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 7 minutes. How large a sample of workers should she take if she wishes to estimate the mean assembly time to within 19 seconds. Assume the confidence level to be 99%.

Answer 1

Solution :

Given that,

standard deviation = = 7 * 60 second = 420

margin of error = E = 19

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Sample size = n = ((Z_/2 * ) / E)²

= ((2.576 * 420) / 19)²

= 3242.34

                              = 3242

Sample size = 3242