I have two samples from two different language corpora: Sample one contains 82 verbs, sample 2 contains 89 verbs. I want to compare the frequencies of a particular verb type, let's call them oral verbs, across both samples and see if they differ significantly from each other (I would have used another verb type in which I don't expect differences as a comparison group for a 4-cell chi square test). Originally, I wanted to do a chi square test but then realized that wouldn't be possible given the different sample sizes. Which test might I be able to apply? Thank you!
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2 Answers
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You can use a chi-squared test in your example with different sample sizes. Your "another verb type" would be verbs that are not oral verbs, i.e. all the other verbs
Suppose in your example, $10$ of the $82$ verbs in sample one were oral verbs and $72$ were not, while $20$ of the $89$ verbs in sample two were oral verbs and $69$ were not. Then the table for your four cell chi-squared test could look like
10 72 | 82
20 69 | 89
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30 141 | 171
and in R you might get
chisq.test(rbind(c(10, 72), c(20, 69)))
# Pearson's Chi-squared test with Yates' continuity correction
#
# data: rbind(c(10, 72), c(20, 69))
# X-squared = 2.4459, df = 1, p-value = 0.1178
so this example would not be statistically significant
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$\begingroup$ Thanks Henry, that was quick and super useful! $\endgroup$– SolesApr 26, 2019 at 8:34
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$\begingroup$ Thank you for this good answer. I am wondering why the Yates Continuity Correction is applied (I checked the docs and figured that in the 2x2 case it is the default? docs ). However I dont really understand why, I thought Yates Correction is necessary for a total N that is smaller than 40? $\endgroup$– BjörnOct 12, 2020 at 12:02
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$\begingroup$ @BjörnB - if a continuity correction is sensible for small samples then it is also sensible for large samples; it will then only make a small difference and so used to be ignored for hand calculations, but that is not a good reason now we use computers $\endgroup$– HenryOct 14, 2020 at 14:58
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Just in case anyone is looking for the Python version of this, you can use scipy ch2_contingency: https://docs.scipy.org/doc/scipy/reference/generated/scipy.stats.chi2_contingency.html
Using the same example as @Henry
import numpy as np
from scipy.stats import chi2_contingency
obs = np.array([[10, 72], [20, 69]])
chi2, p, dof, ex = chi2_contingency(obs)
print(chi2, dof, p)
> 2.44591778277931 1 0.11783094937852609
Which is the same result as R chisq.test
