calculating population variance from sampling distribution variance

[babucher]

Hi all,

Let's say that I want to estimate the population mean and variance of a large bag of popcorn (unpopped). I have a scale, but it's not sensitive enough to weigh individual popcorn kernels. What I propose is to weigh groups of 5 popcorn kernels at a time, then calculate the population mean and variance from this data. For example:

1. Weigh 20 sets of 5 popcorn kernels = 20 data points
2. Calculate the sum of the data points
3. Calculate the sample average by dividing the total sum by 100
4. Assume the sample average is an estimator of the population mean

(so far, so good I think)

5. Calculate the variance of the data points, giving the _sampling_ (not sample) distribution variance.
6. From the equation VAR_population = n * VAR_sampling_distribution, calculate VAR_population

This I think is correct, because #6 is simply a rearrangement of the typical Standard Error of the Mean equation.

My primary question is, in calculating the variance of the sampling distribution, do I divide the sum of squares by n or by (n-1)?

Thanks in advance.

[babucher]

Perhaps to make this more clear, since I've recently learned that "sampling distribution variance" might be confusing, I should note that by that phrase I mean the Variance of the Mean (the average of the samples).

So, using the equation VAR_population = n * VAR_mean
Calculate the variance of the population.

My primary question is, in calculating the variance of the sampling distribution, do I divide the sum of squares by n or by (n-1)?

I believe I've figured this out. I need to divide by (n-1) in order to calculate VAR_mean.

If anyone can, please point me to any references that discuss this, either in textbooks or journal articles.

Thanks

[zmogggggg]

For any random sample {X1, ..., Xn}, the sample variance is defined by S = sxx / (n - 1), where sxx = sum((Xi - Xbar)^2) = sum(Xi^2) - 1/n*[sum(Xi)^2], for i = 1, ..., n.

Keep up the good work

[babucher]

Thanks zmog.

Any references that discuss getting an estimate of the population standard deviation by generating the sampling distribution/distribution of the mean?

Brian

[zmogggggg]

If you wish to estimate the population variance, the sample variance, S^2, is a great unbiased estimate. Then of course sqrt(S^2) = S yields an estimator for the population standard deviation. However, this is a negatively biased estimator of sigma, although extremely commonly used. You can correct this, as Wikipedia demonstrates:

The good stuff

[osamahamza]

hello all
my question is not about how to calculate the sampling distribution of the variance.
i am asking you about how to evaluate the standard error of the sampling distribution of the varince.
does the bootsrab do that? or not?
i want to know the methods if there.
thank you all.

[osamahamza]

hello again
my next question is about what is the difference between the two folloing formulas of the sample varince
[s^2]=[sum(xi-m)^2/n-1]
and the formul
[s^2][sum(xi-m)^2/n]
i know the difference is the bias,but, how?

[osamahamza]

hello all
my question is not about how to calculate the sampling distribution of the variance.
i am asking you about how to evaluate the standard error of the sampling distribution of the varince.
does the bootsrab do that? or not?
i want to know the methods if there.
thank you all.

[Dragan]

hello all
my question is not about how to calculate the sampling distribution of the variance.
i am asking you about how to evaluate the standard error of the sampling distribution of the varince.
does the bootsrab do that? or not?
i want to know the methods if there.
thank you all.

Certainly, the bootstrap will produce bootstrap confidence intervals and an estimate of the standard error for the variance of a set of data. S-Plus will easily do this.

Now, what I don't understand is what you mean by evaluate the standard error of the sampling distribution....can you explain further.