Question

Q.3. In a management study of workers’ wages, a simple random sample of 196 employees was taken from a production line work force. The mean of the sample was found to be Rs. 10000 per annum and the standard deviation Rs. 760. (i) Determine a 95% C-I for the work force, (ii) Determine a 99% C-I for the work force.?

solve the above problem step by step in proper format

Answer 1

a)

95% Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 196- 1 ) = 1.972
10000 ± 1.972 * 760/√(196)
Lower Limit = 10000 - 1.972 * 760/√(196)
Lower Limit = 9893
Upper Limit = 10000 + 1.972 * 760/√(196)
Upper Limit = 10107
95% Confidence interval is ( 9893 , 10107 )

b)

99% Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 196- 1 ) = 2.601
10000 ± 2.601 * 760/√(196)
Lower Limit = 10000 - 2.601 * 760/√(196)
Lower Limit = 9859
Upper Limit = 10000 + 2.601 * 760/√(196)
Upper Limit = 10141
99% Confidence interval is ( 9859 , 10141 )