Statistical Methods in Clinical Trials

II Categorical Data

Ziad Taib

Biostatistics

AstraZeneca

March 7, 2012

Types of Data

ContinuousBlood pressureTime to event

Ordered CategoricalPain level

DiscreteNo of relapses

Categoricalsex

quantitative qualitative

Types of data analysis (Inference)

ParametricVs

Non parametric

FrequentistVs

Bayesian

Model basedVs

Data driven

Inference problems

1. Binary data (proportions)• One sample• Paired data

2. Ordered categorical data

3. Combining categorical data

4. Logistic regression

5. 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

Randomexperience

Failure0

Success1

Hole in one?

With probability p

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 H0: P=0.5

against 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 (cont’d)

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 n responders*rt-PA 312 147 (47.1%)placebo 312 122 (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.

Failure Success Total

Drug Y10 Y11 Y1.

Placebo Y20 Y21 Y2.

Total Y.0 Y.1 N=Y..

R E S P O N S E

TREATMENT

2 x 2 Contingency table

Failure Success Total

Drug 165 147 312

Placebo 190 122 312

Total 355 462 N=624

R E S P O N S E

TREATMENT

2 x 2 Contingency table

Hyper geometric distribution

Urn containing W white balls and R red balls: N=W+R

•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)

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 Fisher’s 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

35 contingency table

The Chi-square test is used for testing the independence

between the two factors

Other factor

A B C D E

i niA niB niC niD niE ni

One Factor ii niiA niiB niiC niiD niiE nii

iii niiiA niiiB niiiC niiiD niiiE niii

nA nB nC nD nE niA

The (Pearson) Chi-square test

The test-statistic is:

i j

ij

2ijij2

E

)E(O

where Oij = observed frequencies

and Eij = expected frequencies (under independence)

the test-statistic approximately follows a chi-square distribution

p

Example 5Chi-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 non-resp responder

rt-PA 165 147 312

placebo 190 122 312

355 269Observed frequencies

non-resp responder

rt-PA 177.5 134.5 312

placebo 177.5 134.5 312

355 269

Expected frequencies

• 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 Fisher’s exact test or using a continuity correction the chi square test the p-value is 0.052.

TABLE OF GRP BY Y

Frequency‚ Row Pct ‚nonresp ‚resp ‚ Total ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ placebo ‚ 190 ‚ 122 ‚ 312 ‚ 60.90 ‚ 39.10 ‚ ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ rt-PA ‚ 165 ‚ 147 ‚ 312 ‚ 52.88 ‚ 47.12 ‚ ƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒˆ Total 355 269 624

STATISTICS FOR TABLE OF GRP BY Y

Statistic DF Value Prob ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ 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

SAS| output

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 testOther factor

A B C D E

i niA niB niC niD niE ni

One Factor ii niiA niiB niiC niiD niiE nii

iii niiiA niiiB niiiC niiiD niiiE niii

nA nB nC nD nE 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 Disease category

dip snip fup bop other

treatment A 33 15 34 26 8 116

group B 28 18 34 20 14 114

61 33 68 46 22 230

Chi-square test: 2 = 3.084 , df=4 , p = 0.544

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 Mantel-Haensel statistic

which is chi square distributed with one df.

• 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

• 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

Age SBP Age SBP Age SBP

22 131 41 139 52 128 23 128 41 171 54 105 24 116 46 137 56 145 27 106 47 111 57 141 28 114 48 115 58 153 29 123 49 133 59 157 30 117 49 128 63 155 32 122 50 183 67 176 33 99 51 130 71 172 35 121 51 133 77 178 40 147 51 144 81 217

Table 1 Age and systolic blood pressure (SBP) among 33 adult women

80

100

120

140

160

180

200

220

20 30 40 50 60 70 80 90

SBP (mm Hg)

Age (years)

adapted from Colton T. Statistics in Medicine. Boston: Little Brown, 1974

Simple linear regression

• Relation between 2 continuous variables (SBP and age)

• 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

y

x

xβαy 11Slope

Multiple linear regression

• Relation between a continuous variable and a set of i continuous variables

• 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

xβ ... xβ xβαy ii2211

Multiple linear regression

Predicted Predictor variables

Response variable Explanatory variables

Outcome variable Covariables

Dependent Independent variables

xβ ... xβ xβα y ii2211

Logistic regressionTable 2 Age and signs of coronary heart disease (CD)

How can we analyse these data?

• Compare mean age of diseased and non-diseased

– 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

0

20

40

60

80

100

0 2 4 6 8

Diseased %

Age group

Logistic function (1)

0.0

0.2

0.4

0.6

0.8

1.0

Probability of disease

x

Transformation

)(

)(

xyP

xyP

1

logit of P(y|x)

{

0.0

0.2

0.4

0.6

0.8

1.0

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

iiiii xyxylL

1

)(1ln)1()(ln)(ln)(

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 …

• 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

ii2211 xβ ... xβ xβαP-1

P ln

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 non-resp responder

rt-PA 165 147 312

placebo 190 122 312

355 269

Observed frequencies

SAS| output

The LOGISTIC Procedure Response Profile Ordered Binary Value Outcome Count 1 EVENT 269 2 NO EVENT 355 Model Fitting Information and Testing Global Null Hypothesis BETA=0 Intercept Intercept and Criterion Only Covariates Chi-Square for Covariates AIC 855.157 853.069 . SC 859.593 861.941 . -2 LOG L 853.157 849.069 4.089 with 1 DF (p=0.0432) Score . . 4.084 with 1 DF (p=0.0433) Analysis of Maximum Likelihood Estimates Parameter Standard Wald Pr > Standardized Odds Variable DF Estimate Error Chi-Square Chi-Square Estimate Ratio INTERCPT 1 -0.4430 0.1160 14.5805 0.0001 . . GRP 1 0.3275 0.1622 4.0743 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)

SAS| output

Response Profile

Ordered Value BI_60 Count

1 1 750 2 0 603

Analysis of Maximum Likelihood Estimates

Parameter Standard Wald Pr > Standardized Odds Variable DF Estimate Error Chi-Square Chi-Square Estimate Ratio

INTERCPT 1 2.4244 0.5116 22.4603 0.0001 TRT 1 0.1299 0.1310 0.9838 0.3213 0.035826 1.139 STRATUM 1 0.1079 0.1323 0.6648 0.4149 0.029751 1.114 AGE 1 -0.0673 0.00671 100.6676 0.0001 -0.409641 0.935 SEVERITY 1 0.0942 0.00642 215.0990 0.0001 0.621293 1.099

Conditional Odds Ratios and 95% Confidence Intervals

Wald Confidence Limits Odds Variable Unit Ratio Lower Upper

TRT 1.0000 1.139 0.881 1.472 STRATUM 1.0000 1.114 0.859 1.444 AGE 1.0000 0.935 0.923 0.947 SEVERITY 1.0000 1.099 1.085 1.113

A Bayesian alternative

•Prior knowledge is part of the Bayesian

approach.

•Prior knowledge matters

Case-Control

• Imagine a randomised clinical trial or a case control study. The analysis uses a chi square test and the corresponding p-values. 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

0

00 BF] H P[

] H P[-11 D] | H P[

]H | P[D

]H | P[D BF

1

0

where

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

1BF

BF

BF

1BF

BF21

21

1 D] | P[H1

1

0

Berger and Selke (1987) have shown that for a

very wide range of cases including the case

control case

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)).

2

1

2

2

BF

e

Conclusion

252.01337.0

0.337 D] | P[H0

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?