A producer of a variety of salty snacks would like to estima

A producer of a variety of salty snacks would like to estimate the average weight of a bag of BBQ potato chips produced during the filling process at one of its plants. Determine the sample size needed to construct a 95% confidence interval with a margin of error equal to 0.003 ounces. Assume the standard deviation for the potato chip filling process is 0.07 ounces. The sample size needed is . (Round up to the nearest integer.)

Solution

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.025  
      
Using a table/technology,      
      
z(alpha/2) =    1.959963985  
      
Also,      
      
s = sample standard deviation =    0.07  
E = margin of error =    0.003  
      
Thus,      
      
n =    2091.460913  
      
Rounding up,      
      
n =    2092   [ANSWER]