1、Chapter 8 Tests of Hypotheses Based on a Single Sample8.1 Hypotheses and Test Procedures8.2 Tests About a Population Mean8.3 Tests Concerning Population Proportion8.4 P-Values8.5 Some Comments on Selecting a Test Procedure Introduction A parameter can be estimated from sample data either by a single
2、 number(a point estimated)or an entire interval of plausible values(a confidence interval).Frequently,however,the objective of an investigation is not to estimate a parameter but to decide which of two contradictory claims about the parameter is correct.Methods for accomplishing this comprise the pa
3、rt of statistical inference called hypothesis testing.In this chapter,we first discuss some of the basic concepts and terminology in hypothesis testing and then develop decision procedures for the most frequently encountered testing problems based on a sample from a single population.8.1 Hypotheses
4、and Test Procedures A statistical hypothesis,or hypothesis,is a claim either about the value of a single population characteristic or about the values of several population characteristics.One example of a hypothesis is the claim=0.75,where is the true average inside diameter of a certain type of PV
5、C pipe.Another example is the statement p5.121212 In any hypothesis-testing problem,there are two contradictory hypotheses under consideration.One hypothesis might be the claim=0.75 and the other .75,or the two contradictory statements might be p.10 and p100.Explain why it might be preferable to use
6、 this Ha rather than:100Test Procedures A test procedure is a rule,based on sample data,for deciding whether to reject.A test of :p=0.10 versus :pza=2.06355.1/15nsxZDont reject H0Example:Individuals filing 1994 federal income tax returns prior to March 31,1995,had an average refund of$1056.Consider
7、the population of“last-minute”filers who mail their returns during the last five days of the income tax period.a.A research suggests that one of the reasons individual wait until the last five days to file their returns is that on average those individuals have a lower refund than early filers.Devel
8、op appropriate hypotheses such that rejection of H0 will support the researchers contention.b.For a sample of 400 individuals who filed a return between April 10 and April 15,the sample mean refund was$910 and the sample standard deviation was$1600.At a=0.05,what is your conclusion?1056:1056:0aHHnsx
9、Z/1056Reject H0 if z-za=-1.645825.1/1056nsxZSo reject H0Solution:Example:New tires manufactured by a company in Findlay,Ohio,are designed to provide a mean of at least 28000 miles.Tests with 30 randomly selected tires showed a sample mean of 27500 miles and a sample standard deviation of 1000 miles.
10、Using a 0.05 level of significance,test whether there is sufficient evidence to reject the claim of a mean of at least 28000 miles.28000:28000:0aHHnsxZ/28000Reject H0 if z2.064 or 30)hypothesis test about a population mean for a one-tailed test of the form000:aHHTest statistic:knownnxz/0 Test statis
11、tic:unknownnsxz/0 Rejection rule at a level of significance of aReject H0 if z30)hypothesis test about a population mean for a one-tailed test of the form000:aHHTest statistic:knownnxz/0 Test statistic:unknownnsxz/0 Rejection rule at a level of significance of aReject H0 if zza000:aHHTest statistic:
12、knownnxz/0Test statistic:unknownnsxz/0Rejection rule at a level of significance of aReject H0 if zza/2Summary:Two-tailed tests about a population meanWhen Large sampleErrors in Hypothesis Testing The basis for choosing a particular rejection region lies an understanding of the errors that one might
13、be faced with in drawing a conclusion.Consider the rejection region x15(n=200)in the circuit board problem.Even when:H0:p=0.10 is true,it might happen that an unusual sample results in x=13,so that H0 is erroneously rejected.Example:A test of H0:p=0.10 versus Ha:p0.10 in the circuit board problem mi
14、ght be based on examining a random sample of n=200 boards.The rejection region consists of x=0,1,2,.,and 15 On the other hand,even when Ha:p70.8 when normal with =72 and 0H)72(72XXX 8.1X7486.025141)67.(1)8.1728.70(1 0174.)67(3300.)8.1708.70(1)70(Example 8.3 Let us use the same experiment and test st
15、atistic X as previously described in the automobile bumper problem,but now consider the rejection region R9=9,10,20.X still has a binomial distribution with parameters n=20 and p,suppose the hypothesis is still 25.0:,25.0:0pHpHaCalculate the probability of type I error and probability of type II err
16、or.Solution:0()(0.25)9(20,0.25)1(8;20,0.25)0.041P type I errorP H is rejected when pP Xwhen XBB The type I error probability has been decreased by using the new rejection.However,a price has been paid for this decrease.0(0.3)(0.3)8(20,0.3)(8;20,0.3)0.887P H is not rejected when pP Xwhen XBB(0.5)(8;2
