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?

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

Re: confident interval question

Quote:

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?

Re: confident interval question

Quote:

Originally Posted by

**wopashui** you meant $\displaystyle z_0.05$, because 1-a is 90%, right?

I get z=1.645

Quote:

Originally Posted by

**wopashui** But how do you know 20/200 is P?

Because there was 20 sucesses out of the 200.