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Chapter 8 Hypothesis Testing ( 假设检验 )

Chapter 8 Hypothesis Testing ( 假设检验 ). The Null and Alternative Hypotheses and Errors in Hypothesis Testing. z Tests about a Population Mean ( known): t Tests about a Population Mean ( unknown). Chapter 8 Hypothesis Testing ( 假设检验 ). z Tests about a Population Proportion

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Chapter 8 Hypothesis Testing ( 假设检验 )

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  1. Chapter 8 Hypothesis Testing(假设检验) • The Null and Alternative Hypotheses and Errors in Hypothesis Testing. • z Tests about a Population Mean ( known): • t Tests about a Population Mean ( unknown)

  2. Chapter 8 Hypothesis Testing(假设检验) • z Tests about a Population Proportion • Type‖Error Probabilities and Sample Size Determination • The Chi-Square Distribution • Statistical Inference for a Population Variance

  3. Learning Objectives In this chapter, you learn: • The basic principles of hypothesis testing • How to use hypothesis testing to test a mean, proportion and population variance. • The assumptions of each hypothesis-testing procedure, how to evaluate them, and the consequences if they are seriously violated

  4. What is a Hypothesis(假设)? • A hypothesis is a claim (assumption) about a population parameter: • population mean • population proportion Example: The mean monthly cell phone bill of this city is μ = $42 Example: The proportion of adults in this city with cell phones is p = 0.68

  5. The Null Hypothesis(零假设), H0 • States the claim or assertion to be tested Example: The average number of TV sets in U.S. Homes is equal to three ( ) • Is always about a population parameter, not about a sample statistic

  6. The Null Hypothesis, H0 (continued) • Begin with the assumption that the null hypothesis is true • Similar to the notion of innocent until proven guilty • Refers to the • Always contains “=” , “≤” or “” sign • May or may not be rejected existing state

  7. The Alternative Hypothesis(备择假设) Ha • Is the opposite of the null hypothesis • e.g., The average number of TV sets in U.S. homes is not equal to 3 ( Ha: μ ≠ 3 ) • Challenges the existing state • Never contains the “=” , “≤” or “” sign • May or may not be proven • Is generally the hypothesis that the researcher is trying to prove

  8. Trash Bag Case Example 8.1 • Tests show that the current trash bag has a mean breaking strength m close to but not exceeding 50 lbs • The null hypothesis H0 is that the new bag has a mean breaking strength that is 50 lbs or less • The new bag’s mean breaking strength is not known and is in question, but it is hoped that the new bag is stronger than the current one • The alternative hypothesis Ha is that the new bag has a mean breaking strength that exceeds 50 lbs • H0:  50 vs. Ha: > 50

  9. Payment Time Case Example 8.2 • With a new billing system, the mean bill paying time m is hoped to be less than 19.5 days • The alternative hypothesis Ha is that the new billing system has a mean payment time that is less than 19.5 days • With the old billing system, the mean bill paying time m was close to but not less than 39 days • The null hypothesis H0 is that the new billing system has a mean payment time close to but not less than 19.5 days • H0:  19.5 vs.Ha: < 19.5

  10. Types of Decisions • As a result of testing H0 vs. Ha, will have to decide either of the following decisions for the null hypothesis H0: • Do not reject H0 OR • Reject H0

  11. Hypothesis Testing(假设检验) Process Claim:the population mean age is 50. (Null Hypothesis: Population H0: μ = 50 ) Now select a random sample X = likely if μ = 50? Is 20 Suppose the sample If not likely, REJECT mean age is 20: X = 20 Sample Null Hypothesis

  12. Reason for Rejecting H0 Sampling Distribution of X X 20 μ= 50 IfH0 is true ... then we reject the null hypothesis that μ = 50. If it is unlikely that we would get a sample mean of this value ... ... if in fact this was the population mean…

  13. Level of Significance(显著水平),  • Defines the unlikely values of the sample statistic if the null hypothesis is true • Defines rejection region(拒绝域) of the sampling distribution • Is designated by , (level of significance) • Typical values are 0.01, 0.05, or 0.10 • Is selected by the researcher at the beginning • Provides the critical value(s) of the test

  14. Level of Significance and the Rejection Region a Level of significance = Represents critical value a a H0: μ = 3 Ha: μ ≠ 3 /2 /2 Rejection region is shaded Two-tail test 3 H0: μ ≤ 50 Ha: μ > 50 a 50 Upper-tail test H0: μ ≥ 19.5 Ha: μ < 19.5 a Lower-tail test 19.5

  15. Errors in Making Decisions • Type I Error(第一类误差) • Rejecting null hypothesis when it is true • Considered a serious type of error The probability of Type I Error is  • Called level of significance of the test • Set by the researcher in advance

  16. Errors in Making Decisions (continued) • Type II Error(第二类误差) • Failing to reject the null hypothesis when it is false The probability of Type II Error is β

  17. Outcomes and Probabilities Possible Hypothesis Test Outcomes Actual Situation Decision H0 True H0 False Do Not No error (1 - ) Type II Error ( β ) Reject Key: Outcome (Probability) a H 0 Reject Type I Error ( ) No Error ( 1 - β ) H a 0

