Question

a previous analysis of paper boxes showed that the standard deviation of the length is 9mm. What is the 98% confidence interval for the average length of a box. How many boxes does he need to measure to be accurate within 3 millimeters.

Answer 1

We cannot calculate the confidence interval without the sample size.  However, for the second question, the sample size needed would be 49.

The formula we use is

$n=(\frac{z_{\frac{\alpha}{2}}\times \sigma}{E})^2$

To find the z-score:
Convert 98% to a decimal:  0.98
Subtract from 1:  1-0.98 = 0.02
Divide 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 is closest to a z-score of 2.33.

Using the z-score, our standard deviation (9) and our margin of error (3), we have:
$n=(\frac{2.33(9)}{3})^2=(6.99)^2=48.86\approx 49$