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- 1. Chap 9-1Copyright ©2012 Pearson EducationBasic Business Statistics12th EditionGlobal EditionChapter 9Fundamentals of HypothesisTesting: One-Sample TestsChap 9-1
- 2. Chap 9-2Copyright ©2012 Pearson Education Chap 9-2Learning ObjectivesIn this chapter, you learn: The basic principles of hypothesis testing How to use hypothesis testing to test a mean orproportion The assumptions of each hypothesis-testingprocedure, how to evaluate them, and theconsequences if they are seriously violated How to avoid the pitfalls involved in hypothesis testing The ethical issues involved in hypothesis testing
- 3. Chap 9-3Copyright ©2012 Pearson Education Chap 9-3What is a Hypothesis? A hypothesis is a claim(assertion) about apopulation parameter: population mean population proportionExample: The mean monthly cell phone billin this city is μ = $42Example: The proportion of adults in thiscity with cell phones is π = 0.68DCOVA
- 4. Chap 9-4Copyright ©2012 Pearson Education Chap 9-4The Null Hypothesis, H0 States the claim or assertion to be testedExample: The average diameter of amanufactured bolt is 30mm ( ) Is always about a populationparameter, not about a samplestatistic30μ:H030μ:H0 30X:H0DCOVA
- 5. Chap 9-5Copyright ©2012 Pearson Education Chap 9-5The Null Hypothesis, H0 Begin with the assumption that the nullhypothesis is true Similar to the notion of innocent untilproven guilty Refers to the status quo or historical value Null always contains “=“sign May or may not be rejected(continued)DCOVA
- 6. Chap 9-6Copyright ©2012 Pearson Education Chap 9-6The Alternative Hypothesis, H1 Is the opposite of the null hypothesis e.g., The average diameter of a manufacturedbolt is not equal to 30mm ( H1: μ ≠ 30 ) Challenges the status quo Alternative never contains the “=”sign May or may not be proven Is generally the hypothesis that theresearcher is trying to proveDCOVA
- 7. Chap 9-7Copyright ©2012 Pearson Education Chap 9-7The Hypothesis TestingProcess Claim: The population mean age is 50. H0: μ = 50, H1: μ ≠ 50 Sample the population and find sample mean.PopulationSampleDCOVA
- 8. Chap 9-8Copyright ©2012 Pearson Education Chap 9-8The Hypothesis TestingProcess Suppose the sample mean age was X = 20. This is significantly lower than the claimed meanpopulation age of 50. If the null hypothesis were true, the probability ofgetting such a different sample mean would bevery small, so you reject the null hypothesis. In other words, getting a sample mean of 20 is sounlikely if the population mean was 50, youconclude that the population mean must not be 50.(continued)DCOVA
- 9. Chap 9-9Copyright ©2012 Pearson Education Chap 9-9The Hypothesis TestingProcessSamplingDistribution of Xμ = 50If H0 is trueIf it is unlikely that youwould get a samplemean of this value ...... then you rejectthe null hypothesisthat μ = 50.20... When in fact this werethe population mean…X(continued)DCOVA
- 10. Chap 9-10Copyright ©2012 Pearson Education Chap 9-10The Test Statistic andCritical Values If the sample mean is close to the statedpopulation mean, the null hypothesis is notrejected. If the sample mean is far from the statedpopulation mean, the null hypothesis is rejected. How far is “far enough” to reject H0? The critical value of a test statistic creates a “line inthe sand” for decision making -- it answers thequestion of how far is far enough.DCOVA
