Nov. 1, 2005 ECON 240A-1 L. PhillipsMidterm1Answer all five questions1. (15 points) The following Box plots describe the midterm scores for Econ 240A for the past three years. The total potential number of points was 75 each year. Note the numbers of students taking the midterm were 45 in 2002, 30 in 2003, and 35 in 2004, so note that the scale differs from box plot to box plot.2002Smallest = 34Q1 = 55Median = 61Q3 = 65.5Largest = 74IQR = 10.5Outliers: 34, 2003Smallest = 49Q1 = 59.75Median = 64Q3 = 67.25Largest = 73IQR = 7.5Outliers: 2004Smallest = 18.75Q1 = 34.5Median = 40.5Q3 = 52.5Largest = 70.5IQR = 18Outliers:Nov. 1, 2005 ECON 240A-2 L. PhillipsMidterm1a. On average, which year’s class appears to do the best? Explain the criterion (ia) that you used. The class of 2002 has the highest median. The question is does this mean they did the best, or are differences among years obscured by the grading of different TAs, etc.?b. Which year’s class was most closely bunched, i.e. had the smallest dispersion?The class of 2003 has the smallest intequartile range.c. Which year’s class(es) did not have any outliers? Would it have been possible,given these distributions (i.e. numbers), to have an outlier at the upper end of the distribution in any of the three years? The classes of 2003 and 2004 had nooutliers. Since Q3 + 1.5* IQR is the potential borderline for outliers, this borderline in the various years was:2002: 65.5 + 1.5* 10.5 = 81.252003: 67.25 + 1.5* 7.5 = 78.52004: 52.5+ 1.5* 18 = 79.5Every year this boundary is above the maximum score possible so no outliers at the upper end.d. How is an outlier calculated in 2002? Q3 + 1.5*IQR, Q1 –1.5*IQRe. Do you think it would be fair to grade each year’s class on an absolute scale, i.e. based on your score as a percent of 75 points, versus grading on a curve? Justify your answer. What can vary from year to year in addition to average student performance? Comparing the median or 2004 with the other years , a curve seems more appropriate. See part a for a possible reason.2. (15 points) You conduct an experiment by throwing a fair die three times. Each throw is independent.a. What is the probability of observing one or more sixes? One or more sixes is the complement of no sixes. The probability of the latter is (5/6)(5/6)(5/6) = 125/216. So the answer is 1-125/216 = 91/216.b. What is the probability of observing exactly one six? Using the binomial, 3!/(1!2!) (1/6)(5/6)2 = 75/216c. What is the probability of observing three sixes? (1/6)(1/6)(1/6) = 1/216.3. (15 points) In last Friday’s Los Angeles Times, Oct. 28, 2005, there was a page one article on the latest poll by the Public Policy Institute of California (PPIC). Governor Schwarzenegger is supporting propositions 74, 75, 76, and 77. The fractions of 1079 likely voters in the sample that support these four propositions are 0.46, 0.46, 0.30, and 0.36, respectively. See the attached page from the report on this poll, available online.a. Calculate a 95% confidence interval for the population proportion of likely voters that will support proposition 74, the “teacher tenure” measure, on election day, Nov. 8th, if this poll is a reliable sample. Using fractions, round off your answer to the third decimal point. Use z =[p-hat-E(p-hat)]/std dev of p-hat. We know p-hat = 0.46, the expected value of p-hat is p, and we know std dev of p-hat= [p-hat(1-p-hat)/1079 = 1.0152. A 95% confidence interval for z is given by theNov. 1, 2005 ECON 240A-3 L. PhillipsMidterm1normal distribution: Prob(-1.96 z 1.96) = 0.95. Substitute in for z from above, Prob(-1.96 [0.46 –p]/1.0152 1-96). Multiply the three parts of the inequality bythe denominator, Prob(-1.96*1.01520.46 – p1.96*1.0152) =0.95 . Multiply the three parts of the inequality by minus one, changing the direction of the inequality, Prob(.0297p – 0.46 -.0297) = 0.95. Lastly add 0.46 to all three partsof the inequality, Prob(0.49p0.43) = 0.95b. This PPIC poll reports a sampling error for likely voters of plus or minus 3%. Does this have any connection to your calculations in part a? This is 0.0297 rounded off and converted to percent.c. If you were to test the null hypothesis that the population of likely voters supporting proposition is 0.5 (or less) against the alternative hypothesis that this population proportion is greater than 0.5, do you think you would reject the null hypothesis? Explain. No, accept since the confidence interval is below 0.5d. Do you think proposition 76, the measure supporting limits on state spending, is likely to pass? Does our finding in class, that the elasticity of state general fund expenditures with respect to personal income is significantly greater than one, shed any light on these poll results for proposition 76? Won’t pass. People like government goods and services but are not keen on paying for them.e. If Governor Schwarzenegger offered you a fair bet, i.e. even odds that proposition 74 will pass, would you take the bet? Yes, with a 95% confidence interval below 0.5 for p, this looks like a good bet.4. (15 points) A random sample of 100 observations was generated using the discretePoisson probability distribution, p(x) = e- x /x!, using a procedure similar to the one in Lab. Three, in which we sampled from the uniform distribution. The expected value of x is , chosen to be three for the random number generation, along with a random seed of 10. A plot of the Poisson probability distribution, given a mean of three, is shown in Figure 4-1. The sample observations are attached. The sample mean is 3.06. The sample standard deviation is 1.99Nov. 1, 2005 ECON 240A-4 L. PhillipsMidterm1a. Test the null hypothesis that the population mean is 3 against the alternative hypothesis that it is not equal to three. Use the sample standard deviation. Select aType I error of size  = 5%. H0 :  = 3, Ha :  3. Use the t-statistic, t = [sample mean- ]/(s/(100)1/2 , so t = [3.06 – 3]/(1.99/10) = 0.06/0.199 = 0.3, a very small t-statistic, so accept the nullb. Report the numerical value of your test statistic in part a. t = 0.3.What is the distribution of this test statistic. Student’s t-distribution.c. What is the distribution of the sample mean? Why is this the case? The sample mean is distributed normal, with expected value equal the population mean and variance equal the population variance divided by n. This is the important result from the central