Note: This document was converted to R-Markdown from this page by M. Drew LaMar.

Download the R code on this page as a single file here

New methods

Hover over a function argument for a short description of its meaning. The variable names are plucked from the examples further below.

Take a random sample of a categorical variable having two possible outcomes from a theoretical population of known probabilities:

sample(c(“L”, “R”), size = 18, prob = c(0.5, 0.5), replace = TRUE)

Create a loop to repeatedly sample a known population and generate an approximate null distribution for a test statistic:

for(i in 1:1000) {
     temporarySample = sample(c(“L”, “R”), size = 18, prob = c(0.5, 0.5), replace = TRUE)
     sum(temporarySample == “R”)
}

Example 6.2. The right hand of toad

Obtain the null distribution for the test statistic (the number of right-handed toads out of 18).

Take a very large number of samples of 18 toads from a theoretical population in which left- and right- handed individuals occur with equal probability (as stipulated by the null hypothesis). For each sample, the number of right-handed toads is calculated and saved in a vector named results18 (the samples themselves are not saved). The results vector is initialized before the loop. The term results18[i] refers to the i-th element of results18, where i is a counter. This many iterations might take a few minutes to run on your computer.

results18 <- vector()
for(i in 1:10000) {
  tempSample <- sample(c("L", "R"), size = 18, prob = c(0.5, 0.5), replace = TRUE)
    results18[i] <- sum(tempSample == "R")
}

Save results18 also as a factor with 19 levels, representing each possible integer outcome between 0 and 18. This step is for tables and graphs only – it makes sure that all 19 categories are included in tables and graphs, even categories with no occurrences.

factor18 <- factor(results18, levels = 0:18)

Tabulate the relative frequency distribution of outcomes in the random sampling process (Table 6.2-1). This is the null distribution. Your results will be similar but not identical to the numbers in Table 6.2-1 because 10,000 times is not large enough for extreme accuracy. The table command calculates the frequencies, and dividing by the length of results18 (i.e., the number of elements in the vector, equal to the number of iterations in the loop) yields the proportions (relative frequencies).

nullTable <- table(factor18, dnn = "nRightToads")/length(factor18)
data.frame(nullTable)
##    nRightToads   Freq
## 1            0 0.0000
## 2            1 0.0000
## 3            2 0.0008
## 4            3 0.0040
## 5            4 0.0131
## 6            5 0.0285
## 7            6 0.0748
## 8            7 0.1229
## 9            8 0.1675
## 10           9 0.1839
## 11          10 0.1648
## 12          11 0.1186
## 13          12 0.0706
## 14          13 0.0348
## 15          14 0.0124
## 16          15 0.0029
## 17          16 0.0004
## 18          17 0.0000
## 19          18 0.0000

Draw a histogram of the null distribution: the number of right-handed toads out of 18. Setting freq = FALSE causes hist to graph density of observations rather than frequency. In this case, height of bars indicates relative frequency if bars are one unit wide (as in the present case).

hist(results18, right = FALSE, freq = FALSE, xlab = "Number of right-handed toads")

We can make a histogram using barplot with the height of the bars as proportions, as given in nullTable.

barplot(height = nullTable, space = 0, las = 1, cex.names = 0.8, col = "white",
    xlab = "Number of right-handed toads", ylab = "Relative frequency")

Calculate the fraction of samples in the null distribution having 14 or more right-handed toads. This won’t be exact because 10,000 times is not large enough for extreme accuracy.

frac14orMore <- sum(results18 >= 14)/length(results18)
frac14orMore
## [1] 0.0157

Calculate the approximate \(P\)-value. This won’t be exact because 10,000 times is not large enough for extreme accuracy.

2 * frac14orMore
## [1] 0.0314

Example 6.4. The genetics of mirror-image flowers

Determine the null distribution for the test statistic, the number of left-handed flowers out of 27.

Obtain a very large number of samples of 27 individuals from a theoretical population in which L and R occur in the ratio 1:3, as specified by the null hypothesis. For each sample, the number of “L” is counted and saved in a vector named results27 (the samples themselves are not saved).

results27 <- vector(mode = "numeric")
for(i in 1:10000) {
    tempSample <- sample(c("L", "R"), size = 27, prob = c(0.25, 0.75), replace = TRUE)
    results27[i] <- sum(tempSample == "L")
}

Save the results also as a factor. This is to ensure that all categories between 0 and 27 are included in tables and graphs, even those that did not occur in the sampling process.

results27fac <- factor(results27, levels = 0:27)

Tabulate the relative frequency distribution of outcomes. To get relative frequencies, use table to get frequencies and then divide by the total number of replicates. Your results will be similar but not identical to the numbers in Table 6.2-1 because 10,000 iterations is not large enough for extreme accuracy.

nullTable <- table(results27fac, dnn = "nRightFlowers")/length(results27fac)
data.frame(nullTable)
##    nRightFlowers   Freq
## 1              0 0.0004
## 2              1 0.0033
## 3              2 0.0191
## 4              3 0.0446
## 5              4 0.0938
## 6              5 0.1413
## 7              6 0.1690
## 8              7 0.1722
## 9              8 0.1417
## 10             9 0.0970
## 11            10 0.0628
## 12            11 0.0319
## 13            12 0.0160
## 14            13 0.0050
## 15            14 0.0011
## 16            15 0.0006
## 17            16 0.0002
## 18            17 0.0000
## 19            18 0.0000
## 20            19 0.0000
## 21            20 0.0000
## 22            21 0.0000
## 23            22 0.0000
## 24            23 0.0000
## 25            24 0.0000
## 26            25 0.0000
## 27            26 0.0000
## 28            27 0.0000

Draw a histogram of the null distribution for the the number of left-handed flowers out of 27. Setting freq = FALSE causes hist to graph density of observations rather than frequency. In this case, height of bars indicates relative frequency if bars are one unit wide (as in the present case).

hist(results27, right = FALSE, freq = FALSE, xlab = "Number of left-handed flowers")

Commands for a fancier histogram:

barplot(height = nullTable, space = 0, las = 1, cex.names = 0.8, col = "white",
    xlab = "Number of left-handed flowers", ylab = "Relative frequency")

Calculate the fraction of results in the null distribution having 6 or fewer left-handed flowers.

frac6orLess <- sum(results27 <= 6)/length(results27)
frac6orLess
## [1] 0.4715

Calculate the \(P\)-value. This won’t be exact because 10,000 times is not large enough for extreme accuracy

2 * frac6orLess
## [1] 0.943