# Thread: Using the normal approximation interval to calculate number of errors

1. ## Using the normal approximation interval to calculate number of errors

Hello I have a bit of a problem how to use the normal approximation interval( Binomial proportion confidence interval - Wikipedia, the free encyclopedia ). I would be very happy if anyone could explain further!

I am trying to calculate the upper probability of an error, say I have a confidence level of 25%. There are 6 people voting for YES and 0 people voting for NO. The result should be 0.206 or the predicted number of errors would be 6 * 0.206 = 1.236.

Could anyone spare the time to explain how this is calculated? I am completely stuck, thank you very much!

2. ## Re: Using the normal approximation interval to calculate number of errors

Originally Posted by elfor
Hello I have a bit of a problem how to use the normal approximation interval( Binomial proportion confidence interval - Wikipedia, the free encyclopedia ). I would be very happy if anyone could explain further!

I am trying to calculate the upper probability of an error, say I have a confidence level of 25%. There are 6 people voting for YES and 0 people voting for NO. The result should be 0.206 or the predicted number of errors would be 6 * 0.206 = 1.236.

Could anyone spare the time to explain how this is calculated? I am completely stuck, thank you very much!
Normal approximation is not appropriate here. You want the probability $p$ such that

$b(0;6,p)=0.25$

where $b(n;N,p)$ denotes the probability mass function of the binomial distribution. This reduces to the requirement that:

$(1-p)^6=0.25$

CB

3. ## Re: Using the normal approximation interval to calculate number of errors

Oh yes you are completely correct Captain, it feels obvious now. Thank you very very much I appreciate it!

4. ## Re: Using the normal approximation interval to calculate number of errors

Hi again! I found that this method is sadly not reasonable for computing larger distributions of maybe some hundred voting YES and some hundred voting NO. For a small example if there are 15 people voting YES and 1 voting NO with a confidence level of 25%. What would be another method to estimate the error in larger cases? And how could this example be calculated?

Thanks for any help!

5. ## Re: Using the normal approximation interval to calculate number of errors

Originally Posted by elfor
Hi again! I found that this method is sadly not reasonable for computing larger distributions of maybe some hundred voting YES and some hundred voting NO. For a small example if there are 15 people voting YES and 1 voting NO with a confidence level of 25%. What would be another method to estimate the error in larger cases? And how could this example be calculated?

Thanks for any help!
For the 100 people in the sample and a reasonable split you can use the normal approximation.

For your sample of 15 you will still have to use an exact method.

If you have $1$ NOs in a sample of $15$ you want the $p$ solving:

$b(0;15,p)+b(1;15,p)=0.25$

If you have access to a modern computational package even the sample of 100 is accessible to the exact method rather than having to resort to the normal approximation.

CB