# Thread: find the sample size required for 95% CI

1. ## 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!

2. Originally Posted by kclark36
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?

3. 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

4. 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

5. Originally Posted by kclark36
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 $z_{0.05/2}=z_{0.025}=1.96$

6. Thank you! That does change the answer quite a bit, from 11289600 to 8643600. Still big, but this time hopefully its right.

7. Originally Posted by kclark36
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)