Question

Statistical estimation

A metalworking company has 192 operators. In a random sample of 50 of them, the average number of overtime hours worked in a week and their standard deviation was 9.6 and 6.2 hours respectively.

a.In order to program the economic resources for all the personnel of operators, the personnel department decides: To estimate with a confidence of 96% the average number of overtime hours worked by each operator during a week.
b. Estimate with a 99% confidence the total number of overtime hours worked by the company's operators for a week.

Answer 1

Solution :

Given that,

sample size = n = 50

Degrees of freedom = df = n - 1 = 50 - 1 = 49

a)

t _/2,df = = 2.1099

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

= 2.010 * (6.2 / 50)

Margin of error = E = 1.84999

#Margin of error = E=1.85

The 96% confidence interval estimate of the population mean is,

- E < < + E

9.6 - 1.85 < < 9.6 + 1.85

7.8 < < 11.45

(7.75 ,11.45)

b)

t _/2,df = 2.680

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

= 2.680 * (6.2 / 50)

Margin of error = E = 2.3

The 99% confidence interval estimate of the population mean is,

- E < < + E

9.6 - 2.3 < < 9.6 + 2.3

7.3 < < 11.9)

(7.3 , 11.9)