To obtain the estimate of time a worker is busy, a manager divides a typical wor
ID: 3227522 • Letter: T
Question
To obtain the estimate of time a worker is busy, a manager divides a typical workday into 480 minutes. Using a random-number table to devise what time to go to an area to observer work occurrences, the manager records observations in a tally sheet like the following. a) Provide the 95% confidence interval of the proportion of time that the worker is busy? b) What is your estimate of the time that the worker is busy? c) The company has a policy for time standards of 90% confidence and 10% accuracy. Based on the available data and your knowledge as a WSU industrial Engineer, do you recommend that the company collect more data? If yes, explain how many more observations are needed and why. If not, explain why no more observations are needed. Circle your answer.Explanation / Answer
By observing tally sheet we can say that tha manager went for inspection for around 20 times, out of which (5+5+5 +1 = 16) times he found worker is busy and production is working and rest 4 times he found worker idle.
(a) Proportion of time worker is busy p^= 16/20 = 0.8
Now we have caclualte 95% confidence interval
so 95% CI = p^ +- Z0.05 * sqrt [p^ * (1-p^)/ n] = 0.8 +- 1.96 * sqrt [ 0.8 * 0.2 /20]
95% CI = 0.8 +- 1.96 * 0.0894 = (0.625, 0.975)
(b) Estimate of time that the worker is busy = Total TIme * proportion = 480 * (0.8) = 384 minutes
(c) Here sample size is n
Here 90% confidence = p^ +- Z0.1 sqrt [ p^ (1-p^)/n] = 0.8 +- 1.645 * sqrt [ 0.8* 0.2/n]
= 0.8 +- 1.645 sqrt (0.16/n)
10% accuracy that means (2 Z0.05 ) times standard deviation will be 10% of the true proportion
=> 0.1 * p^ = 2Z0.1 sqrt [ p^ (1-p^)/n]
=> 0.1 * 0.8 = 2 * 1.645 * sqrt [ 0.8 * 0.2 /n]
=> 0.08 = 3.29 * sqrt (0.16/n)
=> sqrt (n/0.16) = 41.125
=> n = 270.6025 = 271
so to have 90% confidence and 10% accuracy the company must collect more data and this must be minimum of 271 times should it be collected to have such accuracy.
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