Wed Final Exam Solutions

Wed Final Exam Solutions - Se (v'i'TlWMS STT 315 (Secs 12 ~...

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Unformatted text preview: Se (v'i'TlWMS STT 315 (Secs 12 ~ 22) Final Exam - FORM D' 4-: 14/ l ed, Dec 15, 2010 ' Gilliland (Closed book - 120 minutes) ' same. 10:00 am - Noon Name _ ‘ P11) 7 I TA 0. You may use a ealeulator. All electronic devices capable of communication anti/or web access must be turned off: I. In the places above, please put your name, your PH), and the name of your TA. 2. Be sure to code your name, section number and PID number correctly on the answer sheet that has been provided Sign the answer sheet. 3. 0n our-answer sheet on must also code the letter I) as the‘FORM 0 our examination in the s ace rovided. 4. After the exam, return our answer sheet and our exam a er as on leave rom the rout o the room. A copy of this final exam will be placed on the class pages. You may pick-up your exam paper from your TA next semester. ‘ 5. There are 48 questions, each worth 1.459 course points. (Total 70.032 course points possible.) Tables are found on the last sheet. - ' 6. Your score is determined from your answer sheet and not from what you have written on your exam paper. X: safe/r0060 no. when ls‘+ 65PM" 1. You start interviews for a job? one interview after the other, until you get an offer for the first time. Use the model where results of interviews (Offer, No Offer) are independent with chance of Ofi'er equal to .30 on each interview to answer these questions. What is the probability that you get need more than 5 interviews to get your firs an offer? ' (a) .33 (b) .24 (c) .76 (d) .83 r -' O‘FPer "fluke; \K a new} 7—" .7§-:.[6S 2 — 3. At the Paris airshow, suppose that the following 80 aircraft sales resulted. Suppose that a sale is chosen at random from the 80 sales. “- _ 2. What is the probability - . - - ale is by Boeing or to TAM? ‘ (a) 50/80 (b) 58/80 (c) 8/80 85+l8*8:50 (d) 45/80 (6) none ofthese 3. Given that a sale is by Boeing, find the probability that it is to TA u I . (a) 8/13 (b) 8/30 (d) 35/65 (e) none of these - 4 - 5. A device has three components C1, C2, C3 and the device works as long as at least one of the components is functional. - - The probabilities of the components being functional are 09,-0.8, and 0.6. Assume independence- (a) .006 (b) .216 A (c) .904 (e) none of these 5. is the probability that the devic (a)_667 9W 1+6 flaw; en needed? 7 (c) .432 (d) .999 (e) none of these g. A bin contains 52 parts of which 12 are defective and 40 are good. Three parts are selected at random from the bin one after the other without replacement. What is the probability of getting at least one defectiVe part in the three selections? _’ 9.2-.3-3333 x: ____ I, filsf'fifv \HD' 1. For a" particularprobability model, events A and B are i a. e . r -. - t and P(A) = .6, P03) = .3. Then P(A u 13) == (a) .75 (b) .55 w (d) .90 7 (6) none of these I .6t-35 ~Z,&J£u3)" 73+ (a) 36 (b) .40 (c) .64 (d) .70 8 - 9. At a certain university, 30% of the students who take basic statistics are first-year students, 40% are sophomores, 20% arejuniors, and 10% are seniors. From records of the statistics department it is found that out of the firstuyear students who take the basic statistic course 20% get A's; out of the sophomores who take the course 30% get A's; out of thejuniors 35% get A's; and out of the seniors who take the course 40% get A's. ' 8. What percentage of all students who take the course ;: = I (a) 25% (b) 16% . (d) 31.25% (e) none of these - - ; .7. .09 _ r: / ,4 I3 _ I g A, y w l 7 So / , _ . a i :1 _ ' .35 '4. ' I“ f 5 0 I" . I . I S I ill-g 9. Chem that a student received a ' a course, what is the probability that 'she or he is a senior? ' r _ ‘ . (b) .90 (c) .25 (d)r.8-0 (e), .349 t9 :"Li '2 9 l3 ’2‘ p, 10. Ten True-False questions are given to a student. Suppose that the student selects his answers by 10 independent tosses of a fair coin with say, H =True, T = False. (This is a model for guessing.) What is the probability that the student gets at least 7 . ht ti o t? - 3.13 ques onsc H60 (a) 1/2 (b) 1/4 (c) 1/3 . (6) none ofthesc ixfmfiuof’sfi i”l3(>€$‘7):: fl-w-Qtff :‘(DSE 1_1_. How many license plate numbers are there starting with 3 letters drawn from a 26 letter alphabet without replacement and followed y 4 digits drawn with replacement from the set of 10 digits? . ‘ ' f ’L , r/DIID-IO I 7' 2g .25 25/ to 2f?) I (b) “P; . mp4 (c) 26C; - 10C: ((1)263 -106 7 (6) none ofthese ' 12 — 13. Here is the probability mass function of a discrete random variable X. 12. What is the value of the cumulative distribution function at x =. 3? 7 (a) .20 .(b) .70 (d) .90 . (e) 1.00 13. P(l ,<_:_ X < 4) = (a) .50 (b) .60 (d) .80 (e) .70 7 14 -‘— 15. A large population of parts is 20% defective. A random sampIe of 20 parts is to be drawn from the population. Let X denote the number of defective parts in the sample. (a) .206 @P (c) 2/20 (a) .137 (e) .235 «:- 5. s i 5,. 2. management — 362’18 Paw, 2; pax L>-$.?.10e,w.z_ r.qu (a) 3.5? 1.56 (5)4,230’ (6) 4,134. @ SUCK): daemons) :2qu 14. P(O<X52)= '15. Determine 1300 and SD(X). Eco :JZWB =99 _l_§_. A random variable X has a mean of 50 and a standard deviation of 10. By (’I‘Zhebyshevra P(25 < X < 75) i ~I t ‘ (a) 8/9 (b) 24/25 (0) .95 (d) 15116 - 25": 50‘2.SC'0) r ‘_ i ' :3 75 = B o +2504» 352-» _I_‘_7_. The time X of arrival of a customer to an appointment is modeled as a continuous random variable with-a uniform distribution on the interval from 10 minutes late to_30 minutes late, say from 10 minutes to 30 minutes. What is the ,.probability that X is less than 25 minutes? " (a) .40 (c) .60 (e) .25 ,Pflexwx e- P61eXn33 Flappflhupajgfld .1_8_. A small population consists of 6 defective parts and 24 good parts (30 parts in total)- A simple random sample of 20 parts is to be drawn from the population. Let X (1 ' u - umber of defective parts in the sample. P(X =_ 3) = t w (c) .66 (d) .15 '(e) .05 g DH 70 . (gym) .. Mi "$55? (so # moms“ 19 - 20. A variable is modeled as a random variable X with a normal distribution with mean 40 and standard deviation 5. (b) .95 .68 (d) .997 (e) .74 . / - ,3Lfl3 ' x is ' 4% Tat/Elf 2. was so M L -.'i 20. Find the point x such that P(X a x) ==' .15. a .. \ - . (a) 44.7 (c) 46.2 i 19. P(35 <§X <50)= . (d) 43 .4 (e) 45.? 5’ if?“ a as win-testis“? 53-” “#571 a. A large population of trees is 50% diseased. The population is sampled with a simple random sample of size 100. What is the probability that the sample of trees has at least 45 diseased trees in it, that is,-that the sample proportion is at least .45? . If you wish, you may use normal approximation to determine P( 13 Z .45). Pick the closest. _ gem) .76 i (b) .65 ,X‘: Nadineasecifl m. saw}: )(n) 854531.50) x ~ -* rI—- {X54qu PM aw} : I fainkmogemg,mwj I so :3. 1.... .nas new #5 saw/:5 22... Suppose xrin units of feet is normally distributed across a large population of interest with unknown mean it and known standard deviation a: 8 feet. Suppose that a random sample of size n = 16 to be selected from a population. Use a normal distribution for a? to determine the probability that the sample average 3? will fall within 4 feet of p to two decimals. (b) .33 (e) .37 (d) .68 (e) .99 _.. "94*":3 ,Z 50(xlffi'; V72“ Em: M awgga SDCX):‘:5" ' Q. manufacturer of a 100W light bulb has data that suggest that the lifetime of the light bulb follows an exponential - distribution with mean-1000 hours. Use this exponential model to answer the following question about a randomly selected light bulb from the manufacturer. What is the-probabiiity that it will fail in the first 1000 .