2

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?

user1729
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Ishihara
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2 Answers2

1

Perhaps a two-way ANOVA? You have two treatments - sex and carcinoma.

Community
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hejseb
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0

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.

heropup
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