Math 2200 Fall 2014, Final Exam You may use any calculator. You may use a 4 6 inch notecard as a cheat sheet.

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1 1 Math 2200 Fall 2014, Final Exam You may use any calculator. You may use a 4 6 inch notecard as a cheat sheet. Warning to the Reader! If you are a student for whom this document is a historical artifact, be aware that the definitions and conventions on which some of the questions on this exam are based may differ from those adopted in your course. For example, you might be accustomed to using a different form of a test statistic in a hypothesis test. 1. Herodes Ficus, a dedicated philanthropist, makes donations to several charities regularly. Two of his favorites are Guiding Eyes for the Blind and Paws with a Cause. In any month, the probability that he will donate to the former is 0.8 and the probability that he will donate to the latter is 0.6. Because he is sufficiently wealthy to indulge his eleemosynary zeal with ease, his gifts to different charities are independent events. What is the probability that he donates to Guiding Eyes for the Blind and/or Paws with a Cause in a month? A 0.81 B 0.82 C 0.83 D 0.85 E 0.86 F 0.88 G 0.89 H 0.91 I 0.92 J 0.94 Solution. Let A be the even that Mr. Ficus donates to GEFTB in a month and let B be the event that Mr. Ficus donates to PWAC in a month. Then P(A = 0.8 and P(B = 0.6. Because A and B are independent events, P(A B = P(A P(B = (0.8(0.6 = Then Answer: I P(A and/or B = P(A B = P(A + P(B P(A B = = Suppose that A and B are events with P(A = 0.6, P(A B = 0.2, and P(A B = 0.4. This problem and the two that follow it concern three events, E, F, G, that are associated with A and B. Let E be the event that exactly one of the two events A and B occurs, let F be the event that at least one of the two events A and B occurs, and let G be the event that neither of the events A and B occurs. For starters, what is P(E? A 0.0 B 0.1 C 0.2 D 0.3 E 0.4 F 0.5 G 0.6 H 0.7 I 0.8 J 0.9 Solution. First, we need P(B (for all three problems. We obtain this value from the equation P(A B = P(A B, P(B which gives up It follows that P(B = P(A B P(A B = = 0.5. Answer: H P(E = P(exactly one of A and B = P(A + P(B 2 P(A B = = What is P(F? A 0.0 B 0.1 C 0.2 D 0.3 E 0.4 F 0.5 G 0.6 H 0.7 I 0.8 J 0.9

2 2 Solution. Using the value of P(B from the last problem, we have P(F = P(at least one of A and B = P(A + P(B P(A B = = 0.9. Answer: J 4. What is P(G? A 0.0 B 0.1 C 0.2 D 0.3 E 0.4 F 0.5 G 0.6 H 0.7 I 0.8 J 0.9 Solution. Using the value of P(F from the last problem, we have Answer: B P(G = P(neither of A and B = 1 P(at least one of A and B = 1 P(F = = A screening test for women over 40 gives a positive result for 95% of screened women who have a particular disease. It gives a positive result for only 3% of women who are free from the disease. Suppose that the prevalence of the disease is 1%. What is the probability that a woman who tests positive actually has the disease? A 0.24 B 0.27 C 0.30 D 0.33 E 0.36 F 0.39 G 0.42 H 0.45 I 0.48 J 0.51 Solution. Let S be the state of having the disease. The sensitivity of the test is P(POS S = We are also given P (POS S c = This value is 1 P(NEG S c, or 1 - specificity. Bayes s Rule tells us how to express P(S POS in terms of the sensitivity of the test, the specificity of the test, and the prevalence P(S of the state: Answer: A P(S POS = P (POS S P (S P (POS S P (S + P (POS S c P (S c = sensitivity prevalence sensitivity prevalence + (1 specificity (1 prevalence = ( = Two spent batteries have been abesent-mindedly returned to a drawer containing four fresh ones. When a caller ID device needs its four batteries replaced, four are selected from the drawer. What is the probability that the batteries inserted into the device are all fresh? A B C D E F G H I J Solution. The required probability is (4/6(3/5(2/4(1/3, or 1/15, or Answer: C

3 3 7. Horton has a box with six donuts. They are identical on the exterior, but precisely two contain jelly inside. Horton s friend Homer has come over, so Horton takes two donuts from the box at random. If X is the number of jelly donuts selected, what is E(X? A 1/3 B 2/5 C 7/15 D 8/15 E 3/5 F 2/3 G 11/15 H 4/5 I 13/15 J 14/15 Solution. This is an opportune time to thank Tim for his fine work and enterprise. The possible values for X are 0, 1, and 2. The probability that Horton takes 0 jelly donuts, not that we really need it, is (4/6(3/5 = 2/5. The probability that Horton takes 1 jelly donut, a probability we really need, is (4/6(2/5 + (2/6(4/5 = 8/15. The probability that Horton takes 2 jelly donuts, a probability we really need, is (2/6(1/5 = 1/15. We have Answer: F E(X = = For the random variable X of the preceding problem, what is Var(X? A 4/45 B 7/45 C 2/9 D 13/45 E 16/45 F 19/45 G 22/45 H 5/9 I 28/45 J 31/45 Solution. OK. In this problem we do need the value P(X = 0 that we found but did not really use in the last problem. We calculate Answer: E ( Var(X = P(X = ( 2 + P(X = P(X = = 2 ( ( ( = ( Suppose that X and Y are independent normal random variables with E(X = 3, E(Y = 2, Sd(X = 4, and Sd(Y = 3. What is P(X Y + 2? A 0.24 B 0.27 C 0.30 D 0.33 E 0.36 F 0.39 G 0.42 H 0.45 I 0.48 J 0.51 Solution. Sorry we were fresh out of watermelons, so there was no story. We first calculate Var(X Y = Var(X + Var(Y = = 25 and Sd(X Y = Var(X Y = 5. We continue as follows: Answer: G P(X Y + 2 = P(X Y 2 ( X Y (E(X - E(Y = P Sd(X Y = P (Z 0.2 = (3 2 5

4 4 10. Suppose that X is a random variable with standard deviation σ X = 5, that Y is a random variable with standard deviation σ Y = 8, that Z is standard normal, and that X, Y, and Z are independent. Suppose that W = X Y 3Z. A random sample of size 16 results in the sample statistic W. What is the standard deviation of W? A B C D E F G H I J Solution. First, Var (W = Var (X + 12 Y 3Z ( 1 = Var (X + Var 2 Y + Var ( 3Z = Var (X + 1 Var (Y + 9 Var (Z 4