9- 1 Business Statistics: - PowerPoint Presentation

1. 9-1Business Statistics: A Decision-Making Approach8th EditionChapter 9Introduction to Hypothesis Testing

2. 9-2Chapter GoalsAfter completing this chapter, you should be able to: Formulate null and alternative hypotheses involving a single population mean or proportionKnow what Type I and Type II errors areFormulate a decision rule for testing a hypothesisKnow how to use the test statistic, critical value, and p-value approaches to test the null hypothesisCompute the probability of a Type II error

3. 9-3What is Hypothesis Testing?A statistical hypothesis is an assumption about a population parameter (assumption may or may not be true). The best way to determine whether a statistical hypothesis is true would be to examine the entire population.Often impractical. So, examine a random sample If sample data are not consistent with the statistical hypothesis, the hypothesis is rejected.We never 100% “prove” anything because of sampling error

4. Types of HypothesesNull hypothesis. Denoted by H0Alternative hypothesis. Denoted by H1 or Ha9-4

5. 9-5The Null Hypothesis, H0States the assumption to be testedExample: 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 (even though sample is used)

6. 9-6The Null Hypothesis, H0Begin with the assumption that the null hypothesis is trueSimilar to the notion of “innocent until proven guilty”Always contains “=” , “≤” or “” signMay or may not be rejectedBased on the statistical evidence gathered(continued)

7. 9-7The Alternative Hypothesis, HAIs the opposite of the null hypothesise.g.: The average number of TV sets in U.S. homes is less than 3 ( HA: µ < 3 )Contains “≠”, “<” or “>” signNever contains the “=” , “≤” or “” signMay or may not be accepted

8. 9-8Example 1: The average annual income of buyers of Ford F150 pickup trucks is claimed to be $65,000 per year. An industry analyst would like to test this claim.What is the appropriate test?H0: µ = 65,000 (income is as claimed)HA: µ ≠ 65,000 (income is different than claimed) The analyst will believe the claim unless sufficient evidence is found to discredit it.Formulating Hypotheses

9. 9-9Formulating HypothesesExample 2: Ford Motor Company has worked to reduce road noise inside the cab of the redesigned F150 pickup truck. It would like to report in its advertising that the truck is quieter. The average of the prior design was 68 decibels at 60 mph.What is the appropriate test?H0: µ ≤ 67 (the truck is quieter: must be less than 68) HA: µ > 68 (the truck is not quieter)

10. Steps for Formulating Hypothesis TestsState the hypotheses. The hypotheses must be mutually exclusive. That is, if one is true, the other must be false. Formulate an analysis plan. The analysis plan describes how to use sample data to evaluate the null hypothesis. The evaluation often focuses around a single test statistic. 9-10

11. Steps for Formulating Hypothesis TestsAnalyze sample data. Find the value of the test statistic (mean score, proportion, t-score, z-score, etc.) described in the analysis plan and interpret result.Apply the decision rule described in the analysis plan. If the value of the test statistic is unlikely, based on the null hypothesis, reject the null hypothesis. 9-11

12. 9-123 outcomes for a hypothesis testNo error (no need to discuss)Type I error (Khan academy: type I error)Type II errorErrors in Making Decisions

13. 9-13Errors in Making DecisionsType I Error Rejecting a true null hypothesisConsidered as a serious errorThe probability of Type I Error is Called level of significance of the testSet by researcher in advance(continued)

14. 9-14Errors in Making Decisions(continued)Type II ErrorFailing to reject (i.e., accept) a false null hypothesisThe probability of Type II Error is ββ is a calculated value, the formula is discussed later in the chapter

15. 9-15Outcomes and ProbabilitiesState of NatureDecisionDo NotRejectH0No error (1 - )aType II Error ( β )RejectH0Type I Error( )aPossible Hypothesis Test Outcomes H0 False H0 TrueKey:Outcome(Probability)No Error ( 1 - β )

16. 9-16Type I & II Error Relationship Type I and Type II errors cannot happen at the same time (mutually exclusive) 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 )

