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Math Help - Statistics, averages and deviations.

  1. #1
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    Statistics, averages and deviations.

    A population consists of the five numbers 2,3,6,8 and 11. Consider all possible samples of 2 that can be drawn with replacement from this population (i.e. as soon as soon as number has been drawn and its value noted it is replaced and can be drawn again). Find

    i) The mean of the population.

    ii) The standard deviation of the population.

    iii) The mean of the sampling distibution of means.

    iv) The standard deviation of the sampling distribution of means.

    ---------------------------------------

    For i) I calculated the mean to be 6.0 however I found the standard deviation for part ii) to be 3.67 to which I am unsure of its reliability.


    Can anyone please help with these please, I am keen to learn the method to solving these.

    Regards
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  2. #2
    Super Member Aryth's Avatar
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    You have found the mean to be:

    \bar{x} = 6.0

    So, the standard deviation is:

    \sigma = \sqrt{\frac{1}{N}\sum_{i=1}^N (x_i - \bar{x})^2}

    Where N is the number of elements in the population, x_i is the i-th element in the set, and \bar{x} is the mean.

    \sigma = \sqrt{\frac{1}{5} \sum_{i = 1}^5 (x_i - 6.0)^2}

    \sigma = \sqrt{\frac{1}{5}\left[ (2 - 6)^2 + (3 - 6)^2 + (6 - 6)^2 + (8 - 6)^2 + (11 - 6)^2 \right] }

    \sigma = \sqrt{\frac{1}{5}\left[(-4)^2 + (-3)^2 + (0)^2 + (2)^2 + (5)^2 \right]}

    \sigma = \sqrt{\frac{1}{5} (16 + 9 + 0 + 4 + 25)}

    \sigma = \sqrt{\frac{1}{5} (54)}

    \sigma = \sqrt{10.8}

    \sigma = 3.29

    And there you go.
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  3. #3
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    Quote Originally Posted by RossBrons View Post
    A population consists of the five numbers 2,3,6,8 and 11. Consider all possible samples of 2 that can be drawn with replacement from this population (i.e. as soon as soon as number has been drawn and its value noted it is replaced and can be drawn again). Find

    i) The mean of the population.

    ii) The standard deviation of the population.

    iii) The mean of the sampling distibution of means.

    iv) The standard deviation of the sampling distribution of means.

    ---------------------------------------

    For i) I calculated the mean to be 6.0 however I found the standard deviation for part ii) to be 3.67 to which I am unsure of its reliability.


    Can anyone please help with these please, I am keen to learn the method to solving these.

    Regards
    For (iii) and (iv) read this: Sampling distribution - Wikipedia, the free encyclopedia
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  4. #4
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    Thanks, for the input Aryth, that was very helpful.

    Is the standard deviation not...

    \sigma = \sqrt{\frac{1}{N-1}\sum_{i=1}^N (x_i - \bar{x})^2}

    And is the standard deviation of \bar{X} = \frac{\sigma}{\sqrt{n}}
    Thereby meaning the standard deviation is not 3.29 but 1.47?
    I'm lost, can you please untangle this mess?

    Thanks

    Mr Fantastic, I can't understand how to apply the math on that wiki article. Kind Regards,
    Ross
    Last edited by RossBrons; January 4th 2009 at 08:51 AM.
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  5. #5
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    the second one in an unbiased estimatior ?
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  6. #6
    Super Member Aryth's Avatar
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    Quote Originally Posted by RossBrons View Post
    Thanks, for the input Aryth, that was very helpful.

    Is the standard deviation not...

    \sigma = \sqrt{\frac{1}{N-1}\sum_{i=1}^N (x_i - \bar{x})^2}

    And is the standard deviation of \bar{X} = \frac{\sigma}{\sqrt{n}}
    Thereby meaning the standard deviation is not 3.29 but 1.47?
    I'm lost, can you please untangle this mess?

    Thanks

    Mr Fantastic, I can't understand how to apply the math on that wiki article. Kind Regards,
    Ross
    Both can be used, but the one I provided has a smaller error in estimation, but using the N-1 formula, the standard deviation is the 3.67 that you had initially, and not 1.47.

    Your sampling distribution formula is the correct one.
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  7. #7
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    Quote Originally Posted by Aryth View Post
    Both can be used, but the one I provided has a smaller error in estimation, but using the N-1 formula, the standard deviation is the 3.67 that you had initially, and not 1.47.

    Your sampling distribution formula is the correct one.
    Yeah, thanks for that. Greatly appreciated.

    Has anyone got any ideas on how to tackle part iii and iv?

    Cheers,
    Ross
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  8. #8
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    Quote Originally Posted by RossBrons View Post
    Thanks, for the input Aryth, that was very helpful.

    Is the standard deviation not...

    \sigma = \sqrt{\frac{1}{N-1}\sum_{i=1}^N (x_i - \bar{x})^2}

    [snip]
    This is an unbiased estimator and is used to estimate the sd of a population from a sample taken from the population. You have only a sample (no mention of where it's come from) and you want the sd of only the sample. So the original formula you were given is the one to use.
    Last edited by mr fantastic; January 4th 2009 at 01:02 PM.
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  9. #9
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    Quote Originally Posted by RossBrons View Post
    And is the standard deviation of \bar{X} = \frac{\sigma}{\sqrt{n}} Mr F says: Yes.

    Mr Fantastic, I can't understand how to apply the math on that wiki article. Kind Regards,
    Ross
    \mu_{\overline{X}} = \mu.

    \sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}}.

    Substitute \mu = 6.0 and \sigma = 3.29 and n = 2.

    It's that simple.
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