Midterm 2007 - Jun 26, 2007

Midterm 2007 - Jun 26, 2007 - PART 1 of 2: E00220 Midterm...

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Unformatted text preview: PART 1 of 2: E00220 Midterm Test (June 28, 2007) Page 2 of 8 For the 20 questions, choose the best answer and mark it on the SCANTRON form. Choice (E) “Don’t know" is never the correct answer, but you obtain partial credit for choosing it. (1) Which of the following is a continuous probability distribption? (D) (A)Bernoulli> dacm {QMOIM Viv/[W (B) Binomial (C) Poisson (D) Uniform (E) Don't know (2) in measuring the strength of a positive linear relationship between X and Y, which of the following affect the magnitude \of the coefficient of correlation? (C) I. How scattered the points are about the least suares line H. How steep the least squares line is ( 111. How X and Y are measured: their units of measurement (A) I. (B) II. (C) I. and II. (D) I., II. and III. (E) Don’t know (3) Which kind of data often causes an endogeneity bias? (C) (A) Experimental (B) Natural experiment (C) Observational (D) Cross sectional (E) Don't know (4) For a sample taken from negatively skewed population, which of the following statements about the sample mean would be expected? (D) (A) The sample mean is larger than the sample median (B) The sample mean is larger than the sample 45Ih Lpercentile ) The sample mean is smaller than the sample 10 percentile (E The sample mean is smallerthan the sample 55th percentile Don’t know PART 1 of 2: E00220 Midterm Test (June 28, 2007) Page 3 of B (5) For a sample of size 100 taken from a Poisson population with it = , what is the distribution ofthe sample mean? (D) l a! 5 t1. I 6:5» S W, Moisson with x1 = 1 1 , i (an/Poisson with ,1 = 0.01 (3“ ~/ (0) Normal with )1 =1 and 02 =1 IOO (0) Normal with )1 =1 and 0'2 = 0.01 (E) Don't know 0.0%) (6) Considering a bell shaped population with a mean of $4 2(2) and a standard deviation of $10,779, what is the population 515l percentile? (A) l 20,‘ - " _ ( -' - q (A) $41,470 0 .0 27 f; ()3: fiw. (B) $43,950 i ‘ 1;» (C) $66,316 ’Ut?}0\ [a 11% (D) $68,957 ,. it; 3:3! . ‘ (E) Don’t know x : 0.0! 7 4H 6.2 . : Z (7) Which of the following describe observational daa? (B) To study the effect that product placement on grocery store shelves has on 1. customer purchasing behavior, a grocery store rotates product placements and records changes in customer behavior. To study the link between herbal supplements and heart disease, a researcher 11. selects a random sample of adults and asks about their intake of herbal supplements and measures their heart health. (A) Only I. (B) Only II. (C) Both 1. and II. (D) Neither I. nor II. (E) Don't know (8) X measures the negotiated selling price in dollars for a new Toyota Camry in Toronto. The population distribution ofX is normal with a mean of $23,000. What is the probability that for a randomly selected deal the buyer pays a price more than one standard deviation above average? (A) (A) 0.1587 (B) 0.1915 (0) 0.2301 (D) 0.3413 (E) Don’t know PART 1 of 2: EC0220 Midterm Test (June 28, 2007) Page 7 of 8 PART 2 of 2: E00220 Midterm Test (June 28, 2007) Page 2 of 6 , , , , _ , , Problems: Write answers clearly, concisely, and completely on these test papers. Show your work. (16) The Width of the confidence Interval estimator of the population mean w1|| Increase With an increase in WhiCh 0fthe fOHOWinQ? (A) (1) [8 points] Consider this tabulation. X | Free! Percent Cum Is it reasonable to infer that this sample ----- *-+ --------------------------------- —— is from a Poisson population? If not. 0 | 67 23 - 59 23 59 - show why not. If so, show why and l | 91 32.04 55 53 11‘ on'samp mg error give reasonable values for the Poisson 2 | 54 22 - 54 7B 17 parameter“). 3 I 41 14 . 44 92 61 (A) I. 4 I 13 4.58 97 18 (5)11. 