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    Well I got bored so i typed out my answers...

    1. X ~ Po(8) Probability = 0.184

    2. i think it was E(X-Y) = 12 - 7 = 5
    Var(X-Y) = Var(X) + (-1^2)Var(Y) = 11.5
    E(3X+2Y) = 3E(X) + 2E(Y) = 50
    Var(3X+2Y) = 9Var(X) + 4Var(Y) = 91

    Need independence = Var(X-Y) and Var(3x + 2Y)

    3. i) xbar = 9.1
    ii) Reason 1: xbar of second test isn't the same as xbar of first test
    Reason 2: the upper value of the interval for the second test is not in the range of the 90% confidence interval (it should be...)

    iii) n=70

    4. chi squared question? most of it was prove stuff. my value in the largest cell was 4.0 something i think. .'. conclude more men teach in higher education than women

    5. very long chi squared question (9marks) Worked out the expected values using Binomial distribution. Then combined rows 3-6 so that E > 5.

    Final value was X^2 = 10.8 or 11.8 ish i think

    chi squared with 3 dof was 7.779 i think if it was the 0.10 sig level. Hence reject H0, binomial is not a good model for the data.

    6. i) f(x) = 2/pi 0 < x < pi/2 (0 otherwise)

    ii) F(x) = 2x/pi 0< x < pi/2 (0 for x<0 and 1 for x>pi/2)

    iii) X = sin^-1(y/300)

    iv) F(Y) = (2/pi)sin^-1(y/300) for 0 < y < 300

    v) p( boat more than 100m away) = 1 - [(2/pi)sin^-1(100/300)] = 0.784

    7. I can't remember the values at all for this one.

    i) The CI was [ -0.10 , 2.4 ish]

    ii) use two sample t test. i think t = 1.998 which was greater than critical value so reject H0, there is evidence that the second sample have lost more weight (mewx < mewy)

    iii) assume equal variance. This is fair as the ppl will be from similar populations/both wanting to lose weight (i kinda blabbled through this one)
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    There was also a 1 mark part to question 5 i don't remember. I think it asked for an assumption but maybe not?
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    Yes! I got your answers (I can't remember the rest). By the way, that question was something like "Is it fair to assume all the students had been guessing their answers?"

    The answer was no, since the test statistic was far too large.

    Edit: I put a different reason two. I said that the confidence interval for the second one should have been larger than the first, since a larger z value would have been used for a greater % confidence value, hence a larger interval.
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    ah yeh i got it the wrong way round lol.
 
 
 
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