(c) P(BC c ) = One point was lost for multiplying. If, however, you made this same mistake in (b) and (c) you lost the point only once.

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1 Solutions to First Midterm Exam, Stat 371, Fall 2010 There are two, three or four versions of each problem. The problems on your exam comprise a mix of versions. As a result, when you examine the solutions below, first make sure you are reading the solution that corresponds to the version on your exam. 1. Problem 1. The total number of females is 510 in Version 1; and 520 in Version 2. Grading: Each part is worth 1.5 points. This problem examines whether you are good at ignoring extraneous information. Event A never appears in any question, so you can ignore it. The Year at school can be condensed to B or not B. Also, this question is far from my best effort. Both parts (b) and (c) require you to compute P(BC c ); this was dumb by me. Also, it wrong to calculate P(BC c ) by multiplying; you multiply when you have i.i.d. trials, not when you have two features. Unfortunately, multiplying almost gives the correct answer; again, my bad planning. I begin by creating the following table of counts. Sex B c B Total Female (C) Male (C c ) Total (a) P(B c ) = (b) P(B or C c ) = = One point was lost for failing to subtract the probability of the intersection. (c) P(BC c ) = One point was lost for multiplying. If, however, you made this same mistake in (b) and (c) you lost the point only once. I begin by creating the following table of counts. Sex B c B Total Female (C) Male (C c ) Total (a) P(B c ) = (b) P(B or C c ) = = One point was lost for failing to subtract the probability of the intersection. (c) P(BC c ) = One point was lost for multiplying. If, however, you made this same mistake in (b) and (c) you lost the point only once. 2. Problem 2. Your version of problem 2 is the same as your version of problem 1. Grading: One-half point per entry in each of the three spaces. Put k 1 = 4 in the top box; leave the left box empty; put 0.10 in the right box. If you put 0.10 in the left box, you lost one point, for having the wrong entry in both the left and right boxes. If you repeat this error in Problem 3, you lost only 1/2 point for the second occurrence of this same error. Put k 1 = 7 in the top box; leave the left box empty; put 0.05 in the right box. If you put 0.05 in the left box, you lost one point, for having the wrong entry in both the left and right boxes. If you repeat this error in Problem 3, you lost only 1

2 1/2 point for the second occurrence of this same error. 3. Problem 3. Your version of problem 3 is the same as your version of problem 1. Grading: The same as problem 2. Put k 1 = 8 in the top box; put in the left box; leave the right box empty. Put k 1 = 11 in the top box; put in the left box; leave the right box empty. 4. Problem 4. Your version of problem 4 is the same as your version of problem 1. Grading: Your answer for each α is worth 1/2 point. Reject for α = 0.05 and reject for α = Reject for α = Problem 5. There are four versions of problem 5. Each version has the same four tables, just in different orders. Thus, in the answers, I will identify the correct table by its total number of successes in its first segment: 100, 120, 150 or 200. Grading: In parts (a) (c) you lose one point for each table that you label incorrectly. Also, part (d) is worth one point. Thus, if everything is wrong, it appears that you would lose 5 points; but, since the problem is worth only 4 points, 4 is the maximum you can lose. (a) Two tables: 100 and 200. (b) One table: 150. (c) One table: 120. (d) Two tables: 150 and in Version 4. Grading: One point for each part. (a) 600(0.18) = 108. (b) (80 108) 2 /108 = (a) 700(0.21) = 147. (b) ( ) 2 /147 = (a) 800(0.23) = 184. (b) ( ) 2 /184 = Version 4: (a) 900(0.27) = 243. (b) ( ) 2 /243 = Problem 7. There are two versions of this problem. The value of p is: 0.62 in Version 1; and 0.67 in Version 2. Grading: Each part is worth 3 points. (a) P(FSS) = qpp = 0.38(0.62)(0.62) = If you put factorials in your answer, you lost 1.5 points. (b) P(Y = 1) = 4! 1!3! (0.1461)(0.8539)3 = If you did not have the factorials, you lost 1.5 points. If you used p = 0.62 instead of you lost 1.5 points. 6. Problem 6. There are four versions of this problem. The value of n is: 600 in Version 1; 700 in Version 2; 800 in Version 3; and 2

