Sample_Exam_2 - MGMT 305 SAIVIPLE EXAM 2 Identify the letter of the Choice that best completes the statement or answers the question 1 a a b 2.5 C

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Unformatted text preview: MGMT 305 SAIVIPLE EXAM 2' Identify the letter of the Choice that best completes the statement or answers the question. 1. a. a b. 2.5 C. d. Suppose we have a lower-tailed test Ho: p 2 p4, and Ha: p < pg. if we reject H0 at 5% level ofsignificance, which one is the value of the 2 test statistic? 1.4 4.4 -2.5 Which of the following is true about the p~value? p-value is large or small depends on whether its a lowor—tailed test or an upper-tailed test. a. b. p—value depends on the level of significance. 0. In an upper—tailed test, if the p—value is larger than the level of significance, then we reject the null hypothesis. (1. Whether its a lower-tailed test or an upper-tailed test, we do not reject the null hypothesis if the p—value is larger than the level of significance. Suppose the mean sales volume of an automobile dealership is 15 automobile per month. If the manager of the dealership wants to conduct a research study to see whether a new bonus plan increases the sales volume, then which of the following are the right hypotheses? a. Ho:u=15 Hippilfi b. Homsls Hazp.>15 c. Ho:p.>i5 Ha:p515 d. Ho:u215 Ha:p.<15 it is known that the mean of a population is 58. A sample of size 50 is taken and the sample mean is computed as 55. You decided to reject the null hypothesis that tr 2 60. a. you have committed a Type I error. b. you have committed a Type II error. e y0u have committed either Type 1 or Type II error. d you have committed neither Type I nor Type 11 error. If the null hypothesis is not rejected at the 2% level of significance, it will always be rejected at the 1.5% level of significance. will never be rejected at the 3% level of significance. may be rejected or not rejected at the 1.5% level of significance. will never be rejected at the 1.5% level of significance. 999‘s 10. 11. Two independent simple random samples are selected to test the difference between the means of two ' populations whose standard deviations are not known. The sample sizes are n; = 20 and n; = 40. The correct distribution to use is the a b. c. (1 Normal distribution t distribution with degrees of freedom of 60 t distribution with degrees of freedom of 59 t distribution with degrees of freedom of 58 A random sample of 100 people was taken. Fifty-two of the people in the sample favored candidate Bush. We are interested in determining whether-the proportion of the population in favor of Bush is significantly less than 55%. What is the value of the test statistic? EDP-99‘!” 0.43 -0.97 0.97 -0.48 -0.60 Which of the following is tr_ug about the p~value? 2199‘s The p-value is usually pro-specified for a hypothesis test. If the test statistic is negative, the p-value is also negative. 7 The p—value is always less than or equal to the level of significance. We do not reject the null hypothesis when the p-value is larger than the level of significance. Which of the following statements is not true about the level of significance for a hypothesis test? 9-923"? The level of significance is denoted by alpha (0t). If the p-value is less than the level of significance, we reject the null hypothesis. The level of significance is the maximum allowable probability of making a Type I error. The level of significance is determined by the value of the test statistic. In a two-tailed hypothesis test, z-statistic is computed as 2 = -2.000. The p—value is: 9-9 53"?” -2.000 Prob(Z < -2) 2xProb(Z > 2) 2xProb(|Zl > 2) Suppose the p—value in a test is equal to 0.0626. The null hypothesis can be rejected a. b. c. d. ifor=.Ol Kev—“.05 ifcc=.10 not enough data to make a decision whether the null hypothesis can be rejected. i2. Ten students are randomly selected from a statistics class and the midterm and the final exam scores of these 13. 