The production manager for the XYZ manufacturing company is

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 1000 orders. The planner randomly selected 1000 orders and found that 200 orders were shipped late. Construct the 99% confidence interval for the proportion of orders shipped late.

PLEASE SHOW WORK to help me understand

Solution

No of orders n = 1000

No of orders that were shipped late = 200

Proportion of orders that were shipped late p = 200/1000 = 0.20

Standard Error = sqrt[p * (1 - p)/n] = sqrt[0.20 * (1 - 0.20)/1000] = 0.0126

alpha = 1- 0.99 = 0.01

Critical value for alpha = 0.01 is +2.575

Margin of Error = Critical value * Std error = +2.575 * 0.0126 = +0.0326

confidence interval for the proportion of orders shipped late = (sample proportion + Margin of error)

                                                                                                  = 0.20 +0.0326

                                                                                                  = (0.1674 , 0.2326)