Quantitative_Reasoning_for_Business_Week_6_Final

Week 6 Learning Team Assignment QRB 501 University of Phoenix May 16, 2010   Topic 18: Exploration 3 a. If a resident of the U.S. is chosen at random, find the probability that he or she is 25 to 44 years old. = (P(25-44)/Total number of people is 4,486,508. = X’s/(1,107,108 + 452,196 + 1,270,419 + 1,068,243 + 588,542)/4,486,508 = 1,270,419/4,486,508 = 0.28316 (0.2832 rounded) b. If a resident is chosen at random, find the probability that he or she is older than 24 years old. = P(24+)/Total number of people older than 24 = (1,270,419 + 1,068,243 + 588,542)/(1,107,108 + 452,196 + 1,270,419 + 1,068,243 + = 588,542)/4,486,508 = 2,927,204/4,486,508 = 0.6524 c. In what age category does the median age fall' The median age falls in the 25 - 44 category.   Chapter 6 problem A6: A6. (Expected portfolio return) Musumeci Capital Management has invested its portfolio as shown here. What is Musumeci’s expected portfolio return' ASSET PORTFOLIO WEIGHT EXPECTED RETURN 1Money market securities 10% 4% 2Corporate bonds 20 8 3Equities 70 12 Rp = .10(.04) +.20(.08) +.70(12)= .004 + .016 + .084=0.104 or 10% Answer: 10.4% Chapter 6: B6 a) Project # Return Variance Standard Deviation 1 13.4% 4.8% 0.22 2 11.7% 30.8% 0.55 3 10.2% 8.2% 0.29 Return Calculations (.1*.09)+(.7*.13)+(.2*.17)=.134*100%=13.4% (.1*.03)+(.7*.10)+(.2*.22)=.117*100%=11.7% (.1*.15)+(.7*.11)+(.2*.05)=.102*100%=10.2% Variance Calculations .1(9-13)2+.7(13-13) 2+.2(17-13) 2=4.8% .1(3-11.66)2+.7(10-11.66) 2+.2(22-11.66) 2=30.8% .1(15-10.33)2+.7(11-10.33) 2+.2(5-10.33) 2=8.2% Standard Deviation Calculations √(.048)=0.22 √(.308)=0.55 √(.082)=0.29 b) Project Expected Return Risk 1 1 3 2 2 1 3 3 2 Project 1 would be the best choice because it has the highest return with the lowest risk. Chapter 6: B10 (A only) a.) Stock Return X 10.8% Y 7.7% Expected Return for Stock X: (.2*.05)+(.2*.10)+(.4*.12)+(.15*.14)+(.05*.18)= 0.108*100%=10.8% Expected Return for Stock Y: (.2*.12)+(.2*.10)+(.4*.08)+(.15*.0)+(.05*.02)= 0.077*100%=7.7%   Ch. 7. Q 5 When sampling, we are selecting some of the elements in a population, as written in chapter 7 (Cooper-Schindler, 2003). Since, we are conducting a sample of a population who attend live concerts of varying musicians and musical groups and do not know the attendees or the characteristics in advance; we are conducting a non probability sampling. Past seating configurations allow us to calculate the number of tickets that will be available for each of the 200 concerts, giving us the number for each type of music. We also know the gender of attendees: For example: Jazz 1000 700 male (70%) 300 female (30%) Country 1500 600 male (40%) 900 female (60%) Pop 1700 900 male (53%) 800 female (47%) Christian 2000 800 male (40%) 1200 female (60%) I would conduct a non probability sampling, which means it will be nonrandom. I would select a sample similar to the 4 music types above, the same % of male and female within each category as attended the concerts in the last 2 years. I would use quota sampling in that I would select the same distribution in the population that we can estimate. First, I would randomly select from the different music groups, artists. Then within each group, I would randomly select the same % of males from the group as attended in the last 2 years and the same % of females that attended in the last 2 years. Depending upon the questions, if we are expecting the responses to vary between genders, then we should seek proportional response from both men and women, as written in chapter seven (Cooper-Schindler, 2003). If we did not have much time or money then I would not seek precision control even though it does give a higher probability that the sample would more closely