Solutions to First Midterm Exam, Stat 371, Spring those values: = Or, we can use Rule 6: = 0.63.

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1 Solutions to First Midterm Exam, Stat 371, Spring 2010 There are two, three or four versions of each question. The questions 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 percentage of females and juniors is 18 in Version 1; 22 in Version 2; and 17 in Version 3. Each part is worth one I begin by creating the following table. Sex Junior Not Junior Total Female Male Total (a) The probability of a female is 0.40 from the table. (b) The event Junior or female corresponds to three cells; so, we can add those values: = Or, we can use Rule 6: = (c) The probability of male and not a junior is 0.43 from the table. I begin by creating the following table. Sex Junior Not Junior Total Female Male Total (b) The event Junior or female corresponds to three cells; so, we can add those values: = Or, we can use Rule 6: = (c) The probability of male and not a junior is 0.37 from the table. I begin by creating the following table. Sex Junior Not Junior Total Female Male Total (a) The probability of a female is 0.36 from the table. (b) The event Junior or female corresponds to three cells; so, we can add those values: = Or, we can use Rule 6: = (c) The probability of male and not a junior is 0.39 from the table. 2. Problem 2. The ratio is: 1:3:6:2 in Version 1; 2:5:4:3 in Version 2; and 4:2:1:6 in Version 3. Let a > 0 be unknown. The probabilities are a, 3a, 6a and 2a. These must sum to one; thus, 12a = 1, which implies that a = 1/12. Thus, the probabilities are 1/12, 3/12, 6/12 and 2/12. Let a > 0 be unknown. The probabilities are 2a, 5a, 4a and 3a. These must sum to one; thus, 14a = 1, which implies that a = 1/14. Thus, the probabilities are 2/14, 5/14, 4/14 and 3/14. (a) The probability of a female is 0.55 from the table. 1

2 Let a > 0 be unknown. The probabilities are 4a, 2a, a and 6a. These must sum to one; thus, 13a = 1, which implies that a = 1/13. Thus, the probabilities are 4/13, 2/13, 1/13 and 6/ Problem 3. The number of categories is: 9 for Version 1; 11 for Version 2; 5 for Version 3. Each box is worth 1/2 points. Put k 1 = 8 in the top box; leave the left box empty; put 0.01 in the right box. Put k 1 = 10 in the top box; leave the left box empty; put 0.05 in the right box. Put k 1 = 4 in the top box; leave the left box empty; put 0.10 in the right box. 4. Problem 4. The number of categories is: 7 for Version 1; and 10 for Version 2. Each box is worth 1/2 points. Put k 1 = 6 in the top box; put in the left box; leave the right box empty. Put k 1 = 9 in the top box; put in the left box; leave the right box empty. 5. Problem 5. The P-value is: for Version 1; and for Version 2. Your answer for each α is worth 1/2 points. We reject for all α s that are larger than the P- value Reject for α = 0.05 and reject for α = Reject for α = Problem 6. Your version of problem 6 is equal to your version of problem 5. Each part is worth one (a) Table 1. (b) Table 3. (c) Tables 2 and 4. (d) Tables 1, 2 and 3. (a) Table 4. (b) Table 2. (c) Tables 1 and 3. (d) Tables 1, 2 and Problem 7. The value of p is: 0.70 for Version 1; 0.40 for Version 2; and 0.80 for Version 3. Each part is worth 3 points. (a) P(SFS) = pqp = 0.7(0.3)(0.7) = If you put factorials in your answer, you lost 1.5 points. (b) P(Y = 2) = 4! 2!2! (0.147)2 (0.853) 2 = If you did not have the factorials, you lost 1.5 points. If you used p = 0.70 instead of you lost 1 (a) P(FSF) = qpq = 0.6(0.4)(0.6) = If you put factorials in your answer, you lost 1.5 points. (b) P(Y = 1) = 4! 1!3! (0.144)(0.856)3 = If you did not have the factorials, you lost 1.5 points. If you used p = 0.40 instead of you lost 1 2

3 (a) P(SFSS) = pqpp = 0.8(0.2)(0.8)(0.8) = If you put factorials in your answer, you lost 1.5 points. (b) P(Y = 1) = 3! 1!2! (0.1024)(0.8976)2 = If you did not have the factorials, you lost 1.5 points. If you used p = 0.80 instead of you lost 1 8. Problem 8. Your version of problem 8 is equal to your version of problem 7. Each part is worth 3 points. (a) P(X 1 = 2X 2 ) = P(0, 0)+P(4, 2) = (0.1)(0.1) + (0.7)(0.2) = = Many people overlooked 0,0; they lost one (b) P(X 1 + X 2 = 4) = P(0, 4) + P(4, 0) + P(2, 2) = 2(0.1)(0.7) + (0.2)(0.2) = = Many people forgot the coefficient of 2 and obtained 0.11; they lost one (a) P(X 1 = 2X 2 ) = P(0, 0)+P(6, 3) = (0.3)(0.3) + (0.2)(0.5) = = Many people overlooked 0,0; they lost one (b) P(X 1 + X 2 = 6) = P(0, 6) + P(6, 0) + P(3, 3) = 2(0.3)(0.2) + (0.5)(0.5) = = Many people forgot the coefficient of 2 and obtained 0.31; they lost one (a) P(X 1 = 2X 2 ) = P(0, 0)+P(8, 4) = (0.5)(0.5) + (0.1)(0.4) = = Many people overlooked 0,0; they lost one (b) P(X 1 + X 2 = 8) = P(0, 8) + P(8, 0) + P(4, 4) = 2(0.5)(0.1) + (0.4)(0.4) = = Many people forgot the coefficient of 2 and obtained 0.32; they lost one 9. Problem 9. The p in the binomial is: 0.30 for Version 1; 0.20 for Version 2; and 0.40 for Version 3. Part (c) is worth 1/2 point; other parts are worth one point each. (a) np = 2100(0.3) = 630. (b) npq = 630(0.7) = 21. (c) Between. (d) 600.5;

