CochranArmitageTest: Cochran-Armitage Test for Trend

Description Usage Arguments Value Author(s) References See Also Examples

View source: R/Tests.r

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

See Also

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 Sept. 13, 2017, 9:04 a.m.