- 11. Chap 9-11Copyright ©2012 Pearson Education Chap 9-11The Test Statistic andCritical ValuesCritical Values“Too Far Away” From Mean of Sampling DistributionSampling Distribution of the test statisticRegion ofRejectionRegion ofRejectionRegion ofNon-RejectionDCOVA
- 12. Chap 9-12Copyright ©2012 Pearson Education Chap 9-12Possible Errors in Hypothesis TestDecision Making Type I Error Reject a true null hypothesis Considered a serious type of error The probability of a Type I Error is Called level of significance of the test Set by researcher in advance Type II Error Failure to reject a false null hypothesis The probability of a Type II Error is βDCOVA
- 13. Chap 9-13Copyright ©2012 Pearson Education Chap 9-13Possible Errors in Hypothesis TestDecision MakingPossible Hypothesis Test OutcomesActual SituationDecision H0 True H0 FalseDo NotReject H0No ErrorProbability 1 - αType II ErrorProbability βReject H0 Type I ErrorProbability αNo ErrorProbability 1 - β(continued)DCOVA
- 14. Chap 9-14Copyright ©2012 Pearson Education Chap 9-14Possible Errors in Hypothesis TestDecision Making The confidence coefficient (1-α) is theprobability of not rejecting H0 when it is true. The confidence level of a hypothesis test is(1-α)*100%. The power of a statistical test (1-β) is theprobability of rejecting H0 when it is false.(continued)DCOVA
- 15. Chap 9-15Copyright ©2012 Pearson Education Chap 9-15Type I & II Error Relationship Type I and Type II errors cannot happen atthe same time A Type I error can only occur if H0 is true A Type II error can only occur if H0 is falseIf Type I error probability ( ) , thenType II error probability (β)DCOVA
- 16. Chap 9-16Copyright ©2012 Pearson Education Chap 9-16Factors Affecting Type II Error All else equal, β when the difference betweenhypothesized parameter and its true value β when β when σ β when nDCOVA
- 17. Chap 9-17Copyright ©2012 Pearson Education Chap 9-17Level of Significanceand the Rejection RegionLevel of significance =This is a two-tail test because there is a rejection region in both tailsH0: μ = 30H1: μ ≠ 30Critical valuesRejection Region/230/2DCOVA
- 18. Chap 9-18Copyright ©2012 Pearson Education Chap 9-18Hypothesis Tests for the MeanKnown UnknownHypothesisTests for(Z test) (t test)DCOVA
- 19. Chap 9-19Copyright ©2012 Pearson Education Chap 9-19Z Test of Hypothesis for theMean (σ Known) Convert sample statistic ( ) to a ZSTAT test statisticXThe test statistic is:nσμXZSTATσ Known σ UnknownHypothesisTests forKnown Unknown(Z test) (t test)DCOVA
- 20. Chap 9-20Copyright ©2012 Pearson Education Chap 9-20Critical ValueApproach to Testing For a two-tail test for the mean, σ known: Convert sample statistic ( ) to test statistic(ZSTAT) Determine the critical Z values for a specifiedlevel of significance from a table orcomputer Decision Rule: If the test statistic falls in therejection region, reject H0 ; otherwise do notreject H0XDCOVA
- 21. Chap 9-21Copyright ©2012 Pearson Education Chap 9-21Do not reject H0 Reject H0Reject H0 There are twocutoff values(critical values),defining theregions ofrejectionTwo-Tail Tests/2-Zα/2 0H0: μ = 30H1: μ 30+Zα/2/2LowercriticalvalueUppercriticalvalue30ZXDCOVA
- 22. Chap 9-22Copyright ©2012 Pearson Education Chap 9-226 Steps inHypothesis Testing1. State the null hypothesis, H0 and thealternative hypothesis, H12. Choose the level of significance, , and thesample size, n3. Determine the appropriate test statistic andsampling distribution4. Determine the critical values that divide therejection and nonrejection regionsDCOVA