14. students in the statistics course are tabulated below. What are the appropriate null and alternative hypotheses if we want to test to determine whether the final exam scores are better than the midterm exam scores? WEE-El mm a. Ho:ud=0,Ha:t1d;é0 b. Ho:udsO,Ha:pd>0 c. d. Ho: [1.1- ].12 2 0, Ha! [41- H2 < 0 Ho: 141' 11-2 5 0, Ha: Hr lJ-z > 0 Answer questions 13-14 based on the information given in Exhibit 1. Exhibit 1 In order to estimate the difference between the average daily sales of two branches of a department store, the following data has been gathered. Assume the two populations are normally distributed and have equal va- riances. Downtown Store North Mall Store Sample size 12 days 14 days Sample mean (in $1,000) 36 32 Sample standard deviation 1.2 l (in $1,000) Refer to Exhibit 1. The 95% confidence interval for the difference between the two population means is (in $1,000) a. 3.11 to 4.89 b. 3.26 to 4.74 c 1.74 to 6.26 d 2.12 to 5.88 Refer to Exhibit 1. We wouid like to test whether the average daily sales of two branches of a department store are different. The value of the test statistic is a. 3.649 b. 9.276 c. 1.645 d. 1.96 Answer questions 15 - 16 based on the information given in Exhibit 2. Exhibit 2 A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 75%. 15. Refer to Exhibit 2. The test statistic equals a. 1.155 b. 1.25 c. —1.155 d. -1.25 16. Refer to Exhibit 2. What is the appropriate conclusion at the 3% level of significance? at. The proportion of the population in favor of Candidate A is significantly more than 75%. b. The proportion of the population in favor of Candidate A is not significantly more than 75%. o The proportion of the population in favor of Candidate A is significantly more than 80%. d The proportion of the population in favor of Candidate A is not significantly more than 80%. 17. Which of the following statements is [12L a required assumption for developing an interval estimate of the difference between two population means when the samples are small and the pooled variance is used? a. Both populations have normal distributions b. 0", = 0-2 = 1 0. Independent random samples are selected from the two populations. d. All of the above are required assumptions. Exhibit 3 ATM machines must be stocked with enough cash to satisfy customers making withdrawals over an entire weekend. On the other band, if too much cash is unnecessarily kept in the ATMs, the bank is forgoing the opportunity of investing the money and earning interest. Suppose that six months back at a particular branch the population average amount of money withdrawn from ATM machines per customer transaction over the weekends was $160. Recently a random sample of 16 customers is examined and it is observed that the with- drawals have a sample mean of $175 and a sample standard deviation of $25. The amount of withdrawals per customer transaction over an weekend can be assumed to be normally distributed. The branch manager wants to determine whether the average amoant of withdrawals per customer transaction has increased recently. 18. Refer to Exhibit 3. The appropriate null and alternate hypotheses are H0: 113 160, Ha: p> 160 Ho: p< 160, Hazp2160 Hg: p S 160, Ha: p< 160 Hozpz 160, Ha: u < 160 9-???» 19. Refer to Exhibit 3. The value of the test statistic is a. t=—24 h t=24 c 2:24 d z=QA 20. Refer to Exhibit 3. The p-value is p-value > 0.025 p—value < 0.010 p-value < 0.025 0.01 < p—value < 0.025 9-957?“ 21. Refer to Exhibit 3. The conclusion at 5% ievel of significance is Reject Ho and avg. amt .of withdrawals per customer transaction has increased. Reject Ho and avg. amt .of withdrawals per custorner transaction has not increased. Do not Reject Ho and avg. amt .of withdrawals per customer transactiOn has increased. Do not Reject He and avg. amt .of withdrawals per customer transaction has not increased. 