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    (Original post by jacobe)
    To clarify, for the 11 marker, you estimated the population variance from the data given to get the population variance was 18, then you divided 18 by 120 to estimate the sample variance?
    Yeah thats what I did
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    (Original post by jacobe)
    To clarify, for the 11 marker, you estimated the population variance from the data given to get the population variance was 18, then you divided 18 by 120 to estimate the sample variance?
    Yup, I think I got it right.

    calculated mean = 11.76
    unbiased variance = 18

    H0: u = 11.0
    H1: u != 11.0 @ 10% significance level

    \frac{11.76-11.0}{\sqrt{\frac{18}{120}}} = 1.962

    1.962 > 1.645 since two tailed test

    Reject H0. There is significant evidence to suggest that the mean consultation time at the doctor's has changed.
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    My answers:
    1: 2, 146
    2: do not occur at predictable intervals, then independent
    0.222
    3: 6, 1.5 x is values taken by X
    4: 0.831, 0.254
    5: do not reject
    6: reject, then no not necessary, CLT can be used since n is large
    7: do not reject, , 37 since apply continuity correction
    8: 0.0315, then 0.347 (I know this is wrong, since I forgot the 1-ANS, so I will have lost 1 mark)
    Is there anything people disagree with apart from 8ii?
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    (Original post by bentuk)
    pretty sure the probability is 0.6533 for 8ii), quite a few of my class got that, using different methods as well! Did anyone else get that?
    I got 0.654 to s.f because I rounded off one of my earlier numbers, but I've noticed that most past mark schemes said a.r.t (whatever) so it doesn't matter much does it? 😅 ._.
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    (Original post by jacobe)
    For 7(ii) i got 37, I don't think it was about late trains though?
    I got 37 too. So did everyone else in my class.
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    (Original post by JustaDreamer)
    I used a Normal approx to the Poisson? Anyone else do this?
    Ya. It wasn't about trains but about the number of points failures that occured in a 3 day period (or weeks, don't know it was 3 smth though). 1 points failure occured every 3 (days) and the question asked for the largest number of points failures that could occur in a 150 day period for H0 to be rejected I think? So X~Po(50), 50>15 so that can be approximated by a normal distribution N(50, 50). Then the P(X</= x +0.5) </= 0.05 so when you standardise it and equate (x+ 0.5- 50)/rt(50) to -1.645 you get x= 37.3...(or smth like that) which rounds down to 37.......... ?

    Someone correct me if I'm wrong =|
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    How many marks for last part of question 7?

    Edit: 7ii)
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    (Original post by tbhlouise)
    Same here! Got n <= 37.665 (smth) so hope n=37 is right!
    (Original post by jacobe)
    To clarify, for the 11 marker, you estimated the population variance from the data given to get the population variance was 18, then you divided 18 by 120 to estimate the sample variance?
    It should be fine. In STEP the examiners reports are very detailed and they award full marks for two different methods that lead to two slightly different answers when approximating so I assume they should follow the same rules here as well
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    (Original post by tokopoko)
    How many marks for last part of question 7?

    Edit: 7ii)
    6 I think

    Posted from TSR Mobile
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    (Original post by NinjaPandaa)
    6 I think

    Posted from TSR Mobile
    Thanks
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    (Original post by gothmog827)
    My answers:
    1: 2, 146
    2: do not occur at predictable intervals, then independent
    0.222
    3: 6, 1.5 x is values taken by X
    4: 0.831, 0.254
    5: do not reject
    6: reject, then no not necessary, CLT can be used since n is large
    7: do not reject, , 37 since apply continuity correction
    8: 0.0315, then 0.347 (I know this is wrong, since I forgot the 1-ANS, so I will have lost 1 mark)
    Is there anything people disagree with apart from 8ii?
    What's the 0.222 I don't remember the question
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    (Original post by Younjaxx)
    What's the 0.222 I don't remember the question
    Its the one with the dead rabbits. Something to do with following a poisson distribution p (1) for 600m and you had to work out the probability of finding exactly 3 rabbits for 1650m I think
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    do you remember the specific probabilities for q5? or the z value you used to show you had to not reject H0? thanks
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    (Original post by nozzelf78)
    do you remember the specific probabilities for q5? or the z value you used to show you had to not reject H0? thanks
    I think the value were n=18 and p=0.25. I can't remember the info about the sample though. Sorry
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    (Original post by totoro1997)
    I think the value were n=18 and p=0.25. I can't remember the info about the sample though. Sorry
    thanks a lot, do you also have any idea for 7i?
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    (Original post by nozzelf78)
    thanks a lot, do you also have any idea for 7i?
    It was something like po (1) every 3 days and you had to look at 15 days so po (5) and work out p (x less than or equal to 2)
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    (Original post by totoro1997)
    It was something like po (1) every 3 days and you had to look at 15 days so po (5) and work out p (x less than or equal to 2)
    thats amazing, also if you have time; in 4ii, do you remember how many of the 90 observations you were trying to estimate, as in what x was >or< than?
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    (Original post by nozzelf78)
    thats amazing, also if you have time; in 4ii, do you remember how many of the 90 observations you were trying to estimate, as in what x was >or< than?
    I left question 4 out but part ii had something to do with finding x <29
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    for q 5, does p=0.0569 > 0.05 therefore do not reject H0 ring any bells? seems like everyone got do not reject H0 but p=0.0569?
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    also for q4i, was the question x< or equal to 6?
 
 
 
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