Question

artocchio studied and estimated the costs of In an article in the Journal of Management employee absences. Assume Infinite Maag of Maniteploriod Jos ulation. The mean amount of paid time lo Part 1 of3 th period was 1.0 days per employee with a f unpaid time lost during a three-month Popula with a standard deviation of 1.4 days. The mean period was 1.2 day per employee with a standard e of 100 blue-colar workers. me lost during a three-month period for time los ose we randomly select a sample of 100 blue- mount o e averag days? Please give a precise explanation- as if you are thing about statistics - for the reasoning behind your answer deviation of 1.6 days. Suppose we randomly select a the 100 blue-collar workers will exceed 1.5 telling someone who doesn't know anything Show steps 15 pts ys? Please sn't know anything about statistics amount of unpaid time lost during a three-month period en 1.4 and 1.5 days? Please give a precise explanation know anything about statistics - for the reasoning behind your (b) What is the probability 1.5 e 100 blue-collar workers wil be between as if you are telling someone who do answer I Show steps 15 pts esn't m sample is needed to construct a 95 percent confidence interval for the mean (c) How large a rando for the population paid time lost with a margin of error equal to .1? Please give a precise explanation as if you are telling someone who doesn't know anything about statistics-for the reasoning behind your answer (15 pts)

Answer 1

(a)

Standard error of mean paid time lost = $\sigma/\sqrt{n} = 1.4 / \sqrt{100}$ = 0.14

We know that the mean of any distribution follow normal distribution.

Let X be the mean paid time lost.

Thus, probability that mean paid time lost exceeds 1.5 days

= P(X > 1.5) = P[Z > (1.5 - 1) / 0.14] = P[Z > 3.57]

= 0.0002   

(b)

Standard error of mean unpaid time lost = $\sigma/\sqrt{n} = 1.6 / \sqrt{100}$ = 0.16

We know that the mean of any distribution follow normal distribution.

Let Y be the mean unpaid time lost.

Thus, probability that mean paid time lost exceeds 1.5 days

= P(1.4 < Y < 1.5) = P(Y < 1.5) - P(Y < 1.4) = P[Z < (1.5 - 1.2) / 0.16] - P[Z < (1.4 - 1.2) / 0.16]

= P[Z < 1.875] - P[Z < 1.25]

= 0.9696 - 0.8944

= 0.0752

(c)

z value for 95% confidence interval is 1.96

Sample size = (z * $\sigma$ / Margin of error)²

= (1.96 * 1.4 / 0.1)²

= 753 (Rounding to nearest integer)