# Math Help - determine sample size

1. ## determine sample size

I hope someone can work out this problem for me to use as a reference for my homework. Thank you so much in advance.

The length in centimeters if n=29 fish yielded an average length of x=16.82 (i don't know how to put the line over the x) and $s^{2}=34.9$. Determine the size of a new sample so that x(+/-)0.5 is an approximate 95% confidence interval for $\mu$.

2. A 95% confidence interval for $\mu$ is $\bar{x} \pm z_{0.025} {\sigma \over \sqrt{n}}$

$\sigma$ is unknown but since the sample size is fairly large, you can replace it with $s$

What you want is to find a new $n$ such that $z_{0.025} {s \over \sqrt{n}} = 0.5$.

Solving for n you get $n= \frac {z_{0.025}^2 s^{2}}{0.25}$

3. I'm glad someone asked this question! I wasn't sure I did it correctly. It really is so simple. Since I got an answer of 536.28, it makes sense to round up to get 534. So the new sample size would be n=534.

4. IF you did get 536.28, why would you use 534?
and you really should round up.
The books are weak on that point, any number above 537 would give at least that precision.

5. I'm sorry, I meant to type 537. The book really did not explain that, but it maked sense. It seems to be weak in other areas as well. thank you

6. What you're doing is setting $z_{0.025} {\sigma \over \sqrt{n}}\le .5$ and solving for n.

And if you don't know $\sigma$, then you try to estimate it somehow.
You may have some prior information, but you really haven't calculated s yet.

7. I am so desperate to learn how to do these types of problems! To get the SD, can't I just take the square root of the variance? So since the square root of 34.9 is 5.9, sigma=5.9. Also isn't sigma the same as s?

$\sigma$ is the standard deviation of the population
$s$ is the standard deviation of the sample