the human resource director of a large corporation wishes to study absenteeism among its mid level...

the human resource director of a large corporation wishes to study absenteeism among its mid level managers at its central office during the year a random sample of 25 mid level managers reveals the following Absenteeism x =6.2 days s= 7 days. 13 mid level managers cite stress as a cause of absence. construct a 95% confidence interval estimate for the mean number of absences for mid level managers during the year

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Expert Solution

Solution :

Given that,

Point estimate = sample mean = = 6.2

sample standard deviation = s = 7

sample size = n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t_/2,df = t_0.025,24 = 2.064

Margin of error = E = t_/2,df * (s / $\sqrt$ n)

= 2.064 * ( 7/ $\sqrt$ 25)

= 2.9

The 95% confidence interval estimate of the population mean is,

- E < < + E

6.2 - 2.9 < < 6.2 + 2.9

3.3 < < 9.1

A 95% confidence interval estimate for the mean is (3.3 , 9.1)

orchestra answered 1 year ago

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