# How do I test difference in Percentages?

I have some survey data and know the number and percentage of "yes" responses. How do I determine if any of the percentage yes regional responses are statistically different from the national yes response percentage.

Use a hypergeometric test to see whether each region's proportion is significantly greater or significantly less than the national proportion.

The hypergeometric test treats the population as a bag with 21568 stones, 10820 of which are white. Each region is then treated as a random sample from that bag. For example the North East is like grabbing 1919 of those stones and getting 1032 white ones. You can calculate how unlikely that sample is in R:

> phyper(1032-1, 10820, 21568-10820, 1919, lower.tail = F)
[1] 0.0004970996


The 0.0004970996 says there's there's a very small chance that randomly drawing from the national votes 1919 times will give you 1032 yeses. I.e. this tests whether the proportion of North East yeses is significantly different than the population average, and it is (p = 0.0004).

However, you're also interested in whether the regional proportion is significantly less than the national average! To calculate that you can use the lower.tail=T option:

> phyper(1031, 10820, 10748, 1919, lower.tail = T)
[1] 0.9995029


So the North East's proportion looks significantly greater than the national average (p = 0.0004) and not significantly less (p = 0.9995).

To answer your question you could repeat this process of testing both tails for each region. Since this means conducting 26 hypothesis tests, you could expect that one or two might be significant just by random chance if you're using a threshold like p<0.05. For that reason I'd finish the analysis off with a multiple comparison correction.

You can apply z test (or t test) of difference in proportions of responses for national and regional area respectively Compute difference in two proportions and then compute standard error of difference in prortions.Divide the difference with S.E. You will get z-statistic. Check table for its significance at 5% or 1% etc.

note that percentage be converted in propotion form eg .60 for 60% before computations.

• This answer is incorrect because it does not account for two important things: (1) the regional and national results will be correlated because the former is a part of the latter; and (2) because multiple comparisons using the same data will be made, some adjustment for that is essential. An ANOVA, using either fixed or random effects, would be the first solution anyone ought to consider, and any reasonable solution will be a version of that (such as a logistic GLM). – whuber May 31 '16 at 16:11