Compute confidence limits for a difference of two independent proportions.

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Description

The usual formula for the c.i. of at difference of proportions is inaccurate. Newcombe has compared 11 methods and method 10 in his paper looks like a winner. It is implemented here.

Usage

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ci.pd(aa, bb=NULL, cc=NULL, dd=NULL,
     method = "Nc",
      alpha = 0.05, conf.level=0.95,
     digits = 3,
      print = TRUE,
detail.labs = FALSE )

Arguments

aa

Numeric vector of successes in sample 1. Can also be a matrix or array (see details).

bb

Successes in sample 2.

cc

Failures in sample 1.

dd

Failures in sample 2.

method

Method to use for calculation of confidence interval, see "Details".

alpha

Significance level

conf.level

Confidence level

print

Should an account of the two by two table be printed.

digits

How many digits should the result be rounded to if printed.

detail.labs

Should the computing of probability differences be reported in the labels.

Details

Implements method 10 from Newcombe(1998) (method="Nc") or from Agresti & Caffo(2000) (method="AC").

aa, bb, cc and dd can be vectors. If aa is a matrix, the elements [1:2,1:2] are used, with successes aa[,1:2]. If aa is a three-way table or array, the elements aa[1:2,1:2,] are used.

Value

A matrix with three columns: probability difference, lower and upper limit. The number of rows equals the length of the vectors aa, bb, cc and dd or, if aa is a 3-way matrix, dim(aa)[3].

Author(s)

Bendix Carstensen, Esa Laara. http://BendixCarstensen.com

References

RG Newcombe: Interval estimation for the difference between independent proportions. Comparison of eleven methods. Statistics in Medicine, 17, pp. 873-890, 1998.

A Agresti & B Caffo: Simple and effective confidence intervals for proportions and differences of proportions result from adding two successes and two failures. The American Statistician, 54(4), pp. 280-288, 2000.

See Also

twoby2, binom.test

Examples

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( a <- matrix( sample( 10:40, 4 ), 2, 2 ) )
ci.pd( a )
twoby2( t(a) )
prop.test( t(a) )
( A <- array( sample( 10:40, 20 ), dim=c(2,2,5) ) )
ci.pd( A )
ci.pd( A, detail.labs=TRUE, digits=3 )

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