# Choosing Sample Size problem on MyOpenMath

Here is the screenshot of my problem from MyOpenMath. I've been following the formula from last Friday's slideshow on how to do this type of problem and I just can't seem to get the right answer. I know that the 90% confidence level would give me a Z* of 1.644, but I'm confused about where to go from there.

## Comments

• You can apply the formulae on this class slide. In particular:

$n> (z^**)^2 / (4 ME^2)$.

Plugging a confidence level of 0.9 into our Normal Calculator, I find that our $z^$-multiplier is $z^ = 1.644854$. Thus, we get

$n> (z^**)^2 / (4 ME^2) = 1.644854^2 / (4 times 0.01^2) = 6763.86$.

Thus, a sample of size 6764 should suffice.

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