MGMT 500 MIDTERM STUDY QUESTIONS_2012(1).doc-solns
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MGMT 500 MIDTERM STUDY QUESTIONS_2012(1).doc-solns

Course Number: ECON 500, Spring 2012

College/University: Sabancı University

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MGMT 500 BUSINESS STATISTICS MIDTERM STUDY QUESTIONS for eight countries during the 1990s Q1. The table shown below contains information technology (IT) investment as a percentage of total investment for eight countries during the 1990s. It also contains the average annual percentage change in employment during the 1990s. Explain how these data shed light on the question of whether IT investment creates or costs...

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500 MGMT BUSINESS STATISTICS MIDTERM STUDY QUESTIONS for eight countries during the 1990s Q1. The table shown below contains information technology (IT) investment as a percentage of total investment for eight countries during the 1990s. It also contains the average annual percentage change in employment during the 1990s. Explain how these data shed light on the question of whether IT investment creates or costs jobs. (Hint: Make use of relevant graphical tools) Country Netherlands Italy Germany France Canada Japan Britain U.S. % IT 2.5% 4.1% 4.5% 5.5% 8.3% 8.3% 8.3% 12.4% % Change 1.6% 2.2% 2.0% 1.8% 2.7% 2.7% 3.3% 3.7% To analyze this we plot each countrys annual percentage change in employment during the 1990s on the vertical axis and the corresponding information technology (IT) investment as a percentage of total investment for eight countries on the horizontal axis. In order to sow the relationship, I created a scatter plot graph by the help of stat tools function of excel. As the points tend to move up and to the right, and as the correlation between these two variables is 0.930, this implies a reasonably strong positive linear relationship between IT investment and the jobs. Briefly we can say that IT investment creates jobs. Q2. 1 The percentage of the US population without health insurance coverage for samples from the 50 states and District of Columbia for both 2003 and 2004 produced the following tables of summary measures and correlations. Summary Measures Table: Count Mean Median Standard deviation Minimum Maximum First quartile Third quartile Skewness Percentage in 2003 51.000 14.455 13.700 3.724 9.100 24.900 11.600 16.800 0.910 Percentage in 2004 51.000 14.855 14.200 4.098 8.000 26.300 12.200 16.500 0.699 Table of Correlations: Percent 2003 Percent 2003 Percent 2004 1.000 0.903 a. Describe the distribution of state percentages of Americans without health insurance coverage in 2004. Be sure to employ both measures of central location and dispersion in developing your characterization of this sample. ( 5 pts) 1.000 The variance is essentially the average of the squared deviations from the mean. A more intuitive measure is the standard deviation, defined as the square root of the variance. The distribution of state percentages of Americans without health insurance coverage in 2004 can be summarized as below; As the mean of the population is 14.855; If the data was approximately symmetric, then the 1st,2nd and the 3rd quartiles had to include these below values; Approximately 68% of the observations are within 1 standard deviation of the mean, that is, within the interval XS which turns this range (10.757,18.953) Approximately 95% of the observations are within 2 standard deviation of the mean, that is, within the interval X2S which turns this range (6.659,23.051) Approximately 99.7% of the observations are within 3 standard deviation of the mean, that is, within the interval X3S which turns this range (2.561,27.149) As the distribution is skewed, we can say that in 2004, the state percentages of Americans without health insurance coverage in 2004 wasnt normally distributed, and there was a huge gap between the health insurance coverages. 2 b. Compare the 2004 distribution of percentages with the corresponding set of percentages taken in 2003. How are these two sets similar? In what ways are they different? (5 pts) When I compare the 2004 distribution of percentages with the corresponding set of percentages taken in 2003, I can see that the 2003 distribution was more skewed compared to the 2004 distribution. Also the standard deviation of the 2003 values is smaller than the 2004 values. The number of the observations is the same in each variable. Also the means are closer, but the variance in the 1 st variable is smaller, therefore the 2003 values exhibit less variability about the mean than do the 2004 values. By the affect of this, range of the distribution in 2003 