Solution to the “Trust in Clergy” Question

R code for the “trust in clergy” problem we discussed in class

#Trust levels in 2010 and 2015
p2010 <- .6
p2015 <- .56

#Sample size for the poll
n <- 1000

#standard erros for the poll in 2010 and 2015
SE2010 <- sqrt((p2010*(1-p2010))/n)
SE2015 <- sqrt((p2015*(1-p2015))/n)

#SE of the difference between polls
SEdiff <- sqrt(SE2010^2 + SE2015^2)

#Z-test
z_test <- (p2015 – p2010) / SEdiff

#P-value
pnorm(z_test)

#Box Model
pop <- 1000000 #number of tickets in the box

#writing 1’s and 0’s on the tickets
box <- c(rep(1, pop*p2010), rep(0, pop*(1-p2010)))

#running the model
diffs <- NA
for (i in 1:10000){
samp1 <- sample(x = box, size = n)
samp2 <- sample(x = box, size = n)
diffs[i] <- mean(samp1) – mean(samp2)
}

#Distribution of the differences
hist(diffs)
abline(v=-0.06, col = “red”)

length(diffs) #total number of differences we calculated

#number of times diff was more than what we observed
length(diffs[diffs >= (p2010 – p2015)])

#p-value value via box model
length(diffs[diffs >= (p2010 – p2015)]) / length(diffs)