1. Statistics for Managers 5th Edition Chapter 9 Fundamentals of Hypothesis Testing: One-Sample Tests

2. Chapter Topics Hypothesis testing methodology Z test for the mean ( known) P-value approach to hypothesis testing Connection to confidence interval estimation One-tail tests T test for the mean ( unknown) Z test for the proportion Potential hypothesis-testing pitfalls and ethical considerations

3. What is a Hypothesis? A hypothesis is a claim (assumption) about the population parameter Examples of parameters are population mean or proportion The parameter must be identified before analysis I claim the mean GPA of this class is 3.5! © 1984-1994 T/Maker Co.

4. The Null Hypothesis, H 0 States the assumption (numerical) to be tested e.g.: The average number of TV sets in U.S. Homes is at least three ( ) Is always about a population parameter ( ), not about a sample statistic ( )

5. The Null Hypothesis, H 0 Begins with the assumption that the null hypothesis is true Similar to the notion of innocent until proven guilty Refers to the status quo Always contains the “=” sign May or may not be rejected (continued)

6. The Alternative Hypothesis, H 1 Is the opposite of the null hypothesis e.g.: The average number of TV sets in U.S. homes is less than 3 ( ) Challenges the status quo Never contains the “=” sign May or may not be accepted Is generally the hypothesis that is believed (or needed to be proven) to be true by the researcher

7. Hypothesis Testing Process Identify the Population Assume the population mean age is 50. ( ) REJECT Take a Sample Null Hypothesis No, not likely!

8. Sampling Distribution of = 50 It is unlikely that we would get a sample mean of this value...... Therefore, we reject the null hypothesis that μ = 50. Reason for Rejecting H 0  20 If H 0 is true... if in fact this were the population mean.

9. Filling Process You are in charge of the filling process for the one pound jars of Cheese Whiz You take a random sample of 36 jars each hour to determine if the filling process is in control

10. Steps: State the Null Hypothesis (H 0 :   ) State its opposite, the Alternative Hypothesis (H 1 :   16) Hypotheses are mutually exclusive & exhaustive Sometimes it is easier to form the alternative hypothesis first. Identify the Problem

11. Decision Rule Accept H 0 if15.8 oz  X  oz Reject H 0 ifX 16.2 oz Assume a sample of 36 jars tested each hour Population standard deviation is.6 oz.

12. Possible Results States of Nature Correct Decision Type II Error (  Correct Decision Type I Error (  Do Not Reject H o Reject H 0 ACTIONSACTIONS H 0 is True  = 16 H 0 is False 

13. Type I Error X Reject H 0 Do Not Reject H 0 16.2 1615.8 - 2 + 2 Z.0228.4772  Z  n X    36 15.8   16   Z  36 16.2   16   

14. Z1Z1  36 15.8   15.9   Z2Z2  36 16.2   15.9   15.815.916.2 X Reject H 0 Do Not Reject H 0.3413.4987  0 Z.3413.4987.8400 = 

15.   Reduce probability of one error and the other one goes up.  &  Have an Inverse Relationship

16. True Value of Population Parameter Increases When Difference Between Hypothesized Parameter & True Value Decreases Significance Level  Increases When  Decreases Population Standard Deviation  Increases When   Increases Sample Size n Increases When n Decreases Factors Affecting Type II Error,       n

17. Hypothesis Testing: Steps 1. State H 0 and H 1 2. Choose  - level of significance 3. Use Information in Steps 1 and 2 to Design a Decision Rule 4. Collect Data 5. Compare Sample Data to Decision Rule and State Conclusion.

18. Setting Up H 0 and H 1 H 0 must contain the equality (=) Either H 0 or H 1 must precisely represent what is being tested H 0 and H 1 must be mutually exhaustive (must cover all possible outcomes)

19. Level of Significance, Defines unlikely values of sample statistic if null hypothesis is true Called rejection region of the sampling distribution Is designated by, (level of significance) Typical values are.01,.05,.10 Is selected by the researcher at the beginning Provides the critical value(s) of the test

20. Level of Significance and the Rejection Region H 0 :   3 H 1 :  < 3 0 0 0 H 0 :   3 H 1 :  > 3 H 0 :   3 H 1 :   3    /2 Critical Value(s) Rejection Regions

21. Example: One Tail Test Q. Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed = 372.5. The company has specified  to be 15 grams. Test at the  0.05 level. 368 gm. H 0 :  368 H 1 :  > 368

22. Finding Critical Value: One Tail Z.04.06 1.6.9495.9505.9515 1.7.9591.9599.9608 1.8.9671.9678.9686.9738.9750 Z 0 1.645. 05 1.9.9744 Standardized Cumulative Normal Distribution Table (Portion) What is Z given  = 0.05?  =.05 Critical Value = 1.645.95

23. Example Solution: One Tail Test  = 0.5 n = 25 Critical Value: 1.645 Test Statistic: Decision: Conclusion: Do Not Reject at  =.05 No evidence that true mean is more than 368 Z 0 1.645.05 Reject H 0 :  368 H 1 :  > 368 1.50

