1. ©2003 Thomson/South-Western 1 Chapter 8 – Hypothesis Testing for the Mean and Variance of a Population Slides prepared by Jeff Heyl, Lincoln University ©2003 South-Western/Thomson Learning™ Introduction to Business Statistics, 6e Kvanli, Pavur, Keeling

2. ©2003 Thomson/South-Western 2 Hypothesis Testing on the Mean  Null hypothesis (H o ): A statement concerning a population parameter  Alternative hypothesis: A statement in contradiction of the null hypothesis

3. ©2003 Thomson/South-Western 3 Type I and Type II Errors  =probability of rejecting the H o when H o is true (Type I error)  =probability of failing to rejecting the H o when H o is false (Type II error) Actual Situation Conclusion H o TrueH o False Fail to Reject H o Correct decision Type II error Reject H o Type I error Correct decision

4. ©2003 Thomson/South-Western 4 Hypothesis Testing Process  Determine the H o and H a  Determine the significance level  Compare the sample mean (variance) to the hypothesized mean (variance)  Decide whether to fail to reject or reject H o  Determine what the decision means in reference to the problem

5. ©2003 Thomson/South-Western 5 Height Example H o :  = 5.9 H a :   5.9  =.05 = P(rejecting H o when H o is true) critical value = ± 1.96 Reject H o if > 1.96 X - 5.9  / n

6. ©2003 Thomson/South-Western 6 Height Example Z.025 -k-k-k-k 0 |Z| > k k.025 Area =.5 -.025 =.475 Figure 8.1

7. ©2003 Thomson/South-Western 7 Height Example Figure 8.2 X distance =.14’ = 2.53 x.055 5.76’  x ≈ =.055’.48 75 75 µ x = 5.9’

8. ©2003 Thomson/South-Western 8 Computed Value Because - 2.53 < - 1.96, we reject H o thus we conclude that the average population male height is not equal to 5.9 Z = ≈ = = -2.53 = Z* X - 5.9  / n X - 5.9 s / n 5.76 - 5.9.48 / 75

9. ©2003 Thomson/South-Western 9 Height Example Figure 8.3 X Area =.5 -.005 =.495 -k-k-k-k 0 Area =.01 k.005

10. ©2003 Thomson/South-Western 10 Height Example Figure 8.4 Z Reject H o if Z* falls here -2.575 Z* = -2.53 2.575 0 Area =.01 Reject H o if Z* falls here

11. ©2003 Thomson/South-Western 11 Hypothesis Testing 5 Step Procedure 3.Define the rejection region 4.Calculate the test statistic 5.Give a conclusion in terms of the problem 1.Set up the null and alternative hypothesis Z = ≈ X - µ o  / n X - µ o s / n 2.Define the test statistic

12. ©2003 Thomson/South-Western 12 Everglo Light Example 1.Define the hypotheses H o : µ = 400 H a : µ ≠ 400 3.Define the rejection region reject H o if Z > 1.645 or Z 1.645 or Z < -1.645 2.Define the test statistic Z = ≈ X - 400  / n X - 400 s / n

13. ©2003 Thomson/South-Western 13 Everglo Light Example Figure 8.5 Z -1.645 Z* = 2.59 1.645 0 Area =  =.1 Area =.5 -.05 =.45 2 =.05

14. ©2003 Thomson/South-Western 14 Everglo Light Example 5.State the conclusion There is sufficient evidence to conclude that the average lifetime of Everglo bulbs is not 400 hours 4.Calculate the value of the test statistic Z* ≈ = = 2.59 114.25 411 - 400 42.5 / 100

15. ©2003 Thomson/South-Western 15 Confidence Intervals and Hypothesis Testing When testing H o : µ = µ o versus H a : µ ≠ µ o using the five-step procedure and a significance level, , H o will be rejected if and only if µ o lies outside the (1 -  ) 100 confidence interval for µ X - k to X + k  n n

16. ©2003 Thomson/South-Western 16 Power of a Statistical Test  = P(fail to reject H o when H o is false) 1-  = P(rejecting H o when H o is false) 1-  = the power of the test

17. ©2003 Thomson/South-Western 17 Probability of Rejecting H o Figure 8.6 391.775400403408.225 A B X Area =  =.10 Area = 1 -  when µ = 403

