Cochran-Armitage Test

Objective

The Cochran-Armitage test is used to test whether there is a linear trend when the response is binary. This test is used with data in the form of a contingency table, such as that described in Independence Testing, where there are only two data rows and we suspect (or hope) that the columns are ordered.

For example, in clinical trials, we might be interested in the relationship between increased dosage and the effect of a drug or treatment under study. The effect of the drug or treatment is binary (e.g. cured not cured) while the dosage levels are ordinal (e.g. 10 mg, 20 mg, 30 mg, etc.).

Example

Example 1. Determine whether there is a linear trend based on the dosages (0, 10, 20, or 30 mg) for the drug treatment illustrated in Figure 1. Here the first data row counts the number of patients who were not cured while the next row counts the number of patients who were cured.

Figure 1 – Drug Treatment data

Before presenting the Cochran-Armitage test results for Example 1, we first provide some theoretical background.

Theoretical Background

Let x_j be the (numeric) labels of the columns (j = 1, 2, …k), n_j = the number of elements in the j^th column, and y_j = the value in the cell in the first row and j^th column. Define p_j = y_j/n_j (the observed  response probability) and

The one-sided hypotheses are:

H₀: π₁ = π₂ = … = π_k

H₁: π₁ ≤ π₂ ≤ … ≤ π_k with at least one strict inequality

Here, π_j is the population probability that corresponds to the sample proportion p_j.

The null hypothesis for the two-sided test is as for the one-sided test and the alternative hypothesis is either π₁ ≤ π₂ ≤ … ≤ π_k or π₁ ≥ π₂ ≥ … ≥ π_k with at least one strict inequality.

The Cochran-Armitage test is

which is equivalent to

Here, c is an optional continuity correction factor. If the x_j are equally spaced then a suitable correction factor is c = (x_j+1–x_j) /2 for any j < k. If the x_j are not equally spaced then a possible correction is

although it is probably better not to use a correction factor at all in this case.

Test

For the one-sided test, the null hypothesis is rejected when z ≥ z_crit where z_crit = NORM.S.INV(1–α), the right side critical value.

For the two-sided test with no continuity correction, the null hypothesis is rejected when |z| ≥ NORM.S.INV(1–α/2). When the continuity correction factor is used then the null hypothesis is rejected when z⁺≥ z_crit+ or z^– ≤ z_crit-. Here, z⁺ is the z-statistic described above and z_crit+ = NORM.S.INV(1–α/2) is the right tail critical value. z^– is the z-statistic described above except that the correction factor is added instead of subtracted and z_crit- = NORM.S.INV(α/2) is the left tail critical value.

Example continued

The analysis for Example 1, where no correction factor is used, is shown in the upper part of Figure 2. Since we obtain a significant result (p-value = .0066 < .05 = alpha in the two-tailed test), we conclude that there is a linear trend.

Figure 2 – Cochran-Armitage Test

If a non-zero correction factor is used in cell I5, the calculations for the left-sided test are identical to those shown in Figure 2. In fact, if we use a correction factor of 5, based on =(C3-B3)/2, then z = -2.81367 and p-value = .002449. For a right-sided test, the formula in cell I6 changes to =I4+I5, which for a correction factor of 5, results in z = -2.61797 and p-value = .004423. The p-value for the two-sided test is then p-value = .002449 + .004423 = .006872.

Worksheet Function

Real Statistics Function: The following array function is provided in the Real Statistics Resource Pack to carry out the Cochran-Armitage test.

CATEST(R1, lab, tails): returns a 2 × 1 array consisting of the z-statistic and p-value for the test. R1 is a 2 × k contingency table with 3 rows, the first of which contains numeric labels in increasing order. tails = 1 (default) or 2. if lab = TRUE then an extra column is appended to the output with labels (default FALSE).

This function does not use a correction factor.

Note that the array formula =CATEST(B3:E5,TRUE,2) in range H14:I15 of Figure 2 yields the same test results as shown in the rest of the figure.

Examples Workbook

Click here to download the Excel workbook with the examples described on this webpage.

Reference

Hintze, J. L. (2008) Cochran-Armitage Test for Trend in Proportions. NCSS
https://www.ncss.com/wp-content/themes/ncss/pdf/Procedures/PASS/Cochran-Armitage_Test_for_Trend_in_Proportions.pdf