Nov. 6, 2007 ECON 140A/240A-1 L. PhillipsMidtermAnswer all five questions. The first is a descriptive statistics question. The second is a conditional probability question given joint probabilities and hence marginal ones.The third is a Binomial distribution problem, the fourth, a rates and proportions problem and the fifth a regression problem.1. (15) The current sub-prime crisis is causing many home-owners to default and, as a consequence, their banks or lenders are stuck with bad loans that are not being paid off. A bank is evaluating a “scorecard” based on each applicant’s credit history, i.e. information obtained by the bank from credit bureaus. The bank is applying this methodology to two groups: (1) applicants that received loans and later repaid, and (2) applicants who received loans and later defaulted. The cutoff score for an applicant to receive a loan is 650 or better. There are box plots of this retrospective score data for two groups of borrowers from the bank: (1) borrowerswho repaid and (2) borrowers who defaulted. The data follow:a. Do you think this scorecard system adequately discriminates between the two groups? Yes, there is a 123 point difference in the medians.b. Is the dispersion of scores about the same for the two groups? Explain. Yes. The inter-quartile ranges are 87 and 85.75, quite close.Nov. 6, 2007 ECON 140A/240A-2 L. PhillipsMidtermc. Of those who repaid and were good risks, would less than one quarter have scored below the cutoff? Yes, the first quartile is 714, well above the cutoff of 650.d. Of those who defaulted and were bad risks, would more than one quarter have scored above the cutoff? Yes, the third quartile is 670.5 and the cutoffof 650 is between the median of 633 and this value of 670.3e. For which group does this methodology predict better: (1) those who repay, or (2) those who default? Those who repay. The cutoff of 650 would have excluded less than a quarter of these good risks, while it would have included more than a quarter of the bad risks.2. (15) The following table displays the joint probabilities associated with smoking and lung disease for men ages 60-65.Smokers Non-SmokersLung Disease 0.12 0.03No Lung Diseas 0.19 0.66a. What percent of men in this age group with lung disease are smokers? Eighty %, 0.12/0.15b. What is the joint probability of having lung disease and being a non-smoker for this age group for men? From the Table, 0.03.c. What percentage of men in this age group are smokers? 31%, 0.12 + 0.19d. What is the conditional probability of having lung disease given you are a smoker for men at these ages? P(LD/SM) = P(LD and SM)/P(SM) = 0.12/0.31 = 0.308e. What is the conditional probability of having lung disease given you are a non-smoker for this age group of men? P(LD/NSM) = P(LD and NSM)/P(NSM) = 0.03/0.69 = 0.043. (15) Suppose in an unnamed university, for every 100 male revelers on Del Playa on a certain holiday night, there are 150 females in party costume. A professor of Sociology figures out how to take a random sample of ten by recording the genderNov. 6, 2007 ECON 140A/240A-3 L. PhillipsMidtermof the next person to come around a blind corner after his random timer sounds off.a. What is the probability that seven or more of this random sample of ten will be female? P(k>=7) = 1 – P(k<=6) where the probability of a success (female) =150/250 = 0.6. From Table 1, p. B-3 in the Text, for a sample size of 10, P(k<=6) = 0.618, so answer is 0.382.b. Should you use the normal distribution for this calculation? Explain. No, the sample size is 10, so n(1-p) =10*0.4 =4<5.c. What distribution should you use? The Binomial.4. (15) The Los Angeles Times/Bloomberg conducted a poll of 1209 adults. The results were reported in the October 25 Business section of the LA Times, pp. C1-C2.One question was: “How likely is it that the nation could face an economicrecession in the next year?” Sixty five percent of the adults polled said likely.Nov. 6, 2007 ECON 140A/240A-4 L. PhillipsMidterma. Is this significantly more than half the adult population thinking a recession is likely? (Hint: You can group don’t know with not likely so that this is a Bernoulli, i.e. yes/no response). H0 : p = 0.5, Ha : P>0.5.With n = 1209, use the Normal distribution, so z = (p-hat – p)/[p(1-p)/n]1/2 , so z = (0.65 – 0.5)[0.5*0.5/1209]1/2 = 0.15/0.0144 = 10.4. For a Type I error of 5% with corresponding critical value of 1.645, the null is easily rejected, so the answer is yes.b. On p. C2, the poll methodology reported “the margin of sampling error forall adults is plus or minus three percent” Show how this was calculated. You can round to the nearest percent.From part a above the standard deviation is 0.0144 and for a confidence interval of 95% we would have 0.65 + or – 1.96*0.0144 or + or - 0.028, i.e. a sampling error of about 3%c. What size type I error is implied by the sampling error reported in part b?From part b, a 95% confidence interval implies a type I error of 5%.5. (15) In 1979 the Gann Initiative was placed on the California ballot and was passed by 75% of the voters. This legislation attempted to limit the growth of government, allowing for inflation and population growth. The question is: has it worked? California General Fund Expenditures in real 2007-08 dollars per capita is plotted against fiscal year in Figure 5-1 for years 1955-56 through 2007-08.Nov. 6, 2007 ECON 140A/240A-5 L. PhillipsMidterma. Did the Gann Initiative make a difference? Explain? Yes the rate of growth, i.e. slope of real General Fund Expenditures per capita was higherbefore 1979, the year the Gann initiative passed.Real general Fund Expenditures per capita were regressed against a time index with the value zero in 1978-79, one in 1979-80, … and 29 in 2007-08.. The results are shown in Table 5-1.b. Was there a significant trend in real California General Fund Expenditures per capita after the passage of the Gann initiative? Explain.Yes, the t-statistic (or F-statistic) are highly significant. The likelihood of getting statistics this large by chance is about one in ten thousand.Lastly, a plot of the dependent variable, its fitted value, and the residual is shown in Figure 5-3. c. Do you see any evidence that one of the assumptions of least squares regression has been violated? Explain. Yes, the residual is auto-correlatedNov. 6, 2007 ECON 140A/240A-6 L. PhillipsMidtermTable 5-1Dependent Variable: CAGFDEXPCMethod: Least
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