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The personnel manager wants to determine if there is a difference in the average time lost due to absenteeism between two plants. From historical data, the estimated standard deviations of lost time are 200 and 250 minutes, respectively for plants 1 and 2.

(a) Assuming that equal sample sizes will be selected from each plant, what should the sample size be if the bound on the error of estimation is 40 minutes with a probability of 0.90

So here the formula:

$n=\frac{\mathbb{Z}_\frac{\alpha}{2}^2 \cdot (\sigma^2_1+\sigma^2_2) }{B^2}$

So getting the information from the question I got:

$n=\frac{\mathbb{Z}_{0.05}^2 \cdot (200+250) }{40^2}=173.35473$

So the sample size would be 173, did i do this correctly?

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I see some quibbles with the way you displayed your work: (1) For 90%, I think $z = 1/645,$ not sure what you used. (2) The number 200 and 250 are standard deviations and need to be squared to get variances. (3) It is customary to round up to the next higher integer. Even so, using R as a calculator, I got about the same answer you did.

1.645^2*(200^2 + 250^2)/40^2
## 173.3547
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