17. 9-17Level of Significance,   defines rejection region (Cutoff) of the sampling distributionIs designated by level of significanceTypical values are 0.01, 0.05, or 0.10Based on COST! Want small probability of Type I error  higher costProvides the critical value(s) of the test The value corresponding to a significance level

18. 9-18Hypothesis Tests for the Meanσ Knownσ UnknownHypothesis Tests for Assume first that the population standard deviation σ is known

19. 9-19Type of Hypothesis TestH0: μ ≥ 3 HA: μ < 30H0: μ ≤ 3 HA: μ > 3H0: μ = 3 HA: μ ≠ 3aa /2Lower tail testLevel of significance = a0 /2aUpper tail testTwo tailed test0a-zαzα-zα/2zα/2Reject H0Reject H0Reject H0Reject H0Do not reject H0Do not reject H0Do not reject H0Example:Example:Example:*comparison operator of each test*

20. 9-20z-units (Not as universal as using p-value):For given , find the critical z value(s): -zα , zα ,or ±zα/2 Convert the sample mean x to a z test statistic:Reject H0 if z is in the rejection region, otherwise do not reject H0Two Equivalent Approaches to Hypothesis Testing

21. 9-21x units – sample mean (Not Recommended):Given , calculate the critical value(s)xα , or xα/2(L) and xα/2(U)The sample mean is the test statistic. Reject H0 if x is in the rejection region, otherwise do not reject H0Two Equivalent Approaches to Hypothesis Testing

22. 9-22Hypothesis Testing ExampleTest the claim that the true mean # of TV sets in US homes is at least 3.(Assume σ = 0.8)1. Specify the population value of interest Testing mean number of TVs in US homes2. Formulate the appropriate null and alternative hypothesesH0: μ  3 HA: μ < 3 (This is a lower (left) tail test)3. Specify the desired level of significanceSuppose that  = 0.05 is chosen for this test: probability is given, need to find z value

23. Formula Table6-23We haveXZp We wantXX=μ+zσNorminv(p,μ,σ)ZNormsinv(p as )pNormdist(x,μ,σ,1)Normsdist(z)

24. 9-24Reject H0Do not reject H04. Determine the rejection region = .05-zα= -1.6450This is a one-tailed test with  = 0.05 Since σ is known, the cutoff value is a z value: Reject H0 if z < z = -1.645 ; otherwise do not reject H0Hypothesis Testing Example(continued)Normsinv(.05) = 1.645

25. 9-255. Obtain sample evidence and compute the test statisticSuppose a sample is taken with the following results: n = 100, x = 2.84 ( = 0.8 is assumed known) Then the test statistic is:Hypothesis Testing Example

26. 9-26Reject H0Do not reject H0 = .05-1.6450 6. Reach a decision and interpret the result-2.0Since z = -2.0 < -1.645, we reject the null hypothesis that the mean number of TVs in US homes is at least 3. There is sufficient evidence that the mean is less than 3.Hypothesis Testing Example(continued)z

27. 9-27Reject H0 = .052.8684Do not reject H03 An alternate way (not recommend) of constructing rejection region:2.84Since x = 2.84 < 2.8684, we reject the null hypothesisHypothesis Testing Example(continued)xNow expressed in x, not z unitsNot enough statistical evidence to conclude that the number of TVs is at least 3

28. 9-28Using p-Valuep-value: The p-value is the probability that your null hypothesis is actually correct.  Simple and easy to apply: always reject null hypothesis if p-value <  .Best and Universal approach: because p-values are usually computed by statistical SW packages (i.e., Excel, Minitab)

29. 9-29p-Value Approach to TestingConvert Sample Statistic ( ) to Test Statistic (a z value, if σ is known)Determine the p-value from a table or computerCompare the p-value with If p-value <  , reject H0If p-value   , do not reject H0 x(continued)

30. 9-30More than just a simple “reject”Can now determine how strongly we “reject” or “accept”The further the p-value is from a, the stronger the decisionJust compare “absolute value” whether it is one-tail or two-tail test. Unlike z value (or t value), no need to worry about “negative” or “positive”p-Value Approach to Testing(continued)

