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For context: I should note in advance I am a relative beginner with this.

Data context

I have data on some 600 000 persons which includes a column of whether these persons took parental leave or not (coded simply as 1 - took parental leave, 0, took no parental leave). I also have a column coding each person as male or female. I want to know whether persons coded as female are more likely to take parental leave than persons coded as male.

So I made a 2x2 table (female/male; no parental leave/parental leave) and applied the chi-square test which is significant (as expected). The residuals + prop table show that indeed women are overrepresented in taking 'parental leave'. So far so good.

Problem statement

However, the effect size is relatively small (Cramer's V about 0,15). For a number of reasons this seems counterintituive - the difference between men and women in the 'parental leave = 1' group seems quite large. I googled/read a bit about effect size & unbalanced groups. In this case there is a large dataset, with a relatively small proportion of the 600 000 persons taking parental leave. Could this affect the effect size, if yes, is there any measure other than Cramer's V that should be used in this regard?

Note: I am not specifically looking for a large effect size, just wondering whether I am applying the right measure.

Own research I have read the post: Chi-square Test with High Sample Size and Unbalanced Data but it didn't quite answer my question (the issue seems similar though).

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  • $\begingroup$ Everything in this question is based on your perception that "the difference between men and women ... seems quite large." Please tell us what that difference is and why it seems large to you. $\endgroup$
    – whuber ♦
    Jan 21, 2022 at 20:25
  • $\begingroup$ As an alternative, odds ratio is easy to calculate for a 2 x 2 table and is easy to understand. But Cramer's V (or phi for the 2 x 2 case) is appropriate as well. Or as mentioned in an answer in the post you cited, simply looking at the proportions of observation in each cell in rows or columns. $\endgroup$
    – Sal Mangiafico
    Jan 21, 2022 at 20:40
  • $\begingroup$ Apologies for not responding sooner - I didn't see that I got comments. I think the difference is quite large as, relative the proportion of men/women in the overall population, women are about twice as likely to take up parental leave. I suppose I am wondering how the odds ratio stacks up to the effect size... $\endgroup$
    – Statsbeginner
    Feb 9, 2022 at 16:05
  • $\begingroup$ It sounds like you are describing the most useful effect size for your situation: "relative to the proportion of men/women in the overall population, women are about twice as likely to take up parental leave". If I understand what you are saying, this is the odds ratio. As an example of a 2 x 2 table with Cramer V = 0.15, and OR = 2, the following is code in R : Matrix = matrix(c(550, 1100, 250, 1000), nrow=2, byrow=TRUE); library(vcd); assocstats(Matrix); oddsratio(Matrix, log=FALSE) $\endgroup$
    – Sal Mangiafico
    Feb 10, 2022 at 0:22
  • $\begingroup$ Thanks for the response; I will just stick with the odds ratio then! $\endgroup$
    – Statsbeginner
    Feb 15, 2022 at 4:32

1 Answer 1

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Answer from comment thread:

It sounds like you are describing the most useful effect size for your situation:

relative to the proportion of men/women in the overall population, women are about twice as likely to take up parental leave.

If I understand what you are saying, this is the odds ratio.

For future readers, as an example of a 2 x 2 table with Cramer V = 0.15, and OR = 2, the following is code in R:

Matrix = matrix(c(550, 1100, 250, 1000), nrow=2, byrow=TRUE)

library(vcd)

assocstats(Matrix)

oddsratio(Matrix, log=FALSE)

   ###                     X^2 df   P(> X^2)
   ### Likelihood Ratio 64.712  1 8.8818e-16
   ### Pearson          63.294  1 1.7764e-15
   ### 
   ### Phi-Coefficient   : 0.148 
   ### Contingency Coeff.: 0.146 
   ### Cramer's V        : 0.148 
   ###
   ### odds ratio
   ###
   ### 2

OR = (550 / 1100) / (250 / 1000)

names(OR) = "Odds ratio"

OR

   ### Odds ratio 
   ###          2
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