values are smaller than the 2004 values. c. What does the table of correlations for the two given sets of percentages tell you in this case? (5 pts) The correlation between the 2003 and the 2004 variables is 0,903.It shows that there is a strong positive relationship between these two variables and these observations are affected by the same impacts/events. d. Based on your answers to Questions b and c, what would you expect to find upon analyzing similar data for 2005? (5 pts) I would expect to find a distribution which would be less skewed, has a wider range, and a bigger st. dev. 3 Q3. The salaries of all Michigan State University business college professors have a mean salary of $79,580, median salary of $79,000, and standard deviation of $13,500. a. If you increased every professors salary by $2,000, what would happen to the mean and median salary? (5 pts) If you dont change the number of the professors during this period of time mean would increase by 2, the median would increase by 2. b. If you increased every professors salary by $2,000, what would happen to the sample standard deviation of the salaries? Why? ( 5 pts) Variance is essentially the average of the squared deviations from the mean. And the std. dev is the square root of this value; the standard deviation wouldnt change, because the observations will also increase by 2, as well as the mean. c. If you increased every professors salary by 5%, what would happen to the sample standard deviation of the salaries? ( 5 pts) If the salaries of each professor are d ifferent from each other, then the standard deviations of the sample would definitely change. As every salary will increase by %5, the differences from the mean will change/increase. Q4. An ice cream vendor sells three flavors: chocolate, strawberry, and vanilla. Forty five percent of the sales are chocolate, while 30% are strawberry, with the rest vanilla flavored. Sales are by the cone or the cup. The percentages of cones sales for chocolate, strawberry, and vanilla, are 75%, 60%, and 40%, respectively. For a randomly selected sale, define the following events: A1 = chocolate chosen = 0,45 A2 = strawberry chosen = 0,30 A3 = vanilla chosen =0,25 B = ice cream on a cone =0,62 B = ice cream in a cup=0,38 a. Find the probability that the ice cream was sold on a cone and was chocolate flavor. (8 pts) P(X=A1 and Y=B)=0,45*0,62=0,279 b. Find the probability that the ice cream was sold in a cup and was chocolate flavor. ( 8 pts) P(X=B and Y=A1)=0,38*0,45=0,171 c. Find the probability that the ice cream was sold on a cone. ( 9 pts) P(B)=0,62 4 Q5. A set of final exam scores in a business statistics course was found to be normally distributed, with a mean of 73 and a standard deviation of 8. a. What is the probability of getting a score between 65 and 89 on this exam? (5 pts) We can easily compute these variables by converting them to the Z value and using the advantages of standard normal distribution. P(65<X<89)=P(65-73/8<Z<89-73/8)=P(-1<Z<2)=P(Z<2)-(1-P(Z<1))=0.97725-0.1586 =0.818 b. What percentage of students scored between 75 and 80 on this exam? (5 pts) P(75<X<80)=P(75-73/8<Z<80-73/8)=(0,25<Z<0,875)=0.5987-0.1908=0.4079 c. Only 5% of the students taking the test scored higher than what value? (5 pts) Z=1.64 Upper limit=+Z. Stdev=73+1.64*8=86.15 d. If the professor grades on a curve (i.e., gives As to the top 10% of the class, regardless of the score), are you better off with a score of 81 on this exam or a score of 68 on a different exam, where the mean is 62 and the standard deviation is 3? Show your answer in detail, and explain. (10 pts) In order to be in the %10 of the class, the minimum note has to be; Z=1.2815 Upper limit= +Z. Stdev= 73+1.2815*8=83,25 With score the of 81, you cant be better off as you have to score higher than 83,25 in order to be in top %10. For the second case; Upper Limit= +Z. Stdev=62+1.2815*3=65.8445 As 68 is higher than 65.84, you would be better off in this case. Q6. A financial analyst collected useful information for 30 employees at Gamma Technologies, Inc. These data include each selected employees gender, age, number of years of relevant 5 work experience prior to employment at Gamma, number of years of employment at Gamma, the number of years of post-secondary education, and annual salary. a. Indicate the type of data for each of the six variables included in this set. (5 pts) A variable is numerical if meaningful arithmetic can be performed on it. Otherwise the variable is categorical. Age, number of years of relevant work experience prior to employment at Gamma, number of years of employment at Gamma, the number of years of post-secondary education, are numerical and discrete