Question

It is desired to know the number of readings to time in production operation with a confidence level of 90% and an error of 8% for which the following times have been recorded at the preliminary level determined.

Lecture	time(min)	Lecture	Time(min)	Lecture	Time(min)	Lecture	Time(min)
1	8.8	4	7.7	7	7.3	10	6.8
2	7	5	7.8	8	6.2	11	6.1
3	6.8	6	6.3	9	8.1	12	8

A) Average time

B) standard deviation

C) number of readings to time to meet the requirements that the study requires

D) error reached with those 12 data

Answer 1

A) Average time, m = (8.8+7+6.8+7.7+7.8+6.3+7.3+6.2+8.1+6.8+6.1+8)/12 = 7.24

B) Standard deviation, s = SQRT(((8.8-7.24)^2+(7-7.24)^2+(6.8-7.24)^2+(7.7-7.24)^2+(7.8-7.24)^2+(6.3-7.24)^2+(7.3-7.24)^2+(6.2-7.24)^2+(8.1-7.24)^2+(6.8-7.24)^2+(6.1-7.24)^2+(8-7.24)^2)/(12-1))

= 0.852

C) Desired confidence level = 90%

z = NORMSINV(1-(1-.9)/2) = 1.645

Margin of error, e = 8% = 0.08

Number of readings required = (z*s/(e*m))^2

= (1.645*0.852/(0.08*7.24))^2

= 6

D) Error reach with those 12 data

e = zs/(m*sqrt(n) )

= 1.645*0.852/(7.24*sqrt(12))

= 0.056 or 5.6 %