<PRE><B>Question 1043988</B>
The lower the confidence the smaller the sample size needed, A.
the standard error is z*s/sqrt(n), and we want that to equal 2
z*s/sqrt(n)=2; z^2*s^2=4*n, by squaring everything and multiplying both sides by n.
n=z^2*s^2/4
z(.995)=2.576
sigma=13.3
n=293.45 or 294
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Everything is the same for 95% confidence except the z value is 1.96
3.8416*(13.3)^2/4=169.88 or 170, rounding to the higher number.

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