- 23. Chap 9-23Copyright ©2012 Pearson Education Chap 9-236 Steps inHypothesis Testing5. Collect data and compute the value of the teststatistic6. Make the statistical decision and state themanagerial conclusion. If the test statistic fallsinto the nonrejection region, do not reject thenull hypothesis H0. If the test statistic falls intothe rejection region, reject the null hypothesis.Express the managerial conclusion in thecontext of the problem(continued)DCOVA
- 24. Chap 9-24Copyright ©2012 Pearson Education Chap 9-24Hypothesis Testing ExampleTest the claim that the true mean diameterof a manufactured bolt is 30mm.(Assume σ = 0.8)1. State the appropriate null and alternativehypotheses H0: μ = 30 H1: μ ≠ 30 (This is a two-tail test)2. Specify the desired level of significance and thesample size Suppose that = 0.05 and n = 100 are chosenfor this testDCOVA
- 25. Chap 9-25Copyright ©2012 Pearson Education Chap 9-252.00.08.161000.83029.84nσμXZSTATHypothesis Testing Example3. Determine the appropriate technique σ is assumed known so this is a Z test.4. Determine the critical values For = 0.05 the critical Z values are 1.965. Collect the data and compute the test statistic Suppose the sample results aren = 100, X = 29.84 (σ = 0.8 is assumed known)So the test statistic is:(continued)DCOVA
- 26. Chap 9-26Copyright ©2012 Pearson Education Chap 9-26Reject H0 Do not reject H0 6. Is the test statistic in the rejection region?/2 = 0.025-Zα/2 = -1.96 0Reject H0 ifZSTAT < -1.96 orZSTAT > 1.96;otherwise donot reject H0Hypothesis Testing Example(continued)/2 = 0.025Reject H0+Zα/2 = +1.96Here, ZSTAT = -2.0 < -1.96, so thetest statistic is in the rejectionregionDCOVA
- 27. Chap 9-27Copyright ©2012 Pearson Education Chap 9-276 (continued). Reach a decision and interpret the result-2.0Since ZSTAT = -2.0 < -1.96, reject the null hypothesisand conclude there is sufficient evidence that the meandiameter of a manufactured bolt is not equal to 30Hypothesis Testing Example(continued)Reject H0 Do not reject H0= 0.05/2-Zα/2 = -1.96 0= 0.05/2Reject H0+Zα/2= +1.96DCOVA
- 28. Chap 9-28Copyright ©2012 Pearson Education Chap 9-28p-Value Approach to Testing p-value: Probability of obtaining a teststatistic equal to or more extreme than theobserved sample value given H0 is true The p-value is also called the observed level ofsignificance H0 can be rejected if the p-value is less than αDCOVA
- 29. Chap 9-29Copyright ©2012 Pearson Education Chap 9-29p-Value Approach to Testing:Interpreting the p-value Compare the p-value with If p-value < , reject H0 If p-value , do not reject H0 Remember If the p-value is low then H0 must goDCOVA
- 30. Chap 9-30Copyright ©2012 Pearson Education Chap 9-30The 5 Step p-value approach toHypothesis Testing1. State the null hypothesis, H0 and the alternativehypothesis, H12. Choose the level of significance, , and the sample size, n3. Determine the appropriate test statistic and samplingdistribution4. Collect data and compute the value of the test statistic andthe p-value5. Make the statistical decision and state the managerialconclusion. If the p-value is < α then reject H0, otherwisedo not reject H0. State the managerial conclusion in thecontext of the problemDCOVA
