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Math Help - unfair die

  1. #1
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    unfair die

    Here is a problem that I have been working on. I am mixed up on the different testing formulas. I included what i did so far after each question. Thanks for looking

    It is suspected that a particular die used in a game of chance was not fair. It was thought that the 1-side was heacy and the 6-side was light. Data was collected concerning the number of observed 1-s when the die was rolled 5000 times. Let p1 equal the probability of rolling a one with that die. We will test the hypothesis H0 p1=1/6 against the alternative hypothesis H1 p1<1/6 (i don't understand why this is < either)

    a. give the test statistic and critical region that has an alpha=0.01 significant level

    test stat Z=\frac{Y/5000-1/6}{\sqrt{(1/6*5/6)/5000}}
    crit. reg. Z \le -z_{0.01}

    b. If y=775 is the observed number of ones in the 5000 rolls, calculate the value of the test statistic and state conclusion about the die

    Z=\frac{775/5000-1/6}{.0052704628}=-2.213594348

    -2.21<-2.36

    since -2.21 is not less than -2.36, H0 fails to be rejected

    c. find the p-value for this test

    P(Z \le z)=P(Z \le -2.21)=1-0.9864=0.0136=? This part confuses me because of the negative test statistic. I don't know how to calculate the p-value and which way the equality should be facing.

    Thanks
    Last edited by kclark36; July 1st 2009 at 08:58 AM. Reason: changed p-value
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by kclark36 View Post
    Here is a problem that I have been working on. I am mixed up on the different testing formulas. I included what i did so far after each question. Thanks for looking

    It is suspected that a particular die used in a game of chance was not fair. It was thought that the 1-side was heacy and the 6-side was light. Data was collected concerning the number of observed 1-s when the die was rolled 5000 times. Let p1 equal the probability of rolling a one with that die. We will test the hypothesis H0 p1=1/6 against the alternative hypothesis H1 p1<1/6 (i don't understand why this is < either)
    I think that since 1-side was heacy, it makes more probable to be 1 on the lower side.
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  3. #3
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    I am wondering if someone can tell me if this is how this problem should be worked out. I'm hoping that I'm finally getting the hang of this. The p-value really confuses me.
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  4. #4
    MHF Contributor matheagle's Avatar
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    since your p-value, .0136, is greater than alpha, you fail to reject the null hypothesis.
    DRAW a st normal curve and look at both your test stat and where the rejection region begins.
    If the p-value is less than alpha, then your test statistic is in your rejection region.
    Hence you reject the null hypothesis is this case.
    Likewise if the p-value is large (in comparison to alpha) you fail to reject the null hypothesis.
    Last edited by matheagle; July 1st 2009 at 12:41 AM.
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  5. #5
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    Is that the correct way to do the P-value? I only found a p-value example for alternative hypothesis H1 where p>p0. I'm also not sure about the negative and how that changes things. Thanks
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by kclark36 View Post
    Is that the correct way to do the P-value? I only found a p-value example for alternative hypothesis H1 where p>p0. I'm also not sure about the negative and how that changes things. Thanks
    If the alternative hypothesis has a less than sign,
    then you accept the alternative hypothesisis for small values (negative in this case).
    That makes the p-value equal to P(Z<number).
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  7. #7
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    So rather than P(Z \le -2.21) it should be P(Z<-2.21)=1-0.9864=0.136?

    The example I have for alternate hypothesis of H_{1}: p_{1}>p_{0} showed a p value of P(Z \ge z), so that is why I assumed to just reverse the equality but the negative confused me. I can't find much material on finding the p-value.

    \Phi 2.21=0.09864
    Last edited by kclark36; July 1st 2009 at 09:46 AM. Reason: changed value
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  8. #8
    MHF Contributor matheagle's Avatar
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    They are the same.
    This is a continuous distribution.

    P(X<a)=P(X\le a)=\int_{-\infty}^a f(x)dx

    Since P(X=a)=0

    P(X<a)+P(X=a)=P(X\le a)
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