1、Chap 9-1Chapter 9Fundamentals of Hypothesis Testing: One-Sample TestsStatistics For ManagersUsing Microsoft Excel6th EditionChap 9-2Learning ObjectivesIn this chapter, you learn: The basic principles of hypothesis testingHow to use hypothesis testing to test a mean or proportionThe assumptions of ea

2、ch hypothesis-testing procedure, how to evaluate them, and the consequences if they are seriously violatedHow to avoid the pitfalls involved in hypothesis testingThe ethical issues involved in hypothesis testingChap 9-3What is a Hypothesis? A hypothesis is a claim (assumption) about a population par

3、ameter: population mean population proportionExample: The mean monthly cell phone bill of this city is = $42Example: The proportion of adults in this city with cell phones is = 0.68Chap 9-4The Null Hypothesis, H0 States the claim or assertion to be testedExample: The average number of TV sets in U.S

4、. Homes is equal to three ( ) Is always about a population parameter, not about a sample statistic 3:H03:H03X:H0Chap 9-5The Null Hypothesis, H0 Begin 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 “=

5、” , “” or “ ” sign May or may not be rejected(continued)Chap 9-6The Alternative Hypothesis, H1 Is the opposite of the null hypothesis e.g., The average number of TV sets in U.S. homes is not equal to 3 ( H1: 3 ) Challenges the status quo Never contains the “=” , “” or “ ” sign May or may not be prov

6、en Is generally the hypothesis that the researcher is trying to provePopulationClaim: thepopulationmean age is 50.(Null Hypothesis:REJECTSupposethe samplemean age is 20: X = 20SampleNull Hypothesis20likely if = 50?IsHypothesis Testing ProcessIf not likely, Now select a random sample H0: = 50 )XChap

7、9-8Sampling Distribution of X = 50If H0 is trueIf it is unlikely that we would get a sample mean of this value . then we reject the null hypothesis that = 50.Reason for Rejecting H020. if in fact this were the population meanXChap 9-9Level of Significance, Defines the unlikely values of the sample s

8、tatistic if the null hypothesis is true Defines rejection region 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 Chap 9-10Level of Significance and

9、 the Rejection RegionH0: 3 H1: 3 Represents critical valueLower-tail testLevel of significance = 0Upper-tail testTwo-tail testRejection region is shaded /20 /2 H0: = 3 H1: 3Chap 9-11Errors in Making Decisions Type I Error Reject a true null hypothesis Considered a serious type of errorThe probabilit

10、y of Type I Error is Called level of significance of the test Set by the researcher in advanceChap 9-12Errors in Making Decisions Type II Error Fail to reject a false null hypothesisThe probability of Type II Error is (continued)Chap 9-13Outcomes and ProbabilitiesActual SituationDecisionDo NotReject

11、H0No error (1 - ) Type II Error ( )RejectH0Type I Error( ) Possible Hypothesis Test Outcomes H0 False H0 TrueKey:Outcome(Probability)No Error ( 1 - )Chap 9-14Type 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

12、error can only occur if H0 is false If Type I error probability ( ) , then Type II error probability ( )Chap 9-15Factors Affecting Type II Error All else equal, when the difference between hypothesized parameter and its true value when when when nChap 9-16Hypothesis Tests for the Mean Known UnknownH

13、ypothesis Tests for (Z test)(t test)Chap 9-17Z Test of Hypothesis for the Mean ( Known) Convert sample statistic ( ) to a Z test statistic XThe test statistic is:nXZ Known UnknownHypothesis Tests for Known Unknown(Z test)(t test)Chap 9-18Critical Value Approach to Testing For a two-tail test for the

14、 mean, known: 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 XChap 9-19Do not reject

15、 H0Reject H0Reject H0 There are two cutoff values (critical values), defining the regions of rejection Two-Tail Tests /2-Z0H0: = 3 H1: 3+Z /2Lower critical valueUpper critical value3ZXChap 9-206 Steps in Hypothesis Testing1. State the null hypothesis, H0 and the alternative hypothesis, H12. Choose t