4 (a) np = 1600(0.2) = 320. (b) npq = 320(0.8) = 16. (c) Between. (d) 300.5; (a) np = 600(0.4) = 240. (b) npq = 240(0.6) = 12. (c) Between. (d) 210.5; Problem 10. Your version of problem 10 is equal to your version of problem 9. First, ˆp = 480/1200 = The 99% CI is ± [0.4(0.60)]/1200 = ± = [0.364, First, ˆp = 720/1600 = The 98% CI is ± [0.45(0.55)]/1600 = ± = [0.421, First, ˆp = 700/2000 = The 95% CI is ± 1.96 [0.35(0.65)]/2000 = ± = [0.329, Problem 11. Your version of problem 11 is equal to your version of problem 9. As noted on the exam, if you used the snc 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. The inequality θ becomes np becomes p /2000 becomes p The inequality θ becomes np becomes p /800 becomes p The inequality θ becomes np becomes p /1200 becomes p Problem 12. Your version of problem 12 is equal to 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 300 ± = 300 ± 28.5 = [271.5, 328.5]. Divide each endpoint by t = 1.5 to obtain [181.0, 219.0]. The CI is 240 ± = 240 ± 30.4 = 4

5 [209.6, 270.4]. Divide each endpoint by t = 2.5 to obtain [83.8, 108.2]. The CI is 600 ± = 600 ± 57.0 = [543.0, 657.0]. Divide each endpoint by t = 3.5 to obtain [155.1, 187.7]. 13. Problem 13. There are four versions of problem 13, but they are just permutations of the tables 5 8. Instead of typing four answers below, I will refer to a table by its value for the grand total, which is 380 for one of the tables. Each part is worth one (a) The table with 519. (b) 556. (c) Two tables: 160 and 380. (d) 380. (e) Two tables: 519 and 556. (f) Three tables: 160, 519 and Problem 14. The rate per hour is: 20 in Version 1; and 30 in Version 2. The θ for the process to be observed is the rate multiplied by the number of hours. First, θ = 20(7.5) = 150. Thus, the PI is 150 ± = 150 ± Thus, the PI is [130, 170]. First, θ = 30(6) = 180. Thus, the PI is 180 ± = 180 ± Thus, the PI is [154, 206]. 15. Problem 15. Your version of problem 15 is equal to your version of problem 14. Part (a) is worth one point and (b) is worth 1.5 points. (a) E 2 = np 20 = 500(0.22) = 110. (b) (90 110) 2 /110 = (a) E 2 = np 20 = 600(0.3) = 180. (b) ( ) 2 /180 = Problem 16. Your version of problem 16 is equal to your version of problem 14. The two versions are permutations of the same four CI s. In the solution below, I identify each CI by its lower bound. The center of each CI is equal to ˆp. Thus, we must find the interval with a unique center; it is the interval ; thus, the other three intervals are the CI s. Next, we remember: the larger the confidence level, the wider the CI. Thus, is 80%; is 90%; and is 99%. 17. Problem 17. The number of rolls on Monday is: 100 for Version 1; 200 for Version 2; and 150 for Version 3. First, m = 240 and p = 2/6; thus, mp = 80. The PI is 80 ± (4/6) = 80 ± 14 = [66, 94]. First, m = 300 and p = 2/6; thus, mp = 100. The PI is 100± (4/6) = 100±13 = [87, 113]. 5

6 First, m = 180 and p = 2/6; thus, mp = 60. The PI is 60 ± (4/6) = 60 ± 8 = [52, 68]. 18. Problem 18. Your version of problem 18 is equal to your version of problem 17. First, m = 600, n = 100, ˆp = 0.44 and mˆp = 264. Thus, the PI is 264± (0.56) 1 + (600/100) = 264 ± 75 = [189, 339]. First, m = 500, n = 200, ˆp = 0.4 and mˆp = 200. Thus, the PI is 200± (0.6) 1 + (500/200) = 200 ± 40 = [160, 240]. First, m = 450, n = 150, ˆp = 0.42 and mˆp = 189. Thus, the PI is 189± (0.58) 1 + (450/150) = 189 ± 34 = [155, 223]. 19. Problem 19. Your version of problem 19 is equal to your version of problem 17. Taken together, the four conditions occur if, and only if: The next two trials yield a total of exactly one S. The last trial is an F. Thus, the probability is: p(3p 2 q)(2pq)q = 6p 4 q 3. Taken together, the four conditions occur if, and only if: The first trial is an F. The next four trials yield a total of exactly three S s. The next two trials yield a total of exactly one S. The last trial is an F. Thus, the probability is: q(4p 3 q)(2pq)q = 8p 4 q 4. Taken together, the four conditions occur if, and only if: The first trial is an S. The next four trials yield a total of exactly two S s. The next three trials yield a total of exactly one S. The last trial is an S. Thus, the probability is: p(6p 2 q 2 )(3pq 2 )p = 18p 5 q 4. The first trial is an S. The next three trials yield a total of exactly two S s. 6