# Clinical Trials A short course - Chalmers Statistical Methods in Clinical Trials II Categorical Data Ziad Taib Biostatistics AstraZeneca March 7, 2012 Types of Data quantitative Continuous Blood pressure Time to event Categorical sex Discrete No of relapses

Ordered Categorical Pain level qualitative Types of data analysis (Inference) Parametric Vs Non parametric Model based Vs Data driven Frequentist Vs Bayesian Inference problems 1. Binary data (proportions)

2. 3. 4. 5. One sample Paired data Ordered categorical data Combining categorical data Logistic regression A Bayesian alternative Categorical data In a RCT, endpoints and surrogate endpoints can be categorical or ordered categorical variables. In the simplest cases we have binary responses (e.g. responders non-responders). In Outcomes research it is common to use many

ordered categories (no improvement, moderate improvement, high improvement). Bernoulli experiment 1 Success With probability p Random experience 0 Failure Hole in one? With probability1-p Binary variables

Sex Mortality Presence/absence of an AE Responder/non-responder according to some pre-defined criteria Success/Failure Estimation Assume that a treatment has been applied to n patients and that at the end of the trial they were classified according to how they responded to the treatment: 0 meaning not cured and 1 meaning cured. The data at hand is thus a sample of n independent binary variables The probability of being cured by this treatment can be estimated by

satisfying Hypothesis testing We can test the null hypothesis Using the test statistic When n is large, Z follows, under the null hypothesis, the standard normal distribution (obs! Not when p very small or very large). Hypothesis testing For moderate values of n we can use the exact Bernoulli distribution of leading to the sum being Binomially distributed i.e. As with continuous variables, tests can be used to build confidence intervals. Example 1: Hypothesis test based on binomial distr. Consider testing

against H0: P=0.5 Ha: P>0.5 and where: n=10 and y=number of successes=8 p-value=(probability of obtaining a result at least as extreme as the one observed)= Prob(8 or more responders)=P8+ P9+ P10= ={using the binomial formula}=0.0547 Example 2 RCT of two analgesic drugs A and B given in a random order to each of 100 patients. After both treatment periods, each patient states a preference for one of the drugs. Result: 65 patients preferred A and 35 B Example (contd) Hypotheses: H0: P=0.5 against H1: P0.5 Observed test-statistic: z=2.90

p-value: p=0.0037 (exact p-value using the binomial distr. = 0.0035) 95% CI for P: (0.56 ; 0.74) Example 3 We want to test if the proportion of patients experiencing an early improvement after some treatment is 0.35. n=312 patients were observed among which 147 experienced such an improvement yielding a proportion of (47.1%). The Z value is 4.3 yielding a p-value of 0.00002. Using the exact distribution 0.00001. Of course n here is large so the normal approximation is good enough. A 95% confidence interval for the proportion is [4.1, 5.2] and does not contain the point 0.35.

Two proportions Sometimes we want to compare the proportion of successes in two separate groups. For this purpose we take two samples of sizes n1 and n2. We let yi1 and pi1 be the observed number of subjects and the proportion of successes in the ith group. The difference in population proportions of successes and its large sample variance can be estimated by Two proportions (continued) Assume we want to test the null hypothesis that there is no difference between the proportions of success in the two groups. Under the null hypothesis, we can estimate the common proportion by Its large sample variance is estimated by Example 4 NINDS trial in acute ischemic stroke Treatment

rt-PA placebo n 312 312 responders* 147 (47.1%) 122 (39.1%) *early improvement defined on a neurological scale Point estimate: 0.080 (s.e.=0.0397) 95% CI: (0.003 ; 0.158) p-value:

0.043 Two proportions (Chi square) The problem of comparing two proportions can sometimes be formulated as a problem of independence! Assume we have two groups as above (treatment and placebo). Assume further that the subjects were randomized to these groups. We can then test for independence between belonging to a certain group and the clinical endpoint (success or failure). The data can be organized in the form of a contingency table in which the marginal totals and the total number of subjects are considered as fixed. 2 x 2 Contingency table RESPONSE T R

E A T M E N T Failure Success Total Drug Y10 Y11 Y1.

