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My data look like this:

UnigueID  Region   Sex   etc.
4567      4        M
3452      2        F
2316     12        F
2347      4        F
3987      7        M
9567      7        M

and so on for 15,000 some obs.

I have multiple regions, different numbers of obs in each region, and I want to know first and foremost are the gender proportions different? And then ideally which are different and by how much (confidence limits, etc). I thought to do an ANOVA test using proportion of females in each group as a gender mean but I wasn't getting an F-value presumably because there is no spread using that idea it is just a single number for each group, so there was no error. I'm not sure what I'm screwing up conceptually.

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  • $\begingroup$ Restated differently: Are you looking for % male for each region (I assume % female is 1 - % male), and then seeing which regions are statistically significantly different than the average and by how much? $\endgroup$ – ilanman Aug 1 '16 at 19:29
  • $\begingroup$ Yes, different from the average and different from each other. That is just right $\endgroup$ – jkh107 Aug 1 '16 at 21:01
  • $\begingroup$ re "... just a single number for each group." Could you clarify this? After all, even your small example shows two individuals in Region 4 and two in Region 7. What distinction are you making, then, between "Region" and "group"? $\endgroup$ – whuber Aug 4 '16 at 18:14
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An ANOVA should work here (assuming regions are independent and variances are equal) You can calculate:

  • Group means - proportion of females per region
  • Group standard deviations - similar as above
  • Grand mean - total proportion of females in the dataset

Using that, you should be able to find an F-statistic and p-value. Now, that will only tell you if each region has the same proportion of females.

If you reject the Null, then there is more work to do. You want to figure out which of the means is actually differen and by how much. There exist many approaches, most basic of them may be Tukey's HSD or Scheffe's method.

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