2011-03_Test_2S_updated.pdf

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Department
Economics for Management Studies
Course
MGEB12H3
Professor
Daga
Semester
Winter

Description
ECMB12H3 Lec30 and Lec60 Quantitative Methods in Economics II Department of Management University of Toronto at Scarborough January – May 2011 Dr. Yu Test 2 Date: Saturday March 19, 2011, 9–11am Time allowed: Two (2) hours Aids allowed: Calculator Notes: • This test consists of 21 questions in 11 pages including this cover page. • It is the student’s responsibility to hand in all pages of this test. Any missing page will get zero mark. • Statistical tables are provided separately. Do not hand in the tabl es. Only hand in your test papers. • Show your work in each question in Part II. • This test is worth 30% of your course grade. Print Last Name Solution Given Name(s) Student Number Circle your Section L30 Wednesday at 7 pm L60 Online Do not write on the space below, for markers only Page Question Max Marks 2-8 1-20 60 9 21 5 10-11 22ab 10 12-14 23abcd 15 15-16 24abc 10 Total 100 1 Part I. Multiple Choice. 3 points in each question. No part mark. Circle only one answer. If there are more than one correct answer, circle the best one. 1. Let X 1 be a normal random variable with mean 6 an d variance 1, X 2 be a normal random variable with mean 7 and variance 1, X 1 and X 2 are independent. The probability P X 1 X 2) is equal to (a) 0.05 (b) 0.1 √ (c) 0.242 (d) 0.258 (e) 0.758 Solution: X − X is normal with mean 6 − 7 = −1 and variance 1+1 = 2 . 1 2  X1− X 2 − ( 1) 0− −  P X 1 X >2 0)= P  2 > 2 = P Z > 0.71 = 0.242   2. Let X 1, X 2, X3 be a random sample of size 3 from a normal population with mean 1 and variance 4. The probability P X 1 2X −22X > 73 )is equal to (a) 0.0005 √ (b) 0.1597 (c) 0.3403 (d) 0.6502 (e) 0.8403 Solution: E X 1 2X − 2X 3) = E X 1 2E X − 22X ( )3 =1+ 2 1 − 2 1 =1 Var X 1 2X −2X 3)=Var X +14Var X + 4V2r X ( 3 = 4+ 4 4 + 4 4 = 36  X + X − X −2 1 7 −  P X 1 2X −22X > 73 )= P 1 2 3 >  = P Z >1 = 0.1597  36 36  3. There are two methods to assemble a computer. The plant manger randomly selected 10 workers to assemble the computer using method A, and independently another 10 workers to assemble the computer using met hod B. The sample mean of the assembly time is 18 minutes for method A and 14 minutes for method B. The sample standard deviation is 3 minutes for method A and also 3 minutes for method B. To test H 0 µ 1 µ =22 ,H 1 µ 1 µ >22 , µ1 is the average assembly time in method A, and µ is the average assembly time in method B. The test statistic is 2 (a) Z =1.49 √ (b) t =1.49 (c) µ1− µ 2 2 (d)µ 1 µ 2 2 (e) α = 0.05 2 4. Let X1 and X2 be the means of two independent random samples, each of size n, from the respective populatioN µ ,σ 2) and N µ ,σ 2 , where the common variance σ 2 is 1 2 known. If the 90 percent confidence interval µ1− µ2 is X1− X 2σ /5 , the value of n is closest to (a) 25 (b) 50 (c) 75 (d) 104 √ (e) 136 5. A simple regression model is assumed for relating the price of grapefruit X (in dollars) to quantities of grapefruit demanded Y . Data for five months is provided below and some calculations are given. X Y X − X X − X )2 Y −Y (X − X Y −Y ) 0.1 50 -0.2 0.04 22 -4.4 0.2 30 -0.1 0.01 2 -0.2 0.3 30 0 0 2 0 0.4 20 0.1 0.01 -8 -0.8 0.5 10 0.2 0.04 -18 -3.6 Total 1.5 140 0 0.10 0 -9.0 Mean 0.3 28 0 0.02 0 -1.8 The slope of the estimated simple regression line is equal to √ (a) –90 (b) 90 (c) –55(d) 55 (e) none of these Questions 6-7, use the following information. If X1,X2,...,50and Y1,Y2,...50be two independent random samples from a population with mean µ and standard deviation 1. 6. Which one of the following has an approximately standard normal distribution? 