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Probability a sample mean will fall in a range

  1. Nov 21, 2015 #1
    1. The problem statement, all variables and given/known data
    A random sample of size n = 81 is taken from an infinite population with the mean μ = 128 and the standard deviation σ = 6.3. With what probability can we assert that the value we obtain for the sample mean X will fall between 126.6 and 129.4?


    3. The attempt at a solution
    z = (x-μ)/(σ/sqrt(n))
    so we have
    z = (126.6-128)/(6.3/9) = -2 and z = (129.4-128)/(6.3/9) = 2
    so the probability it will fall in the range is
    F(2) - F(-2) = .9772 - .0228 = .9544

    is this correct?
     
  2. jcsd
  3. Nov 21, 2015 #2

    Orodruin

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    This depends on the actual distribution in the population. You can only do what you did if this distribution is assumed to be Gaussian.
     
  4. Nov 21, 2015 #3
    Gaussian means "normal" right? I am confused a bit about that. In my book they seem to use "z" for the test statistic and use "t" when the population is known to be normal. From what I can tell they are the same thing except that with z you use the standard normal table and with t you use a different table with a certain amount of degrees of freedom. I don't think I fully get it.
     
  5. Nov 21, 2015 #4

    Ray Vickson

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    I do not actually believe you; I think you are mis-reading your book (although, to be honest, I am making this judgement sight-unseen). Typically, for an independent random sample from an underlying normal (=Gaussian) distribution with mean ##\mu## and variance ##\sigma^2##: (1) we use ##z## and normal tables when we KNOW the value of ##\sigma##; but (2) use ##t## and t-tables when we do not know ##\sigma##, but have estimated it from the sample data itself.

    In case (2), we estimate
    [tex] \text{estimator of }\: \sigma^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2 [/tex]
    where the sample values are ##x_1, x_2, \ldots, x_n## and ##\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i ## is the sample mean. In that case the jargon is that there are ##n-1## "degrees of freedom".

    In the limit as ##n \to \infty## the t-distribution with (n-1) degrees of freedom becomes the standard normal, so using ##z## is like having infinitely many degrees of freedom.
     
  6. Nov 21, 2015 #5
    so for either of the two statistics to work, the distribution must be normal?
     
  7. Nov 21, 2015 #6

    Ray Vickson

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    Theoretically, yes, but for a large sample-size, using the "normal" results give a "reasonably accurate" approximation. This is based on the so-called Central Limit Theorem; see, eg.,
    https://en.wikipedia.org/wiki/Central_limit_theorem
    or http://davidmlane.com/hyperstat/A14043.html
    or http://www.statisticalengineering.com/central_limit_theorem.htm .

    For a "reasonable" non-normal underlying distribution, a sample size of n = 81 is likely large enough that normal-based estimates will be informative, if not absolutely accurate.
     
  8. Nov 21, 2015 #7
    ahh ok what my book actually says is use z for samples of n>30 with σ known and if σ is not known replace σ with s and if the sample is n<30 And the population is normal use t. so since my sample is large enough, my solution to this problem should be close enough?
     
  9. Nov 21, 2015 #8

    Ray Vickson

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    Asked and answered.
     
  10. Nov 21, 2015 #9
    thank you
     
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