```
# Compare 2 proportions using the var stabilizing transformation
# arcsin(sqrt((x+3/8)/(n+3/4))) (Anscombe), which has variance
# 1/[4(n+.5)]
m1 <- 100; m2 <- 150
deaths1 <- 10; deaths2 <- 30
f <- function(events,n) asin(sqrt((events+3/8)/(n+3/4)))
stat <- f(deaths1,m1) - f(deaths2,m2)
var.stat <- function(m1, m2) 1/4/(m1+.5) + 1/4/(m2+.5)
cat("Test statistic:",format(stat)," s.d.:",
format(sqrt(var.stat(m1,m2))), "\n")
#Use unbiased prior with variance 1000 (almost flat)
b <- gbayes(0, 1000, m1, m2, stat, var.stat, 2*m1, 2*m2)
print(b)
plot(b)
#To get posterior Prob[parameter > w] use
# 1-pnorm(w, b$mean.post, sqrt(b$var.post))
#If g(effect, n1, n2) is the power function to
#detect an effect of 'effect' with samples size for groups 1 and 2
#of n1,n2, estimate the expected power by getting 1000 random
#draws from the posterior distribution, computing power for
#each value of the population effect, and averaging the 1000 powers
#This code assumes that g will accept vector-valued 'effect'
#For the 2-sample proportion problem just addressed, 'effect'
#could be taken approximately as the change in the arcsin of
#the square root of the probability of the event
g <- function(effect, n1, n2, alpha=.05) {
sd <- sqrt(var.stat(n1,n2))
z <- qnorm(1 - alpha/2)
effect <- abs(effect)
1 - pnorm(z - effect/sd) + pnorm(-z - effect/sd)
}
effects <- rnorm(1000, b$mean.post, sqrt(b$var.post))
powers <- g(effects, 500, 500)
hist(powers, nclass=35, xlab='Power')
describe(powers)
# gbayes2 examples
# First consider a study with a binary response where the
# sample size is n1=500 in the new treatment arm and n2=300
# in the control arm. The parameter of interest is the
# treated:control log odds ratio, which has variance
# 1/[n1 p1 (1-p1)] + 1/[n2 p2 (1-p2)]. This is not
# really constant so we average the variance over plausible
# values of the probabilities of response p1 and p2. We
# think that these are between .4 and .6 and we take a
# further short cut
v <- function(n1, n2, p1, p2) 1/(n1*p1*(1-p1)) + 1/(n2*p2*(1-p2))
n1 <- 500; n2 <- 300
ps <- seq(.4, .6, length=100)
vguess <- quantile(v(n1, n2, ps, ps), .75)
vguess
# 75%
# 0.02183459
# The minimally interesting treatment effect is an odds ratio
# of 1.1. The prior distribution on the log odds ratio is
# a 50:50 mixture of a vague Gaussian (mean 0, sd 100) and
# an informative prior from a previous study (mean 1, sd 1)
prior <- function(delta)
0.5*dnorm(delta, 0, 100)+0.5*dnorm(delta, 1, 1)
deltas <- seq(-5, 5, length=150)
plot(deltas, prior(deltas), type='l')
# Now compute the power, averaged over this prior
gbayes2(sqrt(vguess), prior, log(1.1))
# [1] 0.6133338
# See how much power is lost by ignoring the previous
# study completely
gbayes2(sqrt(vguess), function(delta)dnorm(delta, 0, 100), log(1.1))
# [1] 0.4984588
# What happens to the power if we really don't believe the treatment
# is very effective? Let's use a prior distribution for the log
# odds ratio that is uniform between log(1.2) and log(1.3).
# Also check the power against a true null hypothesis
prior2 <- function(delta) dunif(delta, log(1.2), log(1.3))
gbayes2(sqrt(vguess), prior2, log(1.1))
# [1] 0.1385113
gbayes2(sqrt(vguess), prior2, 0)
# [1] 0.3264065
# Compare this with the power of a two-sample binomial test to
# detect an odds ratio of 1.25
bpower(.5, odds.ratio=1.25, n1=500, n2=300)
# Power
# 0.3307486
# For the original prior, consider a new study with equal
# sample sizes n in the two arms. Solve for n to get a
# power of 0.9. For the variance of the log odds ratio
# assume a common p in the center of a range of suspected
# probabilities of response, 0.3. For this example we
# use a zero null value and the uniform prior above
v <- function(n) 2/(n*.3*.7)
pow <- function(n) gbayes2(sqrt(v(n)), prior2)
uniroot(function(n) pow(n)-0.9, c(50,10000))$root
# [1] 2119.675
# Check this value
pow(2119.675)
# [1] 0.9
# Get the posterior density when there is a mixture of two priors,
# with mixing probability 0.5. The first prior is almost
# non-informative (normal with mean 0 and variance 10000) and the
# second has mean 2 and variance 0.3. The test statistic has a value
# of 3 with variance 0.4.
f <- gbayesMixPost(3, 4, mix=0.5, d0=0, v0=10000, d1=2, v1=0.3)
args(f)
# Plot this density
delta <- seq(-2, 6, length=150)
plot(delta, f(delta), type='l')
# Add to the plot the posterior density that used only
# the almost non-informative prior
lines(delta, f(delta, mix=1), lty=2)
# The same but for an observed statistic of zero
lines(delta, f(delta, mix=1, x=0), lty=3)
# Derive the CDF instead of the density
g <- gbayesMixPost(3, 4, mix=0.5, d0=0, v0=10000, d1=2, v1=0.3,
what='cdf')
# Had mix=0 or 1, gbayes1PowerNP could have been used instead
# of gbayesMixPowerNP below
# Compute the power to detect an effect of delta=1 if the variance
# of the test statistic is 0.2
gbayesMixPowerNP(g, 1, 0.2, interval=c(-10,12))
# Do the same thing by simulation
gbayesMixPowerNP(g, 1, 0.2, interval=c(-10,12), nsim=20000)
# Compute by what factor the sample size needs to be larger
# (the variance needs to be smaller) so that the power is 0.9
ratios <- seq(1, 4, length=50)
pow <- single(50)
for(i in 1:50)
pow[i] <- gbayesMixPowerNP(g, 1, 0.2/ratios[i], interval=c(-10,12))[2]
# Solve for ratio using reverse linear interpolation
approx(pow, ratios, xout=0.9)$y
# Check this by computing power
gbayesMixPowerNP(g, 1, 0.2/2.1, interval=c(-10,12))
# So the study will have to be 2.1 times as large as earlier thought
```

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