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    (Original post by ResultInExcellence)
    Does anyone remember what they wrote for that 1 marker along the lines of whether it was necessary to have assumed that it was normally distributed?
    Anyone get like 0.653 or something for the 8(ii)? My logic seems sound but I haven't seen anyone else get that.

    Ok, first, what I did was work out the probability of a type 2 error given that p was 0.7/0.5. The probability of a type 2 error if p was 0.7 was 0.0933 (worked out using binomial, n=14,p=0.7, worked out probability that it was less than or equal to 7), and the probability of p being 0.7 was 1/3, so I did 1/3*0.0933 to get the probability of it being 0.7 and a type 2 error. I then did the same for 0.5 and got the probability of a type 2 error given that it was 0.5 to be 0.6047, then did 1/3*0.6047 to work out the probability of it being 0.5 and a type 2 error if it was 0.5. I then added those 2 values up to get the total probability of it being a type 2 error. So, the probability of a type 2 error was 349/1500. I then used another binomial and did n=4, and p =349/1500 and worked out probability of it being greater than or equal to 1 (that was the question).


    Where did I go wrong (if at all)?
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    (Original post by tokopoko)
    * 8 ii) in full from memory* (checking my answer so might as well post it here)

    Find probability that out of 4 tests at least 1 results in a Type 2 error. Probability of p=0.3 or 0.5 or 0.7 is 1/3.

    P(type 2 when p=0.3) = 0
    P(type 2 when p=0.5) = 0.6047 (using tables) NB (from part i) we know critical region X >= 8
    P(type 2 when p=0.7) = 0.0933 (using tables)


    P(Type 2 error) = 1/3 x 0.6047 + 1/3 x 0.0933 = 0.23267

    B(4 , 0.23267)
    P(X at least 1) = 1 - P(X=0) = 1 - (1 - 0.23267)^4

    Ans = 0.6533
    I agree, although I did slightly differently. Instead I found the probability on no type II error for each, I.e for p=0.3 its 1, p=0.5 its 0.3953 and p=0.7 it's 0.9067, so since they're all equally likely you can average them to get (1+0.3953+0.9067)/3= 0.7673 hence probability = 1- 0.7673^4= 0.6533

    Although effectively it's exactly the same method
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    (Original post by tokopoko)
    * 8 ii) in full from memory* (checking my answer so might as well post it here)

    Find probability that out of 4 tests at least 1 results in a Type 2 error. Probability of p=0.3 or 0.5 or 0.7 is 1/3.

    P(type 2 when p=0.3) = 0
    P(type 2 when p=0.5) = 0.6047 (using tables) NB (from part i) we know critical region X >= 8
    P(type 2 when p=0.7) = 0.0933 (using tables)


    P(Type 2 error) = 1/3 x 0.6047 + 1/3 x 0.0933 = 0.23267

    B(4 , 0.23267)
    P(X at least 1) = 1 - P(X=0) = 1 - (1 - 0.23267)^4

    Ans = 0.6533
    **** YES! LOOK AT NEXT PAGE! THATS WHAT I DID!

    I think this is right. Logically, it looks right at least.
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    (Original post by rich1334)
    Hmmm, but then again a lot of people completely skipped out Q4 (10 marks) and Q8ii (5 marks)
    Yep that was what I was thinking plus the continuity correction in q7 threw a few people

    Posted from TSR Mobile
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    Ok, did anyone else get 37 for the points on a rail way?
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    (Original post by rich1334)
    Hmmm, but then again a lot of people completely skipped out Q4 (10 marks) and Q8ii (5 marks)
    what do you think itll be then cus im normally quite pessimistic about things anyways
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    (Original post by tokopoko)

    P(type 2 when p=0.3) = 0
    Pretty sure this isn't right, as we were told in the question that there is an even chance that the true p is 0.3, 0.5, 0.7, therefore if p = 0.5 or 0.7, there is a possibility of a Type II error when p=0.3, as 0.3 is not necessarily the true p. Just my opinion
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    (Original post by dathtom)
    Pretty sure this isn't right, as we were told in the question that there is an even chance that the true p is 0.3, 0.5, 0.7, therefore if p = 0.5 or 0.7, there is a possibility of a Type II error when p=0.3, as 0.3 is not necessarily the true p. Just my opinion
    Definition of Type 2 error : Accept Ho (p=0.3). when in fact it is not true.

