Type of Data: Discrete, categorical (counts or frequencies)
Use: testing counts, proportions, and ratios; plotting barplots and histograms.
In this section, we are concerned with comparing counts, proportions, and ratios.
In all the examples below, we are presented with the counts. But, how do we arrive at these values from our data?
For small data sets, it is straightforward to keep a running total of the counts in each group.
For larger data, or for data where we have only the raw data, we will need to extract this summary information ourselves.
R has several functions available for this purpose.
Consider a sample factor:
x <- c(rep('A', times = 10), rep('B', times = 20), rep('C', times = 15)), and a simple integer vector
y <- 1:length(x).
length() function returns the number of elements in a vector.```
We can calculate the total number of elements in x:
Or, subsets of x:
length(x[x == 'A'])
Note that without subsetting, the length of x remains at the total length, because the logical statement returns a vector of equal length to x, of TRUE or FALSE, depending how each element of x satisfies that condition.
length(x == 'A')
We can also calculate the length of one column that satisfies logical conditions of any number of other columns.
length(x[y < 10])
length() returns just one value.
We can use
table() to count up all instances of each level in a factor.
x A B C 10 20 15
Or, the parts of x that match a logical condition (compare this to the similar use of
table(x == 'A')
FALSE TRUE 35 10
Similarly, for a numeric vector.
table(y > 20)
FALSE TRUE 20 25
If we subset, we can again return a single value.
table(x[x == 'A'])
We can also create multi-dimensional tables.
table(x, y > 15)
x FALSE TRUE A 10 0 B 5 15 C 0 15
hist() is used to create histograms.
However, we can also use it to create a table of the number of counts in each of the bins (what
hist() and also
Let’s generate a vector of 45 numbers from a Normal distribution:
z <- rnorm(n= 45).
The usual use of
hist() is to make a plot. However, we can change a default argument to
plot = FALSE.
hist(z, plot= FALSE)
$breaks  -2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 $counts  2 1 5 8 5 10 7 3 3 1 $density  0.08888889 0.04444444 0.22222222 0.35555556 0.22222222 0.44444444  0.31111111 0.13333333 0.13333333 0.04444444 $mids  -2.25 -1.75 -1.25 -0.75 -0.25 0.25 0.75 1.25 1.75 2.25 $xname  "z" $equidist  TRUE attr(,"class")  "histogram"
Above, you can see all the parts that are usually unseen in a call to
$counts section contains the number of elements in each bin. These are used to plot the histogram.
We can access them directly.
hist(z, plot= FALSE)$counts
 2 1 5 8 5 10 7 3 3 1
We can modify the default
breaks = argument to either set the number of bins …
hist(z, plot= FALSE, breaks = 5)$counts
 2 6 13 17 6 1
… or the specific break points.
hist(z, plot= FALSE, breaks = c(-3,0, 3))$counts
 21 24
In contrast to
table(), hist requires a numeric vector.
Error in hist.default(x) : 'x' must be numeric
|No. Variables||No. levels||Test|
The binomial test is a test of the statistical significance of deviations from a theoretically expected distribution of observations into two categories: e.g., does our count of heads:tails differ from a 1:1 ratio, when tossing a coin?
For example, if we toss a coin 10 times and get 8 heads, is it likely that this coin is biased?
We can specify x the number of failures (or successes) out of n number of trials.
binom.test(x = 8, n = 10)
## Exact binomial test ## ## data: 8 and 10 ## number of successes = 8, number of trials = 10, p-value = 0.1094 ## alternative hypothesis: true probability of success is not equal to 0.5 ## 95 percent confidence interval: ## 0.4439045 0.9747893 ## sample estimates: ## probability of success ## 0.8
With such a small sample, it’s hard to tell. What if we tossed the coin 10 more times?
## Exact binomial test ## ## data: 16 and 20 ## number of successes = 16, number of trials = 20, p-value = 0.01182 ## alternative hypothesis: true probability of success is not equal to 0.5 ## 95 percent confidence interval: ## 0.563386 0.942666 ## sample estimates: ## probability of success ## 0.8
For an alpha of 0.05, it suggests that this coin, or the way it is being flipped, needs some attention!
We can also use this test to ask other questions about the probabilty of two outcomes (including counts of things that do not at first seem to be counts of only two things). For example, we can ask if a six-sided die is biased to throw more 6’s than we would expect.
If fair, the dice would roll a 6 one in six times (1/6). If we rolled 51 6’s in a total of 235 rolls, what can we say?
To fit this into a binomial test, we do the following.
binom.test(x = 51, n = 235, p = 1/6, alternative = 'greater')
## Exact binomial test ## ## data: 51 and 235 ## number of successes = 51, number of trials = 235, p-value = ## 0.02654 ## alternative hypothesis: true probability of success is greater than 0.1666667 ## 95 percent confidence interval: ## 0.1735253 1.0000000 ## sample estimates: ## probability of success ## 0.2170213
The number of successes (
x) is 51. The number of trials (
n) is 235. The probability (null hypothesis) we expect (
p) is 1/6. And the alternative hypothesis is that we roll 51 or more 6’s (
alternative = 'greater')—this is a one-tailed test.
We would use a two-tailed test if we wanted to know whether the die rolled too many or too few 6’s (
alternative = 'two.sided').