- 31. Chap 9-31Copyright ©2012 Pearson Education Chap 9-31p-value Hypothesis TestingExampleTest the claim that the true meandiameter of a manufactured bolt is 30mm.(Assume σ = 0.8)1. State the appropriate null and alternativehypotheses H0: μ = 30 H1: μ ≠ 30 (This is a two-tail test)2. Specify the desired level of significance and thesample size Suppose that = 0.05 and n = 100 are chosenfor this testDCOVA
- 32. Chap 9-32Copyright ©2012 Pearson Education Chap 9-322.00.08.161000.83029.84nσμXZSTATp-value Hypothesis TestingExample3. Determine the appropriate technique σ is assumed known so this is a Z test.4. Collect the data, compute the test statistic and thep-value Suppose the sample results aren = 100, X = 29.84 (σ = 0.8 is assumed known)So the test statistic is:(continued)DCOVA
- 33. Chap 9-33Copyright ©2012 Pearson Education Chap 9-33p-Value Hypothesis Testing Example:Calculating the p-value4. (continued) Calculate the p-value. How likely is it to get a ZSTAT of -2 (or something farther from themean (0), in either direction) if H0 is true?p-value = 0.0228 + 0.0228 = 0.0456P(Z < -2.0) = 0.02280-2.0Z2.0P(Z > 2.0) = 0.0228DCOVA
- 34. Chap 9-34Copyright ©2012 Pearson Education Chap 9-34 5. Is the p-value < α? Since p-value = 0.0456 < α = 0.05 Reject H0 5. (continued) State the managerial conclusionin the context of the situation. There is sufficient evidence to conclude the average diameter ofa manufactured bolt is not equal to 30mm.p-value Hypothesis TestingExample(continued)DCOVA
- 35. Chap 9-35Copyright ©2012 Pearson EducationConnection Between Two-Tail Testsand Confidence Intervals For X = 29.84, σ = 0.8 and n = 100, the 95%confidence interval is:29.6832 ≤ μ ≤ 29.9968 Since this interval does not contain the hypothesizedmean (30), we reject the null hypothesis at = 0.051000.8(1.96)29.84to1000.8(1.96)-29.84DCOVA
- 36. Chap 9-36Copyright ©2012 Pearson Education Chap 9-36Do You Ever Truly Know σ? Probably not! In virtually all real world business situations, σ is notknown. If there is a situation where σ is known then µ is alsoknown (since to calculate σ you need to know µ.) If you truly know µ there would be no need to gather asample to estimate it.DCOVA
- 37. Chap 9-37Copyright ©2012 Pearson Education Chap 9-37Hypothesis Testing:σ Unknown If the population standard deviation is unknown, youinstead use the sample standard deviation S. Because of this change, you use the t distribution insteadof the Z distribution to test the null hypothesis about themean. When using the t distribution you must assume thepopulation you are sampling from follows a normaldistribution. All other steps, concepts, and conclusions are the same.DCOVA
- 38. Chap 9-38Copyright ©2012 Pearson Education Chap 9-38t Test of Hypothesis for the Mean(σ Unknown)The test statistic is:HypothesisTests forσ Known σ UnknownKnown Unknown(Z test) (t test) Convert sample statistic ( ) to a tSTAT test statisticThe test statistic is:HypothesisTests forσ Known σ UnknownKnown Unknown(Z test) (t test)XThe test statistic is:nSμXtSTATHypothesisTests forσ Known σ UnknownKnown Unknown(Z test) (t test)DCOVA
- 39. Chap 9-39Copyright ©2012 Pearson Education Chap 9-39Example: Two-Tail Test( Unknown)The average cost of a hotelroom in New York is said tobe $168 per night. Todetermine if this is true, arandom sample of 25 hotelsis taken and resulted in an Xof $172.50 and an S of$15.40. Test the appropriatehypotheses at = 0.05.(Assume the population distribution is normal)H0: ______H1: ______DCOVA
