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Group assigment statistic group3

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Group assigment statistic group3

  1. 1. Stung Treng Campus, Cambodia MBA Program, Promotion I ASSIGNMENT Subject: Business Statistics Lectured by TepVuthy Group Members: 1. Por Narith 2. Chor Menglon 3. Loeur Sopheaktra Due date: 21st August 2011Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  2. 2. ACADEMIC YEAR 2010– 2011 Layout of AssignmentThe code numbers and page numbers of the problems are extracted from Business Static Bookauthored by AMIR D. ACZEL. The problems or exercises are for group 3. Group Student’s Name Exercise Page Contact Number Number Number 1-Por Narith 1-6 20 Phone : 012 371 003 1-63 63 E-mail : 3. 2-10 78 narith_por01@yahoo.com 2-23 83 3-9 127 3-14 135 4-3 186 4-17 187 4-38 198 2-Chor Menglon 4-47 199 Phone : 017 920 000 4-54 201 5-1 215 5-15 228 5-22 228 6-20 261 6-28 261 Phone : 077 333 262 3- Loeur Sopheaktra 6-44 265 6-48 307 7-22 261 7-16 305Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  3. 3. SolutionProblem 1.6What is difference between a qualitative and a quantitative variable?Resolution 1.6The difference between a qualitative and a quantitative variable included Qualitative Variable Quantitative Variable A qualitative or categorical variable simply A quantitative variable can be described by a record a quality. numbers of which arithmetic operations such as averaging making sense. If a number is used for distinguishing members of different category of a qualitative variable, the number assignment is arbitrary.Problem 1.63The following data are the numbers of tons shipped weekly across the pacific by shippingcompany.398, 412, 560 476, 544, 690, 587, 600, 613, 457, 504, 477, 530, 641, 359, 566, 452, 633, 474499, 580, 606, 344, 455, 505, 396, 347, 441, 390, 632, 400, 582Assume these data represent an entire population. Find the population mean and populationstandard deviation.Resolution 1.63 a. Find the population mean X Observation X-X bar X1 398 11382.2 X2 412 8590.97 X3 560 3059.47 X4 476 822.973 X5 544 1545.47 X6 690 34340.7 X7 587 6775.35 X8 600 9084.47 X9 613 11731.6 X10 457 2274.1 X11 504 0.47266 X12 477 766.598 X13 530 640.723 X14 641 18581.1 X15 359 21224.8 X16 566 3759.22 X17 452 2775.97Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  4. 4. X18 633 16464.1 X19 474 941.723 X20 499 32.3477 X21 580 5671.97 X22 606 10264.2 X23 344 25820.5 X24 455 2468.85 X25 505 0.09766 X26 396 11813 X27 347 24865.3 X28 441 4056.1 X29 390 13153.2 X30 632 16208.5 X31 400 10959.5 X32 582 5977.22 Total 16150 286053 N 32 Sum(X1: X32) 16150Mean= --------------------------------=-----------------= 504.689 N 32 Σ(X- )2 286053Std= √---------------= √--------------= √9227.51= 94.547 N 32Problem 2.10The European version of roulette is difference from the US version in that the Europeanroulette wheel doesn’t have 00. How does this change the probability of winning when you beton a single numbers? European casino charge a small administration fee, which is not the casein US casinos. Does this make sense to you based on your answer to the earlier question?Resolution 2.10Given that a US casino roulette wheel has 38 slots (00 and 0.36), the odds of landing on a betfrom a single number comes out to be 1 in 38 (or about 2.6%).Percentage is found by dividing 1 (the numbers of chances you have to get your numbers) bythe numbers of outcome possible (in this case 38)= 1/38= 0.026316Problem 2.23A firm has 550 employees; 380 of them have had at least some college education, and 412 ofthe employees underwent a vocational training program. Future more, 357 employees both arecollege-educated and have had the vocation training educated or has had the training. Ifemployee is chosen at random, what probability that he or she is college educated or has hadthe trainings or both?Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  5. 5. Resolution 2.23n= 550P(A): the probability of employees who have had at least some college education. 380P(A)= ---------------=69.09% 550The probability of employees who have had at least some college education is 65%.P(B)= the probability of the employees who underwent a vocational training program. 