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Math Help - find the sample size required for 95% CI

  1. #1
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    find the sample size required for 95% CI

    Given a random variable X with distribution N(mu,225) how can I find the sample size n required for a 95% confidence interval for the mean mu with an error of at most epsilon=0.01?

    The only formula I can find is n=(z^2*s^2)/.25 Can I somehow use the population variance (take the square root of 225) and substitute it into an equation? I hope someone can help me. My search through websites has only made it worse. THANKYOUUUU!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by kclark36 View Post
    Given a random variable X with distribution N(mu,225) how can I find the sample size n required for a 95% confidence interval for the mean mu with an error of at most epsilon=0.01?

    The only formula I can find is n=(z^2*s^2)/.25 Can I somehow use the population variance (take the square root of 225) and substitute it into an equation? I hope someone can help me. My search through websites has only made it worse. THANKYOUUUU!
    Since variance is given to you, you can apply the following formula n=\frac{z_{\alpha/2}^2\sigma^2}{\varepsilon^2}, where \sigma=15 and \varepsilon=0.01. Can you find z_{0.025} and take it from here?
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    thankyou! I knew it was something like that. For the 0.025, I would get than answer from the table, but rather than the \alpha answer, it is the \alpha/2 answer (2.240)? I'm just so amazed they even have that on the table.

    \frac{2.24^{2}*225}{.0001}=\frac{1128.96}{.0001}=1  1289600=n?

    Did I mess up somewhere? Than number seems way too huge. Thankyou
    Last edited by kclark36; June 30th 2009 at 07:51 AM. Reason: i forgot the math symbols
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    since the n formula divides \alpha by 2, for 95% would it be 0.025/2 for a z value of 2.240? or is it 0.05/2 for a z value of 1.96? Thanks
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by kclark36 View Post
    since the n formula divides \alpha by 2, for 95% would it be 0.025/2 for a z value of 2.240? or is it 0.05/2 for a z value of 1.96? Thanks
    Since the you're looking at a 95% CI, it follows \alpha=0.05. So when calculating the z value, you use z_{0.05/2}=z_{0.025}=1.96
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    Thank you! That does change the answer quite a bit, from 11289600 to 8643600. Still big, but this time hopefully its right.
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by kclark36 View Post
    Thank you! That does change the answer quite a bit, from 11289600 to 8643600. Still big, but this time hopefully its right.
    Don't be scared of the size of the result...

    As you make \varepsilon\to 0, n\to\infty (i.e. the smaller you make the error, the larger the sample size will be)
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