The operations manager of a large production plant would like to estimate the me

Question

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 plusminus 15 seconds with 92% confidence? c) 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 with a standard deviation of 4.0 minutes.

Explanation / Answer

a) here std error =std deviation/(n)1/2 =3.6/(120)1/2 =0.3286

for 92% confidence ; z=1.7507

therefore 92% confidence interval =sample mean -/+ z*std error =15.62 ; 16.78

b) for margin of error E =15

std deviaiton =3.6 minute =3.6*60=216 seconds

for 92% CI; z=1.7507

therefore sample size n=(z*std deviation/E)2 =~636

c)

here std error =std deviation/(n)1/2 =4/(25)1/2 =0.80

for 92% confidence ; t=1.828

therefore 92% confidence interval =sample mean -/+ t*std error =14.74 ; 17.66

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