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statisticsbe @163.com Password: shnu2010. Statistics for Business and Economics. Chapter 9 Hypothesis Testing. What is a Hypothesis? （假设）. A hypothesis is a claim (assumption) about a population parameter : population mean population proportion. - PowerPoint PPT Presentation

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• Chap 10-*Chapter 9

Hypothesis TestingStatistics for Business and Economics

• Chap 10-*What is a Hypothesis? A hypothesis is a claim (assumption) about a population parameter:

population mean

population proportion

Example: The mean monthly cell phone bill of this city is = \$42Example: The proportion of adults in this city with cell phones is p = .68

• Chap 10-*

• Chap 10-*The Null Hypothesis, H0 States the assumption (numerical) to be testedExample: The average number of TV sets in U.S. Homes is equal to three ( )Is always about a population parameter, not about a sample statistic

• Chap 10-*The Null Hypothesis, H0Begin with the assumption that the null hypothesis is trueSimilar to the notion of innocent until proven guiltyRefers to the status quoAlways contains = , or signMay or may not be rejected(continued)

• Chap 10-*

• Chap 10-*The Alternative Hypothesis, H1Is the opposite of the null hypothesise.g., The average number of TV sets in U.S. homes is not equal to 3 ( H1: 3 )Challenges the status quoNever contains the = , or signMay or may not be supportedIs generally the hypothesis that the researcher is trying to support

• PopulationClaim: 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 )X

• Chap 10-*Sampling 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 meanX

• Chap 10-*Level of Significance, Defines the unlikely values of the sample statistic if the null hypothesis is trueDefines rejection region of the sampling distributionIs designated by , (level of significance)Typical values are .01, .05, or .10Is selected by the researcher at the beginningProvides the critical value(s) of the test

• Chap 10-*Level of Significance and the Rejection RegionH0: 3 H1: < 30H0: 3 H1: > 3aa Represents critical valueLower-tail testLevel of significance = a0Upper-tail testTwo-tail testRejection region is shaded /20a /2aH0: = 3 H1: 3

• Chap 10-*Errors in Making DecisionsType I Error ---- Reject a true null hypothesisConsidered a serious type of error

The probability of Type I Error is Called level of significance of the testSet by researcher in advance

• Chap 10-*Errors in Making DecisionsType II Error ---- Fail to reject a false null hypothesis

The probability of Type II Error is (continued)

• Chap 10-*

• Chap 10-*Outcomes and ProbabilitiesActual SituationDecisionDo NotRejectH0No error (1 - )aType II Error ( )RejectH0Type I Error( )aPossible Hypothesis Test Outcomes H0 False H0 TrueKey:Outcome(Probability)No Error ( 1 - )

• Chap 10-*Reject H0: 52Do not reject H0 : 52Type II Error ExampleType II error is the probability of failing to reject a false H05250Suppose we fail to reject H0: 52 when in fact the true mean is * = 50

• Chap 10-*Reject H0: 52Do not reject H0 : 52Type II Error ExampleSuppose 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)

• Chap 10-*Reject H0: 52Do not reject H0 : 52Type II Error ExampleSuppose we do not reject H0: 52 when in fact the true mean is * = 505250Here, = P( x ) if * = 50(continued)

• Chap 10-*

• Chap 10-*

• Chap 10-*Hypothesis Tests for the Mean Known UnknownHypothesis Tests for

• Chap 10-*

• Chap 10-*

• Chap 10-*

• Chap 10-*

• Chap 10-*Test of Hypothesisfor the Mean ( Known)Convert sample result ( ) to a z value

The decision rule is:Hypothesis Tests for Consider the test(Assume the population is normal)

• Chap 10-*Reject H0Do not reject H0Decision Ruleaz00H0: = 0 H1: > 0 Critical valueZAlternate rule:

• Chap 10-*p-Value Approach to Testingp-value: Probability of obtaining a test statistic more extreme ( or ) than the observed sample value given H0 is trueAlso called observed level of significanceSmallest value of for which H0 can be rejected

• Chap 10-*p-Value Approach to TestingConvert sample result (e.g., ) to test statistic (e.g., z statistic )Obtain the p-valueFor an upper tail test:

