A previous analysis of paper boxes showed that the standard

A previous analysis of paper boxes showed that the standard deviation of their lengths is 15 millimeters. A packer wishes to find the 99% confidence interval for the average length of a box. How many boxes does he need to measure to be accurate within 5 millimeters?

Solution

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    15  
E = margin of error =    5  
      
Thus,      
      
n =    59.71406941  
      
Rounding up,      
      
n =    60   [ANSWER]