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.

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.

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?

Answer 1

Sample size = n = 120

Sample mean = = 16.2

Population standard deviation = = 3.6

We have to construct 92% confidence interval for the population mean.

Here population standard deviation is known so we have to use one sample z-confidence interval.

z confidence interval

Here E is a margin of error

Zc = 1.75   ( Using z table)

So confidence interval is ( 16.2 - 0.5753 , 16.2 + 0.5753) = > ( 15.6247  , 16.7753)

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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?

Confidence level = 0.92

Zc = 1.75    ( Using z table)

Margin of error = E = 15/60 = 0.25

Population standard deviation = = 3.6

We have to find sample size (n)