- 40. Chap 9-40Copyright ©2012 Pearson Education Chap 9-40 = 0.05 n = 25, df = 25-1=24 is unknown, souse a t statistic Critical Value:t24,0.025 = ± 2.0639Example Solution:Two-Tail t TestDo not reject H0: insufficient evidence that truemean cost is different from $168Reject H0Reject H0/2=.025-t 24,0.025Do not reject H00/2=.025-2.0639 2.06391.462515.40168172.50nSμXSTATt1.46H0: μ = 168H1: μ 168t 24,0.025DCOVA
- 41. Chap 9-41Copyright ©2012 Pearson Education Chap 9-41Example Two-Tail t Test Using Ap-value from Excel Since this is a t-test we cannot calculate the p-valuewithout some calculation aid. The Excel output below does this:t Test for the Hypothesis of the MeanNull Hypothesis µ= 168.00$Level of Significance 0.05Sample Size 25Sample Mean 172.50$Sample Standard Deviation 15.40$Standard Error of the Mean 3.08$ =B8/SQRT(B6)Degrees of Freedom 24 =B6-1t test statistic 1.46 =(B7-B4)/B11Lower Critical Value -2.0639 =-TINV(B5,B12)Upper Critical Value 2.0639 =TINV(B5,B12)p-value 0.157 =TDIST(ABS(B13),B12,2)=IF(B18<B5, "Reject null hypothesis","Do not reject null hypothesis")DataIntermediate CalculationsTwo-Tail TestDo Not Reject Null Hypothesisp-value > αSo do not reject H0DCOVA
- 42. Chap 9-42Copyright ©2012 Pearson EducationExample Two-Tail t Test Using Ap-value from MinitabDCOVAOne-Sample TTest of mu = 168 vs not = 168N Mean StDev SE Mean 95% CI T P25 172.50 15.40 3.08 (166.14, 178.86) 1.46 0.157p-value > αSo do not reject H01234
- 43. Chap 9-43Copyright ©2012 Pearson EducationConnection of Two-Tail Tests toConfidence Intervals For X = 172.5, S = 15.40 and n = 25, the 95%confidence interval for µ is:172.5 - (2.0639) 15.4/ 25 to 172.5 + (2.0639) 15.4/ 25166.14 ≤ μ ≤ 178.86 Since this interval contains the hypothesized mean(168), we do not reject the null hypothesis at = 0.05DCOVA
- 44. Chap 9-44Copyright ©2012 Pearson Education Chap 9-44One-Tail Tests In many cases, the alternative hypothesisfocuses on a particular directionH0: μ ≥ 3H1: μ < 3H0: μ ≤ 3H1: μ > 3This is a lower-tail test since thealternative hypothesis is focused onthe lower tail below the mean of 3This is an upper-tail test since thealternative hypothesis is focused onthe upper tail above the mean of 3DCOVA
- 45. Chap 9-45Copyright ©2012 Pearson Education Chap 9-45Reject H0 Do not reject H0 There is only onecritical value, sincethe rejection area isin only one tailLower-Tail Tests-Zα or -tα 0μH0: μ ≥ 3H1: μ < 3Z or tXCritical valueDCOVA
- 46. Chap 9-46Copyright ©2012 Pearson Education Chap 9-46Reject H0Do not reject H0Upper-Tail TestsZα or tα0μH0: μ ≤ 3H1: μ > 3 There is only onecritical value, sincethe rejection area isin only one tailCritical valueZ or tX_DCOVA
- 47. Chap 9-47Copyright ©2012 Pearson Education Chap 9-47Example: Upper-Tail t Testfor Mean ( unknown)A phone industry manager thinks thatcustomer monthly cell phone bills haveincreased, and now average over $52 permonth. The company wishes to test thisclaim. (Assume a normal population)H0: μ ≤ 52 the average is not over $52 per monthH1: μ > 52 the average is greater than $52 per month(i.e., sufficient evidence exists to support themanager’s claim)Form hypothesis test:DCOVA
- 48. Chap 9-48Copyright ©2012 Pearson Education Chap 9-48Reject H0Do not reject H0 Suppose that = 0.10 is chosen for this test andn = 25.Find the rejection region:= 0.101.3180Reject H0Reject H0 if tSTAT > 1.318Example: Find Rejection Region(continued)DCOVA
- 49. Chap 9-49Copyright ©2012 Pearson Education Chap 9-49Obtain sample and compute the test statisticSuppose a sample is taken with the followingresults: n = 25, X = 53.1, and S = 10 Then the test statistic is:0.5525105253.1nSμXtSTATExample: Test Statistic(continued)DCOVA
