Sp09_FinalA_key

Sp09_FinalA_key - EC 41, UCLA Spring 2009 Name (print) F...

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Unformatted text preview: EC 41, UCLA Spring 2009 Name (print) F inal Exam - Version A — 6/8/09 TA: Name & Section Tim - Only pens, pencils and erasers may be used, this is a closed book, closed note, exam. - Students may use a calculator, but nothing that can access the internet. - Write non—integer answers to 3 significant digits, e.g., 333, or 3.33 or .0333 — This exam consists of 20 True/False (20 points), 10 short answer (30 points) - Clearly write answers on this exam. No points are awarded for illegible — Be prepared to show a photo ID during the exam (e.g., UCLA JD) - You may leave when finished. Do not disrupt those still taking th xam. ) I. Circle T for True or F for Fal e (1 point each 1) T 0167:) F latland is divided into three regions: st, Central, and East. If the unemployment rate in the three regions is 3%, 6%, and 12% respect" ely, then the unemployment rate in all Flatland is 7% 10 longer questions (50 points) ers. [email protected] F Evaluated at x* = f the sta ard error of the “estimated mean response for the subpopulation, ,1} ” is smaller than for a future observation )3 .” 36% F A larger regression slope estimate may be less statistically significant than a smaller slope estimate. .m 4Q” F The primary benefit of a well-diversified portfolio is that the standard deviation of its return will typically be smaller than that of a less—diversified portfolio. 5) T o@ Rejecting Ho when CL = .05 provides stronger evidence against Ho than rejecting Ho when or = .01 6©0r F If two assets have identical mean returns and identical standard deviations, it is still possible to benefit by forming a portfolio with both assets, rather then just investing in one of the assets. 76)» F A 95% confidence interval means that percentage of intervals calculated using this method which contain the true population mean will approach 95% as more and more intervals are calculated. [email protected] F When calculating 95% confidence intervals with a given known population standard deviation and different samples of the same size, the intervals will differ, but their size lor width) will be identical. 9) T o@ A particular)? may reject H0: p=0 against H,: p ¢ 0; but not against Ha: u >0. 10) T o@ A “two—sample t-test” is an example of a “non-parametric test.” 11) T o® In the demand for murder example, cross section data for the U.S. suggested that increascd use of capital punishment would reduce murders per capita. 12) T oré'r‘DData from a back to back stemplot can be used to perform a regression b Beam F In a standard OLS regression of Y on X, we assume X is measured without error but that there may be errors in the measurement of Y. [email protected] F An event A and its complement (AC ) can not be independent if P(A) is not equal to zero or one. 15) T 016‘:an statistic has one true value, but each parameter has a distribution of possible values. [email protected] F If the random variable Z = X+Y, then the variance of Z is the sum of the variances of X and Y (oz2 = 0x2 + 0Y2) if the correlation between X and Y is zero. \—__J xi 17g)!“ F If the random variable Z = X+Y, then standard deviation of Z is the sum of the standard deviations of X and Y (oz = ex + 0y) if the correlation between X and Y is one. [email protected] F A lower p-value will provide better evidence that the null hypothesis is false. r F A confidence interval gets wider (larger) as the number of observations in a sample is decreased. fl 2 F Foraleasts uares regression, if f = ,u and T: ,u ;then b > im liesb < q ’x y y o 0 p 1 1 II. Briefly, clearly, and correctly answer the following 10 questions (3 pts. eac ‘ 1) You must decide which of two possible distributions a random variable X has, p0 or p1. The probabilities for specific values of X are: x l 2 3 pg .5 .3 .2 The Null Hypothesis is: Ho: p0 is correct The Alt ive H othesis is: HA: p1 is correct Suppose you follow the rule: accept H0 if X = :' 2 and rejec H0 if X = 3, - a) What is the probability of Type I error? b) What is the probability of Type H error? a c) What is the power of this testing procedure 2) Consider the following 4 (x,y) pairs: (2,0); (2,2); (4,4); (4,6). calculate the sum of the squared {psiduals for the following lines: 1' 1: =—1+1.5 " ' e - z “‘1 a)1ne y ( )X J‘s—2574 Ll‘HJH-ll b 22 2 a as H »i Li; bl .4 b)line2:y=—2+(2)X . 