The operations manager of a large production plant would lik

The operations manager of a large production plant would like to estimate the mean amount of time a worker takes to assemble a new electronic component. Assume that the standard deviation of this assembly time is 3.6 minutes and is normally distributed. After observing 120 workers assembling similar devices, the manager noticed that their average time was 16.2 minutes. Construct a 92% confidence interval for the mean assembly time. How many workers should be involved in this study in order to have the mean assembly time estimated up to x 15 seconds with 92% confidence? Construct a 92% confidence interval if instead of observing 120 workers assembling similar devices, rather the manager observes 25 workers and notice their average time was 16.2 minutes w ith a standard deviation of 4.0 minutes.

Solution

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 =    0.04          
X = sample mean =    16.2          
z(alpha/2) = critical z for the confidence interval =    1.750686071          
s = sample standard deviation =    3.6          
n = sample size =    120          
              
Thus,              
Margin of Error E =    0.575334151          
Lower bound =    15.62466585          
Upper bound =    16.77533415          
              
Thus, the confidence interval is              
              
(   15.62466585   ,   16.77533415   ) [ANSWER]

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b)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.04  
      
Using a table/technology,      
      
z(alpha/2) =    1.750686071  
      
Also,      
      
s = sample standard deviation =    3.6  
E = margin of error =    15/60 = 0.25  
      
Thus,      
      
n =    635.5380207  
      
Rounding up,      
      
n =    636   [ANSWER]

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c)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.04          
X = sample mean =    16.2          
t(alpha/2) = critical t for the confidence interval =    1.828051172          
s = sample standard deviation =    4          
n = sample size =    25          
df = n - 1 =    24          
Thus,              
Margin of Error E =    1.462440938          
Lower bound =    14.73755906          
Upper bound =    17.66244094          
              
Thus, the confidence interval is              
              
(   14.73755906   ,   17.66244094   ) [ANSWER]