ieor_hw2_sol - IEOR 165: Engineering Statistics, Quality...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: IEOR 165: Engineering Statistics, Quality Control and Forecasting, Spring 2008 Homework 2 Solution Chapter 7 Question 8 Noting X = 3.1502 (a) 95 percent CI: X 1.96 n = 3.1502 1.96(.1)/ 5 = (3.0625, 3.2379) (b) 99 percent CI: X z.005 n = 3.1502 12.58(.1)/ 5 = (3.0348, 3.2656) Question 13 99 percent CI: X z.005 n = 1.2 z.005 0.2/ 20 = (1.0848, 1.3152) Question 14 S 99 percent CI: X t.005,n-1 n = 1.2 t.005,19 0.2/ 20 = (1.0720, 1.3280) Question 17 S Using the two-sided CI on mean with unknown variance X t/2,n-1 n (a) 95 percent CI: (331.0572,336.9345) (b) 99 percent CI: (330.0082,337.9836) Question 18 Noting X = 133.22, S = 10.2127 S (a) 95 percent CI: X t.025,n-1 n = (128.14, 138.30) S (b) 95 percent lower CI: (-, X + t.05,n-1 n ) = (-, 137.41) S (c) 95 percent upper CI: (X - t.05,n-1 , ) = (129.03, ) n Question 22 S (a) 95 percent CI: X t.025,n-1 n = 330.2 2.094(15.4)/ 20 = (332.9892, 337.4108) S (b) 99 percent CI: X t.05,n-1 n = 330.2 2.861(15.4)/ 20 = (320.3480, 340.0520) Question 23 S 95 percent CI: X t.025,n-1 n = 1220 1.968(840)/ 300 = (1124.5571, 1315.4429) Question 26 Using the same CI calculation method as question 17 (a) (2013.9, 2111.6) (b) (1996.0, 2129.6) (c) This is just the left hand side of a 95 percent upper CI: 2022.4 Question 30 S Noting X = 11.5667, S = 3.98, 95 percent CI: X t.025,n-1 n = (10.08, 13.05) Question 31 S 95 percent CI: X t.025,n-1 n = (85442.15, 95457.85) 1 ...
View Full Document

This note was uploaded on 05/07/2008 for the course IEOR 165 taught by Professor Shanthikumar during the Spring '08 term at University of California, Berkeley.

Ask a homework question - tutors are online