Question

"a pollster wishes to estimate the number of left-handed scientists. how large of a sample is needed in order to be 98% confident that the sample proportion will no differ from the mean by more than 4%. also suppose we know a previous study concluded that the proportion of left-handed scientists is 8%"

Answer 1

The sample size should be 250.

Our margin of error is 4%, or 0.04.  We use the formula

$\text{ME}=z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

To find the z-score:
Convert 98% to a decimal:  0.98
Subtract from 1:  1-0.98 = 0.02
Divide both sides by 2:  0.02/2 = 0.01
Subtract from 1:  1-0.01 = 0.99

Using a z-table (http://www.z-table.com) we see that this value has a z-score of approximately 2.33.  Using this, our margin of error and our proportion, we have:

$0.04=2.33\sqrt{\frac{0.08(1-0.08)}{n}} \\ \\0.04=2.33\sqrt{\frac{0.08(0.92)}{n}}$

Divide both sides by 2.33:
$\frac{0.04}{2.33}=\sqrt{\frac{0.08(0.92)}{n}}$

Square both sides:
$(\frac{0.04}{2.33})^2=\frac{0.08(0.92)}{n}$

Multiply both sides by n:
$(\frac{0.04}{2.33})^2n=0.08(0.92)$

Divide both sides to isolate n:
$\frac{0.08(0.92)}{(\frac{0.04}{2.33})^2}=n \\ \\249.73=n \\n\approx 250$