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Official Edexcel S3 thread Wednesday 20 May 2015 Morning [6691] Watch

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    (Original post by V0ldemort17)
    Thanks
    It's ok
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    Some of the chi squared mark schemes are a bit dodgy, any advice? E.g for a binomial fit, they had all their E(X) to 3dp, asked you to calculate r and s to 3dp then used 4dp for the test stat! Mark scheme said 4.2-4.4 but if you used 3dp as in question you got 4.524 or something.... no idea what to think


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    (Original post by chughes17)
    Some of the chi squared mark schemes are a bit dodgy, any advice? E.g for a binomial fit, they had all their E(X) to 3dp, asked you to calculate r and s to 3dp then used 4dp for the test stat! Mark scheme said 4.2-4.4 but if you used 3dp as in question you got 4.524 or something.... no idea what to think
    You might as well express the test stat to 3 dp all the time since the chi squared values in the tables are to 3dp. Keeps it consistent.
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    please can someone explan 2014 Q 1 c:
    State the assumption necessary for a product moment correlation coefficient to bevalid in this case.
    ans: an underlying (bivariate) normal distribution
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    (Original post by DCMed96)
    A quick question:
    what would you do if there is tied ranks? (2 marks) how would you answer that?
    You take the average value between the current rank and the next rank, and give them both that rank.
    For example, 2+3/2 = 2.5. The tied values are both assigned the rank 2.5.
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    sorry to ask again but i would really appreciate it if you could help me with june 2011 7d? thanks!
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    Hi guys. Please help me with June 2013 Q7 b. I don't have a clue. Cant even break it down using the mark scheme


    Thanks
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    are there anymore IAL papers apart from June 2014?
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    (Original post by pol234)
    please can someone explan 2014 Q 1 c:
    State the assumption necessary for a product moment correlation coefficient to bevalid in this case.
    ans: an underlying (bivariate) normal distribution
    I think it just means that both of the data should be normally distributed
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    (Original post by em4dy)
    Hi guys. Please help me with June 2013 Q7 b. I don't have a clue. Cant even break it down using the mark scheme


    Thanks
    To estimate standard error first calculate the standard deviation. So find, combined sum of x, and then the combined mean. then find combined sum of x^2 then the standard deviation (sigma/root 40)
    the mark scheme is a bit confusing, write out the standard deviation formula and work backwards.
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    Basically what are people rounding to for chi squared tests? 4 dp is what i will be doing!
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    (Original post by tazza ma razza)
    Basically what are people rounding to for chi squared tests? 4 dp is what i will be doing!
    the mark scheme says anything from 2d.p or more is fine, but i usually do 3 d.p, might end up doing 4 d.p in the exam tho :P just in case
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    (Original post by mmms95)
    To estimate standard error first calculate the standard deviation. So find, combined sum of x, and then the combined mean. then find combined sum of x^2 then the standard deviation (sigma/root 40)
    the mark scheme is a bit confusing, write out the standard deviation formula and work backwards.
    i'm struggling to understand how to combine x^2, please mate if you can do a worked solution for me, i'd be immensely indebted!
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    (Original post by em4dy)
    i'm struggling to understand how to combine x^2, please mate if you can do a worked solution for me, i'd be immensely indebted!
    on a paper maybe email me a picture on [email protected]. Thanks
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    (Original post by mmms95)
    the mark scheme says anything from 2d.p or more is fine, but i usually do 3 d.p, might end up doing 4 d.p in the exam tho :P just in case
    You know when you do

    ∑ (Oi2 / Ei ) - N

    i ended up with a answer to the above for my test statistic that was 0.01 off the markscheme answer, the awrt one, yet my Ei and Oi were exactly the same in the markscheme for the method marks.

    do you round answers mid way through or???
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    (Original post by tazza ma razza)
    You know when you do

    ∑ (Oi2 / Ei ) - N

    i ended up with a answer to the above for my test statistic that was 0.01 off the markscheme answer, the awrt one, yet my Ei and Oi were exactly the same in the markscheme for the method marks.

    do you round answers mid way through or???
    that's rly strange, do you remember which paper? i round my O^2/E answers to 3 d.p then my final answer is also 3d.p and just leave it like that. sometimes they tell us to calculate a missing frequency, and i just keep it to the d.p they ask in for the question. (e.g when they say r to 2d.p, i use the 2d.p value as expected frequency)
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    (Original post by tazza ma razza)
    You know when you do

    ∑ (Oi2 / Ei ) - N

    i ended up with a answer to the above for my test statistic that was 0.01 off the markscheme answer, the awrt one, yet my Ei and Oi were exactly the same in the markscheme for the method marks.

    do you round answers mid way through or???
    That method is a bit less accurate cause the numbers are bigger but it should be fine I think? Maybe you can write more decimal places
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    (Original post by Karoel)
    That method is a bit less accurate cause the numbers are bigger but it should be fine I think? Maybe you can write more decimal places
    is it worth doing both methods then just to check?

    ∑(Oi-Ei)2 / Ei
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    (Original post by em4dy)
    i'm struggling to understand how to combine x^2, please mate if you can do a worked solution for me, i'd be immensely indebted!
    No problem!
    okay so you have
    sum of x which is 32.29 + (32 x 4.55) = 178.89

    sum of x^2 is 141.4035 (from prev sample of x^2) + ((0.25x39)/32 + (4.55^2 x32)
    =811.6335

    therefore

    varianceof both samples=

    ((811.6335/40) - (178.89/40)^2)) x 40/39
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    (Original post by mmms95)
    No problem!
    okay so you have
    sum of x which is 32.29 + (32 x 4.55) = 178.89

    sum of x^2 is 141.4035 (from prev sample of x^2) + ((0.25x39)/32 + (4.55^2 x32)
    =811.6335

    therefore

    varianceof both samples=

    ((811.6335/40) - (178.89/40)^2)) x 40/39

    Thankyou! so in hindsight, S1 letting me down!
 
 
 
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