In: Statistics and Probability
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
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 = t0.025,24 = 2.064
Margin of error = E = t/2,df * (s /n)
= 2.064 * ( 7/ 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)