Thread: Prediction interval for normal distribution

My textbook only has formulas for prediction intervals of t-distributed variables. If I have a binomial distribution, with an estimated p, how can I create a prediction interval that with 95% probability will contain the number of successes of the next 10000 trials? Thanks for any help with this!

with large n you need the CLT.
you need to tell me the TWO sample sizes.
You will take the difference between the two p hat's.
The mean is zero, since p-p=0
and you will just sum the two variances.
I've never seen this before, but it's just like the usual prediction situtaion.

All right, so n = 10000. I've estimated the p to be between 0,0015 and 0,0035. And what I want to do is to find a 95% confidence interval for the number of successes on the next sample of 10000 trials. Thanks for helping out.

Ok, let's get this straight.
A confidence interval is for estimating a parameter.
A prediction interval is for predicting a future random variable.

That is correct and I meant a prediction interval!

well I know I'm right
but you need to be clear on what you want.
The idea is the same as a prediction of a new observation.
You will pivot on the difference between two P hats and then apply the CLT.



where
since I assume we are sampling from the same underlying binomial distribution.
Otherwise this doesn't make sense.

And



But here we have some options, since we don't know p, we need to use the p hat's but one
has been oberved while the other hasn't.
Using Slutsky's theorem I'm sure we can use the known one say,

So

Now pivot and solve for and then multiply by m to obtain the estimate of number of successes in this second future sample.

All right, so I use p=0,0025 as my p. But I'm not sure which formula to put it in. I know this will be a standard normal distribution so anything within 95% would mean a value between two standard deviations. If I knew the standard deviation I could just write

25 - 2 standard deviations < 25 < 25 + 2 standard deviations

I might be way off, but if I'm on to something, how do I find 1 standard deviation here?

you pivot and solve for the second p hat
you want 95 percent, so you place that mess I created between -1.96 and 1.96 and solve for p hat m.
Look at your book where they pivoted and solved for the parameter or future observation as I am doing.