- 50. Chap 9-50Copyright ©2012 Pearson Education Chap 9-50Reject H0Do not reject H0Example: Decision= 0.101.3180Reject H0Do not reject H0 since tSTAT = 0.55 ≤ 1.318There is insufficient evidence that themean bill is over $52.tSTAT = 0.55Reach a decision and interpret the result:(continued)DCOVA
- 51. Chap 9-51Copyright ©2012 Pearson Education Chap 9-51Example: Utilizing The p-valuefor The Test Calculate the p-value and compare to (p-value belowcalculated using Excel spreadsheet on next page)RejectH0= .10Do not rejectH01.3180Reject H0tSTAT = .55p-value = .2937Do not reject H0 since p-value = .2937 > = .10DCOVA
- 52. Chap 9-52Copyright ©2012 Pearson Education Chap 9-52Excel Spreadsheet Calculating Thep-value for The Upper Tail t Testt Test for the Hypothesis of the MeanNull Hypothesis µ= 52.00Level of Significance 0.1Sample Size 25Sample Mean 53.10Sample Standard Deviation 10.00Standard Error of the Mean 2.00 =B8/SQRT(B6)Degrees of Freedom 24 =B6-1t test statistic 0.55 =(B7-B4)/B11Upper Critical Value 1.318 =TINV(2*B5,B12)p-value 0.2937 =TDIST(ABS(B13),B12,1)=IF(B18<B5, "Reject null hypothesis","Do not reject null hypothesis")DataIntermediate CalculationsUpper Tail TestDo Not Reject Null HypothesisDCOVA
- 53. Chap 9-53Copyright ©2012 Pearson EducationUsing Minitab to calculate The p-valuefor The Upper Tail t TestDCOVA1234One-Sample TTest of mu = 52 vs > 5295% LowerN Mean StDev SE Mean Bound T P25 53.10 10.00 2.00 49.68 0.55 0.294p-value > αSo do not reject H0
- 54. Chap 9-54Copyright ©2012 Pearson Education Chap 9-54Hypothesis Tests for Proportions Involves categorical variables Two possible outcomes Possesses characteristic of interest Does not possess characteristic of interest Fraction or proportion of the population in thecategory of interest is denoted by πDCOVA
- 55. Chap 9-55Copyright ©2012 Pearson Education Chap 9-55Proportions Sample proportion in the category of interest isdenoted by p When both X and n – X are at least 5, p canbe approximated by a normal distribution withmean and standard deviationsizesamplesampleininterestofcategoryinnumbernXppμn)(1σ p(continued)DCOVA
- 56. Chap 9-56Copyright ©2012 Pearson Education Chap 9-56 The samplingdistribution of p isapproximatelynormal, so the teststatistic is a ZSTATvalue:Hypothesis Tests for Proportionsn)(1pZSTATπππX 5andn – X 5HypothesisTests for pX < 5orn – X < 5Not discussedin this chapterDCOVA
- 57. Chap 9-57Copyright ©2012 Pearson Education Chap 9-57 An equivalent formto the last slide,but in terms of thenumber in thecategory ofinterest, X:Z Test for Proportion in Terms ofNumber in Category of Interest)(1nnXZSTATX 5andn-X 5HypothesisTests for XX < 5orn-X < 5Not discussedin this chapterDCOVA
- 58. Chap 9-58Copyright ©2012 Pearson Education Chap 9-58Example: Z Test for ProportionA marketing companyclaims that it receivesresponses from 8% ofthose surveyed. To testthis claim, a randomsample of 500 weresurveyed with 25responses. Test at the= 0.05 significance level.Check:X = 25n-X = 475DCOVA
- 59. Chap 9-59Copyright ©2012 Pearson Education Chap 9-59Z Test for Proportion: Solution= 0.05n = 500, p = 0.05Reject H0 at = 0.05H0: π = 0.08H1: π 0.08Critical Values: ± 1.96Test Statistic:Decision:Conclusion:z0Reject Reject.025.0251.96-2.47There is sufficientevidence to reject thecompany’s claim of 8%response rate.2.47500.08).08(1.08.05n)(1pSTATZπππ-1.96DCOVA