  18. Type I & II Error Relationship • Type I and Type II errors cannot happen at the same time • Type I error can only occur if H0 is true • Type II error can only occur if H0 is false If Type I error probability (  ) , then Type II error probability ( β )

  19. Hypothesis Tests for the Mean Hypothesis Tests for   Known  Unknown (Z test) (t test)

  20. Z Test of Hypothesis for the Mean (σ Known) X • Convert sample statistic ( ) to a Z test statistic(Z检验统计量) Hypothesis Tests for   Known σ Known  Unknown σ Unknown (Z test) (t test) The test statistic is:

  21. Critical Value Approach to Testing • Convert sample statistic ( ) to test statistic (Z statistic ) • Determine the critical Z values for a specifiedlevel of significance  from a table or computer • Decision Rule: If the test statistic falls in the rejection region, reject H0 ; otherwise do not reject H0

  22. Two-Tail Tests(双边检验) H0: μ = 3 Ha: μ¹ 3 • There are two cutoff values (critical values), defining the regions of rejection /2 /2 X 3 Reject H0 Do not reject H0 Reject H0 Z -Z +Z 0 Lower critical value Upper critical value

  23. One-Tail Tests(单边检验) • In many cases, the alternative hypothesis focuses on a particular direction H0: μ ≥ 19.5 Ha: μ < 19.5 This is a lower-tail test since the alternative hypothesis is focused on the lower tail below the mean of 19.5 This is an upper-tail test since the alternative hypothesis is focused on the upper tail above the mean of 50 H0: μ ≤ 50 Ha: μ > 50

  24. Lower-Tail Tests H0: μ ≥ 19.5 Ha: μ < 19.5 • There is only one critical value, since the rejection area is in only one tail a Reject H0 Do not reject H0 Z -Z 0 μ X Critical value

  25. Upper-Tail Tests H0: μ ≤ 50 Ha: μ > 50 • There is only one critical value, since the rejection area is in only one tail a Do not reject H0 Reject H0 Z Zα 0 _ μ X Critical value

  26. 6 Steps in Hypothesis Testing • State the null hypothesis, H0 and the alternative hypothesis, Ha • Choose the level of significance, , and the sample size, n • Determine the appropriate test statistic and sampling distribution • Determine the critical values that divide the rejection and non-rejection regions

  27. 6 Steps in Hypothesis Testing (continued) • Collect data and compute the value of the test statistic • Make the statistical decision and state the managerial conclusion. If the test statistic falls into the non-rejection region, do not reject the null hypothesis H0. If the test statistic falls into the rejection region, reject the null hypothesis. Express the managerial conclusion in the context of the problem

  28. Hypothesis Testing Example Example 8.3 Test the claim that the true mean # of TV sets in US homes is equal to 3. (Assume σ = 0.8) 1. State the appropriate null and alternative hypotheses • H0: μ = 3 Ha: μ ≠ 3 (This is a two-tail test) 2. Specify the desired level of significance and the sample size • Suppose that  = 0.05 and n = 100 are chosen for this test

  29. Hypothesis Testing Example (continued) 3. Determine the appropriate technique • σ is known so this is a Z test. 4. Determine the critical values • For  = 0.05 the critical Z values are ±1.96 5.Collect the data and compute the test statistic • Suppose the sample results are n = 100, X = 2.84 (σ = 0.8 is assumed known) So the test statistic is:

  30. Hypothesis Testing Example (continued) • 6. Is the test statistic in the rejection region?  = 0.05/2  = 0.05/2 Reject H0 if Z < -1.96 or Z > 1.96; otherwise do not reject H0 Reject H0 Do not reject H0 Reject H0 -Z= -1.96 0 +Z= +1.96 Here, Z = -2.0 < -1.96, so the test statistic is in the rejection region

  31. Hypothesis Testing Example (continued) 6(continued). Reach a decision and interpret the result  = 0.05/2  = 0.05/2 Reject H0 Do not reject H0 Reject H0 -Z= -1.96 0 +Z= +1.96 -2.0 Since Z = -2.0 < -1.96, we reject the null hypothesis and conclude that there is sufficient evidence that the mean number of TVs in US homes is not equal to 3

  32. Trash Bag Case ( is known) Example 8.4 Tests show that the current trash bag has a mean breaking strength μ close to but not exceeding 50 lbs, The new bag’s mean breaking strength is not known and is in question, but it is hoped that the new bag is stronger than the current one Form hypothesis test: H0: μ ≤ 50 the average breaking strength is not over 50 lbs Ha: μ > 50 the average breaking strength for new trash bag is greater than 50 lbs.