Placebo Y20 Y21 Y2. Total Y.0 Y.1 N=Y.. 2 x 2 Contingency table RESPONSE T R E

A T M E N T Failure Success Total Drug 165 147 312 Placebo

190 122 312 Total 355 462 N=624 Hyper geometric distribution n balls are drawn at random without replacement. Y is the number of white balls (successes) Y follows the Hyper geometric Distribution with parameters (N, W, n)

Urn containing W white balls and R red balls: N=W+R Contingency tables N subjects in total y.1 of these are special (success) y1. are drawn at random Y11 no of successes among these y1. Y11 is HG(N,y.1,y 1.) in general Contingency tables The null hypothesis of independence is tested using the chi square statistic

Which, under the null hypothesis, is chi square distributed with one degree of freedom provided the sample sizes in the two groups are large (over 30) and the expected frequency in each cell is non negligible (over 5) Contingency tables For moderate sample sizes we use Fishers exact test. According to this calculate the desired probabilities using the exact Hyper-geometric distribution. The variance can then be calculated. To illustrate consider: Using this and expectation m11 we have the randomization chi square statistic. With fixed margins only one cell is allowed to vary. Randomization is crucial for this approach. The (Pearson) Chi-square test

One Factor A i niA ii niiA iii niiiA nA Other factor B C D niB niC niD niiB niiC niiD niiiB niiiC niiiD nB nC nD E niE niiE niiiE nE

ni nii niii niA 35 contingency table The Chi-square test is used for testing the independence between the two factors The (Pearson) Chi-square test 2 (O E ) ij ij 2

The test-statistic is: p i j E ij where Oij = observed frequencies and Eij = expected frequencies (under independence) the test-statistic approximately follows a chi-square distribution Example 5 Chi-square test for a 22 table Examining the independence between two treatments and a classification into responder/non-responder is equivalent to comparing the proportion of responders in the two groups NINDS again Observed frequencies

Expected frequencies rt-PA placebo rt-PA placebo non-resp 165 190 responder 147 122 355 269 non-resp 177.5

177.5 responder 134.5 134.5 355 269 312 312 312 312 p0=(122+147)/(624)=0.43 v(p0)=0.00157 which gives a p-value of 0.043 in all these cases. This implies the drug is better than placebo. However when using Fishers

exact test or using a continuity correction the chi square test the p-value is 0.052. TABLE OF GRP BY Y S A S | o u t p u t Frequency Row Pct nonresp resp placebo 190

122 60.90 39.10 rt-PA 165 147 52.88 47.12 Total 355 269 Total 312 312 624 STATISTICS FOR TABLE OF GRP BY Y Statistic

DF Value Prob Chi-Square 1 4.084 0.043 Likelihood Ratio Chi-Square 1 4.089 0.043 Continuity Adj. Chi-Square 1 3.764 0.052 Mantel-Haenszel Chi-Square 1 4.077 0.043 Fisher's Exact Test (Left) 0.982

(Right) 0.026 (2-Tail) 0.052 Phi Coefficient 0.081 Contingency Coefficient 0.081 Cramer's V 0.081 Sample Size = 624 Odds, Odds Ratios and relative Risks The odds of success in group i is estimated by The odds ratio of success between the two groups i is estimated by Define risk for success in the ith group as the proportion of cases with success. The relative risk between the two groups is estimated by

Categorical data Nominal E.g. patient residence at end of follow-up (hospital, nursing home, own home, etc.) Ordinal (ordered) E.g. some global rating Normal, not at all ill Borderline mentally ill Mildly ill Moderately ill Markedly ill Severely ill

Among the most extremely ill patients Categorical data & Chi-square test One Factor A i niA ii niiA iii niiiA nA Other factor B C D niB niC niD niiB niiC niiD niiiB niiiC niiiD nB nC nD E

niE niiE niiiE nE ni nii niii niA The chi-square test is useful for detection of a general association between treatment and categorical response (in either the nominal or ordinal scale), but it cannot identify a particular relationship, e.g. a location shift. Nominal categorical data treatment A group B dip 33

