Correct equation for Breslow-Day statistic in homogeneity test of odds ratio

Question

In Statistical Methods of Cancer Research; Volume 1 - The analysis of case-control studies the authors Breslow and Day derive a statistic to test for the homogeneity of combining strata into an odds ratio (equation 4.30). Given the value of the statistic, the test determines if it is appropriate to combine strata together and compute a single odds ratio.

For example, if we have only one 2x2 contingency table:

_{(source: kean.edu)}

the odds ratio for getting a disease with a risk factor compared to not having the risk factor is:

$$ \psi = (A*D)/(B*C) $$

if we have multiple contingency tables (for example, we stratify by age group), we can use the Mantel-Haenzel estimate to compute the odds ratio across all $I$ strata:

$$ \psi_{mh} = \frac{\sum_{i=1}^{I}A_i D_i / N_i}{\sum_{i=1}^{I}B_i C_i / N_i} $$

For each contingency table we have $R1=A+B$, $R2=C+D$ and $C1=A+C$ so we can express the expected odds ratio for that table in terms of the totals:

$$ \psi_{mh} = \frac{A D}{B C} = \frac{A (R2-C1+A)}{(R1-A)(C1-A)} $$

which gives a quadratic equation for A. Let $a$ be the solution to this quadratic equation (only one root gives a reasonable answer).

Thus a reasonable test for the adequacy of the assumption of a common odds ratio is to sum up the squared deviation; of observed and fitted values, each standardized by its variance:

$$ \chi^2 = \sum_{i=1}^{I}\frac{(a_i - A_i)^{2}}{V_i} $$

where the variance is:

$$ V_i = \left( \frac{1}{A_i} + \frac{1}{B_i} + \frac{1}{C_i} + \frac{1}{D_i} \right)^{-1} $$

If the homogeneity assumption is valid, and the size of the sample is large relative to the number of strata, this statistic follows an approximate chi-square distribution on $I-1$ degrees of freedom and thus a p-value can be determined.

If instead we divide the $I$ strata into $H$ groups and we suspect the odds ratios are homogeneous within groups but not between them, Breslow and Day give an alternative statistic (equation 4.32):

$$ \chi^2 = \sum_{h=1}^{H}\frac{\left( \sum_i a_i - A_i \right)^{2}}{\sum_i V_i} $$

where the $i$ summations are over strata in the $h^{th}$ group with the statistic being chi-square with only $H-1$ degrees of freedom (I assume a different Mantel-Haenzel estimate is computed within each group).

My question is equation 4.32 does not seem right to me. If anything, I'd expect it to be of the form:

$$ \chi^2 = \sum_{h=1}^{H}\frac{ \sum_i \left(a_i - A_i\right)^{2} }{\sum_i V_i} $$

or:

$$ \chi^2 = \sum_{h=1}^{H}\sum_{i}\frac{(a_i - A_i)^{2}}{V_i} $$

with the latter equation approximating a chi-square distribution on $I-1$ degrees of freedom.

Which of these equations should I be using?

Answer 1

This is more directly and more accurately handled through the use of a binary logistic regression model with an interaction term. The usually-best test is the likelihood ratio $\chi^2$ test from such a model. The regression context also allows one to test continuous variables, adjust for other variables, and a host of other extensions.

General comment: I think we spend too much time teaching special cases and would do well to use general tools so that we have more time to deal with complications such as missing data, high dimensionality, etc.

Answer 2

Answer is to use Tarone's Test for Heterogeneity see article by I-Ming Liu.

Liu states: Tarone noted that by replacing the MH estimator equation image by the conditional maximum likelihood estimator, the Breslow–Day test statistic becomes the conditional likelihood score test. Since the MH estimator is inefficient, Tarone and Breslow noted that the test statistic is stochastically larger than a χ2 random variable (see Chi-Square Distribution) under the homogeneity hypothesis. The correct form for the test was derived by Tarone.

So both Tarone and Breslow agree that Tarone's test should be used instead of the above statistic.

Answer 3

It seems like you have the formula for the Cochran–Mantel–Haenszel test wrong. (Disclaimer: I am not a specialist in these type of tests)

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Normal distribution approximation

Basic background. For a single 2x2 contingency table you have effectively one degree of freedom.

$$\begin{array}{cc|c} x_{11}-y & x_{12}+y & R_1 \\ x_{21}+y & x_{22}-y & R_2 \\ \hline C_1 & C_2 & N \\ \end{array}$$

and $$x_{ij} = \frac{R_iC_j}{N}$$

What these type chi-squared type of tests do is approximate the value of $y$ as a normal distributed variable (with a similar covariance as compared to when we consider the distribution as a multinomial distribution) and look at the distribution of the squared value (or a sum of squared values if there are multiple cells).

Cochran–Mantel–Haenszel test

Then for $I$ different tables with $I$ different $y_i$ and possibly different variances $V_i$, the test for the hypothesis $E[y_1] = E[y_2] = \dots = E[y_I] = 0$ (which corresponds to risk ratio's equal to 1) can be tested with the statistic

$$\chi^2_a = \frac{\left(\sum_{i=1}^I y_i\right)^2}{\sum_{i=1}^I V_i}$$

According to the null hypothesis the variable $\sum_{i=1}^I y_i$ is approximately normal distributed with zero mean and variance equal to $\sum_{i=1}^I V_i$.

So this is different from your statistic

$$\chi^2_b = \sum_{i=1}^I \frac{ y_i^2}{ V_i}$$

The difference is in the power.

• The $\chi^2_a$ is more sensitive to risk ratio's that differ from 1 and point in the same direction.
• The $\chi^2_b$ also picks up a difference when the the effects are in opposite directions.

Breslow-Day test

to be continued...

^{I haven't heard of this test before and a quick search shows that it is not widely know or at least not well described in simple sources.}

^{What is seems like is that the test is a sort of mix between $\chi^2_a$ and $\chi^2_b$ and looking for the variability in the risk ratio between different groups assuming that the risk ratio is the same. In the expression $$\chi^2 = \sum_{h=1}^{H}\frac{\left( \sum_i a_i - A_i \right)^{2}}{\sum_i V_i}$$ the term $\frac{\left( \sum_i a_i - A_i \right)^{2}}{\sum_i V_i}$ seems a lot like a CMH test statistic like the $\chi^2_a$ computed above. But now it is done for different groups $\sum_{h=1}^{H}$ like the statistic $\chi^2_b$. Another difference is probably also that the $A_i$ relates to the average risk ratio rather than the assumption that the ratio should be equal to 1.}