Question

The production manager for the XYZ manufacturing company is concerned that the customer orders are being shipped late. He asked one of his planners to check the timeliness of shipments for 900 orders. The planner randomly selected 900 orders and found that 100 orders were shipped late. Construct the 99% confidence interval for the proportion of orders shipped late.

Answer 1

Let X : number of success , number of orders that shipped late

So sample proportion of success , P = X/N

P = 100/900 = .11

so 100(1-a)% confidence interval for proportion of success

[ P - Z_(1-a/2)*√P(1-P)/N , P+Z_(1-a/2)*√P(1-P)/N ]

so a = .01, so a/2 = .005

(1-a/2 ) = .995

Now Z.995= 2.575

Then (1-P) = .89

So CI

2.575*√(.11*.89)/900 = .0268

So CI : ( .11-.0268 , .11+.0268 )

= ( .0832, .1368)