Transcript Chapter09
Chapter 9 Hypothesis Testing with Single Samples © 2002 Thomson / South-Western Slide 9-1 Learning Objectives • Understand the logic of hypothesis testing, and know how to establish null and alternate hypotheses. • Understand Type I and Type II errors. • Use large samples to test hypotheses about a single population mean and about a single population proportion. • Test hypotheses about a single population mean using small samples when is unknown and the population is normally distributed. © 2002 Thomson / South-Western Slide 9-2 Method of Indirect Proof X Either X or Y is true but not both Y X X is demonstrated not to be true Y Y is true by default Y © 2002 Thomson / South-Western Slide 9-3 Hypothesis Testing A process of testing hypotheses about parameters by setting up null and alternative hypotheses, gathering sample data, computing statistics from the samples, and using statistical techniques to reach conclusions about the hypotheses. © 2002 Thomson / South-Western Slide 9-4 Steps in Testing Hypotheses 1. Establish hypotheses: state the null and 2. 3. 4. 5. 6. 7. 8. alternative hypotheses. Determine the appropriate statistical test and sampling distribution. Specify the Type I error rate ( State the decision rule. Gather sample data. Calculate the value of the test statistic. State the statistical conclusion. Make a managerial decision. © 2002 Thomson / South-Western Slide 9-5 Null and Alternative Hypotheses • The Null and Alternative Hypotheses are mutually exclusive. Only one of them can be true. • The Null and Alternative Hypotheses are collectively exhaustive. They are stated to include all possibilities. (An abbreviated form of the null hypothesis is often used.) • The Null Hypothesis is assumed to be true. • The burden of proof falls on the Alternative Hypothesis. © 2002 Thomson / South-Western Slide 9-6 Null and Alternative Hypotheses: Example • A soft drink company is filling 12 oz. cans with cola. • The company hopes that the cans are averaging 12 ounces. Ho: 12 oz Ha: 12 oz © 2002 Thomson / South-Western Slide 9-7 Rejection and Nonrejection Regions Rejection Region Rejection Region Nonrejection Region =12 oz Critical Value © 2002 Thomson / South-Western Critical Value Slide 9-8 Type I and Type II Errors • Type I Error – Rejecting a true null hypothesis – The probability of committing a Type I error is called , the level of significance. • Type II Error – Failing to reject a false null hypothesis – The probability of committing a Type II error is called . – Power is the probability of rejecting a false null hypothesis, and equal to 1- © 2002 Thomson / South-Western Slide 9-9 Decision Table for Hypothesis Testing Null True Null False Fail to reject null Correct Decision Type II error ( ) Reject null Type I error () Correct Decision (Power) © 2002 Thomson / South-Western Slide 9-10 One-tailed and Two-tailed Tests • One-tailed Tests Ho: 12 Ho: 12 Ha: 12 Ha: 12 • Two-tailed Test Ho: 12 Ha: 12 © 2002 Thomson / South-Western Slide 9-11 One-tailed Tests Ho: 12 Ho: 12 Ha: 12 Ha: 12 Rejection Region Nonrejection Region =12 oz Critical Value © 2002 Thomson / South-Western Rejection Region Nonrejection Region =12 oz Critical Value Slide 9-12 Two-tailed Tests Ho: 12 Ha: 12 Rejection Region Rejection Region Nonrejection Region =12 oz Critical Values © 2002 Thomson / South-Western Slide 9-13 CPA Net Income Example: Two-tailed Test Ho : $74,914 If Z Zc 196 . , reject Ho. If Z Zc 196 . , do not reject Ho. Ha : $74,914 .025 2 .025 2 Rejection Region Nonrejection Region Rejection Region © 2002 Thomson / South-Western X n 78, 646 74, 914 2.75 14,530 112 Z = 2.75 Zc = 1.96, reject Ho =0 Zc 196 . Z Zc 196 . Slide 9-14 CPA Net Income Example: Critical Value Method (Part 1) Lower Ho: $74,914 X Z n Ha: $74,914 14,530 c c .025 2 .025 2 Rejection Region Nonrejection Region 72,223 Zc 196 . 74,914 196 . =0 © 2002 Thomson / South-Western 72,223 Rejection Region Upper X 77,605 Zc 196 . 