Dec. 14, 2006 ECON 240A-1 L. PhillipsFinalAnswer all five questions. They are weighted equally.1. (30) For a regression using ordinary least squares, OLS,y = b0 + b1 x1 + b2 x2 …+ bn xn + e, we make certain assumptions about the properties of the error term, e.a. List five assumptions about ei. _E(e) = 0, expected value of error equals zero___________________ii. _Cov(xe) = 0, error and explanatory variable independent______iii. _ Cov(ejek) =0, j ≠k, errors are independent____________________iv. _ Var(ej) =2 , all j, errors are homoskedastic__________________v. _e~N(0, 2), error is normally distributed_____________________b. For one-way analysis of variance, using regression of a quantitative variable against binary dummy explanatory variables (zero/one) we used one of these assumptions to interpret the meaning of the regression coefficients, b0 , b1 etc. Which assumption did we use? __E(e) = 0_______________c. Which assumption is frequently violated in time series regressions? ___ Cov(ejek) =0, j ≠k ________________d. Explain the difference between homoskedasticity and heteroskedasticity. _homeskedastic: errors have same variance;_heteroskedastic: error variance varies across observations__________e. One can obtain estimates of the OLS parameters by minimizing the sum of squared residuals with respect to each regression parameter without making any assumptions about the error term e. So why are these assumptions about the error term important? _The properties of the parameter estimates such as _maximum likelihood estimators depends on this assumption as well as hypothesis tests using Student’s t-distribution and the calculation of confidence intervals____2. (30) The number of days spent recovering from a heart attack was studied for a randomsample of 300 patients in the US. The duration of days recovering was used to calculate the Kaplan-Meier estimates of (1) the hazard function, (2) the cumulative hazardDec. 14, 2006 ECON 240A-2 L. PhillipsFinalfunction, and (3) the survivor function, as displayed in Table 2-1. These Kaplan-Meier estimates for the hazard rate and the cumulative hazard rate are plotted in Figures 2-1 and2-2.Table 2-1: Kaplan-Meier Estimates of Days Recovering from a Heart Attack, USUS days # ending# atrisk interval hazard ratecumulative hazard rate ratio Survivor Function(# ending/# at risk) (# at risk - # ending)/# at risk8 1 300 0.0033 0.0033 0.997 0.9979 1 299 0.0033 0.0066 0.997 0.99412 4 298 0.0134 0.0201 0.987 0.98013 1 294 0.0034 0.0235 0.997 0.97714 4 293 0.0137 0.0371 0.986 0.96415 10 289 0.0346 0.0717 0.965 0.93016 3 279 0.0108 0.0825 0.989 0.92017 8 276 0.0290 0.1115 0.971 0.89418 8 268 0.0299 0.1413 0.970 0.86719 13 260 0.0500 0.1913 0.950 0.82420 12 247 0.0486 0.2399 0.951 0.78421 11 235 0.0468 0.2867 0.953 0.74722 14 224 0.0625 0.3492 0.938 0.70023 13 210 0.0619 0.4111 0.938 0.65724 9 197 0.0457 0.4568 0.954 0.62725 16 188 0.0851 0.5419 0.915 0.57426 19 172 0.1105 0.6524 0.890 0.51027 11 153 0.0719 0.7243 0.928 0.47328 15 142 0.1056 0.8299 0.894 0.42329 16 127 0.1260 0.9559 0.874 0.37030 11 111 0.0991 1.0550 0.901 0.33331 11 100 0.1100 1.1650 0.890 0.29732 12 89 0.1348 1.2998 0.865 0.25733 13 77 0.1688 1.4686 0.831 0.21334 15 64 0.2344 1.7030 0.766 0.16335 6 49 0.1224 1.8255 0.878 0.14336 10 43 0.2326 2.0580 0.767 0.11037 11 33 0.3333 2.3914 0.667 0.07338 6 22 0.2727 2.6641 0.727 0.05339 5 16 0.3125 2.9766 0.688 0.03740 2 11 0.1818 3.1584 0.818 0.03041 1 9 0.1111 3.2695 0.889 0.02742 1 8 0.1250 3.3945 0.875 0.02343 4 7 0.5714 3.9659 0.429 0.01044 1 3 0.3333 4.2993 0.667 0.00747 2 2 1.0000 5.2993 0.000 0.000Dec. 14, 2006 ECON 240A-3 L. PhillipsFinalDec. 14, 2006 ECON 240A-4 L. PhillipsFinalThe exponential distribution is often used for duration studies. The density function, f(t), for the exponential is f(t) =  e-t , where the reciprocal of  would be the mean recovery time. The cumulative distribution function, F(t), or probability the recovery time lasts up to time t* is F(t*) = 1 - e-t* . The survivor function, S(t*), i.e. the probability that recovery time is longer than t*, is S(t*) = 1 –F(t*) = e-t* . The hazard rate,h(t*), or conditional probability of a recovering heart attack patient returning to work after recovering for t* days is h(t*) = f(t*)/ S(t*), which for the exponential is h(t*) = .The cumulative hazard function, H(t*) = *0)(tdtth, and for the exponential is a linear function of recovery time: H(t*) =  t* .a. From Table 2-1 and Figure 2-1, is the conditional probability of a heart attack patient returning to work given they have been recovering for t* days constant, decreasing, or increasing? _increasing______________________Dec. 14, 2006 ECON 240A-5 L. PhillipsFinalb. From Table 2-1 and Figure 2-2, does the cumulative hazard function look like it is a linear function of recovery time? _no, an exponential or power function_______c. Does the exponential appear to be the function to use to fit this US recovery time data? _no, h(t) is not constant, and H(t) is not linear_______________d. If you use the estimated slope from the linear fit of the cumulative hazard function in Figure 2-2 to estimate the mean recovery time in the US, what value do you get, rounded to the nearest day? _1/0.1223 ~ 8 days______e. From Table 2-1, does this estimate in part d of the average number of recoverydays make any sense? _No_____. Is it too high, too low, or just right? _too low__________________3. (30) The number of days spent recovering from a heart attack before returning to work was also collected for a random sample of 300 patients in Canada. The author of the 7th edition of the text (also in the 5th and 6th editions) analyses the data, and asks the question “ Can we conclude that recovery is faster in the United States?” He proceeds by using an equal variances t-test of the differences in the means using the following data. You can round to the second decimal place.Table 3-1 Recovery Days From a Heart Attack Before Returning To WorkSample: 1 300CANADA US Mean 29.43667 26.98333 Median 29.00000 27.00000 Maximum 52.00000 47.00000 Minimum 8.000000 8.000000 Std. Dev. 7.537597 7.476813 Skewness 0.130607 0.066386 Kurtosis 3.050148 2.525550 Jarque-Bera 0.884350 3.034146 Probability 0.642637 0.219353Dec. 14, 2006 ECON 240A-6 L. PhillipsFinal Observations 300 300a. Is the data distributed normally for both countries? _yes, Jarque-Bera insignificant; skewness near zero, kurtosis