Question

How many observations should a time study analyst plan for in an operation that has a standard deviation of 3.23 minutes per piece if the goal is to estimate the mean time per piece to within .16 minute with a confidence of 98.36 percent?

Hint: When rounding your answer follow exactly the same steps as what the equivalent problem set from your homework. Canvas will only accept the whole number. No decimal places.

Answer 1

standard deviation, σ	3.23	m/pc
mean time per piece to within	0.16	m
confidence	0.9836
z value	2.134523
Let the number of observations be n.
sample std deviation, s = σ/√n
Upper Limit = mean + z*s
Lower limit = mean - z*s
upper limit - lower limit = 2*z*s
confidence interval = upper limit - lower limit
hence,
0.16=2*z*s
0.16=2*z*σ/√n
√n=2*z*σ/0.16
n=(2*z*σ/0.16)^2
n=(2*2.13*3.23/0.16)^2
n=	7427.227
hence, the number of observations should be 7428

formulas used:

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