18. ©2003 Thomson/South-Western 18 Power of Tests 2.Power of test = P(Z > z 1 ) + P(Z z 1 ) + P(Z < z 2 ) 1.Determine z 1 = Z  /2 - µ - µ o  / n z 1 = -Z  /2 - µ - µ o  / n

19. ©2003 Thomson/South-Western 19 Power Curve Power curve for test of H o versus H a using five-step procedure Power curve for test of H o versus H a using any other procedure µ 400 403 1.0.1617.10 1 -  = P(rejecting H o ) Figure 8.7

20. ©2003 Thomson/South-Western 20 One Tailed Test for µ H o : µ ≥ 32.5H a : µ < 32.5 1.Define H o and H a prior to observation Z = ≈ X - µ o  / n X - µ o s / n 2.Define the test statistic 3. Reject if H o if Z ≈ < -1.645 X - 32.5 s / n Example 8.4

21. ©2003 Thomson/South-Western 21 One Tailed Test for µ 5.Study supports the claim that the average mileage for the Bullet is less than 32.5 mpg – supports claim of false advertising Z* ≈ = -2.80 30.4 - 32.5 5.3 / 50 4.The value of the test statistic is Example 8.4

22. ©2003 Thomson/South-Western 22 One Tailed Test for µ Figure 8.8 -1.645 0 Area =.5 -.05 =.45 Area =  =.05 Z

23. ©2003 Thomson/South-Western 23 Excel Example Z Z* = 1.54 Area =.49 Area =  =.01 2.33 0 Figure 8.9

24. ©2003 Thomson/South-Western 24 Excel Example Figure 8.10

25. ©2003 Thomson/South-Western 25 Excel Example Figure 8.11

26. ©2003 Thomson/South-Western 26 Power of Tests 2.Power of test = P(Z > z 1 ) 1.Determine z 1 = Z  /2 - µ - µ o  / n H o : µ ≤ µ o verses H a : µ > µ o

27. ©2003 Thomson/South-Western 27 Power of Tests 2.Power of test = P(Z < z 2 ) 1.Determine z 1 = -Z  /2 - µ - µ o  / n H o : µ ≥ µ o verses H a : µ < µ o

28. ©2003 Thomson/South-Western 28 Power of Tests Large-Sample Tests on a Population Mean Two-Tailed Test H o : µ = µ o H a : µ ≠ µ o reject H o if |Z*| > Z  /2 One-Tailed Test H o : µ ≤ µ o H a : µ > µ o reject H o if Z* > Z  H o : µ ≥ µ o H a : µ < µ o reject H o if Z* < -Z 

29. ©2003 Thomson/South-Western 29 Determining the p-Value The p-value is the value of  at which the hypothesis test procedure changes conclusions based on a given set of data. It is the largest value of  for which you will fail to reject H o

30. ©2003 Thomson/South-Western 30 Determining the p-Value Figure 8.12 -1.96 0 Area =.025 Area =.005 Z -2.575 Z* = -2.53 Area =.025 Area =.005 1.96 2.575

31. ©2003 Thomson/South-Western 31 Determining the p-Value Figure 8.13 0 Z -2.53 Area =.4943 (Table A-4) Area=.5 -.4943 =.0057 2.53 Area = p value

32. ©2003 Thomson/South-Western 32 Procedure for Finding the p-Value For H a : µ ≠ µ o p = 2 * (area outside Z * ) For H a : µ > µ o p = area to the right of Z * For H a : µ < µ o p = area to the left of Z * For H a : µ < µ o p = area to the left of Z *

33. ©2003 Thomson/South-Western 33 Determining the p-Value Figure 8.14 0 Z Area =.025 p= area =.5 -.4382 =.0618 Z* = 1.54

34. ©2003 Thomson/South-Western 34 Interpreting the p-Value Classical Approach reject H o if p-value <  fail to reject H o is p-value ≥  General rule of thumb reject H o if p-value is small (p <.01) fail to reject H o is p-value is large (p >.1)

35. ©2003 Thomson/South-Western 35 Interpreting the p-Value Small p Reject H o.01 Large p Fail to reject H o p.1 Inconclusive