24. p -Value Solution Z 0 1.50 P-Value =.0668 Z Value of Sample Statistic From Z Table: Lookup 1.50 to Obtain.9332 Use the alternative hypothesis to find the direction of the rejection region. 1.0000 -.9332.0668 p-Value is P(Z  1.50) = 0.0668

25. p -Value Solution (continued) 0 1.50 Z Reject (p-Value = 0.0668)  (  = 0.05) Do Not Reject. p Value = 0.0668  = 0.05 Test Statistic 1.50 is in the Do Not Reject Region 1.645

26. Example: Two-Tail Test Q. Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed = 372.5. The company has specified  to be 15 grams. Test at the  0.05 level. 368 gm. H 0 :  368 H 1 :  368

27.  = 0.05 n = 25 Critical Value: ±1.96 Example Solution: Two-Tail Test Test Statistic: Decision: Conclusion: Do Not Reject at  =.05 No Evidence that True Mean is Not 368 Z 0 1.96.025 Reject -1.96.025 H 0 :  368 H 1 :  368 1.50

28. p-Value Solution (p Value = 0.1336)  (  = 0.05) Do Not Reject. 0 1.50 Z Reject  = 0.05 1.96 p Value = 2 x 0.0668 Test Statistic 1.50 is in the Do Not Reject Region Reject

29. Connection to Confidence Intervals

30. t Test: Unknown Assumption Population is normally distributed If not normal, requires a large sample T test statistic with n-1 degrees of freedom

31. Example: One-Tail t Test Does an average box of cereal contain more than 368 grams of cereal? A random sample of 36 boxes showed X = 372.5, and  s  15. Test at the  0.01 level. 368 gm. H 0 :  368 H 1 :  368  is not given

32. Example Solution: One-Tail  = 0.01 n = 36, df = 35 Critical Value: 2.4377 Test Statistic: Decision: Conclusion: Do Not Reject at  =.01 No evidence that true mean is more than 368 t 35 0 2.437 7.01 Reject H 0 :  368 H 1 :  368 1.80

33. p -Value Solution 0 1.80 t 35 Reject (p Value is between.025 and.05)  (  = 0.01). Do Not Reject. p Value = [.025,.05]  = 0.01 Test Statistic 1.80 is in the Do Not Reject Region 2.4377

34. PHStat | one-sample tests | t test for the mean, sigma known … Example in excel spreadsheet t Test: Unknown in PHStat

35. Involves categorical variables Fraction or % of population in a category If two categorical outcomes, binomial distribution Either possesses or doesn’t possess the characteristic Sample proportion ( p ) Proportions

36. Example: Z Test for Proportion Problem: AstraZeneca claims that less than 5% of patients taking Nexium experience an upset stomach. Approach: To test this claim, a random sample of 500 patients were interviewed. 15 of the patients experienced stomach pain. Solution: Test at the  =.05 significance level.

37.  =.05 n = 500 p = 15/500 =.03 Z Test for Proportion: Solution H 0 :  .05 H 1 :  .05 Critical Value:  Reject at Reject at  =.05 Decision: Conclusion: We do have sufficient evidence to support the claim that less than 5% of patients experience an upset stomach. Z0 Reject.05 Test Statistic: Z  p -   (1 -  ) n =.03 -.05.05 (1 -.05) 500 = -2.05

38.  =.05 n = 500 p = 15/500 =.03 Z Test for Proportion: Solution H 0 :  .05 H 1 :  .05 Critical Value:  Reject at Reject at  =.05 Decision: Conclusion: We do have sufficient evidence to support the claim that less than 5% of patients experience an upset stomach. Z0 Reject Ho.05 Test Statistic: Z  p -   (1 -  ) n =.03 -.05.05 (1 -.05) 500 = -2.05

39. Z Test for Proportion in PHStat PHStat | one-sample tests | z test for the proportion … Example in excel spreadsheet

40. Controlling  and  A bank wishes to open a new branch if average monthly income is at least 4,000. It does not want to open the branch if average monthly income is less than or equal to 3,800. Assume that  If the bank wants:   What sample size should be used? What should the Decision Rule be?

41. H 0 :   4,00  H 1 :  ≤ 3,800 n = 99 Do not Reject H 0 if X c  $   Reject H 0 if X c < $3,917 2.33 = X c - 3,800 500 n -1.645 = X c - 4,000 500 n 3,800 4,000 XZXZ XZXZ X c -1.645 X c 2.33 0 0 Do not reject H o Do not reject H 0 Reject H 0 reject H 0.05 = .01 = 

42. Potential Pitfalls and Ethical Considerations Randomize data collection method to reduce selection biases Do not manipulate the treatment of human subjects without informed consent Do not employ “data snooping” to choose between one-tail and two-tail test, or to determine the level of significance

43. Potential Pitfalls and Ethical Considerations Do not practice “data cleansing” to hide observations that do not support a stated hypothesis Report all pertinent findings (continued)

44. Chapter Summary Addressed hypothesis testing methodology Performed Z Test for the mean ( Known) Discussed p –Value approach to hypothesis testing Made connection to confidence interval estimation

45. Chapter Summary Performed one-tail and two-tail tests Performed t test for the mean ( unknown) Performed Z test for the proportion Discussed potential pitfalls and ethical considerations (continued)