representative of the population. I would use quota sampling because it is cost effective and is widely used by opinion pollsters and in marketing, as written in our chapter. Reference: Cooper, D.R. , Schindler, P.S. (2003) Business Research Methods (8th Edition) New York: McGraw-Hill. P.201   Ch.8 Ex. 31 The Sony Corporation produces a Walkman that requires two AA batteries The mean life of these batteries in this product is 35.0 hours. The distribution of the battery lives closely follows the normal probability distribution with a standard deviation of 5.5 hours. As a part of their testing program Sony tests samples of 25 batteries. a. What can you say about the shape of the distribution of sample mean' This would reflect a normal distribution under the curve. b. What is the standard error of the distribution of the sample mean' standard error=(standard deviation)/√(number of observations) 5.5/5=1.1 1.1 is the standard error of distribution of the sample mean. c. What proportion of the samples will have a mean useful life of more than 36 hours' z=((36-35))/1.1=0.909 P1 = 0.3186 (up to 36 hours) P2 = 0.50 P= P2-P1 P=0.50-0.3186=0.1814 d. What proportion of the sample will have a mean useful life greater than 34.5 hours' z=((34.5-35))/1.1=0.455 P1=0.1736+(0.455-.45)*(0.1772-0.1736)/(0.46-0.45)=0.1754 by interpolation in Appendix D P2 = 0.50 P = P1 + P2 = 0.50-0.1754=0.675 e. What proportion of the sample will have a mean useful life between 34.5 and 36.0 hours' .3186+ .1736= .4922  Topic 18: Exploration 12 Use the Web to find how this index is calculated. The Purchasing Managers Index (PMI) indicates a relative change over time and measure factors such as new orders, production, supplier delivery times, backlogs, inventories, prices, employment, and import and exports. A score greater than 50 indicates the economy is expanding. A score below 50 forecasts a sluggish economy for the next three to six months. Taken from the website: http://www.missourieconomy.org. The PMI is an indicator for economic activity. Roughly speaking it reflects the percentage of purchasing managers in a certain economic sector that reported better business conditions than in the previous month. Taken from the website: www.investopedia.com Describe what the graph shows in this context. The graph shows from August to Sept. 2003 the PMI declined but then rose up 10% to a 60% PMI and stayed steady until February of 2004, at which time the economy expanded another 10% up to >70% PMI. This lasted for 3 months; until April of 2004 at which point it has steadily been declining with only a slight rise in Dec. 2005 and then continued downward to the base line of 50, which it hit in July 2005. If it continues to decline in Aug. 2005, the economy will be forecasted as sluggish for the next quarter or two.   Ch. 18: Question 27 & 28 27) Margarine: (.89)/(.81)*100=109.88 Shortening: (.94)/(.84)*100=111.90 Milk: 1.43/1.44*100=99.31 Potato chips: 2.07/2.91*100=105.50 28) P2004=6.33 p2000=6.00 P=6.33/6.00*100=105.5 Ch.18. ex. 33 & 34 RC-33: (.60)/(.50)*100=120.00 SM-12: (.90)/1.20*100=75.00 WC50: 1.00/(.85)*100=117.65 34) P2004=2.50 p2000=2.55 P=2.50/2.55*100=98.04 Ch.18. ex.56 Table 1: CPI Data Year CPI % change 1967 33.4 0.00% 1988 118.3 254.19% 1999 166.6 40.83% 2001 177.1 6.30% 2002 179.9 1.58% Table 2: TV Commercial Data Year TV Commercial % change 1967 42,000 0.00% 1988 525,000 1150.00% 1999 1,600,000 204.76% 2001 2,100,000 31.25% 2002 1,900,000 -9.52% Table 3: Game Ticket Data Year Game Ticket % change 1967 8 0.00% 1988 100 1150.00% 1999 325 225.00% 2001 325 0.00% 2002 400 23.08% Comparison Summary: The trend of the CPI is similar to the trends for commercials and tickets. However, the rate of change for commercials and tickets is higher.