I have asked people which food they prefer:
choice group apple orange pizza beer A 374 63 216 101 B 510 65 125 76
Apparently group B prefers fruit and group A prefers pizza and beer, and a chi-square test shows that the overall differences between groups are significant. But how can I test for which individual choice there is a significant difference between groups?
For example, I want to know whether there is a significant difference in the preference for oranges. But I cannot, I believe, just subset the orange choices, because that way I wouldn't consider the total number of participants per group. I mean, a difference between 1 from A and 2 from B will be significant if I have only sampled three people, but not if those are three in a million.
Participants were asked to choose one from the four foods. They could not select multiple answers.
How can I test this?
My hunch would be to either add up the non-orange answers and test the resulting 2×2 table with a chi-square test:
choice group orange not orange A 63 691 B 65 711 orange <- matrix(c(63, 691, 65, 711), 2, 2, TRUE, list(group = c("A", "B"), choice = c("orange", "not orange")) ) chisq.test(orange, correct = FALSE) # p = .9883
or to calculate the percentage of orange answers in each group, consider the two numbers as counts in a binomial distribution and test that with a binomial test:
a <- 63 / (63 + 691) b <- 65 / (65 + 711) all <- 63 + 691 + 65 + 711 binom.test(c(round(a * all / (a + b)), round(b * all / (a + b)))) # p = .9796 # just checkin' all == sum(c(round(a * all / (a + b)), round(b * all / (a + b))))  TRUE
Or is there a better, maybe more common way?
food <- c("apple", "orange", "pizza", "beer") dat <- data.frame( group = rep(c("A", "B"), c(754, 776)), choice = c( rep(food, c(374, 63, 216, 101)), rep(food, c(510, 65, 125, 76)) ) ) tab <- table(dat)
Explanation of second procedure
We want to compare the orange answers between groups. But if we only look at the orange answers themselves, we disregard the fact that other answers could be given. So instead of comparing the absolute numbers of orange answers, what we do is weigh the absolute number of orange answers by their proportion within all the answers in each group. Or in other words, we test if there is a significant difference between the percentages of orange answers in both groups.
Given this contingency table:
choice group orange not orange A 63 691 B 65 711
for group A, the percentage of orange answers is:
a <- 63 / (63 + 691) # 0.08355438 * 100 = 8.36%
and for group B it is:
b <- 65 / (65 + 711) # 0.08376289 * 100 = 8.38%
We can already tell that the difference in percentages is minimal, but this is only an example, so let's continue.
To compare the percentages, we are going to consider them as two categories (A and B) in a binomial distribution. For a binomial test, we need a vector of the same length as the overall number of answers. The overall number of answers in my study is:
all <- 63 + 691 + 65 + 711
To calculate the proportion of the binomial distribution that corresponds to the percentages of orange answers in each group, we simply "scale" (i.e. multiply by the same factor) both percentages to add up to 100% (of all observations); that is, we resolve the calculation:
a * x + b * x = all
The resolution, of course is:
x = all / (a + b)
Now we can calculate the number of observations for each category:
# for A: a * all / (a + b) # for B: b * all / (a + b)
Finally we round the possibly fractional numbers to integers and perform the binomial test:
binom.test(c(round(a * all / (a + b)), round(b * all / (a + b))))
number of successes = 764, number of trials = 1530, p-value = 0.9796 alternative hypothesis: true probability of success is not equal to 0.5 95 percent confidence interval: 0.4739876 0.5247077 sample estimates: probability of success 0.4993464