16、he level of significance, , and the sample size, n3. Determine the appropriate test statistic and sampling distribution4. Determine the critical values that divide the rejection and nonrejection regionsChap 9-216 Steps in Hypothesis Testing5. Collect data and compute the value of the test statistic6

17、. Make the statistical decision and state the managerial conclusion. If the test statistic falls into the nonrejection 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

18、the problem(continued)Chap 9-22Hypothesis Testing ExampleTest the claim that the true mean of TV sets in US homes is equal to 3.(Assume = 0.8)1. State the appropriate null and alternative hypotheses H0: = 3 H1: 3 (This is a two-tail test)2. Specify the desired level of significance and the sample si

19、ze Suppose that = 0.05 and n = 100 are chosen for this testChap 9-23 2.0.08.161000.832.84nXZHypothesis Testing Example3. Determine the appropriate technique is 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

20、 statistic Suppose the sample results are n = 100, X = 2.84 ( = 0.8 is assumed known)So the test statistic is:(continued)Chap 9-24Reject H0Do not reject H0 6. Is the test statistic in the rejection region? = 0.05/2-Z= -1.960Reject H0 if Z 1.96; otherwise do not reject H0Hypothesis Testing Example(co

21、ntinued) = 0.05/2Reject H0+Z= +1.96Here, Z = -2.0 -1.96, so the test statistic is in the rejection regionChap 9-256(continued). Reach a decision and interpret the result-2.0Since Z = -2.0 -1.96, we reject the null hypothesis and conclude that there is sufficient evidence that the mean number of TVs

22、in US homes is not equal to 3 Hypothesis Testing Example(continued)Reject H0Do not reject H0 = 0.05/2-Z= -1.960 = 0.05/2Reject H0+Z= +1.96Chap 9-26p-Value Approach to Testing p-value: Probability of obtaining a test statistic more extreme ( or ) than the observed sample value given H0 is true Also c

23、alled observed level of significance Smallest value of for which H0 can be rejected Chap 9-27p-Value Approach to Testing Convert Sample Statistic (e.g., ) to Test Statistic (e.g., Z statistic ) Obtain the p-value from a table or computer Compare the p-value with If p-value , reject H0 If p-value , d

24、o not reject H0 X(continued)Chap 9-280.0228/2 = 0.025p-Value Example Example: 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?-1.960-2.00.02282.0)P(Z0.02282.0)P(ZZ1.962.0X = 2.84 is translated to a Z score of Z = -2.0

25、p-value = 0.0228 + 0.0228 = 0.04560.0228/2 = 0.025Chap 9-29 Compare the p-value with If p-value , reject H0 If p-value , do not reject H0 Here: p-value = 0.0456 = 0.05Since 0.0456 0.05, we reject the null hypothesis(continued)p-Value Example0.0228/2 = 0.025-1.960-2.0Z1.962.00.0228/2 = 0.025Connectio

26、n to Confidence Intervals For X = 2.84, = 0.8 and n = 100, the 95% confidence interval is: 2.6832 2.9968 Since this interval does not contain the hypothesized mean (3.0), we reject the null hypothesis at = 0.051000.8 (1.96) 2.84 to 1000.8 (1.96) - 2.84Chap 9-31One-Tail Tests In many cases, the alter

27、native hypothesis focuses on a particular directionH0: 3 H1: 3This is a lower-tail test since the alternative hypothesis is focused on the lower tail below the mean of 3This is an upper-tail test since the alternative hypothesis is focused on the upper tail above the mean of 3Chap 9-32Reject H0Do no

28、t reject H0 There is only one critical value, since the rejection area is in only one tailLower-Tail Tests -Z0H0: 3 H1: 3 There is only one critical value, since the rejection area is in only one tailCritical valueZX_Chap 9-34Example: Upper-Tail Z Test for Mean ( Known) A phone industry manager thin