--‘ .12 .50 '.45 ,- (a)_ “M (b) (c) L “My “3 :‘“'-'-€ I PKX6—’W)-fi I- Uer 3, l we 35,. A survey bylan electric company contains questions on the foilowing: Ageofhousehoid head, Sex of househoid head, and Use of electric heating (yes or no). What type of variable is "Age of household head"? I(e) .77 (a) categorical @ 15.. The following data are annualized returnsin % for 14 stocks. If} E— —1.2 3.9 8.2 9.0 10.0 11.0 ll.5 12-5 13.0 14.8 15.5. 16.3' 16.7 18.0 4» 5.5 has? its-1:273) Use the formula presented in the textbook a I- - -- a the 75th percentile return for these stocks. 3 {5:7 ' (a) 15.5% - ’(c) 16.3% " (d) 11.25 (e) noneof these (Ln-HEP :7 ES (We): :: {LLS' (d) 2.326 (e) 2.761 . r / - in {awe 3 ,- ea/ WS 27- 28. Below is a boxplot of the seal strengths of n = 100 sample potato chip bags. 0 1 o 20 3o 40 so 60 7o 30 90 1 00 Seal Strength 27. The median seal strength i ' ' r . (h) 10 ‘ (c) 3 (d) 30 (e) 20 - Qt ' G3 28. Approximately, what perce .. ; A'e'uag ad= seal strengths between 3 and 15‘? _ _ (a) 50% ' (b) 75% (c) 10% (d) 25% (e) 95% Q. A sample of 100 Weights has Q, = 150 'lbs, Q2 = 165 lbs and Q3 «'1 190 lbs. 'The three largest weights in the sample are 220 lbs, 253 lbs and 275 lbs. The upper” (right) whisker extends to what value? ' (a) 230 (b) 253 (c) 250 (d) 190 mt; Flo-H5690) :: £56? . _I I‘- ‘ " r - I I ' ' Q. A confidence interval estimate for the mean ll of a large, h1gth skewed Population 15- needed. A random sample of srze 'n == 3 is den and the t—interval estimate 2? [J3 is calculated. The t—interval estimate may not deliver the designed confidence level 95% in this case. (b) False ' ‘ §_1_. A random sample of n = 1-9 from a normal population has mean A7 = 40.6 and standard deviation 5 m 5 The 98% confidenCe interval estimate of the pOpuiation mean is 40.6 :t ' (a) 1.99 (b) 4.74 - (d) 2.99 (e) 5.60 / 5c; . :é‘ 2.56.1 ‘—-' ‘40 é VT; 32 - 33. The Michigan Bureau of Elections believes that about proportion p = .80 of the signatures in a'Iarge petition drive are valid. ' ‘ 32. Using the value p = .80 for planning purposes, what is the minimum sample size required so that the 90% sat-interval estimate of p has a margin of error of i .03? - ' ' (a) 1,083 ‘ (c) 724- (d) 378 (e) 597 n _, was} (3036-203 b” .023 33. Now supposé‘that the population size is N = 1,000. Using the value p = .80 for planning purposes, what is the minimum sample size required so that the 90% z—interval estimate of p has a margin of error of i .03? Pick the closest. (a) 844 _(b)2'85 (c) 667 @7525 “La w 2" MM 6 _ Hm m—i N. ‘3_4. Consider the confidence interval estimate f i za ,20/ J; of the mean p of a large population. Which case has the - L571, .. .1571. 1:15.: use _3§. The mean u of a large pepulation is the parameter of interest. A simple random sample of size n = 100- is drawn; the sample average is 55 = 30.0 and the sample standard deviation is 5 m 1011 Determine the 90% upper estimate for P. PiCk the closest. I 7 (3331.7 7 (0)310 (d) 30.5 (c) 32.3 ‘0 _:_‘“‘\ 7+»;a‘mi. 30.04— :29 ~—-- ... 31.3 W " T 0130 62 gig. A random sample of 200 accountsis selected from a population of N = 800 accounts and audited. Forty of the sample accounts are found to have errors. Give a 95% confidence interval for the population proportion of accounts with errors. (b)’.20 :l: .043 r (c) .20 i .038 (a) .20 3: .060 (d) .20 :1: .073 (e) .20 i .086 A ... i 10 (I39 p:,i‘3:«2/° .anrmw, Wu” 63%;} 150 - g as: «i _ . . W’ . 