Hi, I'm having some problems trying to figure out exactly what a question is asking, regarding how sampling distribution statistics relate to population statistics. I get that the means should be equivalent with sufficient sample size, and I understand that variance of the sample distribution is equal to population variance divided by the sample size. So I thought I was doing fine with my homework, until I found the following question:
c) What is the mean of the sampling distribution of unbiased sample variance?
It feels like the question should be asking about either the mean of the sampling distribution OR the unbiased sample variance? Would the mean of the sampling distribution of unbiased sample variance be ... the mean?
I found some mention of mu[sub:1q491tpp]s[sup:1q491tpp]2[/sup:1q491tpp][/sub:1q491tpp], which would make sense as the mean of the sample variance, but I don't see any formulas for finding it. Since it's the mean, and we're to assume that the sample n = 36, and there are infinitely many samples, will it be equal to the S[sup:1q491tpp]2[/sup:1q491tpp] of the population?
If so, do I find it by Sum of Squares divided by degrees of freedom? If that's the case, is it just working backwards from the variance value that I have? I.E. N(sigma[sup:1q491tpp]2[/sup:1q491tpp])? But if that's the case, wouldn't it have to be the population N, rather than the sample size N?
Okay, I've now typed out WAY too long of a post, but thank you in advance for anyone who has time to read it and help me out!
c) What is the mean of the sampling distribution of unbiased sample variance?
It feels like the question should be asking about either the mean of the sampling distribution OR the unbiased sample variance? Would the mean of the sampling distribution of unbiased sample variance be ... the mean?
I found some mention of mu[sub:1q491tpp]s[sup:1q491tpp]2[/sup:1q491tpp][/sub:1q491tpp], which would make sense as the mean of the sample variance, but I don't see any formulas for finding it. Since it's the mean, and we're to assume that the sample n = 36, and there are infinitely many samples, will it be equal to the S[sup:1q491tpp]2[/sup:1q491tpp] of the population?
If so, do I find it by Sum of Squares divided by degrees of freedom? If that's the case, is it just working backwards from the variance value that I have? I.E. N(sigma[sup:1q491tpp]2[/sup:1q491tpp])? But if that's the case, wouldn't it have to be the population N, rather than the sample size N?
Okay, I've now typed out WAY too long of a post, but thank you in advance for anyone who has time to read it and help me out!