How many observations should a time study analyst plan for in an operation that

Question

How many observations should a time study analyst plan for in an operation that has a standard deviation of 1.2 minutes per piece if the goal is to estimate the mean time per piece to within .6 minute with a confidence of 95.5 percent? (Do not round intermediate calculations. Round up your final answer to the next whole number.)

How many observations should a time study analyst plan for in an operation that has a standard deviation of 1.2 minutes per piece if the goal is to estimate the mean time per piece to within .6 minute with a confidence of 95.5 percent? (Do not round intermediate calculations. Round up your final answer to the next whole number.)

Explanation / Answer

Basically n = (zs/e)^2

Here z= Number of normal std. deviations for desired confidence

s= Sample std. deviation

e= Max acceptable

n= no. of observations needed

Given:

z for confidence level 95.5 % = 1.967

s= Sample std. deviation = 1.2

e= Max acceptable amount of time error = 0.6

n = ((1.967*1.2)/0.6)^2 = (2.3604/0.6)^2 = (3.934)^2 = 15.4763= 15.5 = 16

So, no of observations needed = 16

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