Midterm2005_answers

Midterm2005_answers - Economics 103 Introduction to...

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Unformatted text preview: Economics 103 Introduction to Econometrics Winter 2005 Professor Sandra Black VERSION A VERSION A VERSION A MIDTERM: VERSION A Instructions: Put your name, ID number, and TA Name (or Section Number) on the Exam. DO NOT USE A CALCULATOR!!!! For complicated math, simply set up the problem but do not solve it. You will be graded on your setup. Total possible points: 132 NAME: STUDENT ID: TA: Some possibly relevant formulas for the bivariate case: fl} = y _ :82 X 23 2 Zoe 3x3,- 47) 2 Z (X,- - X)2 XX? 2 varwl) = n2“. $2 a 2 “ 0' Varwz) - PART 1A: TRUE/FALSE/EXPLAIN YOU ARE GRADED ON YOUR EXPLANATION (5 Points Each) 1. Ordinary Least Squares (OLS) estimators minimize the difference between the actual value (Y1) and the predicted value (Yhat). flléf’ » A/ WMA : (wk ,9 \2' :L WNW 2' @132 ‘ 2. One assumption of the Classical Linear Regression Model is that the sum of the OLS residuals will equal zero. 1 gr ‘ 0 =5} OLS 3. If you use the Minimum Variance Unbiased Estimator (MVUE), your estimate will equal the true population parameter. '3: My, gm mud. um t l m (Hm/E i ON AVEQAQE ) am am TY MM. burl; (Lt/k3 pm \j~-\'CL\,\ (M JLS‘l \WKCCU L3? 4. Var(4X+3Y)=4Var(X)+3Var(Y) <‘-*><**"w Wm MW \/ + g02}cw L><L 5. In the case of a continuous random variable, the area under the cumulative density function will total 1. ( W3”? Vt \ 33‘ Olive: (til idwc.,t“tm ) r: mg was»: \l aw C 0” ‘" l 6. If two random variables have zero covariance, then their joint distribution is equal to the product of their marginal distributions. “puma ft? 2 random ’Vavlavbmg are \M‘le’ENbEHT :9}? itkitji‘ilKlqfijl ’iv‘deLpXMdXt-t‘t a) C: w (11 VD ; 0 BM” ‘ V cw (xg‘j’l "1" \VkCLQtPfil/Wivtwtf 7. An estimator and an estimate are the same thing. A“ at,“ \Wfii’b‘f LS wth m M My hwmmcm CL cumch Vov‘wdg M JXWWOCV \E m \Auwmy we; y! A WWW M 0W" 3 WM? LL dot 0t, twin GUY wgktvuloumv, 8. Let X and Y be two discrete random variables whose marginal distributions are given. Based on this information we can construct the joint distribution of X and Y. gm. we coax CMLLLLOLU it“ \WWXN \Aff \ 9. In the population regression function: Y = ,6, + £2 X + 8 the standard deviation of Y and the standard deviation of e are the same thing. 10. Suppose that X1 ,...,Xn are i.i.d. random variables from a normal distribution N (11,02). If one rejects a null hypothesis in favor of the alternative hypothesis H1 zu<p0, then one will necessarily reject the null hypothesis H0:u=u0 against H1 :u #10 at a given significance level a . (8 points) S G W “a li “film” “it CLXLLE , ,1 Cl" iii V NM W‘I‘Kév‘? "1" GR 5W; Slab/Ci a! watt not Us \A, y” {i 2‘ «2 Part 1B: Questions What role does the central limit theorem play in hypothesis testing and the construction of confidence