# Look at how the required sample size of the one-sample
# proportion test with a two-sided alternative and Type I error
# set to 5\% increases with increasing power:
seq(0.5, 0.9, by = 0.1)
#[1] 0.5 0.6 0.7 0.8 0.9
propTestN(p.or.p1 = 0.7, p0.or.p2 = 0.5,
power = seq(0.5, 0.9, by = 0.1))
#[1] 25 31 38 47 62
#----------
# Repeat the last example, but compute the sample size based on
# the exact test instead of the approximation. Note that because
# we require the actual Type I error (alpha) to be within
# 10\% of the supplied value of alpha (which is 0.05 by default),
# due to the discrete nature of the exact binomial test
# we end up with more power then we specified.
n.list <- propTestN(p.or.p1 = 0.7, p0.or.p2 = 0.5,
power = seq(0.5, 0.9, by = 0.1), approx = FALSE)
lapply(n.list, round, 3)
#$n
#[1] 37 37 44 51 65
#
#$power
#[1] 0.698 0.698 0.778 0.836 0.910
#
#$alpha
#[1] 0.047 0.047 0.049 0.049 0.046
#
#$q.critical.lower
#[1] 12 12 15 18 24
#
#$q.critical.upper
#[1] 24 24 28 32 40
#----------
# Using the example above, see how the sample size changes
# if we allow the Type I error to deviate by more than 10 percent
# of the value of alpha (i.e., by more than 0.005).
n.list <- propTestN(p.or.p1 = 0.7, p0.or.p2 = 0.5,
power = seq(0.5, 0.9, by = 0.1), approx = FALSE, tol.alpha = 0.01)
lapply(n.list, round, 3)
#$n
#[1] 25 35 42 49 65
#
#$power
#[1] 0.512 0.652 0.743 0.810 0.910
#
#$alpha
#[1] 0.043 0.041 0.044 0.044 0.046
#
#$q.critical.lower
#[1] 7 11 14 17 24
#
#$q.critical.upper
#[1] 17 23 27 31 40
#----------
# Clean up
#---------
rm(n.list)
#==========
# Look at how the required sample size for the two-sample
# proportion test decreases with increasing difference between
# the two population proportions:
seq(0.4, 0.1, by = -0.1)
#[1] 0.4 0.3 0.2 0.1
propTestN(p.or.p1 = seq(0.4, 0.1, by = -0.1),
p0.or.p2 = 0.5, sample.type = "two")
#[1] 661 163 70 36
#Warning message:
#In propTestN(p.or.p1 = seq(0.4, 0.1, by = -0.1), p0.or.p2 = 0.5, :
# The computed sample sizes 'n1' and 'n2' are too small,
# relative to the given values of 'p1' and 'p2', for the normal
# approximation to work well for the following element indices:
# 4
#----------
# Look at how the required sample size for the two-sample
# proportion test decreases with increasing values of Type I error:
propTestN(p.or.p1 = 0.7, p0.or.p2 = 0.5,
sample.type = "two",
alpha = c(0.001, 0.01, 0.05, 0.1))
#[1] 299 221 163 137
#==========
# Modifying the example on pages 8-5 to 8-7 of USEPA (1989b),
# determine the required sample size to detect a difference in the
# proportion of detects of cadmium between the background and
# compliance wells. Set the complicance well to "group 1" and
# the backgound well to "group 2". Assume the true probability
# of a "detect" at the background well is 1/3, set the probability
# of a "detect" at the compliance well to 0.4 and 0.5, use a 5%
# significance level and 95% power, and use the upper
# one-sided alternative (probability of a "detect" at the compliance
# well is greater than the probability of a "detect" at the background
# well). (The original data are stored in EPA.89b.cadmium.df.)
#
# Note that the required sample size decreases from about
# 1160 at each well to about 200 at each well as the difference in
# proportions changes from (0.4 - 1/3) to (0.5 - 1/3), but both of
# these sample sizes are enormous compared to the number of samples
# usually collected in the field.
EPA.89b.cadmium.df
# Cadmium.orig Cadmium Censored Well.type
#1 0.1 0.100 FALSE Background
#2 0.12 0.120 FALSE Background
#3 BDL 0.000 TRUE Background
# ..........................................
#86 BDL 0.000 TRUE Compliance
#87 BDL 0.000 TRUE Compliance
#88 BDL 0.000 TRUE Compliance
p.hat.back <- with(EPA.89b.cadmium.df,
mean(!Censored[Well.type=="Background"]))
p.hat.back
#[1] 0.3333333
p.hat.comp <- with(EPA.89b.cadmium.df,
mean(!Censored[Well.type=="Compliance"]))
p.hat.comp
#[1] 0.375
n.back <- with(EPA.89b.cadmium.df,
sum(Well.type == "Background"))
n.back
#[1] 24
n.comp <- with(EPA.89b.cadmium.df,
sum(Well.type == "Compliance"))
n.comp
#[1] 64
propTestN(p.or.p1 = c(0.4, 0.50), p0.or.p2 = p.hat.back,
alt="greater", sample.type="two")
#[1] 1159 199
#----------
# Clean up
#---------
rm(p.hat.back, p.hat.comp, n.back, n.comp)Run the code above in your browser using DataLab