1 1 (a)X −Y (b) (X −Y ) (c)2 X −Y ) (d) (X −Y ) √ (e)5 X −Y ) 2 5 Solution:X isN mean = µ, variance = 1 ; Y is N mean = µ, variance = 1   50  50  X −Y is N ean = 0, variance =1 1 = N ean = 0, variance = 1   50   25 Z = X −Y − mean = X − Y−0 = 5(X −Y )is the standard normal distribution. StdDev 1/25 7. The probabilityP X −Y > 0.2)is closest to (a) 0.001 (b) 0.005 (c) 0.0062 √ (d) 0.1587 (e) 0.4287 Solution:P X −Y > 0.2 )=P 5 X −Y >1 = P Z >1 = 0.1587 3 Questions 8-10, use the following information. In a Gallup poll of 1600 Canadians in year 2006, 720 of them answered “yes” to the question, “Is there any area in the city – where you would be afraid to walk alone at night?” In an earlier poll of 1600 Canadians in year 2001, only 672 had answered “yes”. 8. The margin of error in calculating a 95% confidence interval for the difference in proportions who are afraid is closest to (a) 0.0175 (b) 0.0288 √ (c) 0.03434 (d) 0.0475 (e) 0.0525 9. The p-value for testing the hypothesis of no change in proportions is closest to (a) 0.0175 (b) 0.0245 (c) 0.0375 (d) 0.0436 √ (e) 0.0872 10. At a significance level 0.05, which one of the following statement is the best conclusion? (a) The change from 42% to 45% is statistically significant. √ (b) The change from 42% to 45% is not statistically significant. (c) The change from 42% to 45% is not conclusive. 4 Questions 11-13, use the following information. The number of auto parts that a company produces follows a normal distribution. The company has two production lines: Line 1 produces an average of 75 parts per day with a standard deviation of 20. Line 2 produces an average of 65 parts per day with a st andard deviation of 21. 11. If Line 1 and Line 2 are independent, the probability that Line 1 produces more parts than Line 2 in any single day is closest to (a) 0.1587 (b) 0.3413 (c) 0.5 √(d) 0.6331 (e) 0.7794 12. If Line 1 and Line 2 are independent, what is the probability that the average number of parts produced by Line 1 is greater than that produced by Line 2 in the next 5 days? (a) 0.1587 (b) 0.3413 (c) 0.5 (d) 0.6331 √ (e) 0.7794 Solution: Let X 1 X 2 follows a normal distribution with mean 75− 65 =10 and standard 2 2 deviation 20 + 21 = 841 . 5 5 5 P X 1 > X 2)=(P X 1− X 2> 0 = P  > 0−10   =P Z > −0.771 = 0.7794  841/5  13. Suppose Line 1 and Line 2 are not independent a nd the correlation coefficient between the numbers of parts produced is 0.5. Let D be the difference between the number of parts produced by Line 1 and Line 2 in a day. The variance of D is closest to (a) 400 (b) 415 √ (c) 421 (d) 441 (e) 460 Questions 14-16, use the following information. A random sample of 4 market survey data is coded and given below: x y 1 7 1 8 2 6 4 3 Total 8 24 Assume a straight line regression modely = β + β x +ε . 0 1 14. The slope of the estimated regression equation is equal to 3 1 1 3 √ (a) − (b) − (c) (d) (e) none of these 2 2 2 2 5 15. In testing the significance of the model, the value of the test statistic is (a) −5.15 √ (b) − 7.35 (c) 5.15 (d) 7.35 (e) none of these 16. The sample correlation coefficient betweenx and y is closest to √ (a) − 0.98 (b) − 0.85 (c) 0.85 (d) 0.98 (e) 0.95 6 Part II. Show your work in all questions. 17. (10 marks) In order to compare the means of two populations, two large independent random samples are selected from each population with the following results: Sample 1 Sample 2 Size n n Mean 5,275 5,240 Standard Deviation 150 200 In testing the hypothesiH 0 µ 1 µ =20 versus H a µ 1 µ >20 , the p- value is 0.025. Find the value of n. (10 marks) X − X 5275 5240 35 35 n 7 n Solution: The test statistiZ =s 1 2 = = = = S 2 S2 1502 2002 62500 250 50 1 + 2 + n n1 n2 n n 2 Since p-value =P Z > 7 n = 0.025, we have7 n =1.96 . n = 98 =196 .  50  50  7    18. (10 marks) Let X and Y be the means of two independent random samples, each sample size n, from a distribution with meµn and standard deviatioσ . Assume n is large. Find the  σ σ  value of n such thP−
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