Cochran-Armitage Test For Trend - Introduction

Introduction

The trend test is applied when the data take the form of a 2 × k contingency table. For example, if k = 3 we have

B = 1 B = 2 B = 3
A = 1 N11 N12 N13
A = 2 N21 N22 N23

This table can be completed with the marginal totals of the two variables

B = 1 B = 2 B = 3 Sum
A = 1 N11 N12 N13 R1
A = 2 N21 N22 N23 R2
Sum C1 C2 C3 N

where R1 = N11 + N12 + N13, and C1 = N11 + N21, etc.

The trend test statistic is

where the ti are weights, and the difference N1iR2 −N2iR1 can be seen as the difference between N1i and N2i after reweighting the rows to have the same total.

The hypothesis of no association (the null hypothesis) can be expressed as:

Assuming this holds, then, using iterated expectation,

The variance can be computed by decomposition, yielding

{\rm Var}(T) = \frac{R_1R_2}{N} \left(\sum_{i=1}^kt_i^2C_i(N-C_i) - 2\sum_{i=1}^{k-1}\sum_{j=i+1}^kt_it_jC_iC_j\right),

and as a large sample approximation,

The weights ti can be chosen such that the trend test becomes locally most powerful for detecting particular types of associations. For example, if k = 3 and we suspect that B = 1 and B = 2 have similar frequencies (within each row), but that B = 3 has a different frequency, then the weights t = (1,1,0) should be used. If we suspect a linear trend in the frequencies, then the weights t = (0,1,2) should be used. These weights are also often used when the frequencies are suspected to change monotonically with B, even if the trend is not necessarily linear.

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