Please show work so I can see how you got to the answer. The operations manager

Question

Please show work so I can see how you got to the answer.

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. Assuming the standard deviation of this assembly time is 3.6 minutes. (i). 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. (ii). How many workers should be involved in this study in order to have the mean assembly time estimated up to 15 seconds with 92% confidence.

Explanation / Answer

(i) Given a=1-0.92=0.08,

                   Z(0.04) =1.75 (from standard normal table)

So the lower bound is

xbar - Z*s/vn =16.2- 1.75*3.6/sqrt(120)

                    =16.2-0.575108685

                    =15.62489131

So the upper bound is

xbar + Z*s/vn =16.2+ 1.75*3.6/sqrt(120)

                     =16.2+0.575108685

                      =16.77510869

(ii) 15 seconds = 15/60

                        =0.25 minutes

    So n=(Z*s/E)^2

             =(1.75*3.6/0.25)^2

             =635.04

            n=636(round upto)

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