概率与统计（英文）chapter 8 Tests of Hypotheses Based on a Single Sample

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

3、 the part 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 Hy

4、potheses 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 cer

5、tain type of PVC 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 mi

6、ght be preferable to use 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 a

7、n average refund of $ 1056. Consider 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 hav

8、e a lower refund than early filers. Develop 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

9、=0.05, what is your conclusion?1056:1056:0aHHnsxZ/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

10、 miles and a sample standard deviation of 1000 miles. 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 tes

11、t of the form000:aHHTest statistic : knownnxz/0 Test statistic : 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

12、a level of significance of aReject H0 if zza000:aHHTest statistic : 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 particu

13、lar rejection region lies an understanding of the errors that one might 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 rej

14、ected . Example: A test of H0: p=0.10 versus Ha : p0.10 in the circuit board problem might 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

15、. 1728 .70(1 0174.)67(3300.)8 . 1708 .70(1)70( Example 8.3 Let us use the same experiment and test statistic 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 hypoth

16、esis is still 25. 0:,25. 0:0pHpHaCalculate the probability of type I error and probability of type II error.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 b

17、een 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;20,0.5)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

18、the new rejection region . Because is still normally distributed with mean value (0.01)X72x XX1.8XandDetermine probability 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 X

19、when XN 7270(70)10.13351.8(67)0.0027 Solution:PropositionSuppose an experiment and a sample size are fixed , and a test 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 H

20、a. This proposition says that once the test statistic and n are fixed, there is no rejection region that will simultaneously 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

21、serious than a type II error (this can always be achieved by proper choice of the hypotheses). The approach adhered to by most 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 m

22、akes as small as possible subject to the bound on The resulting value of is often referred to as the significance level of the 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

23、 this error, the smaller should be the signoficance level. The corresponding test procedure is called a level test ( e.g. a level 0.05 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 den

24、ote the true average radioactivity level (picocuries per liter). The value 5 PCi/L is considered the dividing line between safe and unsafe 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 err

25、or would you consider more serious? Explain. Exercise P320 8 A regular type of laminate is currently being used by a manufacture of circuit boards. 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 anothe

26、r 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 can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate. State the relevant hypotheses, and d

27、escribe the type I and type II errors in the context of this situation.Exercise P320 9 Two different companies have applied to provide cable television service in a certain region. Let p denote the probability of all potential subscribers who favor the first company over the second. Consider testing

28、 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 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 si

29、tuation, 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 type 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 conclu

30、de 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 strength of more than 1300 KN/m2,. The mixture will not be used unless experimental evidence indicates conclusively that the strength sp

31、ecification has been met. Suppose compressive strength for specimens of this mixture is normally distributed with =60. Let denote the true average compressive strength.a. What are the appropriate null and alternative hypotheses?b. Let X denote the sample average compressive strength for n=20 randoml

32、y 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 true? What is the probability of a type I error for the test procedure?c. What is the probability distribution of the test stat

33、istic 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 change the test procedure of part(b) to obtain a test with significanc level 0.05? What impact would this change have on

34、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 The calibration of a scale is to be checked by weighing a 10-kg test specimen 25 times. Suppose that the results of

35、 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 hypotheses should be tested?b. Suppose the scale is to be recalibrated if either x10.1032 or x9.8968. what is

36、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 value c is the rejection region e. of part(b) equivalent to the “two-tailed” region either zc or zc?nXZ/

37、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 10.006 9.857 10.107 9.888 9.728 10.439 10.214 10.190 9.793g. Reexpress the test procedure of

38、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 probability that the new design is not implemented when its true average braking distance is

39、 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 system on a certain type of car has been proposed. For the current system, the true

40、 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 for the new design.a. Define the parameter of interest and state the relevant hy

41、potheses.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 appropriate :80.124:1 xxR20.115:2 xxR87.11413.125:3 xorxeitherxRExercise 6: T

42、he 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 label.a. Develop the appropriate null and alternative hypotheses.b. Wha

43、t 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 implemented if a hypothesis test supports the conclusion that the ne

44、w 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 consequences of making this error? c. What is the Type error in th

45、is 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 these same three cases. Case : A Normal Population wit

46、h 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 :0H0 nxZ0 )testtailedtwo(zzorzzeither)testta

47、iledlower(-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 context of the problem situation.2.Determine the

48、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 quantities, substitute into the formula for the test

49、 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-activation temperature is 130. A sample o

50、f 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 of interest: =true average activ

51、ation 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 we reject if ei

52、ther z2.58 or z -2.58.6. Substituting n=9 and =131.08, 16. 25 . 008. 195 . 113008.131z005.005.zzorzz 005.z0Hx7. The computed value z=2.16 does not fall in the rejection region (-2.582.1620000 based on a sample of size n=16 from a normal population distribution with =1500. A test with =0.01 requires

53、Example 8.7The probability of making a type II error when =21,000 is 0H :aH 33. 2zz01. 3669 . 0) 0.34- ( )16150021000-20000 2.33 ( (21000)29.32) 5.42- (21000-200001.38)1500(2.33n22Case II: Large-Sample TestsWhen is known, we choose the random variable nXZ/ When is unknown, we choose the random varia

54、ble nsXZ/ Example 8.8A dynamic cone penetrometer(DCP) is used for measuring material resistance to penetration (mm/blow) as a cone is driven into pavement or subgrade. Suppose that for a particular application, it is required that the true average DCP value for a certain type of pavement be less tha

55、n 30. The pavement will not be used unless there is conclusive evidence that the specification has been met. Lets state and test the appropriate hypotheses using the following data14.1 14.5 15.5 16.0 16.0 16.7 16.9 17.1 17.5 17.8 17.8 18.1 18.2 18.3 18.3 19.0 19.2 19.4 20.0 20.0 20.8 20.8 21.0 21.5

56、23.5 27.5 27.5 28.0 28.3 30.0 30.0 31.6 31.7 31.7 32.5 33.5 33.9 35.0 35.0 35.0 36.7 40.0 40.0 41.3 41.7 47.5 50.0 51.0 51.8 54.4 55.0 57.01.= true average DCP value2. :=303. -1.645, cannot be rejected. We do not have compelling evidence for concluding that 25 0HaH2.132tt tif HReject 5.25 t4.05,41 -

57、n , 0nsx1.0445. 254. 2547. 52554.27t and5.47,s27.54,x whichfrom 3911.97,x and 137.7x. 62ii 7. Since 1.04 does not fail in the rejection region (1.042.132),H0 cannot be rejected at significance level .05.It is still plausible that is (at most) 25.Exercise P320 11 The calibration of a scale is to be c

58、hecked 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 hypotheses should be tested?b. S

59、uppose the scale is to be recalibrated if either 10.10329.8968xor xWhat is the probability that recalibration is carried out when it is actually unnecessary?c. What is the probability that recalibration is judged unnecessary when in fact =10.1? When =9.8?d. Let (10)/(/)zxnFor what value c is the rej

60、ection region of Part(b) equivalent to the “two-tailed” region either ?zc or zc e. If the sample size were only 10 rather 25, how should the procedure of part (d) be altered so that 0.05Exercise P332 19 The melting point of each of 16 samples of a certain brand of hydrogenated vegetable oil was dete