412P(B)= ---------------=74.91% 550The probability of the employees who underwent a vocational training program is 75%.The probability that he or she is college educated or has had the trainings both include 357P(A B)=--------------------= 64.91% 550P(AUB)=(P(A)+P(B))- P(A B)= ((69.09%+74.91%)-64.91%)= 79.09%The probability that he or she is college educated and has had the trainings includes 79.09%.Problem 3.9Return on investment oversee, especially in Europe and the Pacific Rim, are expected to behigher than those of US market in the near term, and analysis are now recommendinginvestment international portfolios. An investment consultant believes that the probabilitydistribution of return (in percent a year) on one such portfolio is as follows: X(%) P(X) 9 0.05 10 0.15 11 0.3 12 0.2 13 0.15 14 0.1 15 0.05 a. Verify that P(x) is a probability distribution b. What is the probability that returns will be at least 12%? c. Find the cumulative distribution of returnsResolution 3.9The probability distribution of return (in percent a year) on portfolio is as follows:Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  6. 6. X(%) 9 10 11 12 13 14 15 P(X) 0.05 0.15 0.30 0.20 0.15 0.10 0.05 a. A distribution is said to be probability distribution if it satisfies the following condition P(X) ≥ 0 and ∑P(X) = 1 x From the above distribution, we can see that are greater than zero and ∑P(X) = 0.05+0.15+0.30+0.20+0.15+0.10+0.05= 1 x Since the above distribution satisfies both conditions, this is probability distribution. b. The probability that returns will be at least 12% is P(X≥12)= P(X=12)+ P(X=13)+ P(X=14)+ P(X=15)= 0.20+0.15+0.10+0.05= 0.5 c. The cumulative distribution of returns is given by X(%) 9 10 11 12 13 14 15 P(X) 0.05 0.15 0.30 0.20 0.15 0.10 0.05 F(X) 0.05 0.20 0.50 0.70 0.85 0.95 1.00Problem 3.14An automobile dealership records the numbers of cars sold each day. The data is used incalculating the following probabilities distribution of daily sales: Find the mean, variance,standard deviation of the numbers soldResolution 3.14 x P(x) xP(x) x2P(x) 0 0.1 0 0 1 0.1 0.1 0.1 2 0.2 0.4 0.8 3 0.2 0.6 1.8 4 0.3 1.2 4.8 5 0.1 0.5 2.5 Mean of x= 2.8 Mean of x2= 10 a. The mean of numbers of the cars sold each dayE(x)= ΣxP(x)= > E(x)= 2.8Average numbers of the cars sole each day is 2.8. b. Find variance V(X) = E(x2)-[E(x)]2 = 10-(2.8)2= 2.16 = > Variance= 2.6Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  7. 7. The square of deviation from the mean of the cars sold per day is 2.16. c. Find standard deviation SD(X) = [V(X)]1/2= [2.6]1/2= 1.47 = > SD(X) = 1.47 The deviation from the mean of the cars sold per day is 1.47.Problem 4.3Find the probability that standard normal random variable will have a value between -2.50 and-0.89.Resolution 4.3P (-2.50<Z<-0.89)P (0<X<2.50) = 0.4938P (0<X<0.89) = 0.3133P (-2.50<X<-0.89) = P (0<Z<2.50) - P (0<Z<0.89)= 0.4938- 0.3133=0.1805 or 18.05% -2.50 -0.89The probability of the Z value is between -0.89 and -2.50 is 18.05%Problem 4.17Find z such that P (Z>z) = 0.12Resolution 4.17P (Z>z) = 0.12So p(Z <z) =0.88z = 1.174987 z=1.174987Problem 4.38If X is a normally distributed random variable with mean 120 and standard deviation is 4.4,find a value x such that the probability that X will be less than x is 0.56.Resolution 4.38  x= µ± Z*σ  µ= 120  σ= 44P(X<x) =0.56P(X<x)=0.50+0.06P(X<x)= 0.06=> Z⋲0.16P(X<x)=0.5= > Z⋲ 6P(X<x)=0.56= > Z⋲ 6+0.16= 6.16x = 120± (6.16)*44= 391.04 0.16Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  8. 8. Problem 4.47The demand for unleaded gasoline at a service station is normally distributed with mean27,009 gallons per day and standard deviation 4,531. Find two values that will give asymmetric 0.95 probability interval for the amount of unleaded gasoline demand daily.Resolution 4.47  µ= 27,009 gallon  σ=4,530 95%  P= 0.95%= >Z= 1.96X= µ± Z*σX=27,009 ± 1.96*4530X=27,009± 8,878.8X= [18,130.2; 35887.8]0.95 probability internal for the amount of unleaded gasoline demand daily is between18,130.2 and 35887.8.Problem 4.54A computer system contains 45 identical microchips. The probability that any microchip willbe in working order at a given time is 0.80. A certain operation requires that at least 30 of thechips be in working order. What is the probability that