SAS: Meta-Analysis CMH Example for Categorical Variable

A. Reference

B. Meta-Analysis

A meta-analysis is a statistical analysis that combines the results of multiple scientific studies. Meta-analysis can be performed when there are multiple scientific studies addressing the same question, with each individual study reporting measurements that are expected to have some degree of error. The aim then is to use approaches from statistics to derive a pooled estimate closest to the unknown common truth based on how this error is perceived.

Wikipedia
  • In meta-analysis, studies become observations.
  • Research collect data for meta-analysis by systematic review of the literature in the field, and compile data directly from the summary statistics in the publication.

C. Problem with simply lumping the data from different studies together

  • Not consider treatment-by-study interaction
  • Assume response rates are the same in all studies.

D. SAS Solution (follow Hamer and Simpson’s paper, but corrected the output from the paper)

  • Create data set with the results of 2 studies. B: Remitted; N:Not remitted; P: Placebo; D: Drug.
  • I have used B (Better) to indicate Remitted cases because Proc Freq test is based on column 1 and row 1 of the 2 by 2 table, so if we code R for Remitted cases then the remitted case will be in column 2 because the table is by alphabetical order and R is after N.
  • The Hamer and Simpson paper actually tested the null hypothesis for the non-effective cases rather than the effective cases.
data chm;
input study $ response $ trt $ cellfreq @@;
datalines;
study1	B	P	24	study1	N	P	3
study1	B	D	58	study1	N	D	30
study2	B	P	16	study2	N	P	57
study2	B	D	2	study2	N	D	10
;
run;
  • Run Cochran-Mantel-Haenszel Statistics using Proc Freq procedure with cmh option.
proc freq data=chm;
tables study*trt*response /cmh;
weight cellfreq;
run;

E. SAS Output

  • SAS chm table
  • Frequency table
  • Cochrane-Mantel-Haenszel test

F. Notes

  • The Mantel-Haenszel estimator of the common odds ratio assumed the estimation to be homogeneous among both studies.
  • The Mentel-Haenszel statistics tests the null hypothesis that the response rate is the same for the two treatments, after adjusting for possible differences in study response rates.
  • For Proc Freq testing options, make sure the group that you want to tested are in row 1 and column 1. It is also important to crosstab treatment as row and response as column, so the interpretation of the relative risk for the risk of improvement make sense. In Hamper and Simpon’s paper the crosstab has been transposed, therefore the relative risk output doesn’t make sense.

G. Interpretation

  • The CMH test statistics is 4.65 with a p-value of 0.03, therefore, we can reject the null hypothesis that there is no association between treatment and response. P-value lower than 0.05 indicates that the association between treatment and response remains strong after adjusting for study.
  • Relative Risk (Column 1) equals to 0.74 which means the probability of the improvement with the drug is 0.74 time the probability of the improvement with the placebo.
  • Relative Risk (Column 2) equals to 1.51 which means the probability of no improvement in the symptoms with the drug is 1.51 times the probability of no improvement with the placebo.
  • The Breslow-Day test has a large p-value of 0.295 which indicates there is no significant difference in the odds ratios among the studies.

* I will show the odds ratio and relative risk calculation in Excel in another post.