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 0.006 ounces. Assume the standard deviation for the potato chip filling process is 0.05 ounces.

The sample size needed is? (round to nearest integrer)

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 =    0.05  
E = margin of error =    0.006  
      
Thus,      
      
n =    266.7679737  
      
Rounding up,      
      
n =    267   [ANSWER]