Nov. 2, 2010 ECON 240A-1 L. PhillipsMidtermAnswer all four questions. No talking. If you have a question raise your hand.1. (20) Following is the data from Lab 5 from the Fortune 500 for 1999 showingthe ranking by revenue of the top 50 firms, in millions of dollars.Rank CompanyRevenue $M1 General Motors1890582 Wal-Mart Stores1668093 Exxon Mobil1638814 Ford Motor1625585 General Electric1116306 Intl. Business Machines875487 Citigroup820058 AT&T623919 Philip Morris6175110 Boeing5799311 Bank of America Corp.5139212 SBC Communications4948913 Hewlett-Packard4825314 Kroger45351.6215 State Farm Insurance Cos44637.2516 Sears Roebuck4107117 American International Group40656.0818 Enron4011219 TIAA-CREF39410.220 Compaq Computer3852521 Home Depot3843422 Lucent Technologies3830323 Procter & Gamble3812524 Albertson's37478.0825 MCI WorldCom3712026 Fannie Mae36968.627 Kmart3592528 Texaco3569029 Merrill Lynch3487930 Morgan Stanley Dean Witter3392831 Chase Manhattan Corp.3371032 Target3370233 Bell Atlantic3317434 Merck3271435 Chevron3267636 J.C. Penney3251037 Motorola3093138 McKesson HBOC30382.339 Intel2938940 Safeway28859.9Nov. 2, 2010 ECON 240A-2 L. PhillipsMidterm41 Ingram Micro28068.6442 E.I. du Pont de Nemours2789243 Johnson & Johnson2747144 Costco Wholesale27456.0345 Time Warner2733346 United Parcel Service2705247 Allstate2695948 Prudential Ins..Co. of America2661849 Aetna26452.750 Bank One Corp.25986 a. What is the value of the median? (Hint: since there are an even number of observations, you need to split the difference between the relevant values,upper part and lower part) With 50 observations there are 25 in the upper half and 25 in the lower half so the median falls between the 25th and 26th observations = (37120 + 36968.6)/2 = 37044.2b. What is the value for the upper quartile? (Pay attention to the hint.) What is the value for the lower quartile? Since there are 25 observations in the upper half the upper quartile is the 13th observation or 48253. Similarly, the lower quartile is the 38th observation or 30382.3c. What is the value of the inter-quartile range? The IQR is the upper quartile minus the lower quartile or 48253 – 30382.3 = 17870.70d. What is the value where the upper whisker ends? What is the value wherethe lower whisker ends? The potential value where the upper whisker ends is the upper quartile + 1.5*IQR = 48253 + 26808.05 = 75061.05. Thelargest observation less than this # is AT&T at 62391., so the whisker ends there. Similarly, the lower whisker ends potentially at the lower quartile – 1.5* IQR = 3574.25. The lowest actual observation is Bank one at 25986 so the whisker ends there. e. Are there any outliers? If so, how many and which firms are outliers? There are 7 outliers at the upper end: Citigroup, IBM, GE, Ford, Exxon, Wal-Mart and GM.2. (20) The California Field Poll shows that one week before the election, the fraction of 1092 likely voters that support Fiorina for the U.S. Senate is 0.41.a. What is an unbiased point estimate of the percentage of votes that Fiorinawill receive in the election? 0.41b. Is the fraction of votes that Fiorina will receive known with certainty? Explain. No. this was the fraction of this sample that supported her one week ago, but campaigns can be all about changing the minds of voters.c. What is the estimated standard deviation for the estimated fraction, 0.41,of votes favoring Fiorina.)/)1(*,(~ˆ,1092),1(**,(~/ˆnpppNpnsobservatioso withppnpBinnwherekkp so the estimated variance of p-hat is 0.41*0.59/1092 = 0,000222 and theNov. 2, 2010 ECON 240A-3 L. PhillipsMidtermestimated standard deviation of p-hat is the square root of this number, 0.015d. Calculate a 95% confidence interval for the unknown fraction of votes that Fiorina will receive. Does this interval include winning, i.e. 0.5? A 95 5 confidence interval is -1.96<=(p-hat – p)/0.015<=1.96 and mutiplyinh through by 0.015, -0.029<=(0.41 –p)<=0.029 and multiplying by minus one and adding 0.41 to all three sections, 0.44>= p >= 0.38 so a 95% confidence interval does not include 0.5e. Test the null hypothesis that the fraction of votes that Fiorina will receive is 0.5 against the alternative hypothesis that this unknown fraction is less than 0.5. What is the numerical value of the test statistic? At the 5% level of significance for a type I error, do you accept or reject the null hypothesis? The null hypothesis, H0 : p =0.5 and the alternative hypothesis , HA : p<0.5, ie a one tailed test. The test statistic is z = (0.41 -0.5)/0.015 = 0.09/0.015 = -6, i.e. highly significant at the 5% level so we reject the null hypothesis that Fiorina will get 50% of the votes.3. (20) These days, some furniture stores no longer advertise in the Los AngelesTimes but use a web page to attract customers. Other furniture stores still advertise in the LA Times. One of the latter decides to evaluate the effectiveness of their newspaper ads by interviewing a random sample of potential customers who walk into the store, perhaps by flipping a coin as to whether to interview or not. The store management found the following joint probabilities:Made a purchase Did not make a purchaseHad seen ad in LA Times (0.18) 0.28 0.42Had not seen ad in LA Times 0.12 0.18a. What is the probability in this sample that the interviewee saw the store’sad in the LA Times? 0.28 + 0.42 = 0.70b. What is the probability in this sample that the interviewee made a purchase? 0.28 + 0.12 = 0.40c. Given that the interviewee saw the ad, what is the probability that he or she made a purchase? P(purchase/saw ad) = p(purchase and saw ad)/p(saw ad) = 0.28/0.70 = 0.4d. Given that the interviewee had not seen the ad, what is the probability that he or she made a purchase? P(purchase/did not see ad) = p(purchase and did not see ad)/p(did see ad) = 0.12/0.30 = 2/5 =0.4e. Based on this sample, should the store management reconsider its advertising strategy? Explain. Advertising in the newspapers does notNov. 2, 2010 ECON 240A-4 L. PhillipsMidtermaffect whether a customer will purchase or not, i.e. seeing or not seeing the ad does not affect the probability of a purchase.4. (20) The California Department of Finance has data for California Personal Income per Capita in 2005 dollars for the calendar years 1969 through 2009. A plot of this data is shown in Figure 4-1. This variable was regressed against a time index that was coded zero for 1969, one for 1970, etc.a. How many observations are there?