as the values can be counted. They are also discrete as the values can be counted. Annual salary is also numerical but it is also continuous as it depends on an essentially continuous measurement. Gender is categorical as it can be defined as male and female on ly. b. Based on the histogram shown below, how would you describe the age distribution for these data? (5 pts) The age distribution is neither skewed nor trendy. It can be said that the distribution of age is symmetric as the age appear to follow the bell shaped normal distribution. Histogram for Age 10 9 8 7 6 5 4 3 2 1 0 <=20 20- 30 30- 40 40- 50 >50 Category c. Based on the histogram shown below, how would you describe the salary distribution for these data? (5 pts) For the salaries which are in between 20K and 50-60K, the distribution seems normal but as there are no variables in 60-70K and 70-80K, and as there are only 6 observations between 80-90K and higher than 90K, I can describe the salary distribution as a whole as Positively skewed? 6 Histogram for Annual Salary 8 7 6 5 4 3 2 1 0 <=20000 2000030000 3000040000 4000050000 5000060000 6000070000 7000080000 8000090000 >90000 Category Q7. State Alabama Alaska Arizona Arkansas California Colorado Connecticut Delaware Dist. of Columbia Florida Georgia Hawaii Idaho Illinois Indiana Iowa Kansas Kentucky Louisiana Maine Maryland Massachusetts Michigan Minnesota Mississippi Missouri Montana Nebraska Verbal 560 518 523 569 501 554 515 500 Math 553 514 524 555 519 553 515 499 Perc_Take 10 53 32 6 49 27 85 73 489 499 494 487 540 585 501 593 584 559 564 505 511 518 563 587 562 587 537 569 476 499 493 514 539 597 506 602 585 557 561 501 515 523 573 593 547 585 539 576 77 67 73 60 20 10 64 5 9 12 8 76 68 85 11 10 5 8 29 8 7 Nevada New Hampshire New Jersey New Mexico New York North Carolina North Dakota Ohio Oklahoma Oregon Pennsylvania Rhode Island South Carolina South Dakota Tennessee Texas Utah Vermont Virginia Washington West Virginia Wisconsin Wyoming 507 522 501 554 497 499 582 538 569 527 501 503 491 594 567 493 565 516 515 528 524 587 551 514 521 514 543 510 507 601 542 566 528 502 502 495 597 557 499 556 512 509 531 514 596 546 40 80 83 14 87 70 5 28 7 56 74 72 62 5 16 52 7 66 71 52 19 7 12 Examine the relationship between the average scores on the verbal and math components of the SAT test across the 50 states and the District of Columbia by generating scatter plots. Explore the relationship between each of these variables and the proportion of high school graduates taking the SAT. Interpret each of these scatter plots. Scatter plot among the countries are created as below. When we compare the District of Columbia with the other countries, we can easily say that, District of Columbias average verbal score is one of the lowest among the other countries. There is only one countrys verbal score is lower than Dist of Columbia and it is Hawaii. 8 9 Q8. A sample of 1000 households was selected in Los Angeles to determine information concerning consumer behavior. Among the questions asked was Do you enjoy shopping for clothing? Of 480 males, 272 answered yes. Of 520 females, 448 answered yes. 2 X 2 Contingency Table is given below. Gender Enjoy Shopping for Clothing Yes No Total Male 272 208 480 Female 448 72 520 Total 720 280 1000 a. What is the probability that a respondent chosen at random is a male and enjoys shopping for clothing? (4 pts) There are 272 people who are male and also enjoy shopping. Therefore the probab ility for a respondent to be chosen at random to be a male and enjoys shopping for clothing is; P(X=Male and X=Yes)=272/1000=0,272 b. What is the probability that a respondent chosen at random is a female and enjoys shopping for clothing? (4 pts) There are 448 people who are female and also enjoy shopping. Therefore the probab ility for a respondent to be chosen at random to be a female and to enjoy shopping for clothing is; P(X=Female and X=Yes)=448/1000=0,448 c. What is the probability that a respondent chosen at random is a female or enjoys shopping for clothing? (4 pts) P(X=Female) + P(X=Yes) - P(X=Female and X=Yes) =(520/1000)+(720/1000) (448/1000) =0,792 d. What is the probability that a respondent chosen at random is a male or a female? (4 pts) P(X=Male)=480/1000=0.48 P(X=Female)= 520/1000=0.52 e. What is the probability that a respondent chosen at random enjoys or does not enjoy shopping for clothing? (4 pts) Q9. The service manager for a new appliances store reviewed sales records of the past 20 sales of new microwaves to determine the number of warranty repairs he will be called on to perform in the 10 next 90 days. Corporate reports indicate that the probability any one of their new microwaves