the operation will be carried outsuccessfully?Resolution 4.54  n= 45  p=0.80  q=1-0.80=0.20  X=30Binominal distributions in which n are large can be approximated using the normalapproximation. So if X is a binominal random variable with parameter p, then: X-npZ=------------------- √np(1-p)Where the mean of a binomial variable is np and variance is np(1-p), so in this problem p=0.8,n= 45, np =36 and np(1-p)=7.2The probability that the operation will be carried out successfully is given by 1 X- np 30+------ - 36 2P (X≤30) = P (-------------------≤------------------------ (for the normal approximation) √np(1-p) √7.2The continuity correction factor of ½ is includedP(X≤30) = 0.0202Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  9. 9. Problem 5.1Discuss the concepts of parameter, a sample statistic, an estimator, and an estimate. What arethe relations among these entities?Resolution 5.1Sample statistic is a numerical measurement of sample. Population parameter is the numericalmeasurement of population. Estimator of population parameter is a sample statistic used toestimate parameter. An estimate of a parameter is a particular numerical value of estimatorobtains by sampling. When a single value is used as estimate, the estimate is called the pointestimate of the population parameter.Problem 5.15Under what conditions is the central limit theorem most useful in sampling to estimate thepopulation mean?Resolution 5.15The conditions that the central limit theorem is most useful in sampling to estimate thepopulation mean includes  We need to know population standard deviation (σ).  If, the population standard deviation (σ) is not known, it is replaced with estimator, sample standard deviation (S).Problem 5.22An economist wishes to estimate the average family income in a certain population. Thepopulation standard deviation is known to be $4,500, and the economist uses a random sampleof size n=225. What is the probability that the sample mean will fall within $800 of thepopulation mean?Resolution 5.22  σ= 4500  n= 225  µ= 800The sample mean will fall within $800 of the population mean 81.648%So, X1= µ-400= 800- 400= 400 X2= µ+400= 800+ 400=1200 X1- µ 400- 800Z1= -------------------= ----------------------------= - 4/3 (σ/√N) 4500/√225 X2- µ 1200- 800Z2=------------------= --------------------------= 4/3 (σ/√n) 4500/√225P (-4/3) = 0.4082P (4/3) = 0.4082= > P (-4/3<Z<4/3) = 0.4082+ 0.4082=0.81648The probability that the sample mean will fall within $800 of the population mean is 81.648%.Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  10. 10. Problem 6.20The manufacture of batteries used in small electrict appliances wants to estimate the averagelife of battery. A random sample of 12 batteries yields = 34.2 hours and s=5.9 hours. Give95% confidence of interval for the average life of batery.Solution 6.20  n= 12  =34.2 95%  s= 5.9Confidential 95%= > Z= 1.96 σ ± Z* ------------- √s 1.96 -1.96 5.9= 34.2 ± 1.96*-----------------------= [30.86, 37.54] √1295% confidence of interval for the average life of batery is between 30.86 and 37.54 hours.Problem 6.28To aid in planning the development of a tourist shopping area, a state agency wants to estimatethe average dollar amount spent by a tourist in an existing shopping area. A random sample of56 tourists gives =$258 and s= $85. Give a 95% confidence interval for the average amountspent by a tourist at the shoping areaResolution 6.28  n= 56  = 258 95%  s= 85Confidence of 95%= > Z= 1.96 σ ±Z* ------------- √s -1.96 1.96 85= 258±1.96*-------------= [235.74, 293.74] √5695% confidence interval for the average amount spent by a tourist at the shoping area isbetween 235.74$ and 293.74$.Problem 6.44A recent article describes the success of business schools in Europe and the demand on thatcontinent for MBA degree. The article reports that a survey of 280 European business positionresulted in the conclusion that only one-seventh of the position of MBAs at European businessare currently filled. Assuming that this numbers are exacted and that the sample was randomlyBuild Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  11. 