Question

Lazurus Steel Corporation produces iron rods that are supposed to be 33 inches long. The machine that makes these rods does not produce each rod exactly 33 inches long. The lengths of the rods vary slightly. It is known that when the machine is working properly, the mean length of the rods made on this machine is 33 inches. The standard deviation of the lengths of all rods produced on this machine is always equal to 0.3 inch. The quality control department takes a sample of 23 such rods every week, calculates the mean length of these rods, and makes a 99% confidence interval for the population mean. If either the upper limit of this confidence interval is greater than 33.10 inches or the lower limit of this confidence interval is less than 32.9 inches, the machine is stopped and adjusted. A recent sample of 23 rods produced a mean length of 33.06 inches. Based on this sample, will you conclude that the machine needs an adjustment? Assume that the lengths of all such rods have a normal distribution. Round to two decimal places.

The confidence interval is...

Answer 1

Level of Significance ,    α =    0.010
population std dev ,    σ =    0.3000
Sample Size ,   n =    23
Sample Mean,    x̅ =   33.0600

Level of Significance ,    α =    0.01          
'   '   '          
z value=   z α/2=   2.576   [Excel formula =NORMSINV(α/2) ]      
                  
Standard Error , SE = σ/√n =   0.3000   / √   23   =   0.0626
margin of error, E=Z*SE =   2.5758   *   0.0626   =   0.1611
                  
confidence interval is                   
Interval Lower Limit = x̅ - E =    33.06   -   0.161129   =   32.8989
Interval Upper Limit = x̅ + E =    33.06   -   0.161129   =   33.2211
99%   confidence interval is (   32.90   < µ <   33.22   )

since upper limit is greater than 33.10 so machine need to be adjusted..