31. 9-31Example: Based on previous example, how likely is it to see a sample mean of 2.84 (or something further below the mean) if the true mean is  = 3.0? p-value =0.0228 = 0.05p-value Example0-1.645-2.0: z value from previous example zReject H0Do not reject H0Normsdist(-2) = 0.0228Normsinv(.05) = 1.645

32. 9-32Compare the p-value with If p-value <  , reject H0If p-value   , do not reject H0 Here: p-value = 0.0228  = 0.05Since 0.0228 < 0.05, we reject the null hypothesis(continued)p-value Examplep-value =0.0228 = 0.05-1.6450-2.0Reject H0

33. 9-33Example: Upper Tail z Test for Mean ( Known) A phone industry manager thinks that customers’ monthly cell phone bills have increased, and now average over $52 per month. The company wishes to test this claim. (Assume  = 10 is known)H0: μ ≤ 52 the average is not over $52 per monthHA: μ > 52 the average is greater than $52 per month (i.e., sufficient evidence exists to support the manager’s claim)Form hypothesis (upper-tail) test:

34. 9-34Reject H0Do not reject H0Suppose  = 0.10 is chosen for this testFind the rejection region: = 0.10zα=1.280Reject H0Reject H0 if z > 1.28Example: Find Rejection Region(continued)Normsinv(0.1) = 1.28

35. 9-35Obtain sample evidence and compute the test statisticSuppose a sample is taken with the following results: n = 64, x = 53.1 (=10 was assumed known) Then the test statistic is:Example: Test Statistic(continued)

36. 9-36Reject H0Do not reject H0Example: Decision = 0.101.280Reject H0Do not reject H0 since z = 0.88 ≤ 1.28i.e.: there is not sufficient evidence that the mean bill is over $52z = 0.88Reach a decision and interpret the result:(continued)

37. 9-37Reject H0 = 0.10Do not reject H01.280Reject H0z = 0.88Calculate the p-value and compare to (continued)p-value = 0.1894p -Value SolutionDo not reject H0 since p-value = 0.1894 >  = 0.101-NORMSDIST(0.88) = 0.1894

38. 9-38Using t valueWhen Pop. Dist. Type is unknown and sample size is smaller, convert sample statistic ( ) to a t – statistic. x Known UnknownHypothesis Tests for The test statistic is:

39. When to apply t-statisticWhen σ is unknown, convert sample statistic (x) to a t – statistic. This is what the textbook said….based on assumption that the population is approximately normal.In reality, we do not know whether the population is almost or even half normal or not. So, forget about what the textbook says. Thus, if n is less than 30, apply t – statistic. Of course, if n is greater than equal to 30, apply z – statistic. 9-39

40. 9-40Hypothesis Tests for μ, σ UnknownSpecify the population value of interestFormulate the appropriate null and alternative hypothesesSpecify the desired level of significanceDetermine the rejection region (critical values are from the t-distribution with n-1 d.f.)Obtain sample evidence and compute the test statisticReach a decision and interpret the result

41. 9-41Example: Two-Tail Test 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 (sample Std). Test at the  = 0.05 level (Assume the population distribution is normal)H0: μ = 168 HA: μ ¹ 168

42. 9-42a = 0.05n = 25Critical Values: t24 = ± 2.0639  is unknown, so use a t statisticExample Solution: Two-Tail TestDo not reject H0: not sufficient evidence that true mean cost is different than $168Reject H0Reject H0a/2=0.025-tα/2Do not reject H00tα/2a/2=0.025-2.06392.06391.46H0: μ = 168 HA: μ ¹ 168

43. TINVNORMSINV(p) gives the z-value that puts probability (area) p to the left of that value of z. TINV(p,DF) gives the t-value that puts one-half the probability (area) to the right with DF degrees of freedom. The reason for this is that TINV is used mainly to give the t-value used in confidence intervals. Remember in confidence intervals we use zα/2 or tα/2. So, to get tα/2, you put in α and TINV automatically splits it when giving the appropriate t-value.9-43

44. TDISTNORMSDIST(z) gives the probability (area) to the left of z. TDIST gives the area to the right of t; and on top of that it only works for positive values of t if you do want the area to the left of t?1 – (area to the right of t).The general form TDIST(t, degrees of freedom, 1 or 2). The last argument (1 or 2)1 for one-tail test or 2 for two-tail test. This is because that, in general, TDIST is used to get the p-values for t-tests.9-44

