R code for example in Chapter 7:
Analyzing proportions

Table 7.1-1 and Figure 7.1-1. Binomial distribution with n = 27 and p = 0.25

dbinom(6, size = 27, prob = 0.25)

## [1] 0.171883

xsuccesses <- 0:27
probx <- dbinom(xsuccesses, size = 27, prob = 0.25) 
probTable <- data.frame(xsuccesses, probx)
probTable

##    xsuccesses        probx
## 1           0 4.233057e-04
## 2           1 3.809751e-03
## 3           2 1.650892e-02
## 4           3 4.585812e-02
## 5           4 9.171623e-02
## 6           5 1.406316e-01
## 7           6 1.718830e-01
## 8           7 1.718830e-01
## 9           8 1.432358e-01
## 10          9 1.007956e-01
## 11         10 6.047736e-02
## 12         11 3.115500e-02
## 13         12 1.384667e-02
## 14         13 5.325641e-03
## 15         14 1.775214e-03
## 16         15 5.128395e-04
## 17         16 1.282099e-04
## 18         17 2.765311e-05
## 19         18 5.120947e-06
## 20         19 8.085705e-07
## 21         20 1.078094e-07
## 22         21 1.197882e-08
## 23         22 1.088984e-09
## 24         23 7.891188e-11
## 25         24 4.383993e-12
## 26         25 1.753597e-13
## 27         26 4.496403e-15
## 28         27 5.551115e-17

barplot(height = probx, names.arg = xsuccesses, space = 0, las = 1, 
    ylab = "Probability", xlab = "Number of left-handed flowers")

Figure 7.1-2. Sampling distribution of a binomial proportion

Compare sampling distributions for the proportion based on n = 10 and n = 100.

Take a large number of random samples of n = 10 from a population having probability of success p = 0.25. Convert to proportions by dividing by the sample size. Do the same for the larger sample size n = 100. The following commands use 10,000 random samples.

successes10 <- rbinom(10000, size = 10, prob = 0.25)
proportion10 <- successes10 / 10
successes100 <- rbinom(10000, size = 100, prob = 0.25)
proportion100 <- successes100 / 100

Plot and visually compare the sampling distributions of the proportions based on n = 10 and n = 100. The par(mfrow = c(2,1)) command sets up a graph window that will plot both graphs arranges in 2 rows and 1 column.

par(mfrow = c(2,1))
hist(proportion10, breaks = 10, right = FALSE, xlim = c(0,1),
    xlab = "Sample proportion")
hist(proportion100, breaks = 20, right = FALSE, xlim = c(0,1),
    xlab = "Sample proportion")

par(mfrow = c(1,1))

Commands for a fancier plot are here.

oldpar <- par(no.readonly = TRUE) # make backup of default graph settings
par(mfrow = c(2,1), oma = c(4, 0, 0, 0), mar = c(1, 6, 4, 1)) # adjust margins
saveHist10 <- hist(proportion10, breaks = 10, right = FALSE, plot = FALSE)
saveHist10$counts <- saveHist10$counts/sum(saveHist10$counts)
plot(saveHist10, col = "firebrick", las = 1, cex.lab = 1.2,
    ylim = c(0,0.3), xlim = c(0,1), ylab = "Relative frequency",
    xlab = "", main = "")
text(x = 1, y = 0.25, labels = "n = 10", adj = 1, cex = 1.1)
saveHist100 <- hist(proportion100, breaks = 40, right = FALSE, plot = FALSE)
saveHist100$counts <- saveHist100$counts/sum(saveHist100$counts)
plot(saveHist100, col = "firebrick", las = 1, cex.lab = 1.2, 
    ylim = c(0,0.1), xlim = c(0,1), ylab = "Relative frequency", 
    xlab = "", main = "")
text(x = 1, y = 0.08, labels = "n = 100", adj = 1, cex = 1.1)
mtext("Proportion of successes", side = 1, outer = TRUE, padj = 2)

par(oldpar) # Revert to backup graph settings

Example 7.3. Sex and the X chromosome

The binomial test, used to test whether spermatogenesis genes in the mouse genome occur with unusual frequency on the X chromosome.

