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**Hu9hD0oE****Member**- Registered: 2021-08-31
- Posts: 1

Suppose forty independent random samples were taken from a large population and a 90% confidence interval for the population proportion was computed from each of them:

How many of the 90% confidence intervals would you expect to contain the population proportion p?

Write down an expression for the probability that at least 39 of the 40 confidence intervals contain the population proportion p.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,278

hi Hu9hD0oE

Welcome to the forum.

It's ages since I did this stuff so there's a reliability warning attached to this answer.

I read your post this morning at around 8.00am and I've been thinking about it for the last 4 hours on and off. I've finally settled on this answer. Accept it at your own risk.

A 90% confidence interval means you are 90% confident that the true population proportion lies within it. So there's a 10% chance it won't.

With 40 trials that means you won't get p in the interval 10% of 40 = 4 times.

LATER EDIT: I don't think the next part can be correct. I'm having a rethink. Back in another 4 hours

LATER LATER EDIT:

I have modified my calculation now.

So let's take 0.1 as the probability that p doesn't lie in the interval.

It now looks like a binomial with n = 40 and success = 90/100. For once in 40 trials we want 39 'successes' and one 'failure' = 40 * (90/100)^39 * (10/100)^1 and for no successes = 1 * (90/100)^0 * (10/100)^40

I seem to remember you can model this with another normal distribution with n, different p = 90/100 and q = 10/100. You put in an x (distance from the mean) of 39.5.

and

How does that sound?

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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