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are106-homework-4-key

# are106-homework-4-key - Psw AK ArelO6 Homework 4 200 180...

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Unformatted text preview: Psw AK ArelO6 Homework 4 200 180 160 140 PATENTS 120 100 80 1960 1965 1970 1975 1980 1985 1990 The simple linear model does not ﬁt this data series very well because the data points don’t not follow a linear pattern. Therefore, when the best ﬁt line is drawn, it will be clear that many of the points deviate from that line. ii. the regression on time would not be my model of choice b‘\ se research and development should be included as well. By not inc '1] just be ushed into the error term and increase the inaccuracy of the regre ‘. - f L [4’ i . iii. Model 2: OLS, using observations 1960-1993 (T = 34) J Dependent variable: PATENTS coefﬁcient std. error t-ratio p-value const -5044.82 473.199 -10.66 4.63e—012*** YEAR 2.61273 0.239410 10.91 2.56e-012*** Mean dependent var 119.2382 S.D. dependent var 29.30583 Sum squared resid 6002.230 SE. of regression 13.69561 R-squared 0.788217 Adjusted R-squared 0.781599 F(l,32) 119.0982 P-value(F) 2.56e-12 Log-likelihood -136.1938 Akaike criterion 276.3877 Schwarz criterion 279.4404 Hannan-Quinn 277.4288 rho 0.990333 Durbin-Watson 0.127221 alpha hat= -5044.82 beta hat= 2.61273 units for alpha hat measured in thousands of patents and beta hat is measured in thousands of patents per year. iv. 2.612 thousands of patents per year In .me, Q; v. estimated standard error for alpha hat = 473.199 estimated variance for alpha hat= 2239 17 .2936 estimated stande error for beta hat = .239410 estimated variance for beta hat= .057317 vi.l'\o'-[5"0 Hq‘ﬁ‘FO n= 4J9 9“ ‘w = '0‘“ :‘27yclLllD +35 2.031 ‘f3-1=9.0’3‘| < To '2 0W3 waIIIe=050e~0Ia 4 ,05 Reject the null hypothesis because beta is only zero 5° of the time. In this case, beta is signiﬁcant since the patents are increasing overtim . vii. t(32, 0.025) = 2.037 VARIABLE COEFFICIENT 95% CONFIDENCE INTERVAL const ~5044.82 -6008.69 -4080.94 YEAR 2.61273 2.12507 3.10039 P LEI-21.0310; 3: I6 I. fir :1 037%) 6'5 P12 (arms-1 037(333‘11032’r5ﬁ 0 arm? I 2 on (13081031 - 39 9 (1.0509 L l?" L . .Ioo‘chQ)‘.OlS .f M. I RSV. ccrIaIvaere, WI“ we 0, I150?» Io "5 Ioouokmma a this PESNN Ia 1.07m H137.\0151)_OL’- 'S'okmg +1. 031mg I060] 615 ‘“ 9W0 '1) (e “002111023. £»-LIoFo.°II%3 .08 “W93. {393.0% 9- 07“ (:1’5‘31‘1101=.“\%7Q27811+ 10mg 340040311 This value is very close to Gretl’s; therefore, I account the minor difference to rounding errors. viii. Covariance matrix of regression coefﬁcients: const YEAR 223917 -113.287 const 0 0.0573169 YEAR const / CW (121%) vartﬁl ~ 223917 2113.287 const cw {1‘ - -113.287 0.0573169 YEAR i ‘m \ Wu?) Mr (.93) =- 13120111 These are the estimated variances and covariance. The variances of the estimated slope and constant in this gretl table and the variances calculated in part v is exactly the same. ix. the Wear” variable would be 2010. The model would predict 206.7673 thousands of patents in the US in 2010. \/ /\ /\,~ ( /\ /\ X- V pawn-tat = l1 var Lot) T Lfemrq‘vm (E) ~+ (New con‘, f3) V Warts“: = 3930 [’11 210101 (.0912) My 10010101131913: (a 726757 v ems:- m .19! so: / xi. Patents hat are in the form of a normal distribution because alpha hat and beta hat are A normal' D” H L¢,[email protected]\ 213:7"(IM C a Bob/001 l’lah 64:. ﬁOrMuS IS “0"qu. ...
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are106-homework-4-key - Psw AK ArelO6 Homework 4 200 180...

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