3 (a) P(SF) = pq = 0.67(0.33) = If you put factorials in your answer, you lost 1.5 points. (b) P(Y = 2) = 5! 2!3! (0.2211)2 (0.7789) 3 = If you did not have the factorials, you lost 1.5 points. If you used p = 0.67 instead of you lost 1.5 points. 8. Problem 8. Your version of problem 8 is the same as your version of problem 7. Grading: Each part is worth 3 points. (a) The answer is: X 1 X X X X X 2 X X X 4 X 6 8 (b) There are three ways to get a total of 18: 8,8,2 or 8,6,4 or 6,6,6. There are 3 arrangements of 8,8,2; 6 arrangements of 8,6,4; and 1 arrangement of 6,6,6. Thus, there are 10 possibilities and the answer is 10/125. (a) The answer is: X 1 X X 2 X 3 X 6 X X 9 X X X (b) There are three ways to get a total of 15: 9,6,0 or 9,3,3 or 6,6,3. There are 6 arrangements of 9,6,0; 3 arrangements of 9,3,3; and 3 arrangements of 6,6,3. Thus, there are 12 possibilities and the answer is 12/ Problem 9. There are three versions of this problem. The p in the binomial is: 0.42 for Version 1; 0.44 for Version 2; and 0.45 for Version 3. Grading: Part (c) is worth 1/2 point; other parts are worth one point each. (a) np = 2300(0.42) = 966. (b) npq = 966(0.58) = (c) Between. (d) 930.5; (a) np = 2500(0.44) = (b) npq = 1100(0.56) = (c) Between. (d) ; (a) np = 2700(0.45) = (b) npq = 1215(0.55) = (c) Between. (d) ; Problem 10. Your version of problem 10 is the same as your version of problem 9. 3

4 First, ˆp = 495/1500 = The 90% CI is ± [0.33(0.67)]/1500 = ± = [0.310, 0.350]. First, ˆp = 455/1300 = The 98% CI is ± [0.35(0.65)]/1300 = ± = [0.319, 0.381]. First, ˆp = 407/1100 = The 99% CI is ± [0.37(0.63)]/1100 = ± = [0.332, 0.408]. 11. Problem 11. Your version of problem 11 is the same as your version of problem 9. As noted on the exam, if you used any normal approximation you received no credit. The binomial can be approximated by the Poisson with parameter θ = np. Thus, a CI for θ is a CI for np. In fact, the last four lines on page 2 of the notes, tells you exactly how to do this problem. Grading: I gave very little partial credit. The inequality θ 7.75 becomes np 7.75 becomes p 7.75/5000 becomes p The inequality θ 6.30 becomes np 6.30 becomes p 6.30/2000 becomes p The inequality θ becomes np becomes p 10.52/4000 becomes p Problem 12. Your version of problem 12 is the same as your version of problem 9. The number of successes during an interval of t hours has a Poisson distribution with parameter θ = tλ, where λ is the rate of the process. First, you use the data to obtain a CI for θ, then you divide by t to obtain the CI for λ. The CI is 250 ± = 250 ± 36.8 = [213.2, 286.8]. Divide each endpoint by t = 2.5 to obtain [85.3, 114.7]. The CI is 240 ± = 240 ± 39.9 = [200.1, 279.9]. Divide each endpoint by t = 3.5 to obtain [57.2, 80.0]. The CI is 230 ± = 230 ± 25.0 = [205.0, 255.0]. Divide each endpoint by t = 4.5 to obtain [45.6, 56.7]. 13. Problem 13. There are four versions of problem 13. Each version has the same four tables, just in different orders. Thus, in the answers, I will identify the correct table by its total number of successes in its first row: 140, 180, 241 or 339. Grading: Parts (a) (c) are worth 3 points total; you lose one point for each table that you label incorrectly, to a maximum of 3 points that can be lost. Parts (d) (f) are the same. 4

