Elevated percentage of standard BMI associated with Carcinoma Percentage of Standard BMI$\hspace{30mm}$Male$\hspace{40mm}$Female
$\hspace{70mm}$Case$\hspace{10mm}$Control$\hspace{20mm}$Case$\hspace{10mm}$Control
<130$\hspace{61mm}$123$\hspace{12mm}$150$\hspace{27mm}$55$\hspace{15mm}$59
$\ge$130$\hspace{61mm}$85$\hspace{13mm}$45$\hspace{29mm}$51$\hspace{15mm}$46
*Standard BMI: male, 22.1; female, 20.6. Percentage of standard BMI=(observed BMI/standard BMI)x100
Is elevated percentage of standard BMI associated with renal cell carcinoma after controlling the effects of sex?
Anyone know what kind of test to run on the data?
 A: Perhaps a two-way ANOVA? You have two treatments - sex and carcinoma.
A: This is a Mantel-Haenszel test, not two-way ANOVA, because you want to test the association after controlling for sex.
The null hypothesis is that there is no association after controlling for the stratified covariate:  $H_0 : OR_i = 1$, for all $i = 1, 2, \ldots, k$, where there are $k$ strata; versus $H_a : OR_i \ne 1$ for at least one such $i$.  Calculate for each stratum $$\begin{align*} {\rm E}[n_{11i}] &= \frac{n_{1*i} n_{*1i}}{n_{**i}}, \\ {\rm Var}[n_{11i}] &= \frac{n_{*1i} n_{*2i} n_{1*i} n_{2*i}}{n_{**i}^2 (n_{**i} - 1)}, \end{align*}$$ and then the test statistic is $$z^2 = \left( \sum_{i=1}^k n_{11i} - {\rm E}[n_{11i}] \right)^{\!2} \left/ \sum_{i=1}^k {\rm Var}[n_{11i}] \right. \sim \chi_1^2.$$  Note this applies to a $2 \times 2$ set of tables, without continuity correction.  Reject $H_0$ at the $100\alpha\%$ significance level if $z^2$ exceeds the upper $100\alpha$ percentile of the chi-squared distribution with 1 degree of freedom.