    when p=0.3 for certain there is no chance of this
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    (Original post by dathtom)
    Pretty sure this isn't right, as we were told in the question that there is an even chance that the true p is 0.3, 0.5, 0.7, therefore if p = 0.5 or 0.7, there is a possibility of a Type II error when p=0.3, as 0.3 is not necessarily the true p. Just my opinion
    H0 is p=0.3

    if p=0.3, the probability of there being a type 2 error is 0. Because, type 2 is accepting a false. So you do 1/3*0, ie, 0.

    edit2 : Type 2 is accepting h0 given h1 is true. If p = 0.3, then h1 isn't true.
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    Unofficial mark scheme anyone?
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    How many marks out of 5 do you think I'll get for the last question? I calculated the probability of a type 2 error for p=0.5 and p=0.7 then found that P(Type 2) = 0.23267... or what ever it was...
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    (Original post by jacobe)
    Anyone get like 0.653 or something for the 8(ii)? My logic seems sound but I haven't seen anyone else get that.

    Ok, first, what I did was work out the probability of a type 2 error given that p was 0.7/0.5. The probability of a type 2 error if p was 0.7 was 0.0933 (worked out using binomial, n=14,p=0.7, worked out probability that it was less than or equal to 7), and the probability of p being 0.7 was 1/3, so I did 1/3*0.0933 to get the probability of it being 0.7 and a type 2 error. I then did the same for 0.5 and got the probability of a type 2 error given that it was 0.5 to be 0.6047, then did 1/3*0.6047 to work out the probability of it being 0.5 and a type 2 error if it was 0.5. I then added those 2 values up to get the total probability of it being a type 2 error. So, the probability of a type 2 error was 349/1500. I then used another binomial and did n=4, and p =349/1500 and worked out probability of it being greater than or equal to 1 (that was the question).


    Where did I go wrong (if at all)?
    I got 0.653 as well
    (Original post by NinjaPandaa)
    Yep that was what I was thinking plus the continuity correction in q7 threw a few people

    Posted from TSR Mobile
    A lot of people seemed to like it from my class. I think the grade boundaries might be what you suggested as well however for the reasons you stated. Some of the questions weren't your everyday stats problems.
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    (Original post by tokopoko)
    * 8 ii) in full from memory* (checking my answer so might as well post it here)

    Find probability that out of 4 tests at least 1 results in a Type 2 error. Probability of p=0.3 or 0.5 or 0.7 is 1/3.

    P(type 2 when p=0.3) = 0
    P(type 2 when p=0.5) = 0.6047 (using tables) NB (from part i) we know critical region X >= 8
    P(type 2 when p=0.7) = 0.0933 (using tables)


    P(Type 2 error) = 1/3 x 0.6047 + 1/3 x 0.0933 = 0.23267

    B(4 , 0.23267)
    P(X at least 1) = 1 - P(X=0) = 1 - (1 - 0.23267)^4

    Ans = 0.6533
    FUUUUUUUUUUUU :') I did that exact method and I'm pretty sure you got it right but I messed up and did 1-0.6047 and 1-0.0933 because I misread the rejection criterion -_-
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    What UMS do people think 67 or 68/72 will be? Thanks
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    (Original post by jacobe)
    H0 is p=0.3

    if p=0.3, the probability of there being a type 2 error is 0. Because, type 2 is accepting a false. So you do 1/3*0, ie, 0.

    edit2 : Type 2 is accepting h0 given h1 is true. If p = 0.3, then h1 isn't true.
    I thought a hypothesis test is never 100% certain
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    (Original post by irbif)
    what do you think itll be then cus im normally quite pessimistic about things anyways
    I'd say 58 for an A
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    (Original post by Rifleboy123)
    FUUUUUUUUUUUU :' I did that exact method and I'm pretty sure you got it right but I messed up and did 1-0.6047 and 1-0.0933 because I misread the rejection criterion -_-
    unlucky, there will be some method marks for sure though
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    By my prediction, either exactly or just under 100UMS but I could be wrong
    (Original post by Avpatil)
    What UMS do people think 67 or 68/72 will be? Thanks
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    What did people get for question 4? Weren't both parts an approximation to binomial, one poisson one normal? I can't remember what probabilties I got however haha.
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    (Original post by totoro1997)
    I thought a hypothesis test is never 100% certain
    8ii) wasnt a hypothesis test. it basically said p=0.3 1/3 of the time. so 1/3 of the time H0 is true.

    ask yourself how can a type 2 error occur when H0 is true?
 
 
 
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