112 c Zc n 14,530 74,914 196 . 112 77,605 Slide 9-15 CPA Net Income Example: Critical Value Method (Part 2) 2 .025 2 Rejection Region Nonrejection Region 72,223 Zc 196 . .025 Rejection Region 77,605 Zc 196 . =0 If X 77,223 or X 77,605, reject Ho. If 77,223 X 77,605, do not reject Ho. Since X 78,646 © 2002 Thomson / South-Western X c 77,605, reject Ho. Slide 9-16 Demonstration Problem 9.1 (Part 1) Ho : 4.30 Ha : 4.30 Rejection Region =.05 Nonrejection Region Zc 1645 . If Z 1645 . , reject H0. If Z 1645 . , do not reject H0. 0 X 4.156 4.30 Z 142 . s 0.574 n 32 Z 142 . 1645 . , do not reject H0. © 2002 Thomson / South-Western Slide 9-17 Demonstration Problem 9.1 (Part 2) Ho: 4.30 Ha: 4.30 =.05 Nonrejection Region Zc 1645 . s XcZ n 4.30 ( 1.645) Rejection Region 0.574 32 xc 4.133 0 4.30 4.133 If X 4.133, reject H0. X 4156 . 4133 . , do not reject H0. If X 4.133, do not reject H0. © 2002 Thomson / South-Western Slide 9-18 Demonstration Problem 9.1 (Part 3) Ho: 4.30 Ha: 4.30 Rejection Region =.05 Nonrejection Region 0 If p - value < , reject Ho. If p - value , do not reject Ho. X 4.156 4.30 1.42 s 0.574 n 32 P ( Z 1.42) .0778 Z © 2002 Thomson / South-Western Since p - value = .0778 > = .05, do not reject Ho. Slide 9-19 Two-tailed Test: Small Sample, Unknown, = .05 (Part 1) Weights in Pounds of a Sample of 20 Plates 22.6 27.0 26.2 25.8 22.2 26.6 25.3 30.4 23.2 28.1 23.1 28.6 27.4 26.9 24.2 23.5 24.5 24.9 26.1 23.6 X 2551 . , S = 2.1933, and n = 20 © 2002 Thomson / South-Western Slide 9-20 Two-tailed Test: Small Sample, Unknown, = .05 (Part 2) Ho: 25 Ha: 25 Rejection Regions 2 df n 1 19 .025 2 .025 Nonrejection Region t c t 2.093 c 2.093 Critical Values © 2002 Thomson / South-Western Slide 9-21 Two-tailed Test: Small Sample, Unknown, = .05 (Part 3) If t 2.093, reject Ho. Rejection Regions If t 2.093, do not reject Ho. .025 2 .025 2 Non Rejection Region t c t 2.093 Critical Values © 2002 Thomson / South-Western c 2.093 X 2551 . 250 . t 104 . S 21933 . n 20 Since t 104 . 2.093, do not reject Ho. Slide 9-22 Demonstration Problem 9.2 (Part 1) Size in Acres of 23 Farms 445 463 466 561 489 466 477 560 474 557 557 505 502 433 553 449 545 477 438 511 545 500 590 X 498.78, S = 46.94,and n = 23 © 2002 Thomson / South-Western Slide 9-23 Demonstration Problem 9.2 (Part 2) Ho : 471 Rejection Region Ha : 471 .05 df n 1 22 Nonrejection Region . t 1717 c Critical Value © 2002 Thomson / South-Western Slide 9-24 Demonstration Problem 9.2 (Part 3) If t 1717 . , reject Ho. If t 1717 . , do not reject Ho. Rejection Region .05 Nonrejection Region . t 1717 c Critical Value © 2002 Thomson / South-Western X 498.78 471 t 2.84 S 46.94 n 23 Since t 2.84 1.717, reject Ho. Slide 9-25 Z Test of Population Proportion p P Z PQ n where: p = sample proportion P = population proportion n P 5, and nQ 5 Q = 1-P © 2002 Thomson / South-Western Slide 9-26 Testing Hypotheses about a Proportion: Manufacturer Example (Part 1) Rejection Regions Ho : P .08 Ha : P .08 .05 2 Nonrejection Region Z c 1645 . Z c .05 2 1645 . Critical Values © 2002 Thomson / South-Western Slide 9-27 Testing Hypotheses about a Proportion: Manufacturer Example (Part 2) If Z 1645 . , reject Ho. Rejection Regions If Z 1645 . , do not reject Ho. .05 2 Nonrejection Region Z c 1645 . Z Critical Values c .05 2 1645 . 33 p .165 200 p P .165.08 Z 4.43 P Q (.08)(.92) n 200 Since Z 4.43 1645 . , reject Ho. © 2002 Thomson / South-Western Slide 9-28 Demonstration Problem 9.3 (Part 1) Ho: P .17 Ha: P .17 Rejection Region .05 Nonrejection Region Z c 1645 . Critical Value © 2002 Thomson / South-Western Slide 9-29 Demonstration Problem 9.3 (Part 2) Rejection Region .05 Nonrejection Region Z Critical Value © 2002 Thomson / South-Western c 1645 . If Z 1645 . , reject Ho. If Z 1645 . , do not reject Ho. 115 p .209 550 p P .209 .17 Z 2.44 P Q (.17)(.83) n 550 Since Z = 2.44 1645 . , reject Ho. Slide 9-30