needs a warranty repair in the first 90 days is 0.05. The manager assumes that calls for warranty repair are independent of one another and is interested in predicting the number of warranty repairs he will be called on to perform in the next 90 days for this batch of 20 new microwaves sold. a. What is the probability that none of the 20 new microwaves sold will require a warranty repair in the first 90 days? (5 pts) Number of microwaves: 20 Probability of failure= 0.05 P(F=0)=0.358 Function: BINOMDIST (0, 20, 0.05, 0) or (20,20,0.95,0) e. What is the probability that at most two of the 20 new microwaves sold will require a warranty repair in the first 90 days? (5 pts) P(F<2)=0,9245 Function: BINOMDIST(2,20,0.05,1) f. What is the probability that only one of the 20 new microwaves sold will require a warranty repair in the first 90 days? (5 pts) P(F=1)=0.7358 Function: BINOMDIST(1,20,0.05,1) g. What is the probability that between two and four (inclusive) of the 20 new microwaves sold will require a warranty repair in the first 90 days? ( 5 pts) P(2<F<4)=P(F<4)-P(F<2)=0.0729 0.9974 4failures 26 11 2 failures 0.92451 6 h. What is the expected number of the new microwaves sold that will require a warranty repair in the first 90 days? ( 5 pts) E(X)=N.P=20*0.05=1 Q10. The height of a typical American male adult is normally distributed with a mean of 68 inches and a standard deviation of 5 inches. We observe the heights of 12 American male adults. X~N (68,25/12=2.08) a. What is the probability that exactly half the male adults will be less than 62 inches tall? (10 pts) =P(X<62)=P(Z<62-68/2.08)=P(Z<-2.88)=1-P(Z<2.88)=1-(0.998)=0.0019 This is for one person only. We have to multiply this probability with the probability that we can get from the sample? Bu soruya bakmak lazm.. 6/12* 0.0019=9.5x10^-4 i. Let Y be the number of the 12 male adults who are less than 62 inches tall. the mean and standard deviation of Y. (10 pts) Determine 12

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INTRODUCTION9-1THE CHANGING INTERNETSoftware-as-a-service(SaaS)Push, not pull, technologies and personalizationF2b2CVoice over Internet Protocol (VoIP)Web 2.09-2How can IT support business processes?Information Technology CapabilitiesBusiness
Arizona State University at the Polytechnic Campus - MATHEMATIC - 101.4
MAJOR BUSINESS INITIATIVES
Arizona State University at the Polytechnic Campus - MATHEMATIC - 101.4
THE PROPOSED DESIGNREPORT #3DUE: NOVEMBER 16, 2004The purpose of this report is to develop an information system model for your application.The new (or proposed) system model should address and resolve the problems anddeficiencies that your team iden
Arizona State University at the Polytechnic Campus - MATHEMATIC - 101.4
REPORT #4PROJECT IMPLEMENTATION AND PRESENTATIONIn Report #4, your team will implement your semester-long project. The implementationis based on the work you have completed on your previous reports. Your implementation will require the team to apply al
Arizona State University at the Polytechnic Campus - MATHEMATIC - 101.4
Practice Test #1 Critical Reasoning (54 Questions)69. (25194!item!188;#058&amp;001228)Which of the following most logically completes the reasoning? When species are extensively hunted, individuals that reach reproductive maturity early make up a larger pro
Arizona State University at the Polytechnic Campus - MATHEMATIC - 101.4
Practice Test #1 Critical Reasoning (53 Questions)73. (25075!item!188;#058&amp;001412)Most of Western music since the Renaissance has been based on a sevennote scale known as the diatonic scale, but when did the scale originate? A fragment of a bone flute e
Arizona State University at the Polytechnic Campus - MATHEMATIC - 101.4
Practice Test #2 Critical Reasoning (68 Questions)1. (24021!item!188;#058&amp;000691)On Pacific islands, a newly arrived gecko species, the house gecko, is displacing the previously established mourning gecko in urban areas, but populations of the two speci
Arizona State University at the Polytechnic Campus - MATHEMATIC - 101.4
135 GMAT LSAT GMAT 135 135 Pace 18 30 ChaseDream GMAT http:/forum.chasedream.com/Test 11. Although 90 percent of the population believes itself to be well informed about health care,only 20 percent knows enough about DNA. So apparently at least 8
Arizona State University at the Polytechnic Campus - MATHEMATIC - 101.4
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CofC - FRENCH-101 - 11012
DR. JOHN WALSHFR. 101/BELL 415OFFICE: 412 J.C. LongAUTOMNE 2010; MWF 9h-9h50TEL: 953-6744FINAL EXAM: Dec. 8, 8h-11hOFFICE HOURS: MW, 10h-11h; 14h-15h and by appt.EMAIL: walshj@cofc.edu; j walshtrois@gmail.com (online office hour, W 20h30-22h on gma