9-9371» 22. Refer to Exhibit 3. The conclusion at 1% level of significance is Reject Ho and avg. amt .of withdrawals per customer transaction has increased. Reject Ho and avg. amt .of withdrawals per customer transaction has not increased. Do not Reject Ho and avg. amt .of withdrawals per customer transaction has increased. Do not Reject Ho and avg. amt .of withdrawals per customer transaction has not increased. sass-1» 23. Refer to Exhibit 3. The 95% confidence interval for the population average amount of withdrawals per customer transaction over the weekends is a. (160.00, 180.00) b. (165.00, 195.00) c. (161.68, 188.32) d. (151.68, 198.32) Exhibit 4 From all the members of a club who played the same game, we randomly chose 8 males and 8 females and recorded their scores as follows: Male: 77 74 82 73 87 69 66 80 (sample variance = 48.0) Female: 72 68 76 68 84 68 61 76 (sample variance = 49.1) Assume that the distributions of the scores of the male members and the female members are independently normally distributed with the same variance. We want to test whether the average score of the male members is difierent from that of the female members in the club. 24. Refer to Exhibit 4. The appropriate null and alternative hypotheses are a. Ho: [Jr-t1; = 0, Ha: til—p2 ¢ 0 13. Ho: til—p2 3% 0, Ha: til-1.1.2 = 0 c. H02 LEI—p2 ?E 0, Ha! tit-1.1.2 #4 0 d. None of the above 25. Refer to Exhibit 4. The point estimate of the difference between the average scores of the male and female members is a. 8.375 b. 6.375 c. 5.375 d. 4.375 ____ 26. Refer to Exhibit 4. The value of the test statistic is a. t= 1.9600 b. t= 1.2557 c. t= 1.9600 d. t 3 2.3300 27. Refer to Exhibit 4. The 99% confidence interval for the difference between the average scores of the male and female members is a. (-5.997, 14.747) (-4.997, 15.747) b. c. (-3.997, 16.747) d. None of the above 28. Refer to Exhibit 4. At 1% level of significance, 3.. there exists a significant difference between the average scores of male and female members b. there exists no significant difference between the average scores of male and female members c. male average score is higher than that of female average score cl. female average score is higher than that of male average score Exhibit 5 A company provides customers Opportunities to buy its products over the Internet, and after 10 months of operation, the company reported that 44% of orders were from the repeat customers. Assume that the company will use a sample of customers orders each quarter to determine whether the proportion of orders from repeat customers changed from the initial 44%. w 29. Refer to Exhibit 5. The appropriate null and alternative hypotheses are He: p < 0.44, Ha: p > 0.44 Ho: p a5 0.44, Ha: p = 0.44 Ho: p = 0.44, Ha: p a5 0.44 Ho: p i 0.44, Ha: p i 0.44- 99 ST?“ ____ 30. Refer to Exhibit 5. During the first quarter a sample of 500 orders showed 205 repeat customers. The p~value and the cenclusion (using 0 =5%) are p~value = 0.177 and Do not reject null hypothesis p-value = 0.177 and Reject null hypothesis p—value = 0.085 and Do not reject null hypothesis None of the above 9‘.“ 13"?" 3]. Refer to Exhibit 5. If you use 0 =1%, the p-value and the conclusion are a. p-value = 0.177 and Do not reject null hypothesis b. p~value = 0.177 and Reject null hypothesis 0. p-value = 0.085 and Do not reject null hypothesis (1. None of the above Exhibit 6 A company contracts with a language institute to provide individualized instruction in foreign lan- guages for its executives who will be posted overseas. Last year, 5 executive studied French. All had some knowledge of French, so they were given the Modern Language Association's listening test of understanding of spoken French before the instruction began. Here are the pretest and posttest scores: (Normal property is assumed.) lHim: Define ud as the difference before and after the instruction, (it. pa = HPosttesnPreteSI) ] 32. Refer to Exhibit 6. The point estimate for the difference before and after the instruction is ll 9*? 23‘?“ Q—lDuID-IQ-I || wn—ANr—t OOOONU’I ll _ 33. Refer to Exhibit 6. The 95% confidence interval for the difference before and after the instruction is a. (—1.9655, 5.5655) b. (—0.9655,6.5655) 0. (1.9655, 5.5655) d. News of the above __ 34. Refer to Exhibit 6. The appropriate hypotheses are a. HozudSO,Ha:ud>-0 b. Ho:udZO,Ha:pd<0 c. Ho:pd=0,Ha:ud>0 d. Hozud<0,Ha:ud>0 35. Refer to Exhibit 6. The value of the test statistic and the conclusion at 5% levei of significance is a. t = 1.327 and Do not reject Hg (Instruction is not effective) b. t = 1.327 and Reject Ho (Instruction is effective) c. t = 2.132 and Do not reject Ho (Instruction is not cfieetive) d. t = 2.132 and Do not reject Ho (Instruction is effective) Solution Key owsgwewwr a 35325.3.5;15' >>>O>>Om>w0>OU>Uw>mm>w>mOOUUmUUUWUU _. :4 WUMHWWNMNNNNN N weyvapwsswewfirgfi #4: la ‘5 «at; 0.05 Balsam}: H‘fldo Half: Ha = ,u 5-, 15’ #6,: ,u 7 15 (EL-reami: 19W:anij 'n, 11:58; m=50' cu. manta T’cd _ ‘7 7%: I We!" 1"! 