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) (5 points) 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) (5 points) How many workers should be involved in this study in order to have the mean assembly
time estimated up to

Solution

confidence interval = mean +- margin of error

margin of error = zscore * standard deviation/sqrt(n)

Zvalue at 92% = 1.75

a)

confidence interval is

16.2 - 1.75 * 3.6/sqrt(120) < u < 16.2 + 1.75 * 3.6/sqrt(120)

=>

15.625 < u < 16.775

b)

margin of error = 15/60 = 0.25

=>

1.75 * 3.6/sqrt(n) = 0.25

number of workers n = 635.04   = 635

c)

confidence interval is

16.2 - 1.75 * 4/sqrt(25) < u < 16.2 + 1.75 * 4/sqrt(25)

=>

14.8 < u < 17.6