Exam_02_Review - W CHAPTER q_ Hygomezsrs TESTS (ONE...

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Unformatted text preview: W CHAPTER q_ Hygomezsrs TESTS (ONE gofiuMTzoN) W- E ALI-I‘Dmofi'uc gradud" I‘exeargé $12.91.” Wanbé f0 fave Mm” Me new fuel W'Ecfi'cm S/erm if”: More 741cm .94 mpg. Ho: J1 a .24 mpg ERRORS TYPE I AND mosfl . C/rz urn/S 3/ (13,6 011‘ least? 2i; mfg. Ho: rd 3 24 {Cb/m) ypoTHESIS 75575 ABOJF POPULATION MEAN flu) — Z DISTRIBWUN H REJECTION Ru L63 cgirrCAL APP ROACH STEPS OF HYPOTHESIS TESTING Step 1. Develop the null and alternative hypotheses. ‘ Step 2. Specify the level of Significance Step 3. Collect the sample data and compute the value of the test statistic. i p-Value Approach Step 4. Use the value of the test statistic to compute the pivalue. Step 5. Reject H0 if the p-value E a. Critical Value Approach Step 4. Use the level of significance to determine the critical value and the re- jection rule Step 5. Use the value of the test statistic and the rejection rule to determine whether to reject H” l l HYPOTHESIS‘ T557: {330m- Popuznwow HEAN{,u)- t DISTRIBUTION H VP 0TH 6555 TEST STHTI ST! C | REJECT Ho IF ‘ p-value 505 i HypoTHESIs TESTS ABOUT A PoPuLnTmN PROPORTION ’ é ! Lower Tail Test Upper Tail Test TWO-Tailed Test H ; 2 H ; s H ; = ? Hypotheses 0 p Po 0 P P0 0 P Po Haip<Po Haip>po Harp qtpa i - , - 1 Test Statistic z : p p” - z : —M— z : e-" p" M3140 lactic! M g V n f1 )1 g Rejection Rule: Reject HEI it' Reject Ho if Reject HfJ if p-value S a p-value S a p-value S a p'Value Approach Reject HO if Z S _Za.'2 or if z 2 gm Reject H0 if z 2 an Rejection Rule: Critical Value Approach E] CHAPTER 10. HYPOWESIS TESTs (mo 0 uLnWoNs) H #1951 7- MRGErsfl'HPLE V5. SMALL—Sfifif’w *2 Va t Dzsm,3m70N 1:1:ng ‘ { Rejecf' H0 If fiejecfion Hale: Reject" Ho 1’ P. yaw Approach P— Value 5 0C P Value é oC Rejecfibn Rule: Rejecv‘ H0 If K78de Ho 1}? 0mm! Value 2 s — ax 2 £- - 2’ z -_’1PPI°EC’1_,,__5_-, @.Ez%:/z,,,._--- Corn-fi'dthc‘e Refec’ Ha .. H76 ’00 (fwd)?- Confi'denve Inherua/ 22955 NOT confoz‘n Do [3qu Ho if? P—Value 6 05 Ra'tcz" Ho £710 1'13" .135 “ti/2 DOES NOT CONTAIN Do 10.3 lnferences About the Difference Between Two Population Means: Matched Samples Suppose employees at a manufacturing company can use two different methods to perform a production task. To maximize production output, the company wants to identify the method with the smaller population mean completion time. Let it, denote the population mean completion time for production method 1 and lit: denote the population mean com? pletion time for production method 2. With no preliminary indication of the preferred prtt duction method. we begin by tentatively assuming that the two production methods have the same population mean completion time. Thus. the null hypothesis is HG: fll i it: 3 or If this hypothesis is rejected, we can conclude that the population mean completion timES differ. in this case, the method prm'iding the smaller mean completion time would be fCC' ommended. The null and alternative hypotheses are written as follows. Hugo, — ‘11:, i 0 Ham”! 7 in: #5 O In choosing the sampling procedure that will he used to collect production time data and test the hypotheses, we consider two alternative designs One is based on independent 53"“ ples and the other is based on matched samples. 