28 61 Disease category snip fup bop 15 34 26 18 34 20 33 68 46 other 8 14 22

Chi-square test: 2 = 3.084 , df=4 , p = 0.544 116 114 230 Ordered categorical data Here we assume two groups one receiving the drug and one placebo. The response is assumed to be ordered categorical with J categories. The null hypothesis is that the distribution of subjects in response categories is the same for both groups. Again the randomization and the HG distribution lead to the same chi square test statistic but this time with (J-1) df. Moreover the same relationship exists between the two versions of the chi square statistic. The Mantel-Haensel statistic The aim here is to combine data from several (H) strata for comparing two

groups drug and placebo. The expected frequency and the variance for each stratum are used to define the MantelHaensel statistic which is chi square distributed with one df. Logistic regression Consider again the Bernoulli situation, where Y is a binary r.v. (success or failure) with p being the success probability. Sometimes Y can depend on some other factors or covariates. Since Y is binary we cannot use usual regression. Logistic regression Logistic regression is part of a category of statistical models called generalized linear models (GLM). This broad class of models includes ordinary regression and ANOVA, as well as multivariate statistics such as ANCOVA and loglinear regression. An excellent treatment of generalized linear models is presented in

Agresti (1996). Logistic regression allows one to predict a discrete outcome, such as group membership, from a set of variables that may be continuous, discrete, dichotomous, or a mix of any of these. Generally, the dependent or response variable is dichotomous, such as presence/absence or success/failure. Simple linear regression Table 1 Age and systolic blood pressure (SBP) among 33 adult women Age SBP Age SBP Age

SBP 22 23 24 27 28 29 30 32 33 35 40 131 128 116 106 114 123 117

122 99 121 147 41 41 46 47 48 49 49 50 51 51 51 139 171 137 111 115

133 128 183 130 133 144 52 54 56 57 58 59 63 67 71 77 81 128 105 145

141 153 157 155 176 172 178 217 SBP (mm Hg) 220 200 180 160 140 120

100 80 20 30 40 50 60 70 Age (years) adapted from Colton T. Statistics in Medicine. Boston: Little Brown, 1974 80 90

Simple linear regression Relation between 2 continuous variables (SBP and age) y Slope y 1x 1 x Regression coefficient 1 Measures association between y and x Amount by which y changes on average when x changes by one unit Least squares method Multiple linear regression Relation between a continuous variable and a set of i continuous variables y 1x1 2 x 2 ... i x i

Partial regression coefficients i Amount by which y changes on average when xi changes by one unit and all the other xis remain constant Measures association between xi and y adjusted for all other xi Example SBP versus age, weight, height, etc Multiple linear regression y 1x1 2 x 2 ... i xi Predicted Predictor variables Response variable Explanatory variables Outcome variable

Covariables Dependent Independent variables Logistic regression Table 2 Age and signs of coronary heart disease (CD) How can we analyse these data? Compare mean age of diseased and nondiseased Non-diseased: 38.6 years Diseased: 58.7 years (p<0.0001) Linear regression? Dot-plot: Data from Table 2 Logistic regression (2) Table 3

Prevalence (%) of signs of CD according to age group Dot-plot: Data from Table 3 Diseased % 100 80 60 40 20 0 0 2 4 Age group 6

8 Logistic function (1) Probability of disease 1.0 0.8 0.6 0.4 0.2 0.0 x Transformation P (y x) 1 P (y x) 1.0 0.8

{ 0.6 logit of P(y|x) 0.4 0.2 0.0 Fitting equation to the data Linear regression: Least squares or Maximum likelihood Logistic regression: Maximum likelihood Likelihood function Estimates parameters and Practically easier to work with log-likelihood n L() ln l () yi ln ( xi ) (1 yi ) ln1 ( xi ) i 1

Maximum likelihood Iterative computing (Newton-Raphson) Choice of an arbitrary value for the coefficients (usually 0) Computing of log-likelihood Variation of coefficients values Reiteration until maximisation (plateau) Results Maximum Likelihood Estimates (MLE) for and Estimates of P(y) for a given value of x Multiple logistic regression More than one independent variable Dichotomous, ordinal, nominal, continuous P ln 1x1 2 x 2 ... i xi 1- P Interpretation of i

Increase in log-odds for a one unit increase in xi with all the other xis constant Measures association between xi and log-odds adjusted for all other xi Statistical testing Question Does model including given independent variable provide more information about dependent variable than model without this variable? Three tests Likelihood ratio statistic (LRS) Wald test Score test Likelihood ratio statistic Compares two nested models Log(odds) = + 1x1 + 2x2 + 3x3 (model 1) Log(odds) = + 1x1 + 2x2

