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Final Examination: Spring 2015 OL4 / US2 Instructor ___Kumud Altmaye__

Answer Sheet

Instructions:

This is an open-book exam. You may refer to your text and other course materials as you work on the exam, and you may use a calculator.

Record your answers and work in this document.

Answer all 25 questions. Make sure your answers are as complete as possible. Show all of your work and reasoning. In particular, when there are calculations involved, you must show how you come up with your answers with critical work and/or necessary tables. Answers that come straight from programs or software packages will not be accepted. If you need to use software (for example, Excel) and /or online or hand-held calculators to aid in your calculation, please cite the source and explain how you get the results.

When requested, show all work and write all answers in the spaces allotted on the following pages. You may type your work using plain-text formatting or an equation editor, or you may hand-write your work and scan it. In either case, show work neatly and correctly, following standard mathematical conventions. Each step should follow clearly and completely from the previous step. If necessary, you may attach extra pages.

You must complete the exam individually. Neither collaboration nor consultation with others is allowed. Your exam will receive a zero grade unless you complete the following honor statement.

Please sign (or type) your name below the following honor statement:

I promise that I did not discuss any aspect of this exam with anyone other than my instructor. I further promise that I neither gave nor received any unauthorized assistance on this exam, and that the work presented herein is entirely my own. Name _Brandy L. Currie_ Date__10 May 2015__

Please sign (or type) your name below the following honor statement:

I promise that I did not discuss any aspect of this exam with anyone other than my instructor. I further promise that I neither gave nor received any unauthorized assistance on this exam, and that the work presented herein is entirely my own. Name _Brandy L. Currie_ Date__10 May 2015__

Record your answers and work. Problem Number | Solution | 1(25 pts) | Answers:(a) True because variance is the average of the squared difference from the mean; a variance of zero means that all values are the same as the mean which indicates they are identical.(b) False because there is no common region and therefore P(A and Ac) = 0, not 1.(c) True because normal distribution is symmetric around the mean; therefore the mean and median are always equal.(d) True because it takes a larger Z value to multiply with standard error to get the confidence interval at 99%.(e) False because a smaller significance level has a larger critical value.Work for (a), (b), (c), (d) and (e): | 2(5 pts) | Answer: Checkout Time (in minutes) | Frequency | Relative Frequency | 1.0 - 1.9 | 4 | 0.16 | 2.0 - 2.9 | 10 | 0.4 | 3.0 - 3.9 | 6 | 0.24 | 4.0 - 5.9 | 5 | 0.20 | Total | 25 | 1 |

