To estimate the average time it takes to assemble a certain

To estimate the average time it takes to assemble a certain computer component, the industrial engineer at an electronics firm timed 64 technicians in the performance of this task, getting a mean of 12.50 minutes and a variance of 4.00 minutes. Construct a 98 Percentage confidence interval for the true average time it takes to assemble the computer component. How large a sample is needed so that the engineer will be able to assert with 99 percentage confidence that the error is at most 0.3 minutes

Solution

2.

A)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    12.5          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    2          
n = sample size =    64          
              
Thus,              
Margin of Error E =    0.411213407          
Lower bound =    12.08878659          
Upper bound =    12.91121341          
              
Thus, the confidence interval is              
              
(   12.08878659   ,   12.91121341   ) [ANSWER]

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b)

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 =    2  
E = margin of error =    0.2  
      
Thus,      
      
n =    384.1458821  
      
Rounding up,      
      
n =    385   [ANSWER]