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Expression for Variance

  1. May 5, 2015 #1

    CAH

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    • Member warned multiple times about not using the homework template
    See the photo of the question+ mark scheme!
    I don't understand a question on finding an expression for the variance of something...


    Attempt at solution: also I worked out c as (3/2) previously, which is correct
    Var(U) = (3/2)^2 x Var(Xbar) = 9/4n^2 x a^2/18

    I'll attach a photo of this too if it's easier to read, my problem is that I thought var(Xbar) = var(x/n) = 1/n^2 x var(x)

    ... But they have done var(Xbar) = 1/n x var(x) ...?
     

    Attached Files:

  2. jcsd
  3. May 5, 2015 #2

    statdad

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    Homework Helper

    Think about this:
    [tex]
    Var\left(\bar X\right) = Var\left(\frac 1 n \sum_{i=1}^n X_i\right) = \frac 1 {N^2} \sum_{i=1}^n Var(X_i)
    [/tex]

    What happens when you simplify the sum?
     
  4. May 5, 2015 #3

    CAH

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    Ah I see so it's just sigma^2/n, is this the same for all cases as in, calculating the variance of a sample and estimating a population variance?.. In other words, kinda, will it ever (a level standard) be 1/n^2 x var(x)

    Probably a stupid question, just checking
     
  5. May 6, 2015 #4

    statdad

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    The only time you would have [itex] n^2 [/itex] in the denominator is if you were (for some reason) considering mathematically a single observation [itex] X_1 [/itex] and calculate
    [tex]
    Var\left( \dfrac{X_1}{n}\right) = \dfrac{Var(X)}{n^2}
    [/tex]

    "is this the same for all cases as in, calculating the variance of a sample and estimating a population variance?"
    I'm not exactly sure what you mean by this, so if my response is off-target that's why.

    If you've talked about sampling distributions for the sample mean, the expression [itex] \frac{\sigma^2}n [/itex] is the population variance for that sampling distribution. It will never have denominator [itex] n^2 [/itex], since, as long as the distribution being sampled has a variance, the steps shown above apply.

    The (sample) variance of a sample is a different beast. Essentially
    * if the population variance is [itex] \sigma^2 [/itex], then the sample variance
    [tex]
    s^2 = \dfrac 1 {n-1} \sum_{i=1}^n \, \left(x_i - \bar x\right)^2
    [/tex]
    is an unbiased estimator of the population variance

    * If you refer to the variance of the sampling distribution of [itex] \bar x [/itex] - which is given above - then to estimate that you have two options
    a) If you have a single sample, use the sample variance [tex] s^2 [/tex] to estimate the sampling distribution's variance by calculating
    [tex]
    \dfrac{s^2}{n}
    [/tex]

    b) If you have a large number of samples, all the same sample size, from the same population, then calculate each sample mean and treat those sample means as a new sample. The sample variance of those (call it [itex] s_{\bar x}^2[/itex]) is the estimate of [itex] \dfrac{\sigma^2}{n}[/itex]

    So there are several subtleties to wade through, but in none but the one unusual and unrealistic comment I made at the start will [itex] \frac{\sigma^2}{n^2} [/itex] play a role.
     
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