than two independent groups Pearson’s correlation coefficient (linear correlation): shows linear correlation between two continuous variables Linear regression: multivariate regression technique used when the oute is continuous。 lnOR = 0 Alternative hypothesis: OR? 。Twosample tests Binary or categorical outes (proportions) Oute Variable Are the observations correlated? Alternative to the chisquare test if sparse cells: independent correlated Binary or categorical (. fracture, yes/no) Chisquare test: pares proportions between two or more groups Relative risks: odds ratios or risk ratios Logistic regression: multivariate technique used when oute is binary。 gives multivariateadjusted odds ratios McNemar’s chisquare test: pares binary oute between correlated groups (., before and after) Conditional logistic regression: multivariate regression technique for a binary oute when groups are correlated (., matched data) GEE modeling: multivariate regression technique for a binary oute when groups are correlated (., repeated measures) Fisher’s exact test: pares proportions between independent groups when there are sparse data (some cells 5). McNemar’s exact test: pares proportions between correlated groups when there are sparse data (some cells 5). Recall: The odds ratio (two samples=cases and controls) Smoker (E) Nonsmoker (~E) Stroke (D) 15 35 No Stroke (~D) 8 42 50 50 8*3542*15???bcadOR?Interpretation: there is a higher odds of stroke in smokers vs. nonsmokers. Inferences about the odds ratio… ? Does the sampling distribution follow a normal distribution? ? What is the standard error? Simulation… ? 1. In SAS, assume infinite population of cases and controls with equal proportion of smokers (exposure), p=.23 (UNDER THE NULL!) ? 2. Use the random binomial function to randomly select n=50 cases and n=50 controls each with p=.23 chance of being a smoker. ? 3. Calculate the observed odds ratio for the resulting 2x2 table. ? 4. Repeat this 1000 times (or some large number of times). ? 5. Observe the distribution of odds ratios under the null hypothesis. Properties of the OR (simulation) (50 cases/50 controls/23% exposed) Under the null, this is the expected variability of the sample OR?note the right skew Properties of the lnOR Normal! Properties of the lnOR From the simulation, can get the empirical standard error (~) and pvalue (~.10) Properties of the lnOR dcba1111 ???Or, in general, standard error = Inferences