HNSKD - 3 months ago 9

R Question

Here is my contingency table:

`X`

# Yes No

# Pre 5 685

# Post 17 1351

Fisher Test

`fisher.test(X)`

# Fisher's Exact Test for Count Data

# data: X

# p-value = 0.3662

# alternative hypothesis: true odds ratio is not equal to 1

# 95 percent confidence interval:

# 0.1666371 1.6474344

# sample estimates:

# odds ratio

# 0.5802157

Calculated Odds Ratio

`P1<-5/(5+685)`

P2<-17/(17+1351)

(P1/(1-P1))/(P2/(1-P2))

# [1] 0.5800773

Why are the values different? How does fisher test function in R calculate the estimated odds ratio?

Answer

Looking at the code underlying `fisher.test`

, I see

```
ESTIMATE <- c(`odds ratio` = mle(x))
```

Immediately above this is

```
mle <- function(x) {
if (x == lo)
return(0)
if (x == hi)
return(Inf)
mu <- mnhyper(1)
if (mu > x)
uniroot(function(t) mnhyper(t) - x, c(0, 1))$root
else if (mu < x)
1/uniroot(function(t) mnhyper(1/t) - x, c(.Machine$double.eps,
1))$root
else 1
}
```

Without exploring all the details of the code above the `mle`

definition, it looks like `fisher.test`

is solving an equation for the odds based on theoretical assumptions defined in `mnhyper`

(another function defined in `fisher.test`

[1]) and not calculating it directly from the data. I suspect if I wanted to get a full answer, I would need to read the references in `?fisher.test`

[1] There are several functions in `fisher.test`

such as `dnhyper`

, `mnhyper`

, and `pnhyper`

which appear to be distribution functions for a non-central hypergeometric distribution.