《统计学基础(英文版·第7版)》课件les7e-ppt-ADA-0703.pptx

1、统计学基础(英文版第7版)课件les7e_ppt_ADA_0703Chapter Outline 7.1 Introduction to Hypothesis Testing 7.2 Hypothesis Testing for the Mean known 7.3 Hypothesis Testing for the Mean Unknown 7.4 Hypothesis Testing for Proportions 7.5 Hypothesis Testing for Variance and Standard DeviationSection 7.3Hypothesis Testing

2、 for the Mean UnknownSection 7.3 Objectives Find critical values in a t-distribution Use the t-test to test a mean when is not known Use technology to find P-values and use them with a t-test to test a mean when is not knownFinding Critical Values in a t-Distribution(1 of 2)1.Identify the level of s

3、ignificance.2.Identify the degrees of freedom d.f.1n.3.Find the critical value(s)using Table 5 in Appendix B in the row with 1ndegrees of freedom.If thehypothesis test isa.left-tailed,use“One Tail,”column with anegative sign,b.right-tailed,use“One Tail,”column with apositive sign,c.two-tailed,use“Tw

4、o Tails,”column with anegative and a positive sign.Finding Critical Values in a t-Distribution(2 of 2)Left-Tailed Test Right-Tailed TestTwo-Tailed TestExample:Finding Critical Values for t(1 of 3)Find the critical value 0tfor a left-tailed test given 0.05and 21n.Solution:The degrees of freedom are 2

5、1 120d.f.1n .Use Table 5.Look at 0.05in the“OneTail,”column.Because the test is left-tailed,the critical value is negative.So,01.725t .5Level of SignificanceExample:Finding Critical Values for t(2 of 3)Find the critical value 0tfor a right-tailed test given 0.01and 17n.Solution:The degrees of freedo

6、m are 17 116d.f.1 n.Use Table 5.Look at 0.01in the“OneTail,”column.Because the test is right-tailed,the critical value is positive.So,02.583t.1Level of SignificanceExample:Finding Critical Values for t(3 of 3)Find the critical value 0tand 0tfor a two-tailed testgiven 0.10and 26n.Solution:The degrees

7、 of freedom are 26 125d.f.1 n.Look at 0.10in the“TwoTail,”column.Because the test is two-tailed,one critical value is negative.and one is positive.01.708 tand 01.708t.10Level of SignificanceSolution:Finding Critical Values for tFind the critical values 0tand 0tfor a two-tailed testgiven 0.10and 26n.

8、Solution:You can check your answer using technology.t-Test for a Mean mu(sigma Unknown)t-Test for a Mean The t-test for a mean is a statistical test for a populationmean.The test statistic is the sample mean x.Thestandardized test statistic is xtsnwhen these conditions are met.1.The sample is random

9、.2.At least one of the following is true:The population is normally distributed or 30n.The degrees of freedom are d.f.1n.Using the t-Test for Mean mu(sigma Unknown)(1 of 3)In WordsIn Symbols1.Verify that is not known,thesample is random,and either the population is normally distributed or 30n.2.Stat

10、e the claim mathematically and verbally.Identify the null and alternative hypotheses.State 0Hand aH.3.Specify the level of significance.Identify.Using the t-Test for Mean mu(sigma Unknown)(2 of 3)In WordsIn Symbols4.Identify the degrees of freedom.d.f.1n5.Determine the critical value(s)Use Table 5 i

11、n Appendix A.6.Determine the rejection region(s).7.Find the standardized test statistic and sketch the sampling distribution.xtsnUsing the t-Test for Mean mu(sigma Unknown)(3 of 3)In WordsIn Symbols8.Make a decision to reject or fail to reject the null hypothesis.If t is in the rejection region,reje

12、ct 0H.Otherwise,fail to reject 0H.9.Interpret the decision in the context of the original claim.Example:Hypothesis Testing Using a Rejection RegionA used car dealer says that the mean price of used cars sold in the last 12 months is at least$21,000.Yoususpect this claim is incorrect and find that a