45. Using Excel….Download and review following Word files from the class websiteImplication of TDIST and TINVUsing TDIST and TINV for HypothesisDownload Finding Critical Value and P-Value Excel file….Last slide for today’s lectureReview example from page 349 to 3659-45

46. 9-46Hypothesis Tests in PHStatOptions

47. 9-47Sample PHStat OutputInputOutput

48. 9-48Hypothesis Tests for ProportionsInvolves categorical valuesTwo possible outcomes“Success” (possesses a certain characteristic)“Failure” (does not possesses that characteristic)Fraction or proportion of population in the “success” category is denoted by π

49. 9-49ProportionsThe sample proportion of successes is denoted by p : When both nπ and n(1- π) are at least 5, p is approximately normally distributed with mean and standard deviation

50. 9-50The sampling distribution of p is normal, so the test statistic is a z value:Hypothesis Tests for Proportionsnπ  5andn(1-π)  5Hypothesis Tests for πnπ < 5orn(1-π) < 5Not discussed in this chapter

51. 9-51Example: z Test for Proportion A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses. Test at the  = 0.05 significance level.Check: n π = (500)(0.08) = 40n(1-π) = (500)(0.92) = 460Both > 5, so assume normal

52. 9-52Z Test for Proportion: Solutiona = 0.05 n = 500, p = 0.05Reject H0 at  = 0.05H0: π = 0.08 HA: π ¹ 0.08Critical Values: ± 1.96Test Statistic:Decision:Conclusion:z0RejectReject0.0250.0251.96-2.47There is sufficient evidence to reject the company’s claim of 8% response rate.-1.96

53. 9-53Do not reject H0Reject H0Reject H0/2 = 0.0251.960z = -2.47Calculate the p-value and compare to  (For a two tailed test the p-value is always two tailed)p-value = .0136:p -Value SolutionReject H0 since p-value = 0.0136 <  = 0.05z = 2.47-1.96/2 = 0.0250.00680.0068

54. 9-54Reject H0: μ  52Do not reject H0 : μ  52Type II ErrorType II error is the probability of failing to reject a false H05250Suppose we fail to reject H0: μ  52 when in fact the true mean is μ = 50

55. 9-55Reject H0:   52Do not reject H0 :   52Type II ErrorSuppose we do not reject H0:   52 when in fact the true mean is  = 505250This is the true distribution of x if  = 50This is the range of x where H0 is not rejected(continued)

56. 9-56Reject H0: μ  52Do not reject H0 : μ  52Type II ErrorSuppose we do not reject H0: μ  52 when in fact the true mean is μ = 505250βHere, β = P( x  cutoff ) if μ = 50(continued)

57. 9-57Steps for Calculating b Specify the population parameter of interestFormulate the hypothesesSpecify the significance levelDetermine the critical valueDetermine the critical value for an upper or lower tail testSpecify the “true” population mean value of interestCompute z-value based on stipulated population meanUse standard normal table to find b

58. 9-58Reject H0: μ  52Do not reject H0 : μ  52Suppose n = 64 , σ = 6 , and  = 0.055250So β = P( x  50.766 ) if μ = 50Calculating β(for H0 : μ  52)50.766

59. 9-59Reject H0: μ  52Do not reject H0 : μ  52Suppose n = 64 , σ = 6 , and  = 0.055250Calculating β(continued)Probability of type II error: β = 0.1539

60. 9-60Would like b to be as small as possibleIf the null hypothesis is FALSE, we want to reject iti.e., we want the hypothesis test to have a high probability of rejecting a false null hypothesisReferred to as the power of the testPower = 1 - bPower Of Hypothesis Testing

61. 9-61Chapter SummaryAddressed hypothesis testing methodologyPerformed z Test for the mean (σ known)Discussed p–value approach to hypothesis testingPerformed one-tail and two-tail tests

62. 9-62Chapter SummaryPerformed t test for the mean (σ unknown)Performed z test for the proportionDiscussed Type II error and computed its probability(continued)