- 60. Chap 9-60Copyright ©2012 Pearson Education Chap 9-60Do not reject H0Reject H0Reject H0/2 = .0251.960Z = -2.47Calculate the p-value and compare to(For a two-tail test the p-value is always two-tail)(continued)0.01362(0.0068)2.47)P(Z2.47)P(Zp-value = 0.0136:p-Value SolutionReject H0 since p-value = 0.0136 < = 0.05Z = 2.47-1.96/2 = .0250.00680.0068DCOVA
- 61. Chap 9-61Copyright ©2012 Pearson Education Chap 9-61Potential Pitfalls andEthical Considerations Use randomly collected data to reduce selection biases Do not use human subjects without informed consent Choose the level of significance, α, and the type of test(one-tail or two-tail) before data collection Do not employ “data snooping” to choose between one-tail and two-tail test, or to determine the level ofsignificance Do not practice “data cleansing” to hide observationsthat do not support a stated hypothesis Report all pertinent findings including both statisticalsignificance and practical importance
- 62. Chap 9-62Copyright ©2012 Pearson Education Chap 9-62Chapter Summary Addressed hypothesis testing methodology Performed Z Test for the mean (σ known) Discussed critical value and p–valueapproaches to hypothesis testing Performed one-tail and two-tail tests
- 63. Chap 9-63Copyright ©2012 Pearson Education Chap 9-63Chapter Summary Performed t test for the mean (σunknown) Performed Z test for the proportion Discussed pitfalls and ethical issues(continued)
- 64. Chap 9-64Copyright ©2012 Pearson EducationBasic Business Statistics12th EditionGlobal EditionOnline TopicPower of A Hypothesis Test
- 65. Chap 9-65Copyright ©2012 Pearson EducationRejectH0: μ 52Do not rejectH0 : μ 52The Power of a Test The power of the test is the probability ofcorrectly rejecting a false H05250Suppose we correctly reject H0: μ 52when in fact the true mean is μ = 50Power= 1-βDCOVA
- 66. Chap 9-66Copyright ©2012 Pearson EducationRejectH0: 52Do not rejectH0 : 52Type II Error Suppose we do not reject H0: 52 when in factthe true mean is = 505250This is the truedistribution of X if = 50This is the range of X whereH0 is not rejectedProb. of type IIerror = βDCOVA
- 67. Chap 9-67Copyright ©2012 Pearson EducationRejectH0: μ 52Do not rejectH0 : μ 52Type II Error Suppose we do not reject H0: μ 52 when in factthe true mean is μ = 505250βHere, β = P( X cutoff ) if μ = 50(continued)DCOVA
- 68. Chap 9-68Copyright ©2012 Pearson EducationRejectH0: μ 52Do not rejectH0 : μ 52 Suppose n = 64 , σ = 6 , and = .055250So β = P( X 50.766 ) if μ = 50Calculating β50.7666461.64552nσZμXcutoff(for H0 : μ 52)50.766DCOVA
- 69. Chap 9-69Copyright ©2012 Pearson EducationRejectH0: μ 52Do not rejectH0 : μ 520.15390.84611.01.02)P(Z6465050.766ZP50)μ|50.766XP( Suppose n = 64 , σ = 6 , and = 0.055250Calculating β andPower of the test(continued)Probability oftype II error:β = 0.1539Power= 1 - β= 0.846150.766The probability ofcorrectly rejecting afalse null hypothesis is0.8641DCOVA
- 70. Chap 9-70Copyright ©2012 Pearson EducationPower of the Test Conclusions regarding the power of the test: A one-tail test is more powerful than a two-tail test An increase in the level of significance ( ) results inan increase in power An increase in the sample size results in anincrease in power
- 71. Chap 9-71Copyright ©2012 Pearson EducationOnline Topic Summary Examined how to calculate and interpretthe power of a hypothesis test.

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