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Thread: 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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    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 $\displaystyle n=\frac{z_{\alpha/2}^2\sigma^2}{\varepsilon^2}$, where $\displaystyle \sigma=15$ and $\displaystyle \varepsilon=0.01$. Can you find $\displaystyle z_{0.025}$ 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 $\displaystyle \alpha $answer, it is the $\displaystyle \alpha/2$ answer (2.240)? I'm just so amazed they even have that on the table.

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

    Did I mess up somewhere? Than number seems way too huge. Thankyou
    Last edited by kclark36; Jun 30th 2009 at 06:51 AM. Reason: i forgot the math symbols
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    since the n formula divides $\displaystyle \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 $\displaystyle \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 $\displaystyle \alpha=0.05$. So when calculating the z value, you use $\displaystyle z_{0.05/2}=z_{0.025}=1.96$
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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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    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 $\displaystyle \varepsilon\to 0$, $\displaystyle n\to\infty$ (i.e. the smaller you make the error, the larger the sample size will be)
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