17、0,0.5)0.252BExample 8.4 The true of cutoff value c=70.8 in the paint-drying example resulted in a very small value of but rather large s.Consider the same experiment and test statistic with the new rejection region .Because is still normally distributed with mean value(0.01)X72x XX1.8XandDetermine p
18、robability of type I error and type II error02()72(75,1.8)7275(1.67)0.04750.051.8P H is rejected when it is trueP Xwhen XN 02(72)(72)72(72,1.8)727211(0)0.51.8P H is not rejected whenP Xwhen XN 7270(70)10.13351.8(67)0.0027 Solution:PropositionSuppose an experiment and a sample size are fixed,and a te
19、st statistic is chosen.Then decreasing the size of the rejection region to obtain a smaller value of results in a larger value of for any particular parameter value consistent with Ha.This proposition says that once the test statistic and n are fixed,there is no rejection region that will simultaneo
20、usly make both and s small.A region must be chosen to effect a compromise between and Because of the suggested guidelines for specifying H0 and Ha,a type I error is usually more serious than a type II error(this can always be achieved by proper choice of the hypotheses).The approach adhered to by mo
21、st statistical practitioners is then to specify the largest value of that can be tolerated and find a rejection region having that value of rather than anything smaller.This makes as small as possible subject to the bound on The resulting value of is often referred to as the significance level of th
22、e test.Traditional levels of significance are 0.10,0.05,and 0.01,though the level in any particular problem will depend on the seriousness of a type I errorthe more serious this error,the smaller should be the signoficance level.The corresponding test procedure is called a level test(e.g.a level 0.0
23、5 test or a level 0.01 test).A test with significance level is one for which the type I error probability is controlled at the specified level.Exercise P319 4 Let denote the true average radioactivity level(picocuries per liter).The value 5 PCi/L is considered the dividing line between safe and unsa
24、fe water.Would you recommend testing H0:=5 versus Ha:5 or H0:=5 versus Ha:150.In the context of this situation,describe type and type errors.Which type of error would you consider more serious?Explain.Exercise P320 8 A regular type of laminate is currently being used by a manufacture of circuit boar
25、ds.A special laminate has been developed to reduce warpage.The regular laminate will be used on one sample of specimens and the special laminate on another sample,and the amount of warpage will then be determined for each specimen.The manufacture will then switch to the special laminate only if it c
26、an be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate.State the relevant hypotheses,and describe the type I and type II errors in the context of this situation.Exercise P320 9 Two different companies have applied to provide cable televisio
27、n service in a certain region.Let p denote the probability of all potential subscribers who favor the first company over the second.Consider testing versus based on a random sample of 25 individuals.Let X denote the number in the sample who favor the first company and x represent the observed value
28、of X.a.Which of the following rejection regions is most appropriate and why?123:718,:8,:17Rx xor xRx xRx xb.In the context of this problem situation,describe what type type errors are.c.What is probability distribution of the test statistic X when H0 is true?Use it to compute the probability of a ty
29、pe error.d.Compute the probability of a type error for the select region when p=0.3.e.Using the selected region,what would you conclude if 6 of the 25 queried favored company 1?Exercise P320 10 A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a compressive str
30、ength of more than 1300 KN/m2,.The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met.Suppose compressive strength for specimens of this mixture is normally distributed with=60.Let denote the true average compressive strength.a.W
31、hat are the appropriate null and alternative hypotheses?b.Let X denote the sample average compressive strength for n=20 randomly selected specimens.Consider the test procedure with test statistic X and rejection region x1331.26.what is the probability distribution of the test statistic when H0 is tr
32、ue?What is the probability of a type I error for the test procedure?c.What is the probability distribution of the test statistic when=1350?Using the test procedure of part(b),what is the probability that the mixture will be judged unsatisfactory when in fact=1350(a type II error)?d.How would you cha
33、nge the test procedure of part(b)to obtain a test with significanc level 0.05?What impact would this change have on the error probability of part?e.Consider the standardized test statistic 42.131300/1300 XnXZ What are the values of Z corresponding to the rejection region of part(b)?Exercise P320 11
