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Likelihood Ratio Statistic & P-value

  1. May 11, 2013 #1
    1. The problem statement, all variables and given/known data
    2127q75.jpg


    2. Relevant equations
    So far I have only worked on question 1, as I was not able to solve it.
    The likelihood ratio test statistic is defined as follows:
    λ = 2 Log(L(theta-hat)/L(theta-hat_0))
    Where L is the likelihood function, the product of all the pdfs/pmfs, and theta-hat is the maximum likelikhood estimator, the value of theta that maximizes the likelihood function. Theta-hat_0 is the same, but it is restricted to be in accordance with the H0 hypothesis.

    I sincerely apologize for the lack of Latex, I am still learning how to work with it, I hope it didn't make it too vague.


    3. The attempt at a solution

    Alright, so what I've done so far is define the likelihood function as the product of the two pmfs, and I took theta to consist of p1 and p2, although I'm not sure that is allowed? Should I instead use theta = p1-p2?
    L(p1,p2) = [tex]\text{p1}^x \binom{m}{x} (1-\text{p1})^{m-x}[/tex] * [tex]\text{p2}^y \binom{n}{y} (1-\text{p2})^{n-y}[/tex]

    Now, finding theta-hat, I just took derivatives with respect to p1 and p2, and set it to zero. It gave me p1 = x / m and p2 = y / n.
    I then did the same for theta-hat_0, but I first set p1 = p2 which gave me theta-hat_0 = (x+y) / (m+n)

    However, plugging this all into the expression for λ doesn't give me a very pretty expression, which tells me that maybe I'm doing something wrong.

    I get λ = [tex]2 \log \left(\left(\frac{x}{m}\right)^x \left(1-\frac{x}{m}\right)^{m-x} \left(\frac{y}{n}\right)^y
    \left(1-\frac{y}{n}\right)^{n-y}\right)-2 \log \left(\left(\frac{x+y}{m+n}\right)^{x+y}
    \left(1-\frac{x+y}{m+n}\right)^{m+n-x-y}\right)[/tex]

    Could anyone indicate where I went wrong, if this is wrong? Maybe I'm not seeing some of the trivial simplifications here. After I figure out how to do this, I'll post where I get stuck with b!

    Thank you

    Edit: If it clarifies, I could post what my lamda is without filling in the values, but its basically 2Log of the pmfs multiplied with p1 = x/m, p2 = y/n, divided by the pmfs multiplied with p1 = p2 = m+n/(x+y)
     
    Last edited: May 11, 2013
  2. jcsd
  3. May 12, 2013 #2
    I suppose it might not be wrong after all. I mean, if I plug the values in, I get something like 2.08, or something of the sort. Not a bad value by any means. However, I don't understand how to compute the P-value of this, sadly. Could anyone help me with that?
     
  4. May 14, 2013 #3
    Ok, so I suppose the answer is just right. I get lamda = 2.04, and the chi square distribution with 1 degree of freedom for 5% gives a higher value, so Ho is not rejected. The P-value I get is also bigger than .1, so it's not rejected. Thanks!
     
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