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 6 minutes. How large a sample of workers should she take if she wishes to estimate the mean assembly time to within 30 seconds with a confidence level of 95%. (Be sure to use plenty of deicmal places throughout the calculation, and then round your final answer up to an integer)

Sample Size =

Answer 1

Solution :

Given that,

standard deviation = = 6 minutes

margin of error = E = 30 seconds = 30 / 60 = 0.5 mminutes

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Sample size = n = ((Z_/2 * ) / E)²

= ((1.96 * 6) / 0.5)²

= 553.19 = 554 minutes

Sample size = 554 minutes