Work: 25*0.4 = 10 25-(4+10+5)= 6 4/25 = 0.16 6/25 = 0.24 5/25 = 0.20 | 3(5 pts) | Answer: 44%Work:0.24+0.20=0.44 | 4(5 pts) | Answer: The class interval of “2.0 - 2.9”. This interval contains the middle value and the median is the middle most value in the set. The middle most value is 13 and would fall in this interval.Work: | 5(5 pts) | Answer: The mean would increase while the median stays the same. The mean is the average of the total and the increase in maximum value would increase the average. The median is the middle number and the number of observations is not changing, just a value.Work: | 6(5 pts) | Answer: 36Work:6*6 = 36. | 7(10 pts) | Answer: 1/3Work:P(X2>4 |X1 = even) = (6/36)/(18/36) = 1/3 | 8(5 pts) | Answer: Yes. A and B are independent because A is associated with one die and B with a completely different die.Work: | 9(5 pts) | Answer: (C) Both quizzes have the same value requestedWork:Quiz 1 = 95-40 = 55 Quiz 2 = 90-35 = 55 | 10(5 pts) | Answer: (A) Quiz 1Work:Q3 is lower for Quiz 2 (90) than Quiz 1 (95) | 11(5 pts) | Answer: (B) Quiz 2Work:The median for Quiz 2 is lower than 60 while Quiz 1 is right at 60, therefore more students must have scored less than 60 | 12(10 pts) | Answer: 0.22Work:P= 200/1000 + 100/1000 – 80/1000 = 0.22 | 13(10 pts) | Answer: 0.40Work:P(Psy|Stat) = (80/1000)/(200/1000) = 0.40 | 14(5 pts) | Answer: 720Work:use the permutation =10P3 = 720 | 15(15 pts) | Answer: (a) 6 (b) 34.26Work for (a) and (b): X | P(X) | X*P(X) | X2*P(X) | 100 | 0.1 | 10 | 1000 | 20 | 0.1 | 2 | 40 | 5 | 0.4 | 2 | 10 | -20 | 0.4 | -8 | 160 | | | | | Total | 1 | 6 | 1210 | a) sum of X*P(X) = 6 b) sum of (X2*P(X)) -Mean2 = 1210-62 = 1174sd = √(variance) = √(1174) = 34.26 | 16(20 pts) | Answer: (a) n = 10; p = 0.3; q = 0.7 (b) 0.9718 (c) 3Work for (a), (b) and (c) : a) q = 1-p =1-.3= 0.7 b) P(X ≥ 1) = 1-P(X=0) =1-0.710 = 0.9718 c) np = 10*0.3 = 3 | 17(10 pts) | Answer: 0.3413Work:P(10< X < 12) = P((10-10)/2 < Z< (12-10)/2) ) = P(0<Z < 1) = P(Z<1) – P(Z<0) = 0.8413 - 0.5000 = 0.3413 | 18(5 pts) | Answer: 11.349 ftWork:The 3rd quartile for a standard normal distribution = 0.6745 10+0.6745*2 = 11.349 | 19(5 pts) | Answer: 0.2Work:sd = 2/√(100) = 0.2 | 20(10 pts) | Answer: (1460.80, 1539.20)Work:Z(0.025) = 1.96 95% CI = Sample mean±Z0.025*SDSample Size=1500±1.96*300225= (1460.80, 1539.20) | 21(15 pts) | Answer: (a) 0.9 (b) 0.1841 (c) No. The P-value is > significance level so failed to reject the null hypothesis. There is insufficient evidence.Work for (a), (b) and (c):. a) t-statistic (Z) = p-p0p0*1-p0n=0.53-0.50.52225 = 0.9 b) P-value = P(Z> test statistic) = P(Z > 0.9) = 1- P(Z< 0.9) = 1-0.8159 = 0.1841 | 22(20 pts) | Answer: (a) H0:µd=0 and Ha:µd<0 (b) -2.7456 (c) 0.0258 (d) Yes. The p-value is < significance level so we reject the null hypothesis. There is sufficient evidence to support the claim.Work for (b) and (c): B) sample mean for difference= -1.4 sample sd of difference = 1.1402 sample size = 5 t-statistic = -1.41.14025 = -2.7456 C) sample size = 5 so df = 5-1 = 4 P-value = Pt4<-2.7456= 0.0258 | 23(10 pts) | Answer: (a) 0.7347 (b) 0.2029 (c) No. The p-value is > significance level so we fail to reject the null hypothesis. There is insufficient evidence to justify the rejection.Work for (a), (b) and (c): a) t-statistic = S12S22=122142= 0.7347 b) df = n1-1=31-1=30 df = n2-1=30-1=29 p-value = P(F(30,29) < 0.7347) = 0.2029 | 24(20 pts) | Answer: (a) Y = 0.5255+1.3220X (b) 5.8135Work for (a) and (b): | X | Y | X2 | Y2 | X*Y | | 0 | 1 | 0 | 1 | 0 | | 1 | 2 | 1 | 4 | 2 | | 3 | 3 | 9 | 9 | 9 | | 5 | 8 | 25 | 64 | 40 | | | | | | | Total | 9 | 14 | 35 | 78 | 51 | Average | 2.25 | 3.5 | 8.75 | 19.5 | 12.75 | a) b=xy-1nxyx2-1nx2=51-14*9*1435-924 =1.3220a=1ny-b1nx=3.5-1.3220*2.25 = 0.5255Y = 0.5255+1.3220X b) Y = 0.5255+1.3220*4 = 5.8135 | 25(15 pts) | Answers: (a) H0:pbrown=0.4; pyellow=porange=0.2; pgreen=ptan=0.1 And Ha:Atleast one proportion differs significantly (b) 10.65 (c) 0.0308 (d) No. P-value is < significance level thus we can reject the null hypothesis. There is sufficient evidence to reject the claim.Work for (b) and (c): B) Chi-sq = 42-40240+21-20220+12-20220+7-10210+18-10210 = 10.65 C) Df = 5-1=4.P-value = P(Chi-sq(4) > 10.65) = 0.0308 |…...

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