Question

The operations manager of a large production plant would like to estimate the average amount of time workers take to assemble a new electronic component. After observing a number of workers assembling similar devices, she guesses that the standard deviation is 10 minutes. How large a sample of workers should she take if she wishes to estimate the mean assembly time to within 20 seconds. Assume the confidence level to be 97%.

Answer 1

Solution :

Given that,

standard deviation = = 10 minutes = 10 * 60 = 600 second

margin of error = E = 20 second

At 97% confidence level the z is ,

= 1 - 97% = 1 - 0.97 = 0.03

/ 2 = 0.03 / 2 = 0.015

Z_/2 = Z_0.015 = 2.17

Sample size = n = ((Z_/2 * ) / E)²

= ((2.17 * 600) / 20)²

= 4238 secons

Sample size = 4238 seconds = 4238 / 60 = 70.63 minutes