34、The calibration of a scale is to be checked by weighing a 10-kg test specimen 25 times.Suppose that the results of different weightings are independent of one another and that the weight on each trial is normally distributed with=0.200kg.Let denote the true average weight reading on the scale.a.What
35、 hypotheses should be tested?b.Suppose the scale is to be recalibrated if either x10.1032 or x9.8968.what is the probability that recalibration is carried out when it is actually unnecessary?c.What is the probability that recalibration is judged unecessary when in fact=10.1?when=9.8?d.Let .For what
36、value c is the rejection region e.of part(b)equivalent to the“two-tailed”region either zc or zc?nXZ/10 e.If the sample size were only 10 rather than 25,how should the procedure of part(d)be altered so that=0.05?f.Using the test of part(e),what would you conclude from the following sample data:9.981
37、10.006 9.857 10.107 9.888 9.728 10.439 10.214 10.190 9.793g.Reexpress the test procedure of part(b)in terms of the standardized test statistic nXZ/10 c.What is the significance level for the appropriate region of part(b)?How would you change the region to obtain a test with=0.001?d.What is the proba
38、bility that the new design is not implemented when its true average braking distance is actually 115 ft and the appropriate region from part(b)is used?e.Let .What is the significance level for the rejection regionz:z-2.33?For the region x:z-2.88nXZ/120 Exercise P321 12 A new design for the braking s
39、ystem on a certain type of car has been proposed.For the current system,the true average braking distance at 40 mph under specified conditions is known to be 120 ft.it is proposed that the new design be implemented only if sample data strongly indicates a reduction in true average braking distance f
40、or the new design.a.Define the parameter of interest and state the relevant hypotheses.b.Suppose braking distance for the new system is normally distributed with=10.let X denote the sample average braking distance for a random sample of 36 observations.Which of the following rejection regions is app
41、ropriate:80.124:1 xxR20.115:2 xxR87.11413.125:3 xorxeitherxRExercise 6:The label on a three-quart container of orange juice indicates that the orange juice contains an average of one gram of fat or less.Answer the following questions for a hypothesis test that could be used to test the claim on the
42、label.a.Develop the appropriate null and alternative hypotheses.b.What is the Type error in this situation?What are the consequences of making this error?c.What is the Type error in this situation?What are the consequences of making this error?Exercise 8:Suppose a new production method will be imple
43、mented if a hypothesis test supports the conclusion that the new method reduces the mean operating cost per hour.a.State the appropriate null and alternative hypotheses if the mean cost for the current production method is$220 per hour.b.What is the Type error in this situation?What are the conseque
44、nces of making this error?c.What is the Type error in this situation?What are the consequences of making this error?8.2 Tests About a Population MeanThe general discussion in Chapter 7 of confidence intervals for a population mean focused on three different cases.We now develop test procedures for t
45、hese same three cases.Case:A Normal Population with Known Although the assumption that the value of is known is rarely met in practice,this case provides a good starting point because of the ease with which general procedures and their properties can be developed.Null hypothesis:Test statistic value
46、:0H0 nxZ0 )testtailedtwo(zzorzzeither)testtailedlower(-zz)testtailedupper(zzTest Level for Region Rejection22 0a0a0a:H:H:H Hypothesise Alternativ Use of the following sequence of steps is recommended in any hypothesis-testing analysis.1.Identify the parameter of interest and describe it in the conte
47、xt of the problem situation.2.Determine the null value and state the null hypothesis.3.State the appropriate alternative hypothesis.4.Give the formula for the computed value of the test statistic5.State the rejection region for the selected significance level .6.Compute any necessary sample quantiti
48、es,substitute into the formula for the test statistic value,and compute that value.7.Decide whether should be rejected and state this conclusion in the problem context.0HExample 8.6 A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average systems-
49、activation temperature is 130.A sample of n=9 systems,when tested,yields a sample average activation temperature of 131.08F.If the distribution of activation times is normal with standard deviation 1.5F,does the data contradict the manufacturers claim at significance level =0.01?Solution:1.Parameter
50、 of interest:=true average activation temperature2.Null hypothesis:=130(null value=0=130)3.Alternative hypothesis:130 4.Test statistic value:nxnxz5.113000H:aH 5.Rejection region:The form of implies use of a two-tailed test with rejection region either From Section 4.3 or Appendix Table A.3,=2.58,so
