Question

How many observations should be taken to be 90% confident of being within + 2% of the actual time a machine is busy if it is believed that the machine is in use 70% of the time? Select one: A. 365 observations B. 1,421 observations C. 2,021 observations D. 789 observations

Answer 1

Solution:

Given that,

= 70% = 0.70

1 - = 1 - 0.70 = 0.30

margin of error = E = 2% = 0.02

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Sample size = n = ((Z _{/ 2}) / E)² * * (1 - )

= (1.645 / 0.02)² * 0.70 * 0.30

= 1420.66

= 1421

n = sample size = 1421

Option B ) is correct.