11. chosen from the entire population of interest, give a 90% confidence interval for the proportionof filled MBA position in Europe.Resolution 6.44 p*qp± Zα/2*√------------ nl 90%90% confidence = > Z= 1.64n= 280 1Success case= -----*280= 40 7 1.64 -1.64 40p= ---------=14% 280q=1-0.14= 86% 0.14*0.860.14± 1.64*√--------------------= = [10.6%, 17.4%] 28090% confidence interval for the proportion of filled MBA position in European is between10.6% and 17.4%.Problem 6.48Before launching its Buyers’ Assurance Program, American Express wanted to estimate theproportion of cardholders who would be interested in this automatic insurance coverage plan.A random sample of 250 American Express Card holders was selected and sent questionnaires.The result was that 121 people expressed interest in the plan. Give a 99% confidence intervalfor the proportion of all interested American Express cardholders.Resolution 6.48  99% confidence= > Z= 2.58  n=250Success case=121 Success Case 121 99%= >p=------------------------------= ---------------=48.4% n 250q=1-p = 1- 0.484= 51.6% p*qp± Zα/2*√------------ n -2.58 2.58 0.484*0.5160.484± 2.58*---------------------------= [40.25%, 56.55%] 250Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  12. 12. 99% confidence interval for the proportion of all interested American Express cardholders isbetween 40.25% and 56.55%.Problem 7.16An automobile manufacturer substitutes a different engine in cars that were known to have anaverage miles-per-gallon ratting of 31.5 on the highway. The manufacturer wants to testwhether the new engine changes the miles-per-gallon rating of the automobile model. Arandom sample of 100 trial runs gives x-bar=29.8 miles per gallon and s=6.6. Using the 0.05level of significance, is the average miles-per-gallon ratting on the highway for cars using thenew engine different from the ratting for cars using the old engine?Resolution 7.16  µ =31.5  n=100  X-bar=29.8  s=6.6  α =0.05An automobile manufacturer substitutes a different engine in cars that were known to have anaverage miles-per-gallon ratting of 31.5 on the highwayWe want to test whether the new engine changes the miles-per-gallon rating of the automobilemodel.Null hypothesisH0= the average miles-per-gallon rating on the highway for cars using the new engine is notsignificantly different from the ratting for cars using the old engine. H0= µ= 31.5 mpgAlternative HypothesisH1= the average miles-per-gallon rating on the highway for cars using the new engine issignificantly different from the ratting for cars using the old engine. H1= µ≠31.5 mpgThe level of significance, α= 0.05A random sample of 100 trial runs give X-bar = 29.8 mpg and s =6.6 α/2= α/2=Under H0, the test statistic is given by 0.025 0.025 X-bar- µ0 29.8-31.5 Z=- Z0=- Z0=1.Z=---------------=-------------------=- 2.58 2.58 1.96 96 σ/√n 6.6/√100As the Z value is in the lower rejection area, H0 is rejected. The given data provides strongevidence to conclude that the average miles-per-gallon ratting on the highway for cars using thenew engine is significantly different from the ratting for cars using the old engine.Problem 7.22Average total daily sales at a small food store are known to be $452.80. The store’smanagement recently implemented some changes in displays of goods, order within aisles, andother changes, and it now wants to know whether average sales volume has changed. ABuild Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra
  13. 13. random sample of 12 days show X-bar= $501.9 and s= $65.00. Using α = 0.05, is the samplingresult significant? ExplainResolution 7.22  µ0=452.80  n=12  X-bar=501.90  s=65  α=0.05It was supposed to test the average total daily sales has changed from $452.8 or not.H0= the average total daily sales has not changed from 452.8 µ0= $452.8H1= the average total daily sales have changed from $452.8. µ0≠ $452.8The level of significance, α= 0.05The test statistic is X-bar - µ t= ------------------------- s√n 501.90-452.80 t =--------------------------------=2.62 65/√12P-value of the test = 2P (T>2.62) = 0.024Since the P-value is less than the significance level, α=0.05, we reject the null hypothesis.Hence, we may conclude that the average total daily sales have changed from $452.8.Build Bright University, Stung Treng Campus, MBA, Promotion I, Group 3: Por Narith, Chor Menglon, Loeur Sopheaktra

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