1. Independent sample design: A simple random sample of workers is selected and each werker in the sample uses method 1. A second independent simple random sample of workers is selected and each worker in this sample uses method 2‘ The test of the difference between population means is based on the procedures in Section 10.2, 2. Matched sample design: One simple random sample of workers is selected. Each worker first uses one method and then uses the other method. The order of the two methods is assigned randomly to the workers, with sotne workers performing method I first and others performing method 2 first. Each worker provides a pair of data values, one value for method 1 and another value for method 2, In the matched sample design the two production methods are tested under similar con ditions (i.e., with the same workers); hence this design often leads to a smaller sampling error than the independent sample design. The primary reason is that in a matched sample design, variation between workers is eliminated because the same workers are used for both production methods. Let us demonstrate the analysis of a matched sample design by assuming it is the method used to test the difference between pOpulation means for the two production methods. A random sample of six workers is used. The data on completion times for the six workers are given in Table 10.2. Note that each worker provides a pair of data values, one for each production method. Also note that the last column contains the difference in completion times d; for each worker in the sample. The key to the analysis of the matched sample design is to realize that we consider only the column of differences. Therefore. we have six data values (.6. -.2, .5. .3, .0. and .6) that will be used to analyze the difference between population means of the two production methods. Let lad = the mean of the dtflerence in values for the population of workers. With this notation, the null and alternative hypotheses are rewritten as follmvs. ; = _ .H new 2“ Him 0 Us)“ 0 new“,ng If H0 is rejected, we can conclude that the population mean completion times differ. Other than the use of the d notation. the formulas for the sampie mean and sample standard deviation are the same ones used previously in the text. The d notation is a reminder that the matched sample provides difference data. The E] sample mean and sample standard deviation for the six difference values in Table 10.2 follow. _ Ed. 1.8 d=—‘:—:. t1 6 30 — i—fld‘iaz— if 335 5d \ rt—l _ 5 —' TABLE 10.2 TASK COMPLETION TIMES FOR A MATCHED SAMPLE DESIGN Completion Time Completion Time Difference in for Method 1 for Method 2 Completion Worker (minutes) (minutes) Times (di) 1 6.0 5.4 .6 2 5.0 5.2 —.2 3 7.0 6.5 .5 4 6.2 5.9 .3 5 6.0 6.0 .0 6 6.4 5.8 .6 it is not necessan- to make the assumption that the population has a normal distribution ifthg sompir’ size is large. Sample size guidelinesfor using the t distribution were presented in Chapters 8 and 9. OnCe the dfierence dam are computed, the t distribution procedure for matched sampies is the same as the one-population estimation and hypothesis testing procedures described in Chapters 8 and 9. With the small sample of n = 6 workers. we need to make the assumption that the popu, lation of differences has a normal distribution. This assumption is necessary so that we may use the tdistribution for hypothesis testing and interval estimation procedures. Based on this assumption, the following test statistic has a t distribution with n 7 1 degrees of freedom. TEST STATISTIC FOR HYPOTHESIS TESTS INVOLVING MATCHED SAMPLES [Mo = Sgt/f” {10.9) Let us use equation ([09) to test the hypotheses H0: ,ud I 0 and H3: ,ud as 0, using a = _05_ Substituting the sample results d : .30, sa. : .335, and it : 6 into equation (10.9), we com- pute the value of the test statistic. d_ _ 4a t:—(&i)0= 30 O=2.20 sd/vfi .335 N8 Now let us compute the prvalue for this two-tailed test. Because t = 2.20 > 0, the test statistic is in the upper tail