# Count data with largely varying sample sizes. Can I get anything from the data?

I only have categorical data, between two groups, with more than three categories per data set.
Here are some actual examples of the data: Comparing (https://www.strawpoll.me/4380945/r) to (https://www.imdb.com/poll/M43-BGuMniY/results?ref_=po_sr) or (https://strawpoll.de/ygx16fa) to (https://www.imdb.com/poll/T8dGOnA-1ck/results?ref_=po_sr). Alternatively here is a much shorter distribution I threw together: (I want to compare group 1 to group 2 in the example)

There are 22 of these surveys, half of them representing opinions on movies and half representing opinions on video games. But all of them vary between the samples and some of them have some very small counts at the very bottom of the distribution (some less than 5, but some larger than 1000). All I want to know is if the two groups have different preferences, or have different distributions. Is there any test I can run?

• Would a Yate's correction work for the data less than 5? Then could I report the discrepancy in sample size and say no strong inference should be made?
– user301453
Nov 13 '20 at 23:53
• Forget Yates' correction as a cure for your low expected counts. See my comment about combining the last three categories into one. // Doing the computations by hand or on a calculator may be a bit tedious, but maybe not a waste of time for a beginner. However, nothing is stopping you from emulating my R code right now. No need to know "all about" R before using needed parts. (The worst that can happen is that you crash R and have to restart your computer.) Nov 14 '20 at 0:52
• I'm trying R right now and I'm getting the hang of it. Thank you!
– user301453
Nov 14 '20 at 1:35
• Glad to hear that! Lots of online tutorials for various procedures in R. Nov 14 '20 at 1:37

In principle, it seems you'd want to use a chi-squared test to see if the two groups tend to have the same distribution of category counts.

In practice, sparse data as in the last few categories of your first dataset make it impossible to do a 'standard' chi-squared test. In particular, several expected cell counts are smaller than five. (Some authors are OK with a few counts as low as three as long as all the rest are higher than five---questionable for your first dataset.)

Fortunately, the implementation of chisq.test in R simulates reasonably accurate P-values for tests in many such problematic situations. The simulation is OK for the table as a whole, but if the null hypothesis of homogeneity is rejected, any ad hoc tests trying to identify specifically which categories differ must be limited to categories with higher numbers of expected counts.

Here is output from chisq.test for your first dataset:

x1 = c(45, 16, 9, 7, 5, 3, 1, 0)
x2 = c(23, 75, 145, 85, 23, 13, 9, 5)
TBL = rbind(x1, x2);  TBL
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
x1   45   16    9    7    5    3    1    0
x2   23   75  145   85   23   13    9    5
chi.out = chisq.test(TBL, sim=T)
chi.out

Pearson's Chi-squared test
with simulated p-value
(based on 2000 replicates)

data:  TBL
X-squared = 127.6, df = NA, p-value = 0.0004998


The simulated P-value is much smaller the 0.05, so there are highly significant differences among categories for the two groups.

The chi-squared statistic $$Q$$ is composed of 16 components as follows:

$$Q = \sum_{i=1}^2\sum_{j=1}^8 \frac{(X_{ij} - E_{ij})^2}{E_{ij}} = 127.6,$$

where the $$X_{ij}$$ are observed counts from the contingency table.

chi.out$obs [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] x1 45 16 9 7 5 3 1 0 x2 23 75 145 85 23 13 9 5  Also, the expected counts, based on the null hypothesis, are computed in terms of row and column totals from the contingency table, approximately as follows: round(chi.out$exp, 2)
[,1]  [,2]   [,3]  [,4]  [,5]  [,6] [,7] [,8]
x1 12.6 16.87  28.54 17.05  5.19  2.97 1.85 0.93
x2 55.4 74.13 125.46 74.95 22.81 13.03 8.15 4.07


Because of the low expected counts in the last two categories, the chi-squared statistic does not necessarily have (even approximately) the distribution $$\mathsf{Chisq}(\nu = (r-1)(c-1) = 7).$$ This is the reason for we needed to simulate the P-value of this test. [A traditional (pre-simulation) approach might be to combine the last three categories into one.]

The Pearson residuals are of the form $$R_{ij}=\frac{X_{ij}-E_{ij}}{\sqrt{E_{ij}}}.$$ That is, $$Q = \sum_{i,j}R_{ij}^2.$$ By looking among the $$R_{ij}$$ with largest absolute values, one can get an idea which categories made the most important contributions to a $$Q$$ large enough to be significant:

round(chi.out\$res, 2)
[,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]  [,8]
x1  9.13 -0.21 -3.66 -2.43 -0.08  0.02 -0.63 -0.96
x2 -4.35  0.10  1.74  1.16  0.04 -0.01  0.30  0.46


So it seems that comparisons involving categories A, C, and E may be most likely to show significance. (A superficial look at the original contingency table shows that these categories have large and discordant counts.)

In order to avoid false discovery from multiple tests on the same data, you should use some method of of choosing significance levels smaller than 5% for such comparisons. (One possibility is Bonferroni's method; perhaps using 1% instead of 5% levels.)

Addendum: Comparison of Cat A with sum of C&D. Output from Minitab.

This is one possible ad hoc test. It uses a simple $$2 \times 2$$ table that you should be able to compute by hand. You can check your expected values in the output below.

Data Display

Row  Cat  Gp1  Gp2
1  A     45   23
2  C&D   16  130

Chi-Square Test for Association: Cat, Group

Rows: Cat   Columns: Group

Gp1     Gp2  All

A         45      23   68
19.38   48.62

C&D       16     130  146
41.62  104.38

All       61     153  214

Cell Contents:      Count
Expected count

Pearson Chi-Square = 69.408, DF = 1, P-Value = 0.000


Very small P-value suggests that Gp 1 prefers Cat A while Gp 2 prefers Cats B & C.

• I'll be honest, I have just learned about the chi-square test. I'll be learning R in a couple of months but this is honestly beyond me right now. I guess I'll stick to descriptives. Thank you anyways! I appreciate the help!
– user301453
Nov 14 '20 at 0:44
• See Addendum with ad hoc test on part of first dataset, as suggested in Answer. Output from Minitab. Enjoy. Nov 14 '20 at 1:20