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

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    For the second part of b I added the U1s to get (4/5)U1 - (U2/5 etc)
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    (Original post by Navo D.)
    Stated hypothesis
    Mu lies within the given CI
    For a 5% sig level (two tailed)
    Did the same thing. At least that's an extra 3 marks


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    (Original post by Navo D.)
    Stated hypothesis
    Mu lies within the given CI
    For a 5% sig level (two tailed)
    It's 5% in each tail but wouldn't you say 10% for the overall significance level?
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    (Original post by xKay)
    Pls tell me this is the answer
    It should be. You get same mean and variance as in C so answer same
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    (Original post by 13 1 20 8 42)
    It's 5% in each tail but wouldn't you say 10% for the overall significance level?
    Yeah I suppose you're right maybe I made a mistake then :P (I did mention 10% overall I think)
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    But the question says to find the ACTUAL probability
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    (Original post by xyzmaster)
    It should be. You get same mean and variance as in C so answer same
    Part d is already tell you that it is NOT the same answer as 0.181
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    (Original post by xKay)
    But the question says to find the ACTUAL probability
    You'd end up finding p(z>root5/6. ) which was same as I'm c therefore actual was 0.1814 and to three sf 0.181
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    Question 5 was a wipe out
    Nice paper apart from that
    Couldn't do b
    Got 0.304 or something for d


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    (Original post by xyzmaster)
    You'd end up finding p(z>root5/6. ) which was same as I'm c therefore actual was 0.1814 and to three sf 0.181
    How sure are you?
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    (Original post by bobo19966)
    Part d is already tell you that it is NOT the same answer as 0.181
    Why not ?
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    (Original post by Mutleybm1996)
    Question 5 was a wipe out
    Nice paper apart from that


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    Are you kidding?
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    (Original post by xyzmaster)
    You'd end up finding p(z>root5/6. ) which was same as I'm c therefore actual was 0.1814 and to three sf 0.181
    but p(z>root5/6) was the probability in the part before was it not?
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    Not claiming I'm right or anything but to those who're saying the last part of 5 is the same as the part right before that;

    Why would it be 6 marks for the same thing?
    Why would they tell us that 0.181 isn't the actual probability
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    the last part of question 5 is definitely the hardest question there has ever been
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    (Original post by xKay)
    How sure are you?
    Pretty sure
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    (Original post by Navo D.)
    Stated hypothesis
    Mu lies within the given CI
    For a 5% sig level (two tailed)
    Isnt it 10% sig level 5% each tail
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    (Original post by Navo D.)
    Not claiming I'm right or anything but to those who're saying the last part of 5 is the same as the part right before that;

    Why would it be 6 marks for the same thing?
    Why would they tell us that 0.181 isn't the actual probability
    Part c used two independent samples. This part used only one sample that's why the wanted us to prove it
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    Not sure my application/final answer was right
    But I think the confusion on the last part of 5 was the expectation algebra we use is generally for independent variables
    U1 and U bar are not independent because U bar contains U1
    So you have to sum up the U1s; it is incorrect to say that the Variance is Var(U1) + Var(Ubar)
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    (Original post by xyzmaster)
    Why not ?
    The part d is asking why the model of part c (0.181) cannot be modelled the last part of q5

    As the part c is independent variable but however any normal distribution may not independent
 
 
 
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