We can also use this same test to compare outcomes that we can see but cannot (or do not) measure, when it is called the sign test. The sign test is essentially a special case of the binomial test where the probability of success under the null hypothesis is p = 0.5.
It is often used to compare paired samples before and after some intervention, for example we could ask whether people felt better or worse after taking some cold medicine.
Say that 8 of 9 people felt better after taking our new medicine.
binom.test(x = 8, n = 9)
## Exact binomial test ## ## data: 8 and 9 ## number of successes = 8, number of trials = 9, p-value = 0.03906 ## alternative hypothesis: true probability of success is not equal to 0.5 ## 95 percent confidence interval: ## 0.5175035 0.9971909 ## sample estimates: ## probability of success ## 0.8888889
The p of 0.04 suggests that this medicine works.
We can use the
chisq.test() function to perform a goodness-of-fit test (testing whether the observed data differs from a theoretical distribution).
If we supply
chisq.test() with a single vector (i.e., a one-dimensional contingency table) of non-negative integers, the hypothesis tested is whether the population probabilities equal those in
p, or are all equal if
p is not given. (You must provide a probabilty for each element of
p must sum to 1).
Is 50:50 different from a 1:1 ratio?
# These two give the same answer chisq.test(x = c(50, 50))
## ## Chi-squared test for given probabilities ## ## data: c(50, 50) ## X-squared = 0, df = 1, p-value = 1
# But here we explicitly state the probability chisq.test(x = c(50, 50), p = c(0.5, 0.5) )
## ## Chi-squared test for given probabilities ## ## data: c(50, 50) ## X-squared = 0, df = 1, p-value = 1
No, 50:50 is not different to 1:1.
Is 50:50 different from a 3:1 ratio?
chisq.test(x = c(50,50), p = c(3/4, 1/4))
## ## Chi-squared test for given probabilities ## ## data: c(50, 50) ## X-squared = 33.333, df = 1, p-value = 7.764e-09
We can compare if the counts of things in different groups is the same between those groups.
For example, do males and females like spinach equally?
prop.test() function requires x a vector of the number of successes in each group,
n a vector of the number of trials in each group.
x could also be a matrix or table of 2 columns with successes and failures (in which case you don’t need
For example, here we measure 4/40 males who like spinach and 196/3270 females who like spinach.
prop.test(c(4, 196), c(40, 3270))
## Warning in prop.test(c(4, 196), c(40, 3270)): Chi-squared approximation may ## be incorrect ## ## 2-sample test for equality of proportions with continuity ## correction ## ## data: c(4, 196) out of c(40, 3270) ## X-squared = 0.52289, df = 1, p-value = 0.4696 ## alternative hypothesis: two.sided ## 95 percent confidence interval: ## -0.06591631 0.14603864 ## sample estimates: ## prop 1 prop 2 ## 0.10000000 0.05993884
So, no. There is no difference in the proportion of males and females who like spinach.
Rather than test whether all groups are equal, we can also specify
p to say what probabilities we expect each group to have.
We have used
chisq.test() to test the goodness-of-fit of a one-dimensional contingency table. If we have a table of at least two rows and two columns, the function interprets this as a two-dimensional contingency table: A Pearson’s chi-squared test is performed of the null hypothesis that the joint distribution of the cell counts in a 2-dimensional contingency table is the product of the row and column marginals.
The data we have are our observed (O) data. The expected (E) data are calculated from the contingency table for each cell using the formula (Row Total x Column Total) / sum of all rows and columns.
The chi-square value is then the sum of [ (O - E)^2 / E ] for each cell.
For example, we might ask whether people who prefer gnocchi to pasta are the same as those who prefer grass to peas
count <- matrix(c(38, 14, 11, 51), nrow = 2, dimnames = list(c('gnocchi', 'pasta'), c('grass', 'peas'))) count
## grass peas ## gnocchi 38 11 ## pasta 14 51
## ## Pearson's Chi-squared test with Yates' continuity correction ## ## data: count ## X-squared = 33.112, df = 1, p-value = 8.7e-09
The results suggest an association between those who like both grass and gnocchi, and those who like both peas and pasta.
What’s the difference between
For a 2x2 table, they are essentially the same. See here.
Fisher’s Exact Test is similarly used in the analysis of contingency tables. Fisher is said to have devised it after hearing that a colleague could tell whether the milk or tea was poured into the cup first.
To test this claim, she was given 8 cups of tea, in four of which milk was added first. The null hypothesis is that there is no association between the true order of pouring and the woman’s guess, the alternative that there is a positive association (that the odds ratio is greater than 1).
TeaTasting <- matrix(c(3, 1, 1, 3), nrow = 2, dimnames = list(Guess = c("Milk", "Tea"), Truth = c("Milk", "Tea"))) TeaTasting
## Truth ## Guess Milk Tea ## Milk 3 1 ## Tea 1 3
fisher.test(TeaTasting, alternative = "greater")
## ## Fisher's Exact Test for Count Data ## ## data: TeaTasting ## p-value = 0.2429 ## alternative hypothesis: true odds ratio is greater than 1 ## 95 percent confidence interval: ## 0.3135693 Inf ## sample estimates: ## odds ratio ## 6.408309
The p = 0.2429 tells us that the association could not be established.
Thompson, L. 2006. S-Plus (and R) *Manual to Accompany Agresti’s Categorical Data Analysis *(2002) 2nd edition.
Crawley, M. The R Book. Ch. 8 Classical Tests