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.

a) 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.

b) How many workers should be involved in this study in order to have the mean assembly time estimated up to

Solution

(a) 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) =15.62489

So the upper bound is

xbar + Z*s/vn =16.2+ 1.75*3.6/sqrt(120) =16.77511

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(b) 15 seconds = 0.25 minutes

So n=(Z*s/E)^2

=(1.75*3.6/0.25)^2

=635.04

Take n=636

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(c) The degree of freedom =n-1=25-1=24

Given a=a=1-0.92=0.08, t(0.04, df=24) = 1.828 (from student t table)

So the lower bound is

xbar - t*s/vn =16.2- 1.828*4/sqrt(25) =14.7376

So the upper bound is

xbar + t*s/vn =16.2+ 1.828*4/sqrt(25) =17.6624