7nd! hypofém': 6'0 W (awn/#:za/ rte/M 6’,- "9 W7” $335M #UI muffin“ [95) f5 [vagina/real and 1': moi“ £7 m {Adm (#3413 {vi—5436:5362 ID’Valw = :Qx P(272)=&[J~P(ee-zfl _ @3 — 3x PC? 4?.) £13» 1&- Maflw 521,712.95 fade d 3 (Frmi— Wham) shim? Had-Tam Fina: df v?o-—--3?' “T? “g. - —.‘+'v----?9---"3” qt- v - :80v- -'33----3 .. - $9.-..qg-..--W ion-~82. ‘f-‘a‘ f9. H9 : é 0 Ha: Md 7 0 WW? WWW") = Lfi O-E‘I‘ = (3-H, £1.27) fins: A f: = W :M swat) f: 7. 51?.5’ [(401 I“! #4”; :0 He‘ “up/“41sz iii. bmfi:/00}__X:gah 91 x F's" 75""3" :30 Ha: P 5 0—?5' Ha: P >025” 15—3 2:”— __ R, (Ha) likfifi} 1’55 06 = 0.03 (1.16) Pr mm = Pl27mss) :: (fl P[2él.f5$)= (1- 0- }P, =o.?5 0.39.93; 0- i‘S‘x 0- 25' ’3 U I00 —"-' 1.155? ' 0.15) awn) ’Prtfm : 0.523, 7 30 NOT 25742717 Ho mearhbn af 1%: MM?» in fiat/em ‘ : . 5 V Hg P 5 o 3 f0? Cmcb'alafe )4 I: 7202‘“ symfkanflf Ha: .P> 0-?5’ fins: 8 fl: #qugfih‘m : in) Bar”: loome (6) Inch,an ran #00 [Dara/(MW (J) fllfiwWWiaf/W’re #81 ‘file Armada no?» adme *0 —_-—--- 7m‘rea5ed rec-wig. Ho: M5160 More titan ?5-/- at Mennr’ne wfielfler 1419. commie Gasman {ransaoh'tm M C Rest-arch Hypo 741 6-5") Ha: I“ '7 160 ask—d- Ped’earch [Weight—11;? EB @ a: 119 1315-160- =§_ =3. (U56 '1: 5+ah'sh‘c because G" Unknawn and 72 .55 So .psma” 50mph) I am. 1: r. _-.--.--—U From 1‘: ab'shr’fiub'on +5416 Since P.— UcLlU-t = 0.01 4 p..qu 40.025 => E31 .551. #310. _.——-—" #1. w Srnoe 12.2.4 > toios=1153 M =? Hf 5'1' of Sigmfi'canre, ‘ecf'Hg Concbde Hun“ Ha average mm of? {05%de pen War «Iranmch‘gm flan fncnwmd flec'enflg- => Em #267. Sinre 15:53.19 4 b.0lza.602 30 no+ gged Ho => M— m twat qf Sign? 'cem, Concludz de— Mz Werafie <2an a1? wc’MaEmzwab feta Mfume}: lmru acb'cm dial no+ imam Mean/4?. :)> @313 15.05 t: 44¢ é.fi-2.soz 45:9.- q5+ c1; 1&7 ,u 2 WSi‘ [2.130(25/3?) : 115:1: 13.32 = ((61.68, [88.32 ‘ f ::7) figs] . +952" 4&5; We wanf 10A wiefizefir #22 MW 5% 52,? fine male membw ("as oéz‘ffanen-Z- 799mm Mad- ?! Hm female member/S in {he club. Ho: ,u' —-J.lz-.=.o Ha: I“, "Jul w: X~E = :16 - :1. 525 2 4.343% => [3:] 133%: s“: Cm-IJS‘HHZ-Dsi = (3-043 + 03-:qu _E] n,+n1—~2 8+8—& " I 5: = 3?- _’__+..L — I I g Y: (n! ’13-) " +3») = 3. t :: (Zinfo a 4'3” — 4.525??? =2.» —____.__m say; 3. 454+ #6211: 93-}. (:1 =7». a: = 0.01 :3,» oz/Z = 0005‘ '9 =+o.O’/2 3 15-005 df= :E d? z n,+n2—2 = 14‘ CI =- X, -72_ t td/z say; = 4.3% i 9.9:? (3.439) .-= 0539?, mazwjlfl [E 3:23" gm“? "0 h is wf'fitm fie ‘7‘?“/a Cmfl'd’ence in'f'cruafl, do 170% rg“€&%. Comluoia film!- W average W qf malt: and #977442 Members are naf Weren't at 1 4/.- lwd of sigmaé'cqnce. fiflm: [a u ; :20. £23. “a P W } ,b,= 0.91! #30: x:- 205; “*1: 5°” d—i—‘d 15*- L : Egg 2 0-”! ' n 500 — -135 — fi—Po M h ‘ 5‘00 {Ll/m = axpfz 4.4.35“) ' :52): 0.0885 : 0-H? wme: 7!;{:0.o5" =7: 90 NGTR’ETECI' He ll '2': S‘DCQ P..UO.LLL¢=O-1?? > 05:0-01 - 3c: No7“ Rey-€67 Ho I115:- LE] £33.; Define [id as #:e ab-fl’emnap £40m? (RE-I Md .:' #Pas+*est-meesf) J's a+o+6+c—z)+3 ___ 1.8 A“: $@ and aalfer ir‘I-L Frxsfiucbbn 5' #33! 56* z: 2d?— anJQ' : 53.. 5-(1.g)’-: 3.0332 21—1 511 75'I‘ CI 3 d—i akin—175% = 13$ ltd-025,1! f? 3-0332 .1 1.3: 3.1555 z: .31‘ 95¢ng 1 J's? ; (4.9655) 5-5655)i&$ :30 EA] #3”: Ho:lud.5.0 £22? t: 5"“) = 1-3“) =1.3&¥l sat/J5 3.0332/[5‘ ,4+ 05: 0.05, Cra'ilfmf Value is 750.05,»? 32.1.32 Conclusran: Do not Wed Ho bemufie 15:13.2; 4 2.13:: The FDS‘I‘WCb'an (76 'I'Jo‘l-L Efif’ea‘fue a} :x’:0.05‘. fly-E ...
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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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Sample_Exam_2 - MGMT 305 SAIVIPLE EXAM 2 Identify the letter of the Choice that best completes the statement or answers the question 1 a a b 2.5 C

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