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Hypothesis testing

Discussion in 'Mathematics' started by jsmith613, Dec 13, 2011.

  1. http://www.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/Subject%20Sorted/S2/S2%202006-01.pdf

    Look at question 7 on the last page

    so I know that for (ai)
    H0: p = 0.2
    H1: p <> 0.2 (does not equal)

    Now when I carry out the test, if it was p>0.2 I would say
    P(X>=9).... and then work out the answer
    equall if it was p<>0.2 how do I carry out the test. Please can you explain as well!

    For (b) I am again faced with the same issue
    The distribution is
    X-N(20,16)
    H1 again is p<>0.2

    so for my normal approx to I use
    P(x>=18) or p(x
     
  2. http://www.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/Subject%20Sorted/S2/S2%202006-01.pdf

    Look at question 7 on the last page

    so I know that for (ai)
    H0: p = 0.2
    H1: p <> 0.2 (does not equal)

    Now when I carry out the test, if it was p>0.2 I would say
    P(X>=9).... and then work out the answer
    equall if it was p<>0.2 how do I carry out the test. Please can you explain as well!

    For (b) I am again faced with the same issue
    The distribution is
    X-N(20,16)
    H1 again is p<>0.2

    so for my normal approx to I use
    P(x>=18) or p(x
     
  3. Sorry?
     
  4. here is the website:


    http://www.xtremepapers.com/Edexcel/Advanced%20Level/Mathematics/Subject%20Sorted/S2/S2%202006-01%20MS.pdf
     
  5. now look at question 7
    for ai) how would I carry out a two-tailed test to at the 5% significance level
     
  6. Assume that the probability of reading the Deano is 0.2.
    We take a random sample of 20 from a large population (this is the first flaw in the question, a school is probably not a large enough population to neglect the fact that we are sampling without replacement, but forget that).
    The probablity of getting zero Deano readers in our sample is 0.8^20.
    The probaility of getting one Deano reader in our sample is
    20 * 0.2 * 0.8^19
    One is a Deano reader, all the other are non-Deano readers (0.8), and the Deano reader could be in any position. So multiply by 20.
    More generally, it's
    p(k) = 20! / (k!(20-k)!) * 0.2^k * (1-0.2)^(20-k)
    The mode of the distribution is at k = 4. There are two tails. So if the actual observation, k = 9, is in the upper tail, we can reject the null, that p = 0.2.
    To reject p = exactly 0.2 at the 5% level, we use a two tailed test. So do the values for k = 9-20 sum to 0.025 or less?
    The other problem with the question is that it's unclear what the teacher is after. 20% of the pupils read the Deano is clearly a ball-park figure. He's unlikely to be interested in the fact that p is a bit different from 0.2, it might be 0.18 for example, and what he's saying is still right. If we make the sample size big enough, we'll eventually reject every value of p except the absolutely accurate one.So what he really wants is confidence bounds for p.


     
  7. ...or just ask your teacher or post your question on the student room maths forum.
     
  8. Would you be kind enough to say a little more about this please? Thanks.
     
  9. Let's say we have a school with 20 pupils in it, and we want to know whether exactly 20% of the pupils are Deano readers or not. Let's say we take a sample of 20 pupils, without replacement. When can we accept or reject this hypothesis?
    Now think about what happens when we have 25 pupils, and take a sample of 20 pupils. When can we accept or reject the 20% hypothesis?
    Now assume that we have 10 million pupils in the school,a nd take another sample of 20 pupils. When can we accept or reject the p = 0.2 hypothesis?

     

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