5 (a) One table: 339. (b) One table: 241. (c) Two tables: 140 and 180. (d) Two tables: 140 and 339. (e) One table: 241. (f) One table: Problem 14. There are three versions of this problem. The number of rolls on Monday is: 120 for Version 1; 130 for Version 2; and 150 for Version 3. Grading: You lost one point if you used p = 1/6 instead of p = 2/6. You lost 1.5 points if you did a CI instead of a PI. You lost 1.5 points if you did the PI for unknown p. First, m = 270 and p = 2/6; thus, mp = 90. The PI is 90 ± (4/6) = 90 ±15 = [75, 105]. First, m = 240 and p = 2/6; thus, mp = 80. The PI is 80 ± (4/6) = 80 ±17 = [63, 97]. First, m = 210 and p = 2/6; thus, mp = 70. The PI is 70 ± (4/6) = 70 ±18 = [52, 88]. 15. Problem 15. Your version of problem 15 is the same as your version of problem 14. First, m = 480, n = 120, ˆp = 0.25 and mˆp = 120. Thus, the PI is 120± (0.75) 1 + (480/120) = 120 ± 49 = [71, 169]. First, m = 520, n = 130, ˆp = 0.20 and mˆp = 104. Thus, the PI is 104± (0.80) 1 + (520/130) = 104 ± 52 = [52, 156]. First, m = 500, n = 150, ˆp = 0.22 and mˆp = 110. Thus, the PI is 110± (0.78) 1 + (500/150) = 110 ± 32 = [78, 142]. 16. Problem 16. There is only one version of this problem. The number of successes on Tuesday is 65 and the number on Tuesday and Wednesday combined is 115. Whether the PI s are correct or not depends on your version of problem 14 (and 15). Grading: One point for each part. Fiona s PI is incorrect b/c it does not contain 65; Bob s is correct b/c it does contain 115. Fiona s PI is correct b/c it contains 65; Bob s is correct b/c it contains 115. Fiona s PI is correct b/c it contains 65; Bob s is correct b/c it contains Problem 17. There are three versions of this problem. Events A and B: both refer to successes in Version 1; refer to a success (A) and a failure (B) in Version 2; and both refer to failures in Version 3. Grading: You lost one-half point for failing to 5

6 evaluate factorials. Otherwise, nearly all answers were either correct or received 0 points. Taken together, the four conditions occur if, and only if: The first trial is an S. The next three trials yield a total of exactly two S s. The next four trials yield a total of exactly two S s. The last trial is an S. Thus, the probability is: p(3p 2 q)(6p 2 q 2 )p = 18p 6 q 3. Taken together, the four conditions occur if, and only if: The first trial is an S. The next three trials yield a total of exactly one S. The next three trials yield a total of exactly two S s. The last trial is an F. Thus, the probability is: The next two trials yield a total of exactly one S. The last trial is an F. Thus, the probability is: q(10p 3 q 2 )(2pq)q = 20p 4 q Problem 18. Your version of problem 18 is the same as your version of problem 17. Grading: One point was lost for confusing the future and past in computing c. First, c = 3/5 = 0.60 and x = 200. Thus, the PI is 0.6(200) = 120± ( ) = 120±23 = [97, 143]. First, c = 6/8 = 0.75 and x = 300. Thus, the PI is 0.75(300) = 225± ( ) = 225±39 = [186, 264]. First, c = 8/10 = 0.80 and x = 250. Thus, the PI is 0.80(250) = 200± ( ) = 200±44 = [156, 244]. p(3pq 2 )(3p 2 q)q = 9p 4 q 4. Taken together, the four conditions occur if, and only if: The first trial is an F. The next five trials yield a total of exactly three S s. 6