22 z o 21%: “g c) What is the least squares regression line and the sum of its squared residuals? A . ' 'A ‘ 'Z \,;.. + x .Zzgzg ? gs. mum/(AD cc rug-fix 01a} 3? 5'4 \f 4 u; e ‘, 3 aWhat'sthesum: 1 ['50 ‘ 2 ~ _L : 'K : : ~3 )) 1 2-0%) q) 44/) - ,Jlj g4]. \ b) A perpetuity bond pays $10 at the end of each year forever] starting in one year) These payments are “discounted back” at an interest rate of 4%, so the discount factor is 1/(1.04) = .9615. What is the present vaéiue of this perpetuity bond (= price of Sue .r @ 4) Suppose the probability of receiving an order from any one particular sales call during a given period is exactly equal to .40 for any sales call. What is the probability that: , a) the first three sales calls result in three orders? ( . 3 l“ “i 0&7 ‘l b) the first three sales calls result in two order ? 2 _ I. ) 3) tutu = $0 (‘3 c) the first order is from the 4th sales call (so the first three 0 n esult in orders)? @fflt/JT 096‘! 5) Consider the random variable X with the density curve below. 0 2 a) What is the distribution ofX is called U mt £0? M b) Suppose random variable Y has the same distribution as X. Draw the distribution of Z = X - Y to the right: c) What is the probability that Z is greater than 1? P(Z > 1) = 6) Consider two independent events, A and B, where P(A) = .4 and P(B) = .1 A B a) What is the probability that A occurs given that B occurred = P(A I B)? ,: I) - ' [ } _ .t 04 Pa} ‘ T b) What is the probability of both A and B occur? @ A Z c) What is the probability that neitherAnorBoccur? I __\ [ LI ‘} l _ l | ‘ ‘tlrt‘lt1 a 7) Suppose a simple random sample, size 25 is taken; the true population standard deviation, 0, is u u r to e ual 3; but the true mean u is unknown; and the sample mean, J? = 40. " 4 1295‘ a) A 90% confidence interval for the sample mean is: X " l {33‘ N ’V b) IfHo: px = 39 and Ha: p. 75 39, the implied -y In is: L “ 3M0 X S‘ H PYBlbffiizs’)’; b 147%”) (‘6th ‘ 3 E; Z ' gfllmf-S c) If H0: ux = 39 and H3: ux > 39, the implied p—v (9w Qlthicl im/ .- 8) X and Y represent the return on two individual investments. ux = o x = 3 M = o y= Z = a portfolio, % invested in X and 3A in Y, so Z = .25X + .75Y . Truefimela ' the returns, pxy = .5 it» 3[ is 7 ~ I 75 az- Li + «L? 7 -/ . L ) a) What is the mean return of the portfolio, #2? ’1‘ / V, .7 , 2,2 -_.._, ___ g3: Lzs)‘[3f%(7zri (U + 265] 355947536 . . . 9 A. ‘ ~ b) Whatisthevarianceofthereturnsontheportfolio. laws 7L H.062)” + Z ‘2’?) \ y . . . . - t L I I c) What IS the standard devration ofthe return on the portfolio? ) 7 5’ I . l 7 5317) 3 9) Consider two events, A = a person has disease X; and B = person tests positive for disease X. P(A) 2.01 (= 1 in 100); P(B|A) =90; P(B|AC) =.40 Find the following: - r -. - an a) Probability of A given B: P(A|B). -: M) Dig!) l] _ N : W8) 3pm)?» l Miame .Ov‘l +3310 T? . 3 p ,4} yfl); wt w j i ‘ 0222 [3(3): .5351 pat“ ; 116,; R116: ‘59"; I ' b) Probability oanndBcomplement = P(AnB°). 11mm M45: {4m}; MAM) (46): X4)» P443) : ,0! ~ ,Lmq : 150‘ c) Probability ofA given 13‘: P<A|Bcl ': KM 96) 00‘ y @el ’ , 5‘4 5‘ 10) Consider the following Excel output for a regression of Y on X. SUMMARY OUTPUT a) Write the e - S: Regression Statistics f2a+bX (It: LL25 ¥|SS X. MultipleR 0.657 R Square 0.432 , Adj R Square 035 b) What is the t—stat for the slope estimate? Std Error 1.85 ' Observations 9 Coefiicients Std Error t Stat P-va/ue Intercept 4.25 13409 0.0157 X Variable l 0.55 ' 0.0544 c) Can we reject a null hypothesis of Ill: 0 against an alternative hypothesis: [31> 0? . ‘ I _ a 0 534/1, @ Mistrial F ‘ ' Z I : ,0272 IH. Clearly answer the following questions. Show your work (5 points each, 50 total) ‘- 1) Use the true percent return for two different