Question

Managers believe that the average time to process a package at the front of the counter of the Mercuric post office is about three minutes. They want to do a work sampling study to estimate this time more accurately. From a small pilot sample, the standard deviation is estimated to be 0.16 minutes. Management wants to find the sample size needed to achieve a maximum allowable sampling error of ±0.025 minutes with 95 percent confidence.

Note: Answer in two decimal places.

Answer 1

Solution :

Given that,

standard deviation=   =0.16

Margin of error = E = 0.025

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

n = ( 1.96*0.16 /0.025 )²

n =157.35

Sample size = n =157