2p £1943 In hypothesis testing as taught i _ ' e 9 .l r of the hypotheses isgiven "favored standing"? -' "I7 ypothesis . . (b) The alternative hypothesis H;. e' wn t i ; . (b) Hg is rejected when it is not true. (0) Ho is retained when it is true _' “I is retained when it is not true. 32. Which of these statements is correct? (a) The p-value of a test statistic is the probability that the null hypothesisHo is true. _ 'The p~value of a test statistic is the probability that the alternative hypothesis H1 is true. he p-value of a test statistic is a kind of “credibility rating” for He in light of the evidence. ' d) A large p-value leads to Rejection of Ho. _ ‘5 7_ 10. A simple random sample n == 150 parts is selected from a bin containing 2,000 parts. The sample contains 12 defective - parts so that the sample proportion is 35 == .08. The 95% z—confidence interval estimate for the proportion of partsin the bin that are defective is 0:08 i 0.04. Which of the following statements is correct? (a) 8% of the parts in the bin are defective b We do not know the percentage of parts in the bin that are defective, but we know that it is between 4% and 12%. or about 95% of all simple random samples of size 150, the interval estimate captures the unknown p- Therefore, we can be about 95% confident that the interval estimate 0.08 i 0.04 contains p. can has” -—.-«cr. 41 - 44. A certain prescription medicine is supposed to contain an average of 200 parts per million (ppm) ofa certain ' . chemical. The manufacturer wants to check whether the average concentration in a large shipment is the required 200 ppm or not. A random sample of 3] portions is tested, and it is found that the sample mean is 196.0 ppm and the sample standard- deviation is 10.91 ppm. Use the East to test the null hypothesis that the average concentration in the entire large shipment is 200 ppm versus the alternative hypothesis thatit is.not_4200 ‘ at level a = 0.01. - .51., 5 m “(495% 41. The value of the test statistic based on the a ‘ u'.’ ; (a) «1.96 _ (c) «2.34 '(d) -1.54 (e) -130 M U I ' L! -é 3’: a 0n " ‘" a '0 05'; ‘ ' /U.3 I 42. What is the approximate p—v . w: n ' test statistic? . , 7 e '3 (b) .10 (c) .20 (a) .025 (e) less than .01 “10‘ ’ 1“” sur- 7 gong 9.9M?” "" '95,“ 43. With the critical value approach to hypothesis testing, what are the critical val _, - ' -test‘? (a)_d:i.960 (mamas (c):l:2.326 (ale 3 ' ‘ 45 - 48. A company’s market share is very sensitive to both its leVel of advertising and the levels of its competitors’ advertising. A firm known to have a 60% market share wants to test whether this value is still valid or has decreased in new of recent adVertising campaigns of its competitors and its owu decreased level of advertising. A random sample of 500 consumers reveals that 275 use the company’s product. Is there evidence to conclude that the company’s market share has _ dropped below 60%, at the 0.05 level of significance? - , +0” I 44. What decision is reached at level a= 0.01? ' (a) Reject Ho (Decide that the population average is different from 200ppm) 45. The value of the test statistic based on the given data is zobs =. - ' .6. i5: :17 0:2: 4 $5 . (ea-L65 -(b).—2.33 - (d)—1.80 (e)-l.96 ’5 5'0 7 a g 't .0 , , £0 (#0 1. 5 a a :46. What is the approaimate p—value of the test statistic? (a) .05 (b) .036 (c) .02 - (e) less than .001 no“ 1 - ' 9|l3 7 7 -2.L? 0 ‘2 47. With the critical value approach to hypothesis testing, what are - = a. l value(s) of the 'z—test? (a) ~ 1 .960 (b) —2.576 ' (d) .2304 ' (e) -1232 ".43. What decision is A at level a: m 0.05? ' r a) Reject Ho (Decide that the company’s market share has dropped below 60% 5mm P—«lahié (atriab Karenina. ...
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This note was uploaded on 02/01/2011 for the course STAT 315 taught by Professor Dennisgilliland during the Summer '10 term at Michigan State University.

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Wed Final Exam Solutions - Se (v'i'TlWMS STT 315 (Secs 12 ~...

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