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 known from historical records to be 3.6 minutes. a. After observing 60 workers assembling similar devices, the manager found 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 with 15 seconds precision with 92% confidence?

Explanation / Answer

( a ) Here the population standard deviation is given so we use z-distribution to find the confidence interval.

x bar (-/+) E

x bar = 16.2

E = zc * ( s / Ön)

s = 3.6

n = 60

zc is the critical value. We use normal table and we get the critical value for 8% level of significance as 1.75.

E = 1.75 * ( 3.6 / Ö60) = 0.8133

x bar (-/+) E

16.2 (-/+) 0.8133

15.39 and 17.01

Answer:

The 92% confidence interval is (15.39 t0 17.01).

( b )

n = (zc s/ E)^2

   = (1.75*3.6)^2 / 0.25^2             [ 15 seconds = 0.25 minutes ]

   = 635.04

We take the next digit as sample size cannot be in fraction.

Answer:

636

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