investments given below. > ’ X=Inv1 Probability Y=Inv2 Pr. bability L :44 L? ’ ' 4 .8 Mpg/22+“ 2 7?, (Y: 6 2 = 1H 6 3 21) Find the standard deviation of the return on X (find variance first) K (2 LI : a : (If 86mm 1(6“ We: Dias hertz : .64 =03? !! . ‘3 '5 ‘: , b) Find the standard deviation of the return on Y (find variance first) 3 » v 5) y i l [email protected] i3(_{7-’§?.U1 ‘— [WR 5L23§Z ; 3-56 if? c) What is the average return (in %) for a po io of 50% of investment 1 and 50% of investment 2? O to): Mali/2023- 2.22 G§=tsflwlrwzeau Nil“ d) Assume the returns are independent so the true correlation be een X and Y, pxy = 0. Find the standard deviation of thereturn on theportfolio: GP. < “3.63)” .S‘G‘Sa 33: L3”); ’Lp‘ls 2) Consider the following sample, (n = 8): 0, 0, 1, 1, l, 3, 3, 7 21) Find the (arithmetic) mean of X and sample standard deviation sx fix h). form weenfidence interval for the true mean value of X: ._ ‘ g E ' l .1 ’"l r X" 151 i? 9 W13 8 LL” 7 0) Suppose Ho: u = 0 Ha: it > 0, What is the tealw- ‘ x ‘0 _ “Z 1 . ‘— ce: it = myfi; — .2227? - 1: Li 29, d) Use table D to find a range for the p—value whenTlE: P; u, H: . _ T/ - ‘ 43, 0H1; 7 2355‘ 25‘ 2.817 .01 ' / ae- 3) Conside/r the following observations 9 scores before training and 9 scores after training: / x1 x2 x1—x2 / after before diff 2 o 2 ' 1 2 f 5 3 2 ,1 s 4 1 i 5 7 -2 3 4 4~ 10 4 6 11 6 5 ~ 12 8 4‘ m ’ Cl ‘vP—r at: a mean 6.778 4.111 2.667 stdev 3.598 2.519 2.398 11124: I emupbfi Assume the same people were given the before and after tests so a matched pair t—procedure is a ro . ' - a) Calculate the 95% confidence interval for the difference (11 - it ) , ' . V 3 ~ ‘ 2 7 .8237J (4.5/0 3? \ Its 2456‘? \U \ 2. WEE/[q c‘ What is -1mplied p value if the alt Fm’ttjé Hui ' d) What is e) What is the p-value using the Sign test with null hypothesis 0 a . an alternative of greater than .5 probability of increaszi score? - v ,i - .-7-i E Maw; 74‘ p value a— $3,; ,5 - Mr 4) Assume the before and after tests were given to different people, so there were 18 individuals given tests. also assume different true standard deviations for the before and after tests. Use the smaller of n1-1 or nz-l for degrees of freedom. A a) Calculatethe95%_conf \ _. ference - 2) 2“?!)7 i 2.335 p 5992 gibiqz .. ’ ' dFsQ slit] ' ' ' 3LT; ; K7. ) hat is the appropriate calculated t test statis 1c 1 \ i ‘ El“? 4 i' a 95 7 if? 75H”; We ~ “Mama 0:: tsu- » ~ 5 :- Table D if the alternative hypothesis is (nl— [121$ 0? 2 [m/fg ~ H c) What is a range for’lthe impl' - — .- . -. Wm, it? [3‘1"] 2 t 95‘ L 3260 d) What is a range for the implie a m Table D if the alternative hypothesis is (llr u;) > 0? e) Does this test give stronger oidence that the training imprOVed scores in the previous problem? V 99 This. M ? 7t 5) Consider the 5 scores of treated individuals and 3 scores of those who did treat x1, notreat x2 not receive treatment to the right. Assume the standard deviation of the 5 4 scores with and without treatment is the same. Use n1 + n2 -2 for degrees of freedom. 7 5 a) Find the pooled estimate of the standard deviation for both the before and 3 £— afterscores. 333' L‘Ggs IL 203 1'56, 6 m , _ ~ ‘— 3‘2 ' ‘5 Mm 7 6“ ‘ l. 2211 7 ' cHLLE 9,9; 0‘1: 15' l b) Calculate the 95% confidence interval for the difference (pr “2) 2 21147 6.1553)‘ t gnaw c) What is the appropriate calculated t-statistic if the null hypothesis is (ul- u;) = 0? d) What is a range for the implied p-value from Table D if the alternative hypothesis is (p.1- pg) 76 0? C. 05 4,34 JD? ?~Li‘/7 tv2§ e) What is a range for the implied p-value from T D if the alternative hypothesis is 6 I D < . 2 D b) The followin ; are data on two variables X and Y: a) Draw a scatter plot by hand above to the right and Fill in the blank cells in the above table. Dr L 3 b) What are be and b1 for the regression )9 = be + bIX (use the formula on the last page, calculator, or both). E : animals) ~ 42:2; (1 L9) ' >;(B)~§7‘_ git-zs‘tm‘) "2” i by. "(r hot: Cg, mfg/é). s 1591015 3' c) What is the sample correlation coefficient (you may use the formula on t e l t e or your calculator, or both). 