Question

A total of 280 observations of Bob Ramos, an assembly-line worker, were made over a 40-hour work week. The sample also showed that Bob was busy working (assembling the parts) during 240 observations. (Round all intermediate calculations to at least two decimal places before proceeding with further calculations.)

a) a) The percent of time Bob is working =85.71

b) The number of observations that need to be taken of Bob to be 95 % confident with 5 % acceptable error = __

c) Was the sample size adequate?

What are the answers for b) and c)?

Answer 1

(a) The percent of time Bob was working = 240/280 = 0.8571 =85.71%

(b) The number of sample needed for 95% confidence limits with 5% acceptable error

n= z² x p x (1-p) / h²

where z =1.96 for 95% confidence limits

p = 0.8571 ( idle or busy time)

h =0.05 ( acceptable error)

n = (1.96)² x 0.8571 x( 1-0.8571) / (0.05)²

= 188.2 =188

(c) Sice the sample size in the question is 280, it is adequate.