1. ## confident interval question

Suppose the number of successes in n=200 binomial trials is 20.

a) Find a 90% confidence interval for p, the probability of success at each trial.

b) Find a 98% CI to p. Why is this interval wider than the previous one?

I know how to construct an confidence interval if given the n, s, and x bar, but how do I do it with this question?

2. ## Re: confident interval question

Try $\displaystyle \displaystyle p\pm Z_{0.9}\times \sqrt{\frac{p(1-p)}{n}}$

where n= 200 and p = 20/200

3. ## Re: confident interval question

Originally Posted by Bushy
Try $\displaystyle \displaystyle p\pm Z_{0.9}\times \sqrt{\frac{p(1-p)}{n}}$

where n= 200 and p = 20/200
you meant $\displaystyle z_0.05$, because 1-a is 90%, right? But how do you know 20/200 is P?

4. ## Re: confident interval question

Originally Posted by wopashui
you meant $\displaystyle z_0.05$, because 1-a is 90%, right?
I get z=1.645

Originally Posted by wopashui
But how do you know 20/200 is P?
Because there was 20 sucesses out of the 200.