of the t distribution. With t = 2.20, the area in the upper tail to the right of the test statistic can be found by using the t distribution table with degrees of freedom = n i l = 6 i l = 5. Information from the 5 degrees offreedorn row ofthe tdis- tribution table is as follows: Area in Upper TaiI‘—; .20 .10 .05 .025 .01 .005 0.920 L476 2015 2.571 3.365 4.032 \ t : 2.20 t Value (5 df) l Thus, we see that the area in the upper tail is between ‘05 and 025‘ Because this testis «1 twortailed test, we double these values to conclude that the p-value is between .i0 and .05- This plvalue is greater than a : .05. Thus? the null hypothesis H0: lad = O is not rejected, Using Excel or Minitab and the data in Table 10.2, we find the exam In addition we can obtain an interval estimate ofthe difference between the two popli' lation means by using the single population methodology of Chapter 8. At 95% confidenCC- the calculation follows. (151/. Cr _—.- (l i [m ‘75-]. CI _— ,3 : 2574-7373:?) ‘iE—j. (:1 z .3 : .35 = f- o. 05, o. 65) Thus, the margin of error is .35 and the 95% confidence interval for the difference betWeen the pOpulation means of the two production methods is ~05 minutes to .65 minutes. Sith 95+ CI : (—0-05, 0-65) Con+a"“5 Waist? DD NOT Rfljf’c‘f' H0 2. The manager of an automobile dealership is considering a new bonus plan desi : increase sales volume. Currently, the mean sales volume is 14 automobiles per mun ' manager wants to conduct a research study to see whether the new bonus plan inc ' sales volume. To collect data on the plan, a sample of sales personnel will be allow sell under the new bonus plan for a one-month period. a. Develop the null and altemativeh'ypotheses most appropriate forthis research situti: c: b. Comment on the conclusion when Ill}J cannot be rejected. c. Comment on the conclusion when Ho can be rejected. 6. The CHHPTER 3 w 50 cameo PEDBLEMS Mf— ‘1 a, .. label on a 3~quart container of orange juice claims that the orange juice contains an average of 1 gram of fat or less. Answar the following questions for a hypothesisfcst that could be used to test the claim on the label. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error? e. What is theType 11 error in this situation? What are the consequences of making this error? ‘ 10. Consider the following hypothesis test: Hazy > 25 . l A sample of 40 provided a sample mean of 26-4. The population standard deviation is 6. a. b. c. d. Compute the value of the test statistic. What is the p—value‘? ' At a = .01, what is your conclusion? What is the rejection rule using the critical value? What is your conclusion? 15. Individuals filing federal income tax returns prior to March 3! rCCeived an average refund the a. of $ l056. Consider the population of “last-minute“ filers who mail their tax return during last five days of the income tax period (typically April 10 to April [5). r A researcher suggests that a reason individuals wait until the last five days is that 0“ 3 average these individuals receive lower refunds than do early filers. Develop appro. priate hypotheses such that rejection of H,, will Support the researchers contention i For a sample 01’4th individuals who filed a tax return between April [0 and 15, the sample mean refund was $9”). Based on prior experience a population standard dc. J viatiorl oftr = $1600 may be asstuned. What is the p-value‘.’ I" At a = .05. what is your conclusion? ‘ Repeat the preceding hypothesis test using the critical value approach. I l l i ! E i l l 1 i 1 : Hu:fl525 i i i 24. Consider the following hypothesis test: " H041 =13 , Ham # 18 A sample of 48 provided a sample mean :2 a l? and a sample standard deviation .9 = 4.5. a. b. C. _ d. 26. Consider the following hypothesis test: A sample of 65 is used. identify the p—value and state your conclusion for each of 5*» V fol lowing sample results. Use a = .05. 