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1、Elementary Statistics,Seventh Edition,Chapter 7,Hypothesis Testing with One Sample,Copyright 2019, 2015, 2012, Pearson Education, Inc.,Chapter Outline,7.1 Introduction to Hypothesis Testing7.2 Hypothesis Testing for the Mean,7.3 Hypothesis Testing for the Mean,7.4 Hypothesis Testing for Proportions7
2、.5 Hypothesis Testing for Variance and Standard Deviation,Section 7.3,Hypothesis Testing for the Mean,Section 7.3 Objectives,Find critical values in a t-distributionUse 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 k
3、nown,Finding Critical Values in a t-Distribution (1 of 2),Identify the level of significance,.,Identify the degrees of freedom,.,Find the critical value(s) using Table 5 in Appendix B in the row with,degrees of freedom. If the,hypothesis test is,left-tailed, use “One Tail,” column with a,negative si
4、gn,right-tailed, use “One Tail,” column with a,positive sign,two-tailed, use “Two Tails,” column with a,negative and a positive sign.,Finding Critical Values in a t-Distribution (2 of 2),Left-Tailed Test,Right-Tailed Test,Two-Tailed Test,Example: Finding Critical Values for t (1 of 3),Find the criti
5、cal value,for a left-tailed test given,and,.,Solution:The degrees of freedom are,.,Use Table 5.Look at,in the “One,Tail,” column.,Because the test is left-tailed, the critical value is negative. So,.,Level of Significance,Example: Finding Critical Values for t (2 of 3),Find the critical value,for a
6、right-tailed test given,and,.,Solution:The degrees of freedom are,.,Use Table 5.Look at,in the “One,Tail,” column.,Because the test is right-tailed, the critical value is positive. So,.,Level of Significance,Example: Finding Critical Values for t (3 of 3),Find the critical value,and,for a two-tailed
7、 test,given,and,.,Solution:The degrees of freedom are,.,Look at,in the “Two,Tail,” column.,Because the test is two-tailed, one critical value is negative. and one is positive.,and,.,Level of Significance,Solution: Finding Critical Values for t,Find the critical values,and,for a two-tailed test,given
8、,and,.,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 population,mean. The test statistic is the sample mean,. The,standardized test statistic is,when these conditions are met.The sample is
9、 random.At least one of the following is true: The population is normally distributed or,.,The degrees of freedom are,.,Using the t-Test for Mean mu (sigma Unknown) (1 of 3),In Words,In Symbols,Verify that,is not known, the,sample is random, and either the population is normally distributed or,.,Sta
10、te the claim mathematically and verbally. Identify the null and alternative hypotheses.,State,and,.,Specify the level of significance.,Identify,.,Using the t-Test for Mean mu (sigma Unknown) (2 of 3),In Words,In Symbols,Identify the degrees of freedom.,Determine the critical value(s),Use Table 5 in
11、Appendix A.,Determine the rejection region(s).,Find the standardized test statistic and sketch the sampling distribution.,Using the t-Test for Mean mu (sigma Unknown) (3 of 3),In Words,In Symbols,Make a decision to reject or fail to reject the null hypothesis.,If t is in the rejection region, reject
12、,.,Otherwise, fail to reject,.,Interpret the decision in the context of the original claim.,Example: Hypothesis Testing Using a Rejection Region,A used car dealer says that the mean price of used cars sold in the last 12 months is at least,. You,suspect this claim is incorrect and find that a random
13、 sample of 14 used cars sold in the last 12 months has a mean price of,and a standard deviation of,. Is there enough evidence to reject the dealers,claim at,? Assume the population is normally,distributed. (Adapted from E),Solution: Hypothesis Testing Using a Rejection Region (1 of 3),Solution:Becau
14、se,is unknown, the sample is random, and the,population is normally distributed, you can use the t-test. The claim is “the mean price is at least,.”,So, the null and alternative hypotheses are,(Claim),and,.,Solution: Hypothesis Testing Using a Rejection Region (2 of 3),The test is a left-tailed test
15、, the level of significance is, and the degrees of freedom are,So, using Table 5, the critical value is,. The,rejection region is,. The standardized test,statistic is,Assume,.,.,Solution: Hypothesis Testing Using a Rejection Region (3 of 3),The figure shows the location of the rejection region and t
16、he standardized test statistic t. Because t is in the rejection region, you reject the null hypothesis.,There is enough evidence at the,level of significance,to reject the claim that the mean price of used cars sold in the last 12 months is at least,.,Level of Significance,Example: Hypothesis Testin
17、g Using Rejection Regions,An 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 and 0.35, respectively. Is there enough evidence to reject the companys
18、 claim at,?,Solution: Hypothesis Testing Using Rejection Regions (1 of 4),Solution:Because,is unknown, the sample is random, and, you can use the t-test. The claim is “the,mean pH level is 6.8.” So, the null and alternative hypotheses are,(Claim) and,.,Solution: Hypothesis Testing Using Rejection Re
19、gions (2 of 4),The test is a two-tailed test, the level of significance is, and the degrees of freedom are,. So, using Table 5, the critical values,are,and,. The rejection regions,are,and,. The standardized test,statistic is,Assume,.,.,Solution: Hypothesis Testing Using Rejection Regions (3 of 4),Th
20、e 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.,Level of Significance,Solution: Hypothesis Testing Using Rejection Regions (4 of 4),You can confirm this decision using techno
21、logy, 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,level of,significance to reject the claim that the mean pH level is 6.8.,Example: Using P-values with t-Tests (1 of 4),A department of motor vehicles o
22、ffice 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, test the offices claim. Assume the population,is normally distributed.,Example: Using P-values with t-Tests (2 of 4),Because,is
23、unknown, the sample is random, and the,population 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,and,. (Claim),Example: Using P-values with t-Tests (3 of 4),The TI-84 Plus display at the far left
24、 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 PLUS,T-Test,Inpt: Data Stats,Sx:3.5,n:10,Calculate Draw,TI-84 Plus,T-Test,Example: Using P-values with t-Tests (4 of 4),From the displays, you can see that,.,Because the P-value is greater than, you fail,to reject the null hypothesis.,There is not enough evidence at the,level of,significance to support the offices claim that the mean wait time is less than 14 minutes.,