29、ks that customer 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: 52 the average is not over $52 per monthH1: 52 the average is greater than $52 per month(i.e., sufficient evidence exists to support the mana

30、gers claim)Form hypothesis test:Chap 9-35Reject H0Do not reject H0 Suppose that = 0.10 is chosen for this testFind the rejection region: = 0.101.280Reject H0Reject H0 if Z 1.28Example: Find Rejection Region(continued)Chap 9-36Review:One-Tail Critical ValueZ.07.091.1 .8790 .8810 .88301.2 .8980.90151.

31、3 .9147 .9162 .9177z01.28.08Standardized Normal Distribution Table (Portion)What is Z given = 0.10? = 0.10Critical Value = 1.280.90.89970.100.90Chap 9-37Obtain sample and compute the test statisticSuppose a sample is taken with the following results: n = 64, X = 53.1 (=10 was assumed known) Then the

32、 test statistic is:0.8864105253.1nXZExample: Test Statistic(continued)Chap 9-38Reject 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:(co

33、ntinued)Chap 9-39Reject H0 = 0.10Do not reject H01.280Reject H0Z = 0.88Calculate the p-value and compare to (assuming that = 52.0)(continued)0.18940.810610.88)P(Z6410/52.053.1ZP53.1)XP(p-value = 0.1894p -Value SolutionDo not reject H0 since p-value = 0.1894 = 0.10Chap 9-40t Test of Hypothesis for th

34、e Mean ( Unknown) Convert sample statistic ( ) to a t test statistic XThe test statistic is:nSXt1-nHypothesis Tests for Known Unknown Known Unknown(Z test)(t test)Chap 9-41Example: Two-Tail Test( Unknown) The average cost of a hotel room in New York is said to be $168 per night. A random sample of 2

35、5 hotels resulted in X = $172.50 and S = $15.40. Test at the = 0.05 level.(Assume the population distribution is normal)H0: = 168 H1: 168Chap 9-42 = 0.05 n = 25 is unknown, so use a t statistic Critical Value: t24 = 2.0639Example Solution: Two-Tail TestDo not reject H0: not sufficient evidence that

36、true mean cost is different than $168Reject H0Reject H0/2=.025-t n-1,/2Do not reject H00/2=.025-2.06392.06391.462515.40168172.50nSXt1n1.46H0: = 168 H1: 168t n-1,/2Connection 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

37、 + (2.0639) 15.4/ 25 166.14 178.86 Since this interval contains the Hypothesized mean (168), we do not reject the null hypothesis at = 0.05Chap 9-44Hypothesis Tests for Proportions Involves categorical variables Two possible outcomes “Success” (possesses a certain characteristic) “Failure” (does not

38、 possesses that characteristic) Fraction or proportion of the population in the “success” category is denoted by Chap 9-45Proportions Sample proportion in the success category is denoted by p When both n and n(1-) are at least 5, p can be approximated by a normal distribution with mean and standard

39、deviation sizesamplesampleinsuccessesofnumbernXppn)(1p(continued)Chap 9-46 The sampling distribution of p is approximately normal, so the test statistic is a Z value:Hypothesis Tests for Proportionsn)(1pZn 5andn(1-) 5Hypothesis Tests for pn 5orn(1-) 5Not discussed in this chapterChap 9-47 An equival

40、ent form to the last slide, but in terms of the number of successes, X:Z Test for Proportionin Terms of Number of Successes)(1nnXZX 5andn-X 5Hypothesis Tests for XX 5orn-X 5Not discussed in this chapterChap 9-48Example: 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 = (500)(.08) = 40n(1-) = (500)(.92) = 460Chap 9-49Z Test for Proportion: Solution = 0.05 n = 500, p = 0.05Reject H0 at = 0.05H0: = 0.08 H1: 0.08Cr