Suppose a computer engineer is interested in determining the

Suppose a computer engineer is interested in determining the average weight of a motherboard manufactured by a certain company. A summary of a large sample provided to the engineer suggest a mean weight of 11.8 ounces and an estimated standard deviation, sigma = 0.75. How large a sample size is required if want a 99% confidence interval, with a tolerable interval width of 0.4? How large a sample would we need if were interested in a 95% confidence interval with a tolerable width of 0.5?

Solution

A)

Note that

margin of error = width/2 = 0.4/2 = 0.2.

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 =    0.75  
E = margin of error =    0.2  
      
Thus,      
      
n =    93.30323345  
      
Rounding up,      
      
n =    94   [answer]

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b)

Margin of error = 0.5/2 = 0.25.

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 =    0.75  
E = margin of error =    0.25  
      
Thus,      
      
n =    34.57312939  
      
Rounding up,      
      
n =    35 [ANSWER]