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Math Help - confident interval question

  1. #1
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    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?
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  2. #2
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    Re: confident interval question

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

    where n= 200 and p = 20/200
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  3. #3
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    Re: confident interval question

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

    where n= 200 and p = 20/200
    you meant z_0.05, because 1-a is 90%, right? But how do you know 20/200 is P?
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  4. #4
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    Re: confident interval question

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

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