36. ©2003 Thomson/South-Western 36 Another Interpretation 1.For a two-tailed test where H o : µ ≠ µ o, the p-value is the probability that the value of the test statistic, Z*, will be at least as large (in absolute value) as the observed Z*, if µ is in fact equal to µ o 2.For a one-tailed test where Ha: µ > µ o, the p-value is the probability that the value of the test statistic, Z*, will be at least as large as the observed Z*, if µ is in fact equal to µ o 3.For a one-tailed test where H a : µ < µ o, the p-value is the probability that the value of the test statistic, Z*, will be at least as large as the observed Z*, if µ is in fact equal to µ o

37. ©2003 Thomson/South-Western 37 Excel Example Figure 8.15

38. ©2003 Thomson/South-Western 38 Excel Example Figure 8.16 Z p value = twice this area 3.06

39. ©2003 Thomson/South-Western 39 Excel Example Figure 8.17 Frequency Histogram 35302520151050 10.12 and10.14 and10.16 and10.18 and10.20 and10.22 and10.24 and10.26 and10.28 and10.30 and underunderunderunderunderunderunderunderunderunder 10.1410.1610.1810.2010.2210.2410.2610.2810.3010.32 Class Limits

40. ©2003 Thomson/South-Western 40 Practical Versus Statistical Figure 8.18 Area= p value =.004 (from Table A-4) Z Z* = -2.65

41. ©2003 Thomson/South-Western 41 Hypothesis Testing on the Mean of a Normal Population: Small Sample  Normal population  Population standard deviation unknown  Small sample  Student t distribution  Nonparametric procedure t =t =t =t = X - µ o s / n

42. ©2003 Thomson/South-Western 42 Small Sample Test A Normal population Symmetric (nonnormal) population Skewed population B Figure 8.19

43. ©2003 Thomson/South-Western 43 Clark Products Example 1. When a question is phrased “Is there evidence to indicate that...,” what follows is the alternative hypothesis H o : µ = 10 and H a : µ ≠ 10 2.The test statistic here is t =t =t =t = X - µ o s / n 3.Using a significance level of.05 and Figure 8.20, the corresponding two-tailed procedure is to reject H o if |t| > t.025,17 = 2.11

44. ©2003 Thomson/South-Western 44 Clark Products Example Area=.025 Z -2.11 Reject H o here 2.11 Area =.025 Reject H o here Figure 8.20

45. ©2003 Thomson/South-Western 45 Clark Products Example Figure 8.21 Z 2.11 Area =.05 Area =.025 1.74 t* = 1.83 0

46. ©2003 Thomson/South-Western 46 Clark Products Example 5. There is insufficient evidence to indicate that the average output voltage is different from 10 volts 4.The value of the test statistic is t* = = 1.83 10.331 - 10.767 / 18

47. ©2003 Thomson/South-Western 47 Clark Products Example Figure 8.22

48. ©2003 Thomson/South-Western 48 Small-Sample Tests on a Normal Population Mean Two-tailed test H o : µ = µ o H a : µ ≠ µ o reject H o if |t*| > t  /2, n-1 One tail test H o : µ = µ o H a : µ < µ o reject H o if t* < t , n-1 H o : µ = µ o H a : µ > µ o reject H o if t* > t , n-1

49. ©2003 Thomson/South-Western 49 Chi-Square Distribution 2222 Area = a  a, df 2 Figure 8.23

50. ©2003 Thomson/South-Western 50 Chi-Square Distribution 2222 Area =.1 18.5493 Figure 8.24

51. ©2003 Thomson/South-Western 51 Chi-Square Distribution Figure 8.25 2222 Area =.025 b Area =.95 Area =.025 a

52. ©2003 Thomson/South-Western 52 Hypothesis Testing on  2 One tail test H o :  2 =  o H a :  2 <  o reject H o if  2* <  , n-1 2 2 2 H o :  2 =  o H a :  2 >  o reject H o if  2* >  , n-1 2 2 2 Two-tailed test H o :  2 =  o H a :  2 ≠  o 2 2 Test statistic:  2 = (n - 1)s 2  2 o

53. ©2003 Thomson/South-Western 53 Chi-Square Distribution Figure 8.26 2222 Area = p value 152.6  2 curve with 14 df

54. ©2003 Thomson/South-Western 54 Chi-Square Distribution Figure 8.27