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    Please could anyone explain exercise c question 7a, how do we know that mu=1 by symmetry? Thanks https://771a1ec81340d97ae9ed29694f73...XZCVmc/CH3.pdf
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    (Original post by economicss)
    Please could anyone explain exercise c question 7a, how do we know that mu=1 by symmetry? Thanks https://771a1ec81340d97ae9ed29694f73...XZCVmc/CH3.pdf
    I don't know whether it's possible to draw a table in LaTeX. But if you lay it out it makes complete sense:

    P(X = 0) = 0.4
    P(X = 1) = 0.2
    P(X = 2) = 0.4

    If it isn't immediately obvious to you, you can do:

    \displaystyle E(X) = \sum_{i=0}^{2} P(X = i) \times i
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    (Original post by L'Evil Wolf)
    how are the observed frequencies calculated in this question.
    They calculate the  \text{Mean Rate of Accidents} = \dfrac{\text{Total Number of Accidents}}{\text{Total Number of Employees}} and use a Poisson distribution to model, as if the null hypothesis is correct this mean is constant and representative of all factories
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    (Original post by Euclidean)
    They calculate the  \text{Mean Rate of Accidents} = \dfrac{\text{Total Number of Accidents}}{\text{Total Number of Employees}} and use a Poisson distribution to model, as if the null hypothesis is correct this mean is constant and representative of all factories
    Sorry I meant to say observed.

    When doing the poisson value of say x=0 it yields: 0.0045

    But then multiplying it by 15 the answer is 0.0677. and not 21.6 :/
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    (Original post by L'Evil Wolf)
    Sorry I meant to say observed.

    When doing the poisson value of say x=0 it yields: 0.0045

    But then multiplying it by 15 the answer is 0.0677. and not 21.6 :/
    Sorry I was mistaken with regards to a Poisson model.

    The mean rate of accident is in accidents per thousand, so in a thousand employees we would expect 5.4 accidents. So for n thousand employees we would expect 5.4(n) accidents. So for 4 thousand employees we expect 4(5.4) = 21.6 accidents.
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    (Original post by Euclidean)
    I don't know whether it's possible to draw a table in LaTeX. But if you lay it out it makes complete sense:

    P(X = 0) = 0.4
    P(X = 1) = 0.2
    P(X = 2) = 0.4

    If it isn't immediately obvious to you, you can do:

    \displaystyle E(X) = \sum_{i=0}^{2} P(X = i) \times i
    Thank you
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    (Original post by Euclidean)
    I asked this earlier, I think 15, 15 IAL and 13 R are the difficult ones

    Goodness of fit are normally at least 10ish
    15 IAL is standard
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    (Original post by Euclidean)
    Sorry I was mistaken with regards to a Poisson model.

    The mean rate of accident is in accidents per thousand, so in a thousand employees we would expect 5.4 accidents. So for n thousand employees we would expect 5.4(n) accidents. So for 4 thousand employees we expect 4(5.4) = 21.6 accidents.
    thanks that makes sense but on page 79 they multiply the probability by n(80) not the mean.

    Quite confusing. Could you clarify this please.
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    (Original post by L'Evil Wolf)
    thanks that makes sense but on page 79 they multiply the probability by n(80) not the mean.

    Quite confusing. Could you clarify this please.
    I assume you're talking about Exercise B, Question 8.

    If so, the null hypothesis is referencing the suitability of a Poisson model (this time!) so the procedure to calculate expected values is a little different. We expect 3.45 cells to occur in each square on average but the probability of say 0 cells in the whole population of squares is given by:

    \text{Number of Squares in Total} \times P(X = 0) = 80 \times P(X = 0) where  X \sim Po(3.45)
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    Thankyou so much !
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    Will examiners mark you down for doing Spearman's and chi-squared working on the given table rather than drawing your own to work on?
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    (Original post by nnnnnnnnnn)
    Will examiners mark you down for doing Spearman's and chi-squared working on the given table rather than drawing your own to work on?
    Doubt it but they say 'parts should be labelled clearly' so i woukd just make a new one just for clarity.


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    do we reckon the paper will be easier than last years?
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    (Original post by tazza ma razza)
    do we reckon the paper will be easier than last years?
    Seriously hoping so, this year's M3 was an alright paper apparently so hoping they've maintained that trend for A2 modules in the form of S3.
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    June 2009, Question 3(c)

    Why does the mark scheme answer mean SRCC is better than PMCC? Why does the finishing position need to be normally distributed?




    Mark Scheme:

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    (Original post by Euclidean)
    June 2009, Question 3(c)

    Why does the mark scheme answer mean SRCC is better than PMCC? Why does the finishing position need to be normally distributed?




    Mark Scheme:

    I think they're implictly saying because position isn't normally distributed then you can't use the PMCC since this requires that both variables are normally distributed, so you use SRCC instead, hope that helps
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    (Original post by economicss)
    I think they're implictly saying because position isn't normally distributed then you can't use the PMCC since this requires that both variables are normally distributed, so you use SRCC instead, hope that helps
    Thanks! I didn't know about the normally distributed thing, had a google and found this for anyone who was wondering the same thing:

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    https://771a1ec81340d97ae9ed29694f73...XZCVmc/CH3.pdf

    For question 17 a exercise I why is n=100? Shouldn't it be 200 especially since in part b it is 100?
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    (Original post by Rkai01)
    https://771a1ec81340d97ae9ed29694f73...XZCVmc/CH3.pdf

    For question 17 a exercise I why is n=100? Shouldn't it be 200 especially since in part b it is 100?
    Because 100 is the number of ball bearings where as 200 is the number of samples are taken, it;s like when you're doing goodness of fit and you divide by the value of n to get the probability then you would multiply by the number of samples to get the expected frequency, so here n is 100 because that;s the quantity in each sample and 200 is the number of samples of 100 repeatedly taken, hope that helps
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    (Original post by Euclidean)
    I assume you're talking about Exercise B, Question 8.

    If so, the null hypothesis is referencing the suitability of a Poisson model (this time!) so the procedure to calculate expected values is a little different. We expect 3.45 cells to occur in each square on average but the probability of say 0 cells in the whole population of squares is given by:

    \text{Number of Squares in Total} \times P(X = 0) = 80 \times P(X = 0) where  X \sim Po(3.45)
    Sorry I was refering to the example on page 79/80
 
 
 
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