Pfthb

I have a question about repeated sampling. Let's say I am interested in the distribution of sample means. So what I would do is generate 10000 times a sample of size 1000 and look at the mean of each sample. Can I instead just take one sample of size 10000*1000 and then look at the mean of the first 1000 elements than from 1001 to 2000 and so on?

If you're controlling for the seed, both approaches should yield identical outcomes:

set.seed(1)
mean(sample(1:9, 3))
#[1] 5.666667
mean(sample(1:9, 3))
#[1] 4
mean(sample(1:9, 3))
# [1] 5.333333

set.seed(1)
x <- sample(1:9)
mean(x[1:3])
#[1] 5.666667
mean(x[4:6])
#[1] 4
mean(x[7:9])
# [1] 5.333333

Here is an example that generates 10,000 sample means of 1,000 items drawn randomly from a uniform distribution. Based on the Central Limit Theorem, we expect these means to be normally distributed with a mean of 0.5.

# set seed to make reproducible 
set.seed(95014)
# generate 10,000 means of 1,000 items pulled from a uniform distribution 
mean_x <- NULL
for (i in 1:10000)
 mean_x <- c(mean_x,mean(runif(1000))) 

hist(mean_x)

...and the output:

@ Len Greski
I can also do it that way right?

a <- runif(10000000)
j <- 1
x <- NULL
while (j <= 10000000)
 x <- c(x,mean(a[j:(j+999)]))
 j <- j + 1000

x
hist(x) 

I would say yes. In taking 10,000,000 samples you've randomly sampled most of the experimental space. If you set.seed the same for both the approaches you mention you get the exact same answer. If you change the seed and run a t-test, the results are not significantly different.

#First Method
seed <- 5554
set.seed(seed)
group_of_means_1 <- replicate(n=10000, expr = mean(rnorm(1000)))
set.seed(seed)
mean_of_means_1 <- mean(replicate(n=10000, expr = mean(rnorm(1000))))

#Method you propose
set.seed(5554)
big_sample <- data.frame(
 group=rep(1:10000, each=1000),
 samples=rnorm(10000 * 1000, 0, 1)
)

group_means_2 <- aggregate(samples ~ group,
 FUN = mean, 
 data=big_sample)

mean_of_means_2 <- mean(group_means_2$samples)

#comparison
mean_of_means_1 == mean_of_means_2

t.test(group_of_means_1, group_means_2$samples)