13、random sample of 14 used cars sold in the last 12 months has a mean price of$19,189and a standard deviation of$2950.Is there enough evidence to reject the dealersclaim at=0.05?Assume the population is normallydistributed.(Adapted from E)Solution:Hypothesis Testing Using a Rejection Region(1 of 3)Sol

14、ution:Because is unknown,the sample is random,and thepopulation is normally distributed,you can use the t-test.The claim is“the mean price is at least$21,000.”So,the null and alternative hypotheses are 0$21,000H(Claim)and$21,000aH.Solution:Hypothesis Testing Using a Rejection Region(2 of 3)The test

15、is a left-tailed test,the level of significance is 0.05,and the degrees of freedom are d.f.14 113 So,using Table 5,the critical value is 01.771t .Therejection region is 1.771t .The standardized teststatistic is 19,18921,000295014xtsnAssume 21,000.2.297.Solution:Hypothesis Testing Using a Rejection R

16、egion(3 of 3)The figure shows the location of the rejection region and the standardized test statistic t.Because t is in the rejection region,you reject the null hypothesis.There is enough evidence at the 5level of significanceto reject the claim that the mean price of used cars sold in the last 12

17、months is at least$21,000.5Level of SignificanceExample:Hypothesis Testing Using Rejection RegionsAn industrial company claims that the mean pH level of the water in a nearby river is 6.8.You randomly select 39 water samples and measure the pH of each.The sample mean and standard deviation are 6.7 a

18、nd 0.35,respectively.Is there enough evidence to reject the companys claim at 0.05?Solution:Hypothesis Testing Using Rejection Regions(1 of 4)Solution:Because is unknown,the sample is random,and 3930n,you can use the t-test.The claim is“themean pH level is 6.8.”So,the null and alternative hypotheses

19、 are 06.8H(Claim)and 6.8aH.Solution:Hypothesis Testing Using Rejection Regions(2 of 4)The test is a two-tailed test,the level of significance is 0.05,and the degrees of freedom are d.f.3938.So,using Table 5,the critical valuesare 02.024t and 02.024t.The rejection regionsare 2.024t and 2.024t.The sta

20、ndardized teststatistic is6.76.80.3539xtsnAssume 6.8.1.784.Solution:Hypothesis Testing Using Rejection Regions(3 of 4)The figure shows the location of the rejection regions and the standardized test statistic t.Because t is not in the rejection region,you fail to reject the null hypothesis.5Level of

21、 SignificanceSolution:Hypothesis Testing Using Rejection Regions(4 of 4)You can confirm this decision using technology,as shown.Note that the standardized statistic t differs from the one found using Table 5 due to rounding.There is not enough evidence at the 5level ofsignificance to reject the clai

22、m that the mean pH level is 6.8.Example:Using P-values with t-Tests(1 of 4)A department of motor vehicles office claims that the mean wait time is less than 14 minutes.A random sample of 10 people has a mean wait time of 13 minutes with a standard deviation of 3.5 minutes.At 0.10,test the offices cl

23、aim.Assume the populationis normally distributed.Example:Using P-values with t-Tests(2 of 4)Because is unknown,the sample is random,and thepopulation is normally distributed,you can use the t-test.The claim is“the mean wait time is less than 14 minutes.”So,the null and alternative hypotheses are 014

24、minutesHand 14minutesaH.(Claim)Example:Using P-values with t-Tests(3 of 4)The TI-84 Plus display at the far left shows how to set up the hypothesis test.The two displays on the right show the possible results,depending on whether you select Calculate or Draw.TI-84 PLUST-TestInpt:Data Stats014x 13Sx:

25、3.5n:10000:Calculate DrawTI-84 PlusT-Test14t.9035079029 p.1948994027x13Sx3.5n10Example:Using P-values with t-Tests(4 of 4)From the displays,you can see that 0.1949P.Because the P-value is greater than 0.10,you failto reject the null hypothesis.There is not enough evidence at the 10level ofsignificance to support the offices claim that the mean wait time is less than 14 minutes.