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    Same as me,

    one question find sigma and mew, i got one as 2.5 and one as 5.

    For that blood one i think i got the 3 as the maximum value for which H0 is rejected.
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    (Original post by starcrossed)
    how did everyone do q6b, did u get lamde as 1.05 and then use the poisson formula to find x=0, x=1 x=2 and then do 1 minus that answer
    Yep, thats exactly how i did it.

    How about the last part of question 5, the caspules (sp?) one. The type II error i calculated was quite large, around 0.9/0.8. did anyone else get something similar
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    I belive Type 2 was about 0.918,
    As it was P(X>3) for X~Po(7)
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    (Original post by starcrossed)
    how did everyone do q6b, did u get lamde as 1.05 and then use the poisson formula to find x=0, x=1 x=2 and then do 1 minus that answer
    Yeah.
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    (Original post by PhilC)
    I belive Type 2 was about 0.918,
    As it was P(X>3) for X~Po(7)
    Wasn't it 1-that?
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    (Original post by PhilC)
    I belive Type 2 was about 0.918,
    As it was P(X>3) for X~Po(7)
    i got the same then.

    did anyone manage the last part of No.7, where we had to prove that T(o) was between 22 and 23. Where we supposed to solve that fat arse equation? i dunno cause it was only worth 2 marks....
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    what was the binomial approx to nomal question again, ino p was 0.25 or sumfin but cant remebr nething else
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    The last question was definitely a two tailed test. As the pyscho wasn't trying to find out whether it improved nor impaired the students ability.
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    (Original post by PhilC)
    I belive Type 2 was about 0.918,
    As it was P(X>3) for X~Po(7)
    excellent, thats exactly what i got...woooo I got a question right...yay!!! sorry im not overly confident about my marks in this paper!
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    oh for t0 to equal 23 and 22, i just subbed in those two values and it came out to be in between, bot sure thts how they wanted it dun tho
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    (Original post by abc123)
    i got the same then.

    did anyone manage the last part of No.7, where we had to prove that T(o) was between 22 and 23. Where we supposed to solve that fat arse equation? i dunno cause it was only worth 2 marks....
    You should have rearranged the equation to =0, and sub the values, you should end up with one being positive, the other negative, hence the root lies between them.
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    (Original post by Bhaal85)
    Wasn't it 1-that?
    i think is was 1- P(x<=3) = 1- 0.0818 = 0.918
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    (Original post by abc123)
    i think is was 1- P(x<=3) = 1- 0.0818 = 0.918
    Isn't the P(x<=3) rejecting H0? Oh wait, yeah I think your right, GOD!!!! Its hard trying to remeber all them numbers from stats, at least in Pure you can rember things. lol.
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    (Original post by abc123)
    so for the last question, some of us accepted Ho, while others rejected it. is anyone 110% of what the anwser was, because this question was worth 10 marks?


    I thought that because it was a two tailed test, the 5% would be split in two, 2.5 at the top, 2.5 at the bottem. leading to Ztest= 1.7.... which was less then top Zcrit (1.96). So accept Ho. although i'm not quite sure now....

    I did the same as you. But I'm not 110% sure, as I'm self-taught and the only candidate who sat it in my school.
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    There was a hell of a lot of continutity(sp) corrections!!! anyone else prove that 0.231 thing?
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    (Original post by sock)
    I did the same as you. But I'm not 110% sure, as I'm self-taught and the only candidate who sat it in my school.
    I actually worked out the X values for which the regions would be rejected. I know that it was definitely two tailed. Thereofre you should have 2.5% at either end.
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    (Original post by Bhaal85)
    Isn't the P(x<=3) rejecting H0? Oh wait, yeah I think your right, GOD!!!! Its hard trying to remeber all them numbers from stats, at least in Pure you can rember things. lol.
    Well for the type 2...we had to find p(accepting Ho, which is p(x>3)) under H1 conditions.

    And then to find P(x>3) is the same as 1- P(x<=3)... all under Po-(7)
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    to prove the .021 thing u had to use continuity correction of 50.5 instead of 50, then it came out to .0213, .021 to 3 dp
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    yeh i proved that, i no all 3 types of approxs were iin the paper, wat other qs was hard, could everyone prve the long eqn, t4 + ............... etc
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    (Original post by abc123)
    Well for the type 2...we had to find p(accepting Ho, which is p(x>3)) under H1 conditions.

    And then to find P(x>3) is the same as 1- P(x<=3)... all under Po-(7)
    Yeah, now that I recall, I think I put down 0.9 odd as my answer, and remeber thinking, why oh why was it so big.
 
 
 
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