Assume that for the existing machine the sample mean bolt di

Assume that, for the existing machine, the sample mean bolt diameter is 5.00 mm from a sample of             100     bolts, the population standard deviation is known to be 0.025 mm, and the bolt diameter measurement is normally distributed. Calculate the following quantities, and include units in your answers:

a) A 95% twosided confidence interval for the true mean bolt diameter, .

b) A 99% twosided confidence interval for the true mean bolt diameter, .

c) Explain the difference between the width of the intervals in part a) and part b).

d) Suppose we wanted to be 90% confident that the error in estimating the mean bolt diameter is less than 0.01 mm. What minimum sample size is required?  

Solution

A)

Note that              
              
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 =    0.025          
X = sample mean =    5          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    0.025          
n = sample size =    100          
              
Thus,              
              
Lower bound =    4.99510009          
Upper bound =    5.00489991          
              
Thus, the confidence interval is              
              
(   4.99510009   ,   5.00489991   ) [ANSWER]

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B)

Note that              
              
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 =    0.005          
X = sample mean =    5          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    0.025          
n = sample size =    100          
              
Thus,              
              
Lower bound =    4.993560427          
Upper bound =    5.006439573          
              
Thus, the confidence interval is              
              
(   4.993560427   ,   5.006439573   ) [ANSWER]

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C)

The larger the confidence level is, the greater the margin of error. You want to be more confident that you include the true mean, so you widen your interval.

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d)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.05  
      
Using a table/technology,      
      
z(alpha/2) =    1.644853627  
      
Also,      
      
s = sample standard deviation =    0.025  
E = margin of error =    0.01  
      
Thus,      
      
n =    16.90964659  
      
Rounding up,      
      
n =    17   [ANSWER]