intervals? (7 points) ’EVL w kow to LNWEC camcgcumfi m;le W Md. tb Wow 4&9 0L,th M b Mme/1 W cm ' is.» we LAM LL? Chit“qu XNN()J&FZ> %“ %){- “:3. N .1. -‘V L 7' Part 2: STATA OUTPUT describe Contains data from C:\DOCUME~1\BLACK\LOCALS~1\TEMP\crime1.dta obs: 2,725 vars: 16 13 Sep 2000 15:28 size: 122,625 (88.3% of memory free) storage display value variable name type format label variable label narr86 byte %9.0g # times arrested, 1986 nfarr86 byte %9.0g # felony arrests, 1986 nparr86 byte %9.0g # property crme arr., 1986 pcnv float %9.0g proportion of prior convictions avgsen float %9.0g avg sentence length, mos. tottime float %9.0g time in prison since 18 (mos.) ptime86 byte %9.0g mos. in prison during 1986 qemp86 float %9.0g # quarters employed, 1986 inc86 float %9.0g legal income, 1986, $100s durat float %9.0g recent unemp duration black byte %9.0g =1 if black hispan byte %9.0g =1 if Hispanic born60 byte %9.0g :1 if born in 1960 pcnvsq float %9.0g pcnv‘2 pt86sq int %9.0g ptime86‘2 inc86sq float %9.0g inc86‘2 Sorted by: sort black by black: sum avgsen ________________l________________________________l___________,__~.______ —> black = 0 Variable l Obs Mean Std. Dev. Min Max avgsen l 1693 .3547549 2.45922 0 40.5 ________.__________________________________l__i_________.____~________________.__~ —> black = 1 Variable ] Obs Mean Std. Dev. Min Max _ _ — _ _ _ _ _ — _ — ——+————_——_——_-.—_——_——.—————_——__—_——_.——__—‘._—_——_____-_____ avgsen ] 439 1.587927 5.856033 0 59.2 These data present the average prison sentence among black and white convicted criminals. (Note: The black and white samples are i.i.d.) 1. How would you test whether the difference between the mean sentence for blacks and the mean sentence for whites is statistically different? To answer this, first set up the null and alternative hypothesis. Then, using either the interval or the point estimate approach, test the null hypothesis with a specified (by you) level of significance. (15 points) D =- \/ t X Emr- )i = EVO- EU) Ht; 1 3 O 4 {i l O p we (we ~.3§)-o (2.th )14 (33%? “a ’3 91363 ‘33 2. Assume now that I told you that the p-value of the difference under the null hypothesis that there is no difference is .027. (NOTE: this is NOT the actual value. . .I made it up!) What would this tell you? (5 points) m p I r), 011;; curt L i? refit/t Wth with he? Witt rtfifittrt L\ D with to Mi 0? >. 02! . reg avgsen narr86 Source SS ’ df MS Number of obs = 2132 ——————————— —-+—---————-—-——————-—--————-———— F( 1, 2130) = 0.47 Model 5.65321912 1 5.65321912 Prob > F = 0.4944 Residual 25777.6762 2130 12.1021954 R—squared = — — - - — — — - - — — --+-———-—-————-————--——-—-—--—-—— Adj R—squared = —0.0003 Total 25783.3294 2131 12.0991691 Root MSE = 3.4788 avgsen Coef. Std. Err. t P>|tl [95% Conf. Interval] . _ _ _ _ _ _ — . _ _ ——+_————_.————._—_—-—————v——————-————_—————————-——————.