CochranArmitageTest: Cochran-Armitage Test for Trend In DescTools: Tools for Descriptive Statistics

Description

Perform a Cochran Armitage test for trend in binomial proportions across the levels of a single variable. This test is appropriate only when one variable has two levels and the other variable is ordinal. The two-level variable represents the response, and the other represents an explanatory variable with ordered levels. The null hypothesis is the hypothesis of no trend, which means that the binomial proportion is the same for all levels of the explanatory variable.

Usage

 `1` ```CochranArmitageTest(x, alternative = c("two.sided", "increasing", "decreasing")) ```

Arguments

 `x` a frequency table or a matrix. `alternative` a character string specifying the alternative hypothesis, must be one of `"two.sided"` (default), `"increasing"` or `"decreasing"`. You can specify just the initial letter.

Value

A list of class `htest`, containing the following components:

 `statistic` the z-statistic of the test. `parameter` the dimension of the table. `p.value` the p-value for the test. `alternative` a character string describing the alternative hypothesis. `method` the character string “Cochran-Armitage test for trend”. `data.name` a character string giving the names of the data.

Author(s)

Andri Signorell <[email protected]> strongly based on code from Eric Lecoutre <[email protected]>
https://stat.ethz.ch/pipermail/r-help/2005-July/076371.html

References

Agresti, A. (2002) Categorical Data Analysis. John Wiley & Sons

`prop.trend.test`

Examples

 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20``` ```# http://www.lexjansen.com/pharmasug/2007/sp/sp05.pdf, pp. 4 dose <- matrix(c(10,9,10,7, 0,1,0,3), byrow=TRUE, nrow=2, dimnames=list(resp=0:1, dose=0:3)) Desc(dose) CochranArmitageTest(dose, "increasing") CochranArmitageTest(dose) CochranArmitageTest(dose, "decreasing") # not exactly the same as in package coin: # independence_test(tumor ~ dose, data = lungtumor, teststat = "quad") lungtumor <- data.frame(dose = rep(c(0, 1, 2), c(40, 50, 48)), tumor = c(rep(c(0, 1), c(38, 2)), rep(c(0, 1), c(43, 7)), rep(c(0, 1), c(33, 15)))) tab <- table(lungtumor\$dose, lungtumor\$tumor) CochranArmitageTest(tab) # but similar to prop.trend.test(tab[,1], apply(tab,1, sum)) ```

Example output

```------------------------------------------------------------------------------
dose (matrix)

Summary:
n: 40, rows: 2, columns: 4

Pearson's Chi-squared test:
X-squared = 6.6667, df = 3, p-value = 0.08332
Likelihood Ratio:
X-squared = 7.2877, df = 3, p-value = 0.06327
Mantel-Haenszel Chi-squared:
X-squared = 3.4667, df = 1, p-value = 0.06262

Warning message:
Exp. counts < 5: Chi-squared approx. may be incorrect!!

Phi-Coefficient        0.408
Contingency Coeff.     0.378
Cramer's V             0.408

dose
0      1      2      3    Sum
resp

0    freq       10      9     10      7     36
perc    25.0%  22.5%  25.0%  17.5%  90.0%
p.row   27.8%  25.0%  27.8%  19.4%      .
p.col  100.0%  90.0% 100.0%  70.0%      .

1    freq        0      1      0      3      4
perc     0.0%   2.5%   0.0%   7.5%  10.0%
p.row    0.0%  25.0%   0.0%  75.0%      .
p.col    0.0%  10.0%   0.0%  30.0%      .

Sum  freq       10     10     10     10     40
perc    25.0%  25.0%  25.0%  25.0% 100.0%
p.row       .      .      .      .      .
p.col       .      .      .      .      .

Cochran-Armitage test for trend

data:  dose
Z = -1.8856, dim = 4, p-value = 0.02967
alternative hypothesis: increasing

Cochran-Armitage test for trend

data:  dose
Z = -1.8856, dim = 4, p-value = 0.05935
alternative hypothesis: two.sided

Cochran-Armitage test for trend

data:  dose
Z = -1.8856, dim = 4, p-value = 0.9703
alternative hypothesis: decreasing

Cochran-Armitage test for trend

data:  tab
Z = -3.2735, dim = 3, p-value = 0.001062
alternative hypothesis: two.sided

Chi-squared Test for Trend in Proportions

data:  tab[, 1] out of apply(tab, 1, sum) ,
using scores: 1 2 3
X-squared = 10.716, df = 1, p-value = 0.001062
```

DescTools documentation built on March 19, 2018, 9:03 a.m.