(model 2) LR statistic -2 log (likelihood model 2 / likelihood model 1) = -2 log (likelihood model 2) minus -2log (likelihood model 1) LR statistic is a 2 with DF = number of extra parameters in model Example 6 Fitting a Logistic regression model to the NINDS data, using only one covariate (treatment group). NINDS again Observed frequencies rt-PA placebo non-resp responder 165

147 190 122 355 269 312 312 S A S | o u t p u t The LOGISTIC Procedure

Response Profile Ordered Value 1 2 Binary Outcome Count EVENT NO EVENT 269 355 Model Fitting Information and Testing Global Null Hypothesis BETA=0 Criterion AIC SC

-2 LOG L Score Intercept Only Intercept and Covariates 855.157 859.593 853.157 . 853.069 861.941 849.069 . Chi-Square for Covariates .

. 4.089 with 1 DF (p=0.0432) 4.084 with 1 DF (p=0.0433) Analysis of Maximum Likelihood Estimates Variable DF INTERCPT GRP 1 1 Parameter Estimate Standard Error Wald

Chi-Square Pr > Chi-Square Standardized Estimate Odds Ratio -0.4430 0.3275 0.1160 0.1622 14.5805 4.0743 0.0001 0.0435

. 0.090350 . 1.387 Logistic regression example AZ trial (CLASS) in acute stroke comparing clomethiazole (n=678) with placebo (n=675) Response defined as a Barthel Index score 60 at 90 days Covariates: STRATUM (time to start of trmt: 0-6, 6-12) AGE SEVERITY (baseline SSS score) TRT (treatment group)

S A S | o u t p u t Response Profile Ordered Value BI_60 Count 1 2

1 0 750 603 Analysis of Maximum Likelihood Estimates Variable DF INTERCPT TRT STRATUM AGE SEVERITY 1 1 1 1 1

Parameter Standard Wald Pr > Standardized Estimate Error Chi-Square Chi-Square Estimate 2.4244 0.1299 0.1079 -0.0673 0.0942 0.5116 0.1310 0.1323 0.00671 0.00642 22.4603 0.9838 0.6648

100.6676 215.0990 0.0001 0.3213 0.4149 0.0001 0.0001 0.035826 0.029751 -0.409641 0.621293 Conditional Odds Ratios and 95% Confidence Intervals Wald Confidence Limits Variable Unit Odds Ratio

TRT STRATUM AGE SEVERITY 1.0000 1.0000 1.0000 1.0000 1.139 1.114 0.935 1.099 Lower Upper 0.881 0.859

0.923 1.085 1.472 1.444 0.947 1.113 Odds Ratio 1.139 1.114 0.935 1.099 A Bayesian alternative B e h t f

o t r a p is e dg e l s w r o e n tt k a r m o

i . r e P oach ledg r w p o p n a k r o i r P n a i

s ay e Case-Control Imagine a randomised clinical trial or a case control study. The analysis uses a chi square test and the corresponding pvalues. If this turns out to be less than 0.05 we assume significance. Example 7: Some studies from the year 1990 suggested that the risk to CHD is associated with childhood poverty. Since infection with the bacterium H. Pylori is also linked to poverty, some researchers suspected H. Pylori to be the missing link. In a case control study where levels of infections were considered in patients and controls the following results were obtained. Case/Control

Case CHD Control High 60% 39% n11+n12 Low 40% 61% n21+n22 n11+n21 n12+n22

1 - P[ H 0 ] P[ H 0 | D] 1 P[ H 0 ] BF where P[D | H 0 ] BF P[D | H1 ] 1 The chi square statistic having, in this case, the value 4.37 yields a p-value of 0.03 which is less than the formal level of significance 0.05. There is, however, no theoretical reason to believe that this result is true. So we take again P(H0)=0.5. This leads to 1 1

1 BF 1 BF P[H 0 | D] 1 2 1 BF BF 1 BF 2 Berger and Selke (1987) have shown that for a very wide range of cases including the case

control case 2 BF e 1 2 2 Using the value 4.73 for the chi square variable leads to a BF value of at least 0.337 (M. A. Mendall et al Relation betweenH. Pylori infection and coronary heart disease. Heart J. (1994)).

Conclusion 0.337 P[H 0 | D] 0.252 0.337 1 Taking another (more or less sceptical) attitude does not change a the conclusion that much: P(H0)=0.75 => P[ H0| D] > (0.5) P(H0)=0.25 => P[ H0| D] > (0.1) Questions or Comments?