—————-———————————— narr86 .0613396 .089748 0.68 0.494 —.1146633 .2373424 cons .5853441 .0827162 7.08 0.000 .4231312 .747557 I run the above regression of the average sentence on the number of times arrested. What does the coefficient on narrat86 tell you? How much longer or shorter is the predicted sentence of someone with 6 previous arrests relative to someone with only 4? (5 points) J\ A0 (gym ( 1':- Co) " {EN 0&de (X: gfiauw + sums (ta) —- 38‘33Wl x21; £523 L 302891610. m M ’lflgg 1,0 gfifes 226W »—~"M;‘ hr” ,fi ‘ I "A Tag” Min 15‘ 4% ' gm “I \A m) much DC tum \} mat: i la imam/cred W} “the X2 \b‘ K - ,9. V,“ f“ \ \\/\.,, 1W}; N 1th m ‘1: t c I ,l 12 Latin 3; \fi gm 3 \ " ZW—sflttavez Tau; \39‘81- ~ \‘ M W Part 3: Problem 1. Consider the random variable X- . You observe a random sample of size n. Assume that X- are i.i.d with mean [1. Consider the following estimators for u. 1 20 M1 = %ZX2' 1:1 1 7'1 #72 = —ZX1 n. 1:1 A A 1 #3 = M2+- n 124 = .35 m; : ($253M ; gem 2. Which estimators are consistent? (Again, show your work, 10 points)) f‘ A n .\o\"0LS/W\ )3 a Lb 'ka \ifi’t/vki \QJQ m; 2: W406 :33 Elk/42):).t Part 4 Consider the two random variables X and Y whose marginal distributions are given by X 2 4 6 8 10 Pr<X=x) % % é %% and Y 1 —1 Pr(Y = y) g % Assume X and Y are independent. 1. Construct the joint PDF for X and Y (5 points) if“??? z 3/31 1 i, H0 31 37/ l i . -\ ¥ / 3 »/ 2/ 32 _ \ i i 41 557' a 2. Compute the covariance for X and Y (5 points) ditional on X=6? Explain. (5 points). Statistical Tables TABLE 6.1 Cumulative Areas Under the Standard Normal Distribution 0 1 2 3 4 5 6 7 8 9 ___4_‘_________._——————————- —3.0 '29 —2.8 '2.7 -2.6 -2.5 -2.4 "2.3 —2.2 —2.1 —2.0 — 1.9 - 1.8 —1.7 —l.6 -' 1.5 —1.4 - 1.3 — 1.2 - 1.1 —1.0 ‘09 —0.8 -0.7 0.0013 0.0013 0.0019 0.0018 0.0026 0.0025 0.0035 0.0034 0.0047 0.0045 0.0062 0.0060 0.0082 0.0080 0.0107 0.0104 0.0139 0.0136 0.0179 0.0174 0.0228 0.0222 0.0287 0.0281 0.0359 0.0351 0.0446 0.0436 0.0548 0.0537 0.0668 0.0655 0.0808 0.0793 0.0968 0.0951 0.11510.1131 0.1357 0.1335 0.1587 0.1562 018410.1814 0.21 19 0.2090 0.2420 0.2389 0.0013 0.0012 0.0018 0.0017 0.0024 0.0023 0.0033 0.0032 0.0044 0.0043 0.0059 0.0057 0.0078 0.0075 0.0102 0.0099 0.0132 0.0129 0.0170 0.0166 0.0217 0.0212 0.0274 0.0268 0.0344 0.0336 0.0427 0.0418 0.0526 0.0516 0.0643 0.0630 0.0778 0.0764 0.0934 0.0918 0.1112 0.1093 0.1314 0.1292 0.1539 0.1515 0.1788 0.1762 0.2061 0.2033 0.2358 0.2327 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010 0.0016 0.0023 0.003 1 0.0041 0.0055 0.0073 0.0096 0.0125 0.0162 0.0207 0.0262 0.0329 0.0409 0.0505 0.061 8 0.0749 0.0901 0. 1075 0. 127 1 0. 1492 0. 1736 0.2005 0.2296 0.0016 0.0022 0.0030 0.0040 0.0054 0.007 1 0.0094 0.0122 0.0158 0.0202 0.0256 0.0322 0.0401 0.0495 0.0606 0.0735 0.0885 0. 1056 0. 125 1 0. 1469 0. 17 l 1 0. 1977 0.0015 0.0021 0.0029 0.0039 0.0052 0.0069 0.0091 0.01 19 0.0154 0.0197 0.0250 0.03 14 0.0392 0.0485 0.0594 0.0721 0.0869 0. 