Sample Size Calculation for the Cochran-Mantel-Haenszel Test

Background

The first method to estimate sample size for the CMH test was introduced by Gail (1973) using a least squares method. Munoz and Rosner (1984) later produced sample size estimators useful in the special case when both margins are fixed. However, this is often not the case in retrospective case-control designs. As such, Woolson, Bean, and Rojas (1986) introduced a calculation using the weighted difference between two binomial distributions, which assumes only one margin to be fixed. This was corroborated by Wittes and Wallenstein (1987) in their paper a year later. Later still, Nam (1992) improved upon that calculation by introducing a coefficient to address continuity corrected CMH tests.

The methods included in the package are based on calculations derived by Woolson, Bean, and Rojas (1986) for uncorrected statistics and Nam (1992) for continuity corrected statistics. A slightly more precise power calculation is used based on Wittes and Wallenstein (1987), but it is functionally equivalent to the calculation by Woolson, Bean, and Rojas (1986).

The Research Question

Starting with the example first used by Woolson, Bean, and Rojas (1986) and then revisited by Nam (1992), we will explore the usage of the central function of the samplesizeCMH package, the power.cmh.test() function.

This was a case–control study which looked at whether there was an association between colon cancer and chlorinated drinking water among males in Iowa, stratified by age. There were four age categories, 20–54 years, 55–69 years, 70–79 years, and 80–84 years; therefore the number of levels of the stratifying variable J = 4. The relative proportion of each age group to the total sample is as follows: t₁ = 0.10, t₂ = 0.40, t₃ = 0.35, and t₄ = 0.15. The National Collaborative Bladder Cancer Study showed that exposure rate for controls for each age stratum, denoted by π_2j (where the subscript ‘2’ indicates second column proportions and j is the stratum), was π₂₁ = 0.75, π₂₂ = 0.70, π₂₃ = 0.65, and π₂₄ = 0.60. Level of significance was set at 0.05 and desired power was 90% with an effect size of 3. They matched equal number of cases and controls in each stratum, i.e., s₁ = s₂ = s₃ = s₄= 0.5.

Calculating Sample Size

library(samplesizeCMH)

sample_size_uncorrected <- power.cmh.test(
  p2 = c(0.75,0.70,0.65,0.60),
  theta = 3,
  power = 0.9,
  t = c(0.10,0.40,0.35,0.15),
  alternative = "greater",
  correct = FALSE
)

print(sample_size_uncorrected, detail = FALSE)

## Power and sample size calculation for the Cochran Mantel Haenszel test
## 
##                  N = 171
##        Effective N = 174
## Significance level = 0.05
##              Power = 0.9
##        Alternative = greater
## 
## CALL: 
## power.cmh.test(p2 = c(0.75, 0.7, 0.65, 0.6), theta = 3, power = 0.9, 
##     alternative = "greater", t = c(0.1, 0.4, 0.35, 0.15), correct = FALSE)

sample_size_corrected <- power.cmh.test(
  p2 = c(0.75,0.70,0.65,0.60),
  theta = 3,
  power = 0.9,
  t = c(0.10,0.40,0.35,0.15),
  alternative = "greater",
  correct = TRUE
)

print(sample_size_corrected, n.frac = TRUE)

## Power and sample size calculation for the Cochran Mantel Haenszel test
## 
##                  N = 191.538
##        Effective N = 196
## Significance level = 0.05
##              Power = 0.9
##        Alternative = greater
## 
## Number of subjects per each group:
## ______________________________________________
## Group   |       1        2        3        4 
## ==============================================
## Case    |    9.58    38.31    33.52    14.37 
## Control |    9.58    38.31    33.52    14.37 
## 
## CALL: 
## power.cmh.test(p2 = c(0.75, 0.7, 0.65, 0.6), theta = 3, power = 0.9, 
##     alternative = "greater", t = c(0.1, 0.4, 0.35, 0.15), correct = TRUE)

power.cmh.test(
  p2 = c(0.75,0.70,0.65,0.60),
  s = 1/3,
  theta = 3,
  power = 0.9,
  t = c(0.10,0.40,0.35,0.15),
  alternative = "greater",
  correct = TRUE
)

## Power and sample size calculation for the Cochran Mantel Haenszel test
## 
##                  N = 217
##        Effective N = 220
## Significance level = 0.05
##              Power = 0.9
##        Alternative = greater
## 
## Number of subjects per each group:
## ______________________________________
## Group   |     1      2      3      4 
## ======================================
## Case    |     8     29     26     11 
## Control |    15     58     51     22 
## 
## CALL: 
## power.cmh.test(p2 = c(0.75, 0.7, 0.65, 0.6), theta = 3, power = 0.9, 
##     alternative = "greater", s = 1/3, t = c(0.1, 0.4, 0.35, 0.15), 
##     correct = TRUE)

A Closer Look at a Partial Table

Let’s take a moment to look at the partial table of stratum j = 1. Given the information above, we can reconstruct this table to better understand the question at hand. We can also use this as an opportunity to use some of the peripheral functions included in the samplesizeCMH package.

The proportion of controls exposed to chlorinated drinking water was π₂₁ = 0.75. It follows then that the complementary column proportion is 0.25. The table below shows column percents.

Using the effect.size() function, we can use the expected exposure rate of control to estimate expected exposure rate of cases.

effect.size(0.75,3)

## [1] 0.9

We can now use that to fill in the remainder of our table.

To determine the odds of exposure for the two groups, we can use prop2odds().

# Control
prop2odds(0.75)

## [1] 3

# Case
prop2odds(0.9)

## [1] 9

We can also use either row or column proportions to calculate the odds ratio for a 2 × 2 table.

props2theta(0.90,0.75)

## [1] 3

We see that this brings us back to our specified effect size, theta = 3, which was used in the sample size calculaion.

There are several other peripheral functions in this package to interconvert between proportions, odds, relative risk, and odds ratios. Use help("odds.and.proportions", "samplesizeCMH") to find out more.