1. ## Okay simple question about the populationstandard deviation

I have the following: the expected value and the variance of a PDF.

I am running 3 simulations with n=10,n=20, and n=30.

I need the population standard deviation of the mean of a sample size n from your distribution

I thought it would be just the square root of the (variance/n), is that not correct? (so thus 3 different sd's.

Or is it simply the square root of the variance?

Further on I am asked to find the table of the proportion of means within one and two stadard deviations of E(X), and I use the following code:
mean(abs(xbar10-mu)<sqrt(var/n))

which would mean the sqrt(var/n) is the standard deviation, correct?

2. Originally Posted by xamius
I have the following: the expected value and the variance of a PDF.

I am running 3 simulations with n=10,n=20, and n=30.

I need the population standard deviation of the mean of a sample size n from your distribution

I thought it would be just the square root of the (variance/n), is that not correct? (so thus 3 different sd's.
Yes

Or is it simply the square root of the variance?

Further on I am asked to find the table of the proportion of means within one and two stadard deviations of E(X), and I use the following code:
mean(abs(xbar10-mu)<sqrt(var/n))
What is that mean?

You should be either using Chebyshev's inequality or treating the sample means as if they are normally distributed with the same mean as the population and standard deviations as above.

which would mean the sqrt(var/n) is the standard deviation, correct?
Now without seeing the actual wording of the question I am suspicious about using the population variance in this work rather than the unbiased estimate of the population variance from your samples, in which case you end up with a t-distribution rather than normal distribution at the end.

CB

3. The code was given to me by the professor (R code.)

Here is what I am doing: Using the quanilte function of a pdf to create 100,000 random draws of sample size 10 (and 20 and 30).
N=100,000
xbar10=rep(0,N)
n=10
for(i in 1:N){
xbar10[i]<-mean(myquant(runif(n)))
}

Which creates xbar10, a sample of means. I take the absolute value of each those means and subtract it fromt he real mean and find the mean of all of those. I then compare it to the standard deviation of the population. Does that make sense? Do you see something wrong with that?

4. Originally Posted by xamius
The code was given to me by the professor (R code.)

Here is what I am doing: Using the quanilte function of a pdf to create 100,000 random draws of sample size 10 (and 20 and 30).
N=100,000
xbar10=rep(0,N)
n=10
for(i in 1:N){
xbar10[i]<-mean(myquant(runif(n)))
}

Which creates xbar10, a sample of means. I take the absolute value of each those means and subtract it fromt he real mean and find the mean of all of those. I then compare it to the standard deviation of the population. Does that make sense? Do you see something wrong with that?
Actually I do, you should not be using your knowledge of the population mean and variance, but should be using your simulated data to obtain estimates.

Except the wording of the question you posted is vague and could mean a number of things.

CB

5. Originally Posted by CaptainBlack
Actually I do, you should not be using your knowledge of the population mean and variance, but should be using your simulated data to obtain estimates.

Except the wording of the question you posted is vague and could mean a number of things.

CB
The assignment is to compare the means of the simulated data and compare it with the variance of the population to see if we beat the Central limit therom. This PDF is really skewed, so with these small samples, I am beating the CLT. Does that make sense?