8. b. i = 96.5 ands = 11.0 C. Compute the value of the test statistic. USe the r distribution table (Table 2 in Appendix B) to compute a range for the p-value. ' At a = .05, what is your conclusion? What is the rejection rule using the critical value? What is your conclusion? How = 100 Hazy a lot) i =103ands= 11.5 i = 102 ands = 10.5 34. 36. Joan's Nursery specializes in custom-designed landscaping for residential areas. The call» matedlabor cost associated with a particular landscaping proposal is based on the number of plantings of trees, shrubs, and so on to be used for the project. For cost-estimating purposes. managers use two hours of labor time for the planting of a medium-sized tree. ? Actual times from a sample of to plantings during the past month follow (times in hours). 9 i 1.7 1.5 ' 2.6 2.2 2.4 2.3 2.6 3.0 1.4 2.3;I With a .05 level of significance, test to see whether the mean tree-planting time differs it from two hours. i a. State the null and alternative hypotheses. 5] b. Compute the sample mean. i c. Compute the sample standard deviation. d. What is the p—value? e. What is your conclusion? a 3 Consider the following hypothesis test: 7 ' 30:13 a .75 Ha:p < -75 A sample of 300 items was selected. Compute the p-value and state your conclusion tag . each of the following sample results. Use a = .05. ' a. p=.68 c. gar—.70 1). 5:32 d. p=.77 ' 38. A study by Consumer Reports showed that 64% of supermarket shoppers believe super- . u on hie Drowsy market brands to be as good as national name brands. To investigate whether this result ap- ] plies to its own product, the manufacturer of a national name—brand ketchup asked a sample of shoppers whether they believed that supermarket ketchup was as good as the national ' brand ketchup. 7 7 a. Formulate the hypotheses that could be used to determine whether the percentage of ‘ supermarket shoppers who believe that the supermarket ketchup was as good as the national brand ketchup differed from 64%. b. If a sample of 100 shoppers showed 52 stating that the supermarket brand was as good - as the national brand, what is the p—value? c. At a = .05, what is your conclusion? :1. Should the national brand ketchup manufacturer be pleased with this conclusion? Explain. E l ; 44. In a cover story, BusinessWeek published information about sleep habits of Ameri : (BusinessWaek, January 26, 2004). The article noted that sleep deprivation causes a nu ber of problems. including highway. deaths. Fifty-one percent of adult drivers admit to I ving while drowsy. A reSearcher hypothesized that this issue was an even bigger proble for night shift workers. a. Formulate the hypotheses that can be used to heIp determine whether more than 51 of the population of night shift Workers admit to driving while drowsy. ' b. A sample of 500 night shift workers identified those 'who admitted to driving w drowsy. What is the sample proportion? What is the p—value? c. At a = .01, what is your conclusion? Pas: CHRPTER ‘1- BNSMJERS T0 SUMESTEZD Pflpgfigfig [W] W . 