  33. Case Study: Find Rejection Region (continued) • Suppose that  = 0.05 is chosen for this test Find the rejection region: Reject H0  = 0.05 Do not reject H0 Reject H0 1.645 0 Reject H0 if Z > 1.645

  34. Review:One-Tail Critical Value Standardized Normal Distribution Table (Portion) What is Z given a = 0.05? 0.45 .04 Z .03 .05 a= 0.05 1.5 .4370 .4382 .4394 .4484 .4495 .4505 1.6 z 0 1.645 .4582 1.7 .4591 .4599 Critical Value = 1.645

  35. Case Study: Test Statistic (continued) Obtain sample and compute the test statistic Suppose a sample is taken with the following results: n = 40, X = 50.575 (=1.65 was assumed known) • Then the test statistic is:

  36. Case Study: Decision (continued) Reach a decision and interpret the result: Reject H0 = 0.05 Do not reject H0 Reject H0 1.645 0 Z = 2.20 Reject H0 since Z = 2.20 ≥1.645 i.e.: there is strong evidence that the mean breaking strength for new trash bag is over 50lbs

  37. Discussion Question • Using Critical Value Approach • In this payment time case, we assume that s is known and s = 4.2 days. A sample of n = 65, = 18.1077 days • Have we strong evidence that the mean payment time for new billing system is less than 19.5 days at =0.01 significance level ?

  38. p-Value (p值) Approach to Testing • p-value: Probability of obtaining a test statistic more extreme ( ≤ or  ) than the observed sample value given H0 is true • Also called observed level of significance • Smallest value of  for which H0 can be rejected

  39. p-Value Approach to Testing (continued) X • Convert Sample Statistic (e.g., ) to Test Statistic (e.g., Z statistic ) • Obtain the p-value from a table or computer based on test statistic. • Compare the p-value with  • If p-value < , reject H0 • If p-value  , do not reject H0

  40. Weight of Evidence Against the Null • Calculate the test statistic and the corresponding p-value • Rate the strength of the conclusion about the null hypothesis H0 according to these rules: • If p < 0.10, then there is some evidence to reject H0 • If p < 0.05, then there is strong evidence to reject H0 • If p < 0.01, then there is very strong evidence to reject H0 • If p < 0.001, then there is extremely strong evidence to reject H0

  41. p-Value Example Example 8.5 • How likely is it to see a sample mean of 2.84 (or something further from the mean, in either direction) if the true mean is  = 3.0? X = 2.84 is translated to a Z score of Z = -2.0 /2 = 0.025 /2 = 0.025 0.0228 0.0228 p-value = 0.0228 + 0.0228 = 0.0456 -1.96 0 1.96 Z -2.0 2.0

  42. p-Value Example (continued) • Compare the p-value with  • If p-value < , reject H0 • If p-value  , do not reject H0 Here: p-value = 0.0456  = 0.05 Since 0.0456 < 0.05, we reject the null hypothesis /2 = 0.025 /2 = 0.025 0.0228 0.0228 -1.96 0 1.96 Z -2.0 2.0

  43. Connection to Confidence Intervals • ForX = 2.84, σ = 0.8 and n = 100, the 95% confidence interval is: 2.6832 ≤ μ≤ 2.9968 • Since this interval does not contain the hypothesized mean (3.0), we reject the null hypothesis at  = 0.05

  44. p -Value Solution for trash bag case Example 8.6 Calculate the p-value and compare to  (assuming that μ = 50) p-value = 0.0139 Reject H0 = 0.05 0 Do not reject H0 Reject H0 1.645 Z = 2.20 Reject H0 since p-value = 0.0139 ＜ = 0.05

  45. Question • Using p-Value Approach • In this payment time case, we assume that s is known and s = 4.2 days. A sample of n = 65, = 18.1077 days • Are there strong evidence that the mean payment time for new billing system is less than 19.5 days at =0.01 significance level ?

  46. t Test of Hypothesis for the Mean (σ Unknown) X • Convert sample statistic ( ) to a ttest statistic Hypothesis Tests for   Known σ Known  Unknown σ Unknown (Z test) (t test) The test statistic is:

  47. The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in X = $172.50 and s = $15.40. Test at the  = 0.05 level. (Assume the population distribution is normal) Two-Tail Test( Unknown) Example 8.7 H0: μ= 168 Ha: μ ¹ 168

  48. a = 0.05 n= 25  is unknown, so use a t statistic Critical Value: t24 = ± 2.0639 Example Solution: Two-Tail Test H0: μ= 168 Ha: μ ¹ 168 a/2=.025 a/2=.025 Reject H0 Do not reject H0 Reject H0 t n-1,α/2 -t n-1,α/2 0 2.0639 -2.0639 1.46 Do not reject H0: not sufficient evidence that true mean cost is different than $168

  49. Connection to Confidence Intervals • For X = 172.5, S = 15.40 and n = 25, the 95% confidence interval is: 172.5 - (2.0639) 15.4/ 25 to 172.5 + (2.0639) 15.4/ 25 166.14 ≤ μ≤ 178.86 • Since this interval contains the Hypothesized mean (168), we do not reject the null hypothesis at  = 0.05

  50. t Test of Hypothesis for the Mean (σ Unknown) Continued • ta, ta/2, and p-values are based on n – 1 degrees of freedom (for a sample of size n) * either t > ta/2 or t < –ta/2

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