Jamie Leigh - 7 months ago 36

R Question

I am trying to do a permutation test for ANOVA on (y_{1},...,y_{N}) with group identifiers

`g`

`## data`

y <- c(6.59491, 6.564573, 6.696147, 6.321552, 6.588449, 6.853832,

6.370895, 6.441823, 6.227591, 6.675492, 6.255462, 6.919716, 6.837458,

6.41374, 6.543782, 6.562947, 6.570343, 6.993634, 6.666261, 7.082319,

7.210933, 6.547977, 6.330553, 6.309289, 6.913492, 6.597188, 6.247285,

6.644366, 6.534671, 6.885325, 6.577568, 6.499041, 6.827574, 6.198853,

6.965038, 6.58837, 6.498529, 6.449476, 6.544842, 6.496817, 6.499526,

6.709674, 6.946934, 6.23884, 6.517018, 6.206692, 6.491935, 6.039925,

6.166948, 6.160605, 6.428338, 6.564948, 6.446658, 6.566979, 7.17546,

6.45031, 6.612242, 6.559798, 6.568082, 6.44193, 6.295211, 6.446384,

6.658321, 6.369639, 6.066747, 6.345537, 6.727513, 6.677873, 6.889841,

6.724438, 6.379956, 6.380779, 6.50096, 6.676555, 6.463236, 6.239091,

6.797642, 6.608025)

## group

g <- structure(c(2L, 2L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 3L, 3L, 3L, 2L,

3L, 2L, 3L, 2L, 3L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 1L, 2L, 2L,

2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 3L,

3L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,

3L, 1L, 3L, 1L, 2L, 2L, 1L, 3L, 2L, 2L, 3L, 1L, 2L, 2L, 2L, 1L,

2L), .Label = c("B1", "B2", "B3"), class = "factor")

This is what I have right now, but it is not working when I change it to test for the sample means instead of the F statistic. I'm pretty sure that I need to change

`T.obs`

`T.perm`

`by(y, g, mean)`

`n <- length(y) #sample size n`

T.obs<- anova(lm(y ~ g))$F[1] #Observed statistic

n.perm <- 2000 # we will do 2000 permutations

T.perm <- rep(NA, n.perm) #A vector to save permutated statistic

for(i in 1:n.perm) {

y.perm <- sample(y, n, replace=F) #permute data

T.perm[i] <- anova(lm(y.perm ~ g))$F[1] #Permuted statistic

}

mean(T.perm >= T.obs) #p-value

Answer

I don't really know what you mean by "**it is not working**". As far as I can see, it works alright, except that it is slightly slow.

```
set.seed(0)
n <- length(y) #sample size n
T.obs <- anova(lm(y ~ g))$F[1] #Observed statistic
n.perm <- 2000 # we will do 2000 permutations
T.perm <- rep(NA, n.perm) #A vector to save permutated statistic
for(i in 1:n.perm) {
y.perm <- sample(y, n, replace=F) #permute data
T.perm[i] <- anova(lm(y.perm ~ g))$F[1] #Permuted statistic
}
mean(T.perm >= T.obs)
# [1] 0.4915
```

This is fairly close to the theoretical value

```
anova(lm(y ~ g))$Pr[1]
# [1] 0.4823429
```

So, yes, you are doing all correct!

From the first paragraph of your question it sounds like we want to compute F-statistic ourselves, so the following function does this. There is a switch `"use_lm"`

. If set `TRUE`

, it uses `anova(lm(y ~ g))`

as what is done in your original code. **This function aims to make computation of F-statistic and p-value transparent**. Also, manual computation is 15 times faster than calling `lm`

and `anova`

(which is an obvious thing...).

```
fstat <- function (y, g, use_lm = FALSE) {
if (!use_lm) {
## group mean (like we are fitting a linear model A: `y ~ g`)
mu_g <- ave(y, g, FUN = mean)
## overall mean (like we are fitting a linear model B: `y ~ 1`)
mu <- mean(y)
## RSS (residual sum of squares) for model A
RSS_A <- drop(crossprod(y - mu_g))
## RSS (residual sum of squares) for model B
RSS_B <- drop(crossprod(y - mu))
## increase of RSS from model A to model B
RSS_inc <- RSS_B - RSS_A
## note, according to "partition of squares", we can also compute `RSS_inc` as
## RSS_inc <- drop(crossprod(mu_g - mu))
## `sigma2` (estimated residual variance) of model A
sigma2 <- RSS_A / (length(y) - nlevels(g))
## F-statistic
fstatistic <- ( RSS_inc / (nlevels(g) - 1) ) / sigma2
## p-value
pval <- pf(fstatistic, nlevels(g) - 1, length(y) - nlevels(g), lower.tail = FALSE)
## retern
return(c(F = fstatistic, pval = pval))
}
else {
anovalm <- anova(lm(y ~ g))
return(c(F = anovalm$F[1L], pval = anovalm$Pr[1L]))
}
}
```

Let's first have a check on the validity of this function:

```
F_obs <- fstat(y, g)
# F pval
#0.7362340 0.4823429
F_obs <- fstat(y, g, TRUE)
# F pval
#0.7362340 0.4823429
```

Don't be surprised that it is insignificant. Your data does not really suggest significant group difference. Look at the boxplot:

```
boxplot(y ~ g) ## or use "factor" method of `plot` function: `plot(g, y)`
```

Now we move on to the permutation. We write another function `perm`

for this purpose. It is actually pretty easy, because we have a nicely defined `fstat`

. All we need to do is to use `replicate`

to wrap up `sample`

+ `fstat`

.

`lm`

is actually very slow:

```
library(microbenchmark)
microbenchmark(fstat(y, g), fstat(y, g, TRUE), times = 200)
#Unit: microseconds
# expr min lq mean median uq max neval cld
# fstat(y, g) 228.44 235.32 272.1204 275.34 290.20 388.84 200 a
# fstat(y, g, TRUE) 4090.00 4136.72 4424.0470 4181.02 4450.12 16460.72 200 b
```

so we write this function using `f(..., use_lm = FALSE)`

:

```
perm <- function (y, g, n) replicate(n, fstat(sample(y), g)[[1L]])
```

Now let's run it with `n = 2000`

(setting random seed for reproducibility):

```
set.seed(0)
F_perm <- perm(y, g, 2000)
## estimated p-value based on permutation
mean(F_perm > F_obs[[1L]])
# [1] 0.4915
```

Note how close it is to the theoretical p-value:

```
F_obs[[2L]]
# [1] 0.4823429
```

As you can see, the result agrees with your original code.

Source (Stackoverflow)