Hypothesis Testing - Binomial Distribution

Apr 2008
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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
 
Apr 2008
748
159
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?
 

pickslides

MHF Helper
Sep 2008
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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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Apr 2008
748
159
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?
 

pickslides

MHF Helper
Sep 2008
5,237
1,625
Melbourne
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 \(\displaystyle np = 4\)


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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