2. a. Hows l4 Ha:y>14 ‘ 36_a_z=fi£p0—:Mzuzgo b. No evidence that the new plan increases [pea _ p0) _75(1 2 '75) sales _ r n 300 c; The research hypothesis [1 > 14 15 SUPPOITCd; the new p-value = .0026 plan increases sates ' p-value S .05; reject HfJ .72 — .75 b. “ m_ = r 1.20 6. a. H020 s 1 W. Z _ u . Ha: 'u > 1 5 .75(1 .75) 300 b. Claimingy > 1 when it is not true I j “51 c. Claiming}; E 1 when it is not true p-va ue _ ' . .1 _ p—value > .05; do not reject H.J 10.a.z=x"“°=M=1.48 j c.z=—M=—2.oo UN; 5/% 1.750 — .75) b. Using normal table with z = 1.48zp-va1ue = 300 1.0000 — .9306 = .0694 p-value = .0228 c. pnvalue > .01, do not reject H0 5' puvalue E .05; reject H0 d. Reject H0 ifz 3 2.33 .77 — .75 d.z= ‘280 175(1 — .75) _ 300 p—value = .7881 p—value > .05; do not reject H0 2.. i—u—————=—1.83 ' 38.3.H:p=.64 ONE WOO/11400 _ H; p 2 .54 p-value = .0336 b - = 52,100 I 2 c. pfvglue :I .05, reject H0; the mean refund of j ' p c _ '5 52 _ 64 “fast-minute" filers is less than $1056 z = —pee£°— - a = —2.50 1pm — p.) _ (.640 — .64) 1.54 n 100 ln-value : 2(.0062) = .0124 1.48 < 2.33. do not reject 1-1O i b. Degre I n — l = 47 c. p-value E .05; reject HO 1 i 1 i 1 E E 15. a. Hazy 21056 M Ha: M < 1056 b _i~,u0_910—1056 ‘ _17—18 ' between .05 and .10 Proportion differs from the reported .64 is between .10 and .20 (1. Yes, because 13 = .52 indicates that fewer believe the supermarket brand is as good as the name brand c. reject H0 d. _ 1‘ 44- 3- H03? 5 '51 eject H0 if: 5 —2.0120 2 2.012 I Ha: p > 51 t = —1.54; do not reject H0 'b. 1'7 = -58:P‘Valuc = ‘0026 c. Reject H0 2.6. a. Bet . 9'7;rejectH0 1 13 Between .01 and .0 —value = .0125; reject Ho 5 c. Between .10 an = .1285; do not "' -1g - -—1 Ll‘ reject HD I (a) z 5”?! gig/fig 34. a. Hnifl=2 l .. H :fl :15 2 4 _, .4. '- ' 13.2.2 (19) P” V411“ " 2x P “‘52 :ZxO-0618 (1. Between .20 and .40 Exact p-value = .2535 5 O ‘ 0 1 e. Do not reject Hu ...———-""' J1 CC) P—Vahu = 0.0335 5 ad 5 0.05“ awe» - 1 (a) 22-2.; eP»Vabu=ZE-P’Zéz"fl Do nor 120‘” H” % =a£1~onszfl Cd) 2 I :Z‘ _..__j,9g - = . scan-as ""L' ’7- i RHCC’LH" “(Mr E Since 2:454 > Ev: J96 ‘ | (b) Z=_2.§7.=> [34mm .-_- 2x0.00$! :o.otaz :po no+ Rged’ Ho . :0-0102. éct=0'”" gawk!" 501(2) ::3 | C9) 2:169! => P-valuz = aka— 0.?382) :- 0.1235 i " 1 at: 0.05‘ $0 "01' Wadi-Lowe '2 04135, > i ' was «1.59 2. co Monastic co . em -;,:. CHaPTER to - Suaaesrw PRO BLEMS‘ Consider the following hypothesis test. Hair“: ‘1‘250 HaZflt —#2>0 The following results are for two independent samples taken from the two populating Sample Satanic 1 “1 =- 40 112 3 it = 25.2 :2 = 223 ‘ or! = 5.2 oz 2 6.0 a. What is the value of the test statistic? What is the p—value? With a -—- ‘05. what is your hypothesis testing conclusion? 5"?" 6. The nation’s 40,000 mortgage brokerages are some of the most profitable small businesses in the United States. These lowuprofile companies find loans for customers in exchange for conunissions. Mortgage Bankers Association of America provides data on the average size | of loans handled by mortgage brokerages (The Wall Street Journal, February 24, 2003). The CD file named Mortgage contains data from a sample of 250 loans made in 2001 and a sample of 270 loans made in 2002 that are consistent with these data. Based on hiStori- i cal loan data the population standard deviations for the loan amounts can be assumed i known at $55,000 in 2002 and $50,000 in 2001. Do the sample data indicate an increase ‘ in the mean loan amount between 2001 and 2002? Use a = .05. 12. The US. Department of Transportation provides the number of miles that residents of the litan areas travel per day in a car. Suppose that for a simple random 75 largest metropo sample of 50 Buffalo residents the mean is 22.5 miles a day and the standard deviation is 8.4 miles a day, and for an independent simple random sample of 40 Boston residen ' mean is 18.6 miles a day and the standard deviation is 7.4 miles a day. a. What is the point estimate of the difference between the mean number of mil Buffalo residents travel per day and the mean number of miles that Boston res travel per day? 1). What is the 95% confidence interval for the difference between the two .4}: .