So, I'm working with GWAS SNP data and want to perform several tests for association between genotype and phenotype. There are two phenotypes (case and control) and 2 or three genotypes. Most of them are Chi-squared tests with different contingency tables, $2 \times 2$ or $2 \times 3$, one of them is the Cochran-Armitage trend test (CATT)

Once I have constructed the contingency table, I can easily get a $p$-value using the Apache commons math library for the Chi-squared tests. No problem.

However, the explanation of the CATT on Wikipedia is not sufficient for me to implement it (my statistics knowledge is limited and I'm still learning).

Like in the example, I suspect a linear trend, so my weights are $t = (0,1,2)$, which make the formula for $T$ to: $$ T \equiv (N_{12}R_2 - N_{22}R_1) + 2(N_{13}R2 - N_{23}R1) $$ and the one for the variance $$ Var(T) = {{R_1 R_2} \over N} ( N(C_2+4C_3) - (C_2 - 2C_3)^2) $$

I checked how the program PLINK does it, since it's already implemented there, but it differs slightly from the above formulas. The C++ source code there would correspond to this: $$ T = {(N_{12}R_2 - N_{22}R_1) + 2(N_{13}R2 - N_{23}R1)\over N} $$ and $$ Var(T) = {{R_1 R_2} \over N} {( N(C_2+4C_3) - (C_2 - 2C_3)^2) \over N^2} $$

Then it does calculates a chi-square value like this $$ \chi^2_{T} = {T^2 \over Var(T)} $$ and calculates the $p$-value like for any other chi-squared value with $df = 1$

I don't need to understand the theory completely, as long as my program calculates correctly, but understanding it would give me additional confidence.

Is this correct or legitimite? Is this how I'll get the $p$-value?

  • 1
    $\begingroup$ Here are some references on linear trend test with application to genetic studies. $\endgroup$ – chl Jun 14 '12 at 13:01
  • $\begingroup$ Thanks, but I've seen this two links before and neither one tells my how to get the p-value afterwards. $\endgroup$ – Sentry Jun 14 '12 at 13:09

This is just a different definition of the statistic $T$. Call your statistic $T_1$ and the other $T_2$. Note the $T_2 = T_1/N$ and that is the reason that the variance of $T_2$ differs from $T_1$ by a factor of $1/N^2$. However you should note that the chi square stitistic is the same in either case. For $T_2$ there is a factor of $1/N^2$ in the numerator and denominator that cancels and does not appear in the formula using $T_1$. You use the same test statistic either way.

  • $\begingroup$ Sounds reasonably. But why does Wikipedia use the Normal distribution instead of the chi-squared one? $\endgroup$ – Sentry Jun 14 '12 at 13:07
  • $\begingroup$ Nevermind, I just got it. For $df = 1$, the Chi-squared distribution is the same as a single squared normal distributed variable, hence the equality. I really should read up more on statistics ;) $\endgroup$ – Sentry Jun 14 '12 at 13:24

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