5 I 7 2 46 99.55 (C) Both 1. and II. 8 l 1 o 35 100 00 (D) Neither I. nor II. ————— ——+ _________________________________ __ (E) Don't know Total l 284 100.00 Solution: > For Questions (17) - (18): Consider the following histogram of a random sample. )7 = 0.2359 * 0 + 0.3204 *1+ 0.225;” 2 + 0.1444 *3 + 0.0458 )l: 4+ 0.0245 * 5 + 0.0035 1: 8 :154 Z /\: Let's see what the probability distribution would be if 1 = 1.54 (by using Poisson probability formula): it seems reasonable to infer that the sample?anfie from a Poisson population with AM .54. The probabilities are not exactly equal to the relative frequencies in the sample, but they are close. We (A) 57.2 6 : know that there will be sampling noise in the sample so it is not reasonable to expect equality. (B) 67.7 a ' (C) 68.3 ; 0 13C? 33 (Note: Some students may calculate the sample variance, which is 1.73, and note that it is close to (D) 68.6 I 5 00‘ W the sample mean of 1.54. However, this is an incomplete answer. Just because the mean and (E) Don’t know 0‘ variance are close does n_ot necessarily mean that the shape is Poisson: it is necessary but not sufficient. Hence, substantial points should be deducted if the above table is missing.) (18) Which is the approximate 95% confidence interval estimator of the population mean? (D) (A) (20. 40) Jr , 5’ B 21,39 v /‘ Eciizmsi 3° ~ HQ) it“ (D) (29, 31) J (E) Don’tknow ’ ,1 0.“ 04/ Page 3 of 6 PART 2 of 2: ECO220 Midterm Test (June 28, 2007) Page 5 of 6 PART 2 of 2: EC0220 Midterm Test (June 28, 2007) (4) [18 points] A population is Bernoulli distributed and the probability that X is equal to one is 0.35 (2) [6 points] Consider this box plot of a random and the probability that X is equal to zero is 0.65. sample drawn from a normal population. Compute the best approximate value of the sample mm (to the neareSt hundre‘fih)' (a) [9 points] For a sample size of 3, find the sampling distribution of the mean. Graph it. Use it to find the probability that the sample mean is greater than 0.25. Solution: One approach is to recognize that we can use the Binomial probability formula. Another is to do this by hand (as we did in Lecture 18), which is the solution given h r : m.- (055)2(035)l = 0.1479 Solution: This can be solved by recognizing that the mean and median (because normal opulation. which is symmetric) are 6 (center line of box) and any of the following facts from the bo < plot (student must understand the meaning of the box edges, 25m and 75‘“ percentiles): P(6 < X < 7) = 0.25 that l (0.65)‘ (0.35)2 = 0.0796 P(S < X <6): 0.25 14664 (055)2(035)l = 0.1479 P(X < 5) = 0.25 (0.65)'(0.35)2 = 0.0796 P(X > 7) = 0.25 I .. V (0.65)‘(0.35)2 = 0.0795 2/3 7 Here is one sample solution making use ofthe fact that P(X > 7,).=-0..25 and the fact that the mean is (03 5)“ = 0,0429 about 6. P(X>7)=0.25 .. . . . . . . , P(Z > ?) = 0.25 We can calculate the requested probability by ustng the sampling distributto weJust found. ? =0.675 P(X>0.25) =1—0.2746 =0.7254 \/ 7 —)7 , q , ‘ , 1 S =, fl ‘2?) OJQQGQJ flth WM 0.4%; .2 2 X “M 1 X70 7 — 6 / —— = 0.675 s s ~‘ 1.48 (Note: The value from the standard normal table, 0.675, does not need to be extrapolated for full marks: 0.68 or 0.67 is good enough.) PART 2 of 2: EC0220 Midterm Test (June 28, 2007) Page 6 of 6 (b) [9 points] For a sample size of 300, find the sampling distribution of the mean. Graph it. Use it to find the probability that the sample mean is greater than 0.25. Solution: A sample size of 300 is sufficiently large (300 > 30) so that we can use the Central Limit Theorem (CLT) to figure out that the sampling distribution of the sample mean is bell shaped (normal). All that remains is to find the mean and standard deviation of the distribution. 2 _ V[X]=°'_ E[X]=;1 n ;z=0.35 HEPM 300 ELY]: 0.35 _ I V[X]= 0.000758 50m = 00275 We can calculate the requested probability by using the sampling distribution we just found: 0.25—0.35 P(X‘> 0.25)=P Z> 0.0275 )=P(Z>—3.64)~t ...
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Midterm 2007 - Jun 26, 2007 - PART 1 of 2: E00220 Midterm...

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