I have several variables that consist of ordinal categories (i.e. no fruit consumption, 1 portion a day, 2 portions, etc.) and I would like to test whether they differ significantly with respect to a binary variable (i.e. stroke and no stroke). Can I use a chi-squared test for this or does it violate the assumptions?
Thank you.
Other methods, such as the one suggested in @Noah's comment, may be preferable. However, to answer your question directly, you could use a chi-squared test, provided that you have enough data for the expected counts to be large enough.
Suppose your counts were as follows for 100 Stroke patients and 1000 patients with no stroke.
Fruits/day 0 1 2 3 4+
Stroke 33 37 22 7 1
NonStr 181 347 305 130 37
Because the expected counts are mostly above 5 and
the remaining one is above 3, you could do a chi-squared
test to see whether the probabilities of various
amounts of fruit per day are the same for the two
groups of patients. (You'd still get a significant
result by combining categories 3 and 4+.) Output from R:
Pearson's Chi-squared test
data: DTA
X-squared = 17.307, df = 4, p-value = 0.001685
Warning message:
In chisq.test(DTA) : Chi-squared approximation
may be incorrect
The warning message is because of the one one expected count below 5.
Notice, however, that this test does not take into account that the Fruit variable is ordinal.
Two additional possibilities for such large numbers of subjects: Using actual counts, both a Welch two-sample t test and a two-sample Wilcoxon test give essentially 0 P-values. Strictly speaking, assumptions for neither test are met, but large sample sizes and tiny P-values show strong evidence of a difference---for my fake data.
Note: Fake data were generated in R as:
x1 = rbinom(100, 5, .2)
x2 = rbinom(1000, 5, .3)
Roughly, the model is that people have five opportunities a day to eat fruit, breakfast, lunch, dinner, and a couple of snacks. At any one opportunity, respective probabilities of doing so are $0.2$ and $0.3.$
