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Edexcel S2 - 27th June 2016 AM watch

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    :woo:

    Solomon papers.

    Physicsandmathstutor paper links.

    Examsolutions.

    Madasmaths - some useful things from S2 there.

    This is a fun exam/module so make the most of it, and good luck! :rave:
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    How would you do this one? 6 c and d

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    (Original post by EricPiphany)
    How would you do this one? 6 c and d
    This video explains how to do this sort of thing well.

    But basically, you need to set up your hypothesis: X \sim \text{Po}(\lambda)

    H0: \lambda = 2
    H1: \lambda \neq 2

    You will reject H0 if \mathbb{P}(X \leq x_{\ell}}) \leq 0.025 so you need to lookup this value of x_{\ell} in your table.

    You will also reject H0 if \mathbb{P}(X \geq x_{u}) \leq 0.025 so you need to re-arrange this and get something like \mathbb{P}(X \leq x_u - 1) \geq 0.0975 and use the tables again to find this value of x_u.


    Your critical value is the interval of X s.t if X = x where x is in your critical interval, you reject H0, i.e: it will be [x_u, x_{\ell}]

    The tables will give approximately 0.025 and approximately 0.0975 so you'll select those but the actual significance level will be the probabilities from the table added together. (The video should explain it really well)
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    (Original post by Zacken)

    Good stuff...
    So is this start correct: 2 per minute, so 60 in half an hour... X~Po(60) \approx ~N(60, 60)...
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    (Original post by EricPiphany)
    So is this start correct: 2 per minute, so 60 in half an hour... X~Po(60) \approx ~N(60, 60)...
    Nah, part (b) is asking you to deal with \lambda = 2 and \lambda \neq 2, not X ~ Po(60), so no need for approximations.
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    (Original post by Zacken)
    Nah, part (b) is asking you to deal with \lambda = 2 and \lambda \neq 2, not X ~ Po(60), so no need for approximations.
    But I need a sample statistic and the question says 'To test this, the number of cups of tea sold during a random 30 minute interval is recorded'. I can't see where else in the question this will be used.
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    (Original post by EricPiphany)
    But I need a sample statistic and the question says 'To test this, the number of cups of tea sold during a random 30 minute interval is recorded'. I can't see where else in the question this will be used.
    Your sample statistic is X \sim \text{Po}(\lambda) where \lambda =2 or \lambad \neq 2.

    Part (b) explicitly states the fact that you're testing the "rate of every 2 minutes".

    You use the 30 minute interval thingy in part (b) as part of your explanation. Nowhere else.
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    (Original post by Zacken)
    Your sample statistic is X \sim \text{Po}(\lambda) where \lambda =2 or \lambad \neq 2.

    Part (b) explicitly states the fact that you're testing the "rate of every 2 minutes".

    You use the 30 minute interval thingy in part (b) as part of your explanation. Nowhere else.
    OK.
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    Part (b) explicitly states the fact that you're testing the "rate of every 2 minutes".

    Still not perfectly convinced. 2 a minute, 60 in half an hour is essentialy the same thing. And if I would want to reject the null hypothesis I would actually have to take a sample, which seems to be implied by the question as a sample over half an hour, unless the question is just presented badly.
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    (Original post by EricPiphany)
    unless the question is just presented badly.
    Bang on the nail there. I've just checked the solution bank for that question and it agrees with me.
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    (Original post by SeanFM)
    :woo:
    Put this thread in the A Level Directory Thread which is here :parrot:
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    Please could someone explain how to do question 7e on this paper? https://57a324a1a586c5508d2813730734...%20Edexcel.pdf Thanks
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    (Original post by economicss)
    Please could someone explain how to do question 7e on this paper? https://57a324a1a586c5508d2813730734...%20Edexcel.pdf Thanks
    You've computed \sigma earlier so you just need to note that 2 - \sigma < X < 2 + \sigma is an interval whose width is larger than the IQR width. The IQR width has probability 0.5 by definition (75% - 25% = 50%) so since the given interval is larger than the IQR then the probability of that interval is larger than the probability of the IQR so the probability is greater than 0.5. Let me know if you need a clearer explanation, seriously - there's no problem if you need to clarify something, you seem to never to and I don't know if that's because you understand it all or... :lol:
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    I'm finding S2 so dull/boring compared to mechanics.

    There's just too many formulas left/right/center to remember... :argh:
    any help guys?
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    (Original post by Zacken)
    You've computed \sigma earlier so you just need to note that 2 - \sigma < X < 2 + \sigma is an interval whose width is larger than the IQR width. The IQR width has probability 0.5 by definition (75% - 25% = 50%) so since the given interval is larger than the IQR then the probability of that interval is larger than the probability of the IQR so the probability is greater than 0.5. Let me know if you need a clearer explanation, seriously - there's no problem if you need to clarify something, you seem to never to and I don't know if that's because you understand it all or... :lol:
    Thank you, I see what you mean how the IQR probability is 0.5 always but how do we work out its width in this question? Haha yeah normally you give me the eureka breakthrough, but this question has got me particularly stumped!
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    (Original post by economicss)
    Thank you, I see what you mean how the IQR probability is 0.5 always but how do we work out its width in this question? Haha yeah normally you give me the eureka breakthrough, but this question has got me particularly stumped!
    In this case the distribution is symmetric - so, since the lower quartile is at 1.41 = (2-0.59) then the upper quartile will be at 2 + 0.59 = 2.59. SO the IQR is [1.41, 2.59]

    Whereas the interval given is is [1.184, 2.816]. i.e: you should be able to see how the intervals are always 2-something to 2 + something given that the graph of the distribution is symmetric about the line x=2. So really you don't need to compute 2 + 0.59 = 2.59; it's enough to note that 2-0.816 < 1.41 so the given interval is larger than the IQR.
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    (Original post by Zacken)
    In this case the distribution is symmetric - so, since the lower quartile is at 1.41 = (2-0.59) then the upper quartile will be at 2 + 0.59 = 2.59. SO the IQR is [1.41, 2.59]

    Whereas the interval given is is [1.184, 2.816]. i.e: you should be able to see how the intervals are always 2-something to 2 + something given that the graph of the distribution is symmetric about the line x=2. So really you don't need to compute 2 + 0.59 = 2.59; it's enough to note that 2-0.816 < 1.41 so the given interval is larger than the IQR.
    Great thanks, forgot it was symmetric!!
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    My teacher told me this is a very short module, but I see so many videos on ExamSolutions

    So is it short or long?
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    (Original post by Psst.)
    My teacher told me this is a very short module, but I see so many videos on ExamSolutions

    So is it short or long?
    It's short, only 7 chapters in the book, 2 of them being very short.
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    Do we get marks for stating

    let x represent blah blah

    followed by the distribution x (squiggle) N or B or lambda (distrubution)

    ?
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    (Original post by tazza ma razza)
    Do we get marks for stating

    let x represent blah blah

    followed by the distribution x (squiggle) N or B or lambda (distrubution)

    ?
    No but I would write down the
    x (squiggle) N or B or Po (distrubution)
    so that they know what you were intending to use in case you look up the wrong number.
    There is often a mark for "evidence of using ...... distribution".
 
 
 
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