‘ tion means? 18. Educational testing companies provide tutoring, classroom learning, and practice tests in an: ' fort to help students perform better on tests such as the Scholastic Aptitude Test (SAT). The: preparation companies claim that their courses will improve SAT score performances by erage of 120 points (The Wall Street Journal. January 23, 2003). Aresearcher is uncertain of _ claim and believes that 120 points may be an overstatement in an effort to encourageituden ' take the test preparation course. in an evaluation study of one test preparation service, th A searcher collects SAT score data for 35 students who took the test preparation course and 48 s dents who did not take the course. The CD file named SAT contains the scares for this size ' a. Formulate the hypotheses that can be used to test the researcher's belief that the ' provement in SAT scores may be less than the stated average of 120 points. b. Using a: = .05, what is your conclusion? c. What is the point estimate of the improvement in the average SAT scores pro ' by the test preparation course? Provide a 95% confidence interval estimate 0 ’ improvement. D d. What advice would you have_for the researcher after seeing the confidence in - i i | | i i “ 22. c» Eamings2005 - Earnings 26. ' Ptovide a 95% confidence interval estimate of the difference between the population I, Per'shm earnings data ComPaIiflg 1316 cunent quarter’s earnings with the previous - ter are in the CD file entitled Earnings 2005 (The Wall Street Journal, January 27, for the current quarter versus the previous quarter. Have earnings increased? Streetlnsidemom reported 2002 earnings per share data for a sample of major companies (Februaty 12, 2003). Prior to 2002. financial analysts predicted the 2002 earnings per share for these same companies (Barron ’3, September 10, 2001). Use the following data to com- ment on differences between actual and estimated earnings per share. Company Actual Predicted AT&T 1.29 ‘ _ 0.33 Americanfixpress‘ 7 2.01 7 ' ' 2.31 Citig'mpp - ' ‘ 2.591 3143 r CocaiCola 1.60 - 1.78 DuPotit_ _ ' . 1.84 - 2.18 ExxonMpbil 2.72 2.197 = Generat- Electric 7 1.51 trit- Johnson & Johnson 2.28 a - 2.13 ' McDonald’s 0.77 1.55 ,wat-Man 7 ‘ _ 1.31 . 1.74 a. Use a = .05 and test for any difference between the population mean actual and popu— ., . lation r'neanestimated earnings per share. What isthepwalue? What is your conclusion; b. What is the point estimate of the difference between the two means? Did the analysts tend to underestimate or overestimate the earnings? _ c. At 95% confidence, what is the margin of error for the estimate in part (b)? What would you recornmend based on this information? - E J l ! —-....._.___——..Mfi__———~M"_______,_H_____n-.mm. ‘ CHAPTER 10 ,- ANSWERS To SUGGESTED PROBLEHS' (ii ‘- fz) - Du (25.2 - 7.2.8) - O 2. a. r. g: + 0—5 16.2): + £6): \ 40 50 n3 n,_ b. p—value : 1.0000 — .9788 = .0212 c. p-value S .05; reject H(] = 0.015- 18. a. Haunt-#1 2 £20 Hazyl — #2 < 120 —2.10 ______—-—--"‘_"-"——____— ! flap—value=£+95‘O-Oqu c. 32113118 d. Larger sample size 22. $0.10 +0 $0.32 26. a. £= —.60 Using t table, p»va‘lue is greater than .40 Exactp-value = .5633 Do not reject HG h. —.103 c. .39; larger sample size b. ...
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This note was uploaded on 03/03/2012 for the course MGMT 305 taught by Professor Priya during the Spring '08 term at Purdue University-West Lafayette.

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Exam_02_Review - W CHAPTER q_ Hygomezsrs TESTS (ONE...

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