# Proportion comparaison between 4 modalities of one factor in R (chi2?) [closed]

I would need help in order to perform a proportion comparison between 4 different modalities of one factor.

Here is an example.

I would like to compare the proportion of mother-to-child smoking between individuals from 4 different countries.

Contry   USA  INDIA  CHINA  ITALIA
Mother   100  2000   1600   200
Child    25   860    64     10
%        25%  43%    4%     5%


So what I would like to compare is then the percentage of smoking child between countries I firts tought to chi2 test of fit to a uniform distribution law such as:

           USA  INDIA  CHINA  ITALIA
observed   25%  43%    4%     5%
expected   25%  25%    25%    25%


But I know chi2 does work only on numbers and not on proportions, right ?

Does someone have an idea of a test to support that USA and INDIA have a significantly higher level of smoking childs regarding their mothers compared to CHINA and ITALIA compared to what is expected if it were homogeneous ?

Thanks a lot for you help and time

Note: keep in mind that it is just a toy data, meaningless without any biological sens, I just want to put a sample example where I would like to compare proportions.

• What do the numbers in your first table represent?
– whuber
Oct 12 at 16:48
• Number of people that smoke, in mother and their childs respectively. Oct 12 at 17:29
• I cannot believe that, because I know far more than 100 mothers in the US smoke. And because most mothers have multiple children, and numbers of children per mother vary among countries, what exactly does the "child" count mean? How to conduct your test depends on what your data mean and how you collected them, making clear descriptions essential.
– whuber
Oct 12 at 18:10
• Hello whuber, it is a toy data, I simply invented the data and the biological question so that I could then report it on a real question. The idea here is to give a simple example so that community members can help. Here is basic idea is to compae the proportion of smoking child between countried, that's all... Oct 13 at 5:57

A 4-sample proportion test. The null being prop.test can be used for testing the null that the proportions (probabilities of success) in several groups are the same, or that they equal certain given values.

> prop.test(c(25,860,64,10),c(100,2000,1600,200))

4-sample test for equality of proportions without continuity correction

data:  c(25, 860, 64, 10) out of c(100, 2000, 1600, 200)
X-squared = 772.76, df = 3, p-value < 2.2e-16
alternative hypothesis: two.sided
sample estimates:
prop 1 prop 2 prop 3 prop 4
0.25   0.43   0.04   0.05


resulting in a rejection of the null (unsurprisingly).

• thanks a lot for your help Oct 13 at 8:38
• As pointed out in comments to the previous version of this question, this answer is not necessarily correct, because it depends (among other things) on whether a unique child is associated with each mother or multiple children might have been observed; and it also might depend on how many children the mothers tend to have.
– whuber
Oct 13 at 13:42