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    For the last question I managed to put the type of error, the value of p(X>=20) and p(X>=19) and I claimed the probability of a type 2 would be the probability of p(x>=19) and didn't take it away from 1. How many method marks do youse think would be given? I was assuming 1 each for the boundary probabilities and 1 for the type of error but not entirely sure.
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    For 8 ii, we didn't have to run a complete test did we and state hypotheses etc? Just find out if it changed the result of the test?


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    (Original post by GM16)
    For 8 ii, we didn't have to run a complete test did we and state hypotheses etc? Just find out if it changed the result of the test?


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    The question just says "Determine whether this makes a difference to theconclusion of the test" so you won't need to state the hypotheses again.
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    (Original post by marioman)
    The question just says "Determine whether this makes a difference to theconclusion of the test" so you won't need to state the hypotheses again.
    Phew, thanks


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    (Original post by xsanda)
    Xsanda's solutions

    1) Unbiased mean and variance of summarised data
    E(V) = 268
    Var(V) = 1912/13 = 147 [4]

    2) Normal distribution, given P(M<1.00) = 5%
    P(M>1.05) = 13.6% [6]

    3) Approximated binomial
    n large, np = 26 > 5, nq = 14 > 5
    use normal distribution
    P(X<=30) = 0.932 [7]

    4) Poisson, P(X=4) = P(X=5)
    λ = 5, P(X=5) = 0.175 [5]

    5) School with 55% girls
    i) P(G≥6) = 22% (above significance level), do not reject H₀ [7]
    ii) Fixed prob of girl chosen, independent of previous Head Students [1]

    6) Cars along a narrow road
    i) Constant probability of cars throughout the minute, cars independent [2]
    ii) λ = 6.5, P(4≤X<7) = 0.561[3]
    iiia) N(30,30)
    P(X>35) = 0.158 [6]
    b) Not independent (narrow, slow road, so affected by car in front) [1]

    7) Probability density function
    i) x is a single sample value of X [1]
    ii) a=1, b=1.5 [7]
    iii) E(X) = 7/4 = 1.75 [3]

    8) Animals dying at zoo
    i) P(L<12.48) = 4.88% < 5%
    reject H₀, substantial evidence zoos limit life expectancy [7]
    ii) s² = 12.5
    5.05% > 5%, do not reject H₀, insufficient evidence zoos limit life expectancy [5]
    iii) yes, as the distribution of the original data is not given [1]

    9) Poisson test, λ=11
    i) reject if P(R≥20)
    92% chance of Type II error [6]
    I concur with all of these.
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    For 8ii I didn't calculate an unbiased variance. I followed through with just the number given and got the correct conclusion. How many marks will I lose out of 5?
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    (Original post by Mr M (jr))
    I concur with all of these.
    I keep thinking about my answer to 5ii) . . . . .

    that the sample must be random for the significance test to be valid, and then going on to explain what a random sample is in this particular context. Xsanda's answer isn't the same. . . . . and so you don't concur with me. . . . .

    To me the question is similar to question 6i) on the June 2014 paper...???????
 
 
 
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