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    This is a necessary condition for PMCC

    (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?




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    In contngency tables I believe gagaface asked something similar, but is H0 There is no association beetween the two variables OR is it The two variables are independent
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    (Original post by L'Evil Wolf)
    In contngency tables I believe gagaface asked something similar, but is H0 There is no association beetween the two variables OR is it The two variables are independent
    Either should be perfectly fine but I generally go with whatever it says in the question. Failing that I normally go dor the association stuff.

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    (Original post by Euclidean)
    Thanks! I didn't know about the normally distributed thing, had a google and found this for anyone who was wondering the same thing:

    No problem, thanks for that
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    https://771a1ec81340d97ae9ed29694f73...XZCVmc/CH5.pdf

    For q 14 c exD shouldn't it be two tailed as it is asking for any agreement meaning either positive or negative agreement?
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    (Original post by Rkai01)
    https://771a1ec81340d97ae9ed29694f73...XZCVmc/CH5.pdf

    For q 14 c exD shouldn't it be two tailed as it is asking for any agreement meaning either positive or negative agreement?
    I see what you mean, but I think it wants you to infer from the previous part of the question that because there was quite strong positive correlation between the results then you should test for positive correlation as we can see that there won't be negative correlation so a two-tailed test would be pointless
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    Why is there only 1 constraint here? Calculating the mean is one, but what about the usual one where the expected frequency is equal to the observed frequency? I have a feeling this was asked before but I can't find it lol.
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    (Original post by paradoxequation)
    Why is there only 1 constraint here? Calculating the mean is one, but what about the usual one where the expected frequency is equal to the observed frequency? I have a feeling this was asked before but I can't find it lol.
    When testing for any kind of uniform distribution- so continuous uniform or discrete uniform, the degrees of freedom are always the number of cells after combining-1, if we had been testing for a Poisson distribution then you would have been right and we would have been subtracting 2, hope that helps
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    (Original post by economicss)
    When testing for any kind of uniform distribution- so continuous uniform or discrete uniform, the degrees of freedom are always the number of cells after combining-1, if we had been testing for a Poisson distribution then you would have been right and we would have been subtracting 2, hope that helps
    So because we are not testing for a poisson, but rather if the rate of accidents is constant, then there is no constraint for the expected f = observed f? I guess it makes sense considering we are not testing a model. Thanks!
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    (Original post by paradoxequation)
    So because we are not testing for a poisson, but rather if the rate of accidents is constant, then there is no constraint for the expected f = observed f? I guess it makes sense considering we are not testing a model. Thanks!
    Yep that's it No problem!
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    (Original post by economicss)
    When testing for any kind of uniform distribution- so continuous uniform or discrete uniform, the degrees of freedom are always the number of cells after combining-1, if we had been testing for a Poisson distribution then you would have been right and we would have been subtracting 2, hope that helps
    is this not poisson?
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    Out of questions to do? Try madasmaths statistics booklets.
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    (Original post by physicsmaths)
    is this not poisson?
    I don't think so, because we're not testing whether the Poisson distribution is a suitable model for the data, we're testing whether the probability is constant
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    Does anyone have an idea why 2015 isn't on physicsandmathstutor anymore?
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    (Original post by Euclidean)
    Does anyone have an idea why 2015 isn't on physicsandmathstutor anymore?
    Nah but I'm pretty sure it'll be up on Edexcel anyway
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    (Original post by Euclidean)
    Does anyone have an idea why 2015 isn't on physicsandmathstutor anymore?
    Anymore? I think he was just too lazy/cba to put it up? It's on the Edexcel website anyhow.
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    (Original post by Zacken)
    Anymore? I think he was just too lazy/cba to put it up? It's on the Edexcel website anyhow.
    its weird week before m3 ial june 15 was up then before the exam it was gone....
    his website is so big i think he has to be very careful on what he can put up and stuff.
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    (Original post by Krollo)
    Either should be perfectly fine but I generally go with whatever it says in the question. Failing that I normally go dor the association stuff.

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    Thanks, checked with the edexcel questions that seems a solid approach.

    Can I ask you, on the issue of the Poisson model testing stuff, on Chapter 4, page 79 Ex7 Edexcel obtain the expected value by multiplying the probability for some r, with the total number of observations,

    yet on Ex4b q7 they incline to multiply the probablity for some r, with the mean, lander.

    So which is it. Essentially when calculating the observed value in a poisson test, what do we multiply the probabiilty with to obtain the expected value.
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    (Original post by L'Evil Wolf)
    When calculating the observed value in a poisson test, what do we multiply the probabiilty with to obtain the expected value?
    Number of observations.
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    (Original post by Zacken)
    Number of observations.
    Thanks
 
 
 
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