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    1. ok one question has asked me to find the product moment correlation coefficient, I have done this, but the next part of the question asks to find Spearmans Rank. However, the question is worth 1 mark, therefore is their a quick method to finding it given ive found the product moment correlation?? As doing all the calculations surely is worth more than 1 mark?

    2. Using { = sum of

    {(x-200) = 17000

    {(x-200)^2 = 86000000

    n = 8

    (a) Find the mean

    (b) Find the Standard Deviation

    (c) Find {(x-300)

    Thanks!
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    There is unfortunately no shortcut. You have to work out (sum over i) d(i)^2, where d(i) is the ith difference in ranks. The PMCC doesn't involve ranks.
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    (Original post by Jonny W)
    There is unfortunately no shortcut. You have to work out (sum over i) d(i)^2, where d(i) is the ith difference in ranks. The PMCC doesn't involve ranks.
    blimey thats alot of workings for 1 mark then, any ideas with the second one jonny?
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    This question is OK provided you are happy taking constants out of sums. For example, if you know

    (sum over i from 1 to 100) [y(i)^3 + 7] = 1000000

    then the sum counts the '7' 100 times. So

    (sum over i from 1 to 100) y(i)^3 = 1000000 - 700.

    (a)
    sum(x - 200) = 17000
    sum(x) - 200n = 17000
    sum(x) = 17000 + 200n = 18600
    mean = sum(x) / n = 18600 / 8 = 2325

    (b)
    sum[(x - 200)^2] = 86000000
    sum[x^2 - 400x + 40000] = 86000000
    sum(x^2) - 400sum(x) + 40000n = 86000000
    sum(x^2)
    = 86000000 - 40000n + 400sum(x)
    = 86000000 - 40000*8 + 400*18600
    = 93120000

    variance = (1/n)sum(x^2) - mean^2 = 6234375
    standard deviation = sqrt(variance) = 2496.87

    (c)
    sum(x - 300)
    = sum(x) - 300n
    = 18600 - 300*8
    = 16200
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    ok im having some troubles with part 3, im not totally sure of what you mean

    I found the mean by 17000/8 + 200

    I found the SD by [root]86000000/8 - (mean-200)^2

    but with the last part I dont understand how you relate {(x-300) with {(x-200)

    the answer claims {(x-200) - 800 how does it get the 800? Is it only possible by doing Jonny's method? Surely its easier somehow? with relation to the number of marks available.

    Any help would be asap would be great as exam is tomorrow morn!

    Thanks
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    Let y(i) = x(i) - 200 (ie, subtract 200 from each of the data values).

    Then

    mean(y) = mean(x) - 200 . . . shifting the data shifts the mean,
    sd(y) = sd(x) . . . shifting the data doesn't change the sd.

    The given sums imply

    mean(y) = (1/8)sum(y) = (1/8)sum(x - 200) = 17000/8 = 2125,
    sd(y)
    = sqrt[(1/8)sum(y^2) - mean(y)^2]
    = sqrt[86000000/8 - (17000/8)^2]
    = 2496.87.

    So

    mean(x) = 2125 + 200 = 2325,
    sd(x) = 2496.87.

    In the last part we want to relate sum(y) with sum(z), where y(i) = x(i) - 200 and z(i) = x(i) - 300. Each item in the second sum is 100 less than the corresponding item in the first sum. Since there are eight items, it follows that sum(z) = sum(y) - 8*100 = sum(y) - 800. We know that sum(y) = 17000, so sum(z) = 17000 - 800 = 16200.
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    ok so if i wanted to find {(x-400)

    id notice a difference of 200

    => 200n
    =>200*8 = 1600

    => {(x-400) = {(x-200) - 1600

    => 17000 - 1600

    = 15400

    Correct?


    [Cheers Jonny you are a huge help]
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    Just furthering this,

    now I find {(x-100) difference is = +100

    => 100*8
    = 800

    But, its + 100

    => 17000+800

    = 17800

    Yes? I think ive got it now
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    Your last two posts are correct.
 
 
 
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Updated: June 10, 2004
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