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Thread: Hypothesis Testing - Binomial Distribution

  1. #1
    Super Member craig's Avatar
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    Hypothesis Testing - Binomial Distribution

    Hi, been looking over my Hypothesis Testing notes, and can't seen to find anything about testing the Binomial. For example a recent question read something like this.

    A coin was tossed 20 times, with 16 tails and 4 heads. Let $\displaystyle Y$ be the number of heads thrown. How would I test this?

    Guessing you'd start off with something like:

    $\displaystyle Y - Bi(20,p)$

    $\displaystyle H_0: p=0.5$

    $\displaystyle H_A: p \neq0.5$

    Not sure how to procede from here, if I was testing a Normal Distribution I'd calculate a p-vaule, just not sure how to do it for the Binomial.

    Thanks in advance
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  2. #2
    Super Member craig's Avatar
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    Just found something online that's similar to what I'm looking for.

    What they did was to calculate $\displaystyle P(Y \leq 4) = 0.00591$.

    How ever because a value of 16 heads would count against $\displaystyle H_0$ just as much, then we must also take into account $\displaystyle P(Y \geq 4)$.

    $\displaystyle P(Y \geq 4) = P(Y \leq 4)$ because of symmetry, so the p-value is $\displaystyle 2(P(Y \geq 4)) = 0.0118$.

    This indicates good evidence against $\displaystyle H_0$, ie. evidence against the coin being fair.

    Does this look ok?
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    I was going to suggest finding $\displaystyle \text{C.I}_{0.95}$ if $\displaystyle p=0.5$ is not in this interval reject $\displaystyle H_0$ . This could be done by using the normal approximation to the binomial. But I think $\displaystyle n$ is too small in your case.
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  4. #4
    Super Member craig's Avatar
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    Quote Originally Posted by pickslides View Post
    I was going to suggest finding $\displaystyle \text{C.I}_{0.95}$ if $\displaystyle p=0.5$ is not in this interval reject $\displaystyle H_0$ . This could be done by using the normal approximation to the binomial. But I think $\displaystyle n$ is too small in your case.
    Yeh I'd done that before with a different question, but as you said $\displaystyle n$ was too small this time.

    Was what I did correct?
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  5. #5
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    Quote Originally Posted by craig View Post
    Yeh I'd done that before with a different question, but as you said $\displaystyle n$ was too small this time.
    To have a normal approximation to the binomial then $\displaystyle np>10$ and $\displaystyle (1-p)n>0$

    In your case $\displaystyle np = 4$


    Quote Originally Posted by craig View Post

    What what I did correct?
    Not sure what you are saying here.

    Interval I had in mind was,

    $\displaystyle \text{C.I}_{0.95} = p\pm Z_{\frac{\alpha}{2}}\times \sqrt{\frac{p(1-p)}{n}}$
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    Super Member craig's Avatar
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    Quote Originally Posted by pickslides View Post
    Not sure what you are saying here.
    Oops sorry typo, fixed that.


    Quote Originally Posted by pickslides View Post
    Interval I had in mind was,

    $\displaystyle \text{C.I}_{0.95} = p\pm Z_{\frac{\alpha}{2}}\times \sqrt{\frac{p(1-p)}{n}}$
    Ahh right thankyou!
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