1‘7 I"'\ :7 W: (301-234 Wt W“ @‘74'llf'S-5’7C7) ‘ d) Find the p—value for the significance test of Hefév=0versus Ha: p¢0 01! F; P: 20g ‘ traitor?— Naqi 25$) -' " e “ @ t- : r - — . ;~ : Hts 4U: “W” . mac @F I 3 (l e) F ind the p—value for the significance test of _ I. twig Watt-Q vgitfdltb er gwuufi a O can 0- w l 7) The NHL claims the average number of teeth players have knocked out per hockey game = 4. You believe the average number knocked out is actually larger and consider a specific alternate value of 6 teeth knocked out per game. You take a SRS, n = 36, of hockey games and compute the sample mean. Assume the distribution of knocked out teeth per hockey game is approximately symmetric with no 0 Ila .filpl 3-1.; .5— 2 “a 4:5 ism-Q; Melt/d. )L vqb ’ ~ [gala-31‘ mildew“ ‘Z/aflfifllww' c) What values of A7 will leaa'to rejection of the null hypothesis? Treat f as the unknown in this case. 3 ._> _ -\ NJJPd Zmlg > Z, X > 4 4 "My/g“) a? as >74! 7 Li +~ 13709 ‘ 751”: > ._ >H>LI§ ‘ W > re d) What is the probability of fin 'ngX far enou above 4 o reject H0 if the true mean is actually 6? What is this probability called? ) ‘3? > 523‘” Ma : b 5 > a 7g) > 37l ‘ : I) . 75% If/r.‘ 9/13 \ P Z < t 8) a) What is the probability of rejecting Ho - n ' ' ' fact true? Mat type 0 I We» I mu}? Clearly draw and label a diagram to the right and illustrate the Type I and Type II errors. Show the lowest value of X that will lead to rejection of the null hypothesis. le’s 7114/2/54 ' @ 9) Compa A Sells term life insurance to 70 year old men (coverage starts when '4 they turn 71) hich pays $500 thousand in the event of death. The probability of death for individuals ' is group is: Age: 71 72 73 74 . ‘ Probability: .04 .05 .06 l rig The insurance compan ollects an annual premium on January 1 of any year in which a ndividual is living. The annual premium is $3 ousand per year. Do not include time value of money nsiderations. q ,_ a) What is the probability that an 'dividual in this group lives to be 75 yea : old? g 8 2 fl 1 How much does Co any A collect .from such an indiv' u ual over the 4 year period? 0 0 9 (Arm mama. b) What is the average earning of Company A per ins u ed individual-over a 4 year period? (this will be the sum of 5 terms,_»for each of the 5 possible outcomes per in ' 1dua ver the entire period). 293 (av-53.03 Jr ,UH @fi'~§‘£’0)‘f~ .1? Giu~~§00 “,L .06 020-~§DO) 4- .92 620) 4411+ “7,5 t .-2 .6: 7L -— 2'19 ' ‘1911 2 23¢, e) Consider the “earnings” of mpany A from one individual. What is the s : dard deviation? 2 ( 1 g , ._ A . z - . 7; a Li. (1-. . -11 «113‘ ,1 ‘95 40.5 ’23“! J- :0 e723 “Ya-Ll) 0 C' i W) 0‘! f b 7J2 ( 6(' 2 ¥.S’162"~23‘bl : sir-27‘? it 612" ‘016~’260§ + mac-m H 698? t 7 Q9: 13% =5 g": 2% 357,“ d) Supposc Comany A insures 10 individuals. What is the standard deviation of average u" ings per individual if all claims are ind endent? 10) X and Y represent the return on two individual investments. W = 2 ox = 4 W = 8 oy= 6 Z = a portfolio, halfinvested in X and Y each, so Z = .SX + .5Y . a) What is the mean return of the portfolio, uz? g (2) .j \ C [Q ) '-‘— H U b) Given true correlation between the returns, pxy = 1; what is the standard deviation of the return on the portfolio? 15?: -?*2+.§2'2} r .s 9 “Mel - 3g: (m a b 22);”! (5 GP: 6 -~ dirt-+12 c) Given true correlation between the returns, pxy = .5 what is the standard d " ‘ I of the return on the portfolio? L - ’ \ '2 2 d) Given true correlation between the returns, pxy = 0 what is the standard devi ' - n the return on the portfolio? e) Given true correlation between the returns, pxy = —1 what is the standard deviation of the re? on the portfolio? .7, > x _\ q? HH‘b—lz 1341,; 03):“, ...
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This note was uploaded on 12/06/2009 for the course ECON 41 taught by Professor Guggenberger during the Spring '07 term at UCLA.

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Sp09_FinalA_key - EC 41, UCLA Spring 2009 Name (print) F...

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