# [R] Find if there is independence

Uwe Ligges ligges at statistik.uni-dortmund.de
Thu May 23 22:49:10 CEST 2002

```Ambrosini Alessandro wrote:
>
> Hello
> I have the matrix
>
> a<-matrix(c(2,1,0,1,2,2,1,5,7,2,5,12),nrow=6)
> a
>
>       [,1] [,2]
> [1,]    2    1
> [2,]    1    5
> [3,]    0    7
> [4,]    1    2
> [5,]    2    5
> [6,]    2   12
>
> Suppose that in the first row we have 3 men of England, 2 with hair, and 1
> no
> In the second we have 6 italian men, 1 with hair and 5 no ...
> I want to find if there is a dependence between men withouth hair and
> nationality.
> By the way, which is the simplest command to use to study the independence
> in a matrix like the one that I wrote?

Let me ask another question: What is the appropriate method to ...

Since the cells of your contingency table are not very well filled, I'd
suggest to use Fisher's (general) exact test, if you want to test on
independency of rows and columns, fortunately you do not need to
calculate it manually, see ?fisher.test .

> Using
> > chisq.test(a)
>
>         Pearson's Chi-squared test with Yates' continuity correction
>
> data:  a
> X-squared = 0, df = 1, p-value = 1

No. In my version of R (1.5.0 on WinNT) it results plausibly in:
X-squared = 6.8155, df = 5, p-value = 0.2347

Are you sure you did the things mentioned above exactly?

> Warning message:
> Chi-squared approximation may be incorrect in: chisq.test(a)
>
> Why does it give me the warning message? What have to add into chisq.test(a)
> ?
> And in which value I can understand the level of the dependence?

What does "level of dependence" mean in your case?
Are you sure you want to quantify it, using such a small number of
observations and such vague defintions (hair / no hair). To you want to
quantify it for all countries or separately? But perhaps I misunderstand
...

Uwe Ligges
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```