1038 0.1230 0. 1446 0. 1685 0.1949 0.0015 0.0014 0.0014 0.0021 0.0020 0.0019 0.0028 0.0027 0.0026 0.0033 0.0037 0.0036 0.0051 0.0049 0.0048 0.0068 0.0066 0.0064 0.0089 0.0087 0.0084 0.0116 0.0113 0.0110 0.0150 0.0146 0.0143 0.0192 0.0188 0.0183 0.0244 0.0239 0.0233 0.0307 0.0301 0.0294 0.0384 0.0375 0.0367 0.0475 0.0465 0.0455 0.0582 0.0571 0.0559 0.0708 0.0694 0.0681 0.0853 0.0838 0.0823 0.1020 0.1003 0.0985 0.1210 0.1190 0.1170 0.1423 0.1401 0.1379 0.1660 0.1635 0.161i 0.1922 0.1894 0.1867 0.2266 0.2236 0.2206 0.2177 0.2148 continuer.‘ Appendix G Statisticai Tables TABLE 6.1 (concluded) ___________________.__._——————————-—————- z 0 1 2 3 4 5 6 7 8 .9 -0.6 0.2743 0.2709 -0.5 0.3085 0.3050 —-0.4 0.3446 0.3409 —-0.3 0.3821 0.3783 -—0.2 0.4207 0.4168 —0.1 0.4602 0.4562 —0.0 0.5000 0.4960 0.0 0.5000 0.5040 0.1 0.5398 0.5438 0.2 0.5793 0.5832 0.3 0.6179 0.6217 0.4 0.6554 0.6591 0.5 0.6915 0.6950 0.6 0.7257 0.7291 0.7 0.7580 0.7611 0.8 0.7881 0.7910 0.9 0.8159 0.8186 1.0 0.8413 0.8438 1.1 0.8643 0.8665 1.2 0.8849 0.8869 1.3 0.9032 0.9049 1.4 0.9192 0.9207 1.5 0.9332 0.9345 1.6 0.9452 0.9463 1.7 0.9554 0.9564 1.8 0.9641 0.9649 1.9 0.9713 0.9719 2.0 0.9772 0.9778 2.1 0.9821 0.9826 2.2 0.9861 0.9864 2.3 0.9893 0.9896 2.4 0.9918 0.9920 2.5 0.9938 0.9940 2.6 0.9953 0.9955 2.7 0.9965 0.9966 2.8 0.9974 0.9975 2.9 0.9981 0.9982 0.2676 0.2643 0.3015 0.2981 0.3372 0.3336 0.3745 0.3707 0.4129 0.4090 0.4522 0.4483 0.4920 0.4880 0.5080 0.5120 0.5478 0.5517 0.5871 0.5910 0.6255 0.6293 0.6628 0.6664 0.6985 0.7019 0.7324 0.7357 0.7642 0.7673 0.7939 0.7967 0.8212 0.8238 0.8461 0.8485 0.8686 0.8708 0.8888 0.8907 0.9066 0.9082 0.9222 0.9236 0.9357 0.9370 0.9474 0.9484 0.9573 0.9582 0.9656 0.9664 0.9726 0.9732 0.9783 0.9788 0.9830 0.9834 0.9868 0.9871 0.9898 0.9901 0.9922 0.9925 0.9941 0.9943 0.9956 0.9957 0.9967 0.9968 0.9976 0.9977 0.9982 0.9983 0.2611 0.2578 0.2546 0.2514 0.2483 0.2451 0.2946 0.2912 0.2877 0.2843 0.2810 0.2776 0.3300 0.3264 0.3228 0.3192 0.3156 0.3121 0.3669 0.3632 0.3594 0.3557 0.3520 0.3483 0.4052 0.4013 0.3974 0.3936 0.3897 0.3859 0.4443 0.4404 0.4364 0.4325 0.4286 0.4247 0.4840 0.4801 0.4761 0.4721 0.4681 0.4641 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857 0.9875 0.9878 0.9904 0.9906 0.9927 0.9929 0.9945 0.9946 0.9959 0.9960 0.9969 0.9970 0.9977 0.9978 0.9984 0.9984 0.9881 0.9884 0.9909 0.9911 0.9931 0.9932 0.9948 0.9949 0.9961 0.9962 0.9971 0.9972 0.9979 0.9979 0.9985 0.9985 0.9887 0.9890 0.9913 0.9916 0.9934 0.9936 0.9951 0.9952 0.9963 0.9964 0.9973 0.9974 0.9980 0.9981 0.9986 0.9986 3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990 ________________________________-————————————"‘ Examples: IfZ — Normal(0,1) then P(Z S -1.32) = .0934 and P(Z 5 1.84) = .9671. Source: This table was generated using the State:0 function normd. ...
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