Read and inspect the data. Each row in the data file represents a different spermatogenesis gene.

mouseGenes <- read.csv("../Data/chapter07/chap07e2SexAndX.csv")
head(mouseGenes)

##   chromosome onX
## 1          4  no
## 2          4  no
## 3          6  no
## 4          6  no
## 5          6  no
## 6          7  no

Tabulate the number of spermatogenesis genes on the X-chromosome and the number not on the X-chromosome.

table(mouseGenes$onX)

## 
##   no  yes 
##   15   10

Calculate the binomial probabilities of all possible outcomes under the null hypothesis (Table 7.2-1). Under the binomial distribution with n = 25 and p = 0.061, the number of successes can be any integer between 0 and 25.

xsuccesses <- 0:25
probx <- dbinom(xsuccesses, size = 25, prob = 0.061)
data.frame(xsuccesses, probx)

##    xsuccesses        probx
## 1           0 2.073193e-01
## 2           1 3.367007e-01
## 3           2 2.624760e-01
## 4           3 1.307255e-01
## 5           4 4.670757e-02
## 6           5 1.274386e-02
## 7           6 2.759585e-03
## 8           7 4.865905e-04
## 9           8 7.112305e-05
## 10          9 8.727323e-06
## 11         10 9.071211e-07
## 12         11 8.035781e-08
## 13         12 6.090306e-09
## 14         13 3.956429e-10
## 15         14 2.203032e-11
## 16         15 1.049510e-12
## 17         16 4.261188e-14
## 18         17 1.465509e-15
## 19         18 4.231266e-17
## 20         19 1.012696e-18
## 21         20 1.973624e-20
## 22         21 3.052667e-22
## 23         22 3.605629e-24
## 24         23 3.055193e-26
## 25         24 1.653947e-28
## 26         25 4.297797e-31

Use these probabilities to calculate the P-value corresponding to an observed 10 spermatogenesis genes on the X chromosome. Remember to multiply the probability of 10 or more successes by 2 for the two-tailed test result.

2 * sum(probx[xsuccesses >= 10])

## [1] 1.987976e-06

For a faster result, try R’s built-in binomial test. The resulting P-value is slightly different from our calculation. In the book, we get the two-tailed probability by multiplying the one-tailed probability by 2. As we say on page 188, computer programs may calculate the probability of extreme results at the “other” tail with a different method. The output of binom.test includes a confidence interval for the proportion using the Clopper-Pearson method, which is more conservative than the Agresti-Coull method.

binom.test(10, n = 25, p = 0.061)

## 
##  Exact binomial test
## 
## data:  10 and 25
## number of successes = 10, number of trials = 25, p-value =
## 9.94e-07
## alternative hypothesis: true probability of success is not equal to 0.061
## 95 percent confidence interval:
##  0.2112548 0.6133465
## sample estimates:
## probability of success 
##                    0.4

Example 7.2. Radiologists’ missing sons

Standard error and 95% confidence interval for a proportion using the Agresti-Coull method for the confidence interval.

Read and inspect the data.

radiologistKids <- read.csv("../Data/chapter07/chap07e3RadiologistOffspringSex.csv")
head(radiologistKids)

##   offspringSex
## 1         male
## 2         male
## 3         male
## 4         male
## 5         male
## 6         male

Frequency table of female and male offspring number.

table(radiologistKids$offspringSex)

## 
## female   male 
##     57     30

Calculate the estimated proportion of offspring that are male, and the total number of radiologists.

n <- sum(table(radiologistKids$offspringSex))
n

## [1] 87

pHat <- 30 / n
pHat

## [1] 0.3448276

Standard error of the sample proportion.

sqrt( (pHat * (1 - pHat))/n )

## [1] 0.0509588

Agresti-Coull 95% confidence interval for the population proportion.

pPrime <- (30 + 2)/(n + 4)
pPrime

## [1] 0.3516484

lower <- pPrime - 1.96 * sqrt( (pPrime * (1 - pPrime))/(n + 4) )
upper <- pPrime + 1.96 * sqrt( (pPrime * (1 - pPrime))/(n + 4) )
c(lower = lower, upper = upper)

##     lower     upper 
## 0.2535425 0.4497542

Agresti-Coull 95% confidence interval for the population proportion using the binom package. To use this package you will need to install it (this needs to be done only once per computer) and load it using the library command (this needs to be done once per R session). The confidence interval from the binom package will be very slightly different from the one you calculated above because the formula we use takes a slight shortcut.

install.packages(“binom”, dependencies = TRUE)

library(binom)
binom.confint(30, n = 87, method = "ac")

##          method  x  n      mean     lower     upper
## 1 agresti-coull 30 87 0.3448276 0.2532164 0.4495625