Hypothesis Testing - Binomial Distribution

**craig** · May 9th 2010, 12:57 PM

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 $Y$ be the number of heads thrown. How would I test this?

Guessing you'd start off with something like:

$Y - Bi(20,p)$

$H_0: p=0.5$

$H_A: p \neq0.5$

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

**craig** · May 9th 2010, 01:08 PM

Just found something online that's similar to what I'm looking for.

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

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

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

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

Does this look ok?

**pickslides** · May 9th 2010, 02:06 PM

I was going to suggest finding $\text{C.I}_{0.95}$ if $p=0.5$ is not in this interval reject $H_0$ . This could be done by using the normal approximation to the binomial. But I think $n$ is too small in your case.

**craig** · May 9th 2010, 02:11 PM

Originally Posted by pickslides

I was going to suggest finding $\text{C.I}_{0.95}$ if $p=0.5$ is not in this interval reject $H_0$ . This could be done by using the normal approximation to the binomial. But I think $n$ is too small in your case.

Yeh I'd done that before with a different question, but as you said $n$ was too small this time.

Was what I did correct?

**pickslides** · May 9th 2010, 02:19 PM

Originally Posted by craig

Yeh I'd done that before with a different question, but as you said $n$ was too small this time.

To have a normal approximation to the binomial then $np>10$ and $(1-p)n>0$

In your case $np = 4$

Originally Posted by craig

What what I did correct?

Not sure what you are saying here.

Interval I had in mind was,

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

**craig** · May 9th 2010, 02:21 PM

Originally Posted by pickslides

Not sure what you are saying here.

Oops sorry typo, fixed that.

Originally Posted by pickslides

Interval I had in mind was,

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

Ahh right thankyou!

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