Decision rule: compare the p-value to If p-value < , reject H0If p-value , do not reject H0 (continued)

• Chap 10-*Example: Upper-Tail Z Test for Mean ( Known) A phone industry manager thinks that customer monthly cell phone bill 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 managers claim)Form hypothesis test:

• Chap 10-*Reject H0Do not reject H0Suppose that = .10 is chosen for this test

Find the rejection region: = .101.280Reject H0Example: Find Rejection Region(continued)

• Chap 10-*Obtain sample and compute the test statistic

Suppose a sample is taken with the following results: n = 64, x = 53.1 (=10 was assumed known) Using the sample results, Example: Sample Results(continued)

• Chap 10-*Reject H0Do not reject H0Example: Decision = .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)

• Chap 10-*Reject H0 = .10Do not reject H01.280Reject H0Z = .88Calculate the p-value and compare to (assuming that = 52.0)(continued)p-value = .1894Example: p-Value SolutionDo not reject H0 since p-value = .1894 > = .10

• Chap 10-*One-Tail TestsIn many cases, the alternative hypothesis focuses on one particular directionH0: 3 H1: < 3H0: 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 3

• Chap 10-*Reject H0Do not reject H0Upper-Tail Testsaz0H0: 3 H1: > 3There is only one critical value, since the rejection area is in only one tailCritical valueZ

• Chap 10-*Reject H0Do not reject H0There is only one critical value, since the rejection area is in only one tailLower-Tail Testsa-z0H0: 3 H1: < 3ZCritical value

• Chap 10-*Do not reject H0Reject H0Reject H0There are two critical values, defining the two regions of rejection Two-Tail Tests/20H0: = 3 H1: 3/2Lower critical valueUppercritical value3zx-z/2+z/2In some settings, the alternative hypothesis does not specify a unique direction

• Chap 10-*

• Chap 10-*

• Chap 10-*

• Chap 10-*

• Chap 10-*Hypothesis Testing ExampleTest the claim that the true mean # of TV sets in US homes is equal to 3.(Assume = 0.8)State the appropriate null and alternativehypothesesH0: = 3 , H1: 3 (This is a two tailed test)Specify the desired level of significanceSuppose that = .05 is chosen for this testChoose a sample sizeSuppose a sample of size n = 100 is selected

• Chap 10-*Hypothesis Testing ExampleDetermine the appropriate technique is known so this is a z testSet up the critical valuesFor = .05 the critical z values are 1.96Collect the data and compute the test statisticSuppose the sample results are n = 100, x = 2.84 ( = 0.8 is assumed known)So the test statistic is:(continued)

• Chap 10-*Reject H0Do not reject H0Is the test statistic in the rejection region? = .05/2-z = -1.960Reject H0 if z < -1.96 or z > 1.96; otherwise do not reject H0Hypothesis Testing Example(continued) = .05/2Reject H0+z = +1.96Here, z = -2.0 < -1.96, so the test statistic is in the rejection region

• Chap 10-* 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 in US homes is not equal to 3 Hypothesis Testing Example(continued)Reject H0Do not reject H0 = .05/2-z = -1.960 = .05/2Reject H0+z = +1.96

• Chap 10-*.0228/2 = .025Example: p-ValueExample: 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?-1.960-2.0Z1.962.0x = 2.84 is translated to a z score of z = -2.0 p-value = .0228 + .0228 = .0456.0228/2 = .025

• Chap 10-*Compare the p-value with If p-value < , reject H0If p-value , do not reject H0 Here: p-value = .0456 = .05Since .0456 < .05, we reject the null hypothesis(continued)Example: p-Value.0228/2 = .025-1.960-2.0Z1.962.0.0228/2 = .025

• Chap 10-*Test of Hypothesisfor the Mean ( Known)Convert sample result ( ) to a z value

The decision rule is:Hypothesis Tests for Consider the test(Assume the population is normal)

• Chap 10-*Test of Hypothesisfor the Mean ( UnKnown)Convert sample result ( ) to a z value

The decision rule is:Hypothesis Tests for Consider the test(Assume the population is normal)

• Chap 10-*t Test of Hypothesis for the Mean ( known)Convert sample result ( ) to a t test statistic