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 180 orders were shipped late. Construct the 95% confidence interval for the proportion of orders shipped late.

Please show all work, please type so it is legible, thank you

Answer 1

Solution :

Given that,

n = 900

x = 180

= x / n = 180 / 180 = 0.2

1 - = 1 - 0.2 = 0.8

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

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 * (((0.2 * 0.8) / 900)

= 0.026

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.2 - 0.026 < p < 0.2 + 0.026

0.174 < p < 0.226

(0.174,0.226)

The 95% confidence interval for the proportion of orders shipped late is from 0.174 to 0.226