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Standard Deviation for Sampling Distribution

  1. Sep 12, 2009 #1
    Okay... this is a really simple question but I am pratically going nuts over it I can't sleep, I've been looking at it for like 12 hours, googling and wiking over it I am going crazy...

    Okie...

    Suppose I have 4 numbers, 0, 2, 4 and 6 and I grab 2 random ones with replacement... so I have

    0 2
    0 4
    0 6
    2 4
    2 6
    4 6
    0 0
    2 2
    4 4
    6 6

    Okie 10 possible samples. I want to find the standard deviation of the sampling distribution X... but I am having trouble getting the population variance...

    I tried this... 1/20 (Sum of xi square - Sum of xi square/2)

    20 becoz' there is 20 numbers, 2 becoz for each set there 2. So I been getting 2...

    I also tried this... 1/2 (Sum of xi square - Sum of xi square/2) and got 20...

    I have the answer for the Sampling Distribution with is 1.581, so working backwards... 1.581square divide by 1/(2^2) I get 10... which means the population variance is 10....

    I am going nuts, can anyone help? I really need to sleep now.
     
    Last edited: Sep 12, 2009
  2. jcsd
  3. Sep 12, 2009 #2
    You need to define the problem better. There are n!/(n-r)!r! ways to choose r things out of n things without repetitions and without regard to order, so for choosing random pairs from the set {0,2,4,6} there are 24/4=6 ways.

    If you are choosing from all possible ordered pairs (a,b) from the set {0,2,4,6}, there are 16 such pairs.

    Assuming this is what you want and a uniform discrete PD, the probability of any pair is 1/16. The variance of the discrete uniform distribution is (n^2-1)/12 although this is not used very much since it's not a function of sampling variation with fixed n.

    Is there any reason why this probability distribution would not be uniform?
     
    Last edited: Sep 12, 2009
  4. Sep 12, 2009 #3
    Oops, my bad I meant with replacement, so why would there be 16 pairs? Could u list them out? Order of number does not matter... I think... and why is it a uniform discrete PD? and why does probability come into place?

    ...

    Oh I think I just found the answer, but I would like to clarify stuff. I think I got the formulas messed up.

    Here's what I did I took the Population Variance : 1/20 (Sum of (xi^2...+ zi^2) - Sum of (xi... + zi)^2/20). Basically, I took 20 because there are a total of 20 numbers in the 10 sets of 2. And I know the entire population so I did not use n-1. Which gives me 5.

    So for Sampling Distribution Variance is Population Variance/n, so n is 2 as each sample consist of 2 numbers. so 5/2 is the variance.

    I probably got confused because I didn't know what number is n... so I would like to clarify am I right to say :

    1) Sampling Distribution Variance is Population Variance/n where n = the sample size(The number of variables in 1 set of sample regardless how many samples you have.) For example, you have 3 samples, Sample 1 {A, B}, Sample 2 {B, C}, Sample 3 {C, D}. the sample size is 2 even though you have 3 sets and a total of 6 variables.

    2) As for the population variance n is the total number of variables you have regardless of Sample Size. For example, for Sample 1 {A, B}, Sample 2 {B, C}, Sample 3 {C, D}, n would be 6 as there is 6 variables, even though we have only 3 sets.

    3) Which means to the n for Population Variance and the Variance of Sampling Distribution is a different n? I got confused because the formula I have had the 2 n in the same formula and I though they have to be the same.

    Once this is clarify, I can move on. :) Am I right?
     
    Last edited: Sep 12, 2009
  5. Sep 13, 2009 #4
    I frankly don't understand what you're talking about. Where are you getting your numbers? If you are working from the set {0,2,4,6} there are:

    6 ways to choose pairs without duplication or specifying order.

    16 ways to choose pairs you allow duplication (0,0) and specify all ordered pairs (0,2),(2,0)...

    If you don't allow duplicates but specify ordered pairs, there are 12 ways.

    So you can have 6, 12 or 16 possibilities. One of these is your population n. Your sample n(s) is 2. You can't calculate a sample variance for a sample of n(s)=2. If you don't follow what I'm saying, I'll let someone else respond.
     
  6. Sep 14, 2009 #5
    Yes, you can also have 6 unordered pairs plus 4 duplicates, as you have, but the same question remains. What are trying to do? What distribution are you talking about? Your population is a population of pairs. How do you define the parameters of a population including unordered pairs? You can do this if you have a vector space of ordered pairs (or any tuples). Your sample space is {0,2,4,6}. Is each number equally likely to be chosen? If so, you have a uniform distribution. Are these numbers being treated as additive quantities or simply as objects like cards? If they are additive and you have a uniform distribution, you must calculate your sample space parameters based on a uniform distribution. (Look up discrete uniform distribution).

    EDIT: Since your sample space consists only of specific even numbers and zero, it seems you must consider the numbers as objects since the probability of odd numbers is zero. I don't know how this distribution lends itself to the usual calculation of statistical moments under a uniform distribution(UD). For example the mean of (0,2,4,6) under a UD is b-a/2. If a=0 and b=6 then the mean is 3. But 3 has probability zero, so one could argue that this statistic doesn't apply if the sample space is a set. On the other hand the mean of four numbers (-3,-2,2,3) is zero even though 0 is not among the numbers. This is because the calculation assumes the set of all integers (actually all rational numbers generally). However, if you define your sample space as a set {0,2,4,6}, you must assume that the numbers are objects, not additive quantities since additive quantities are only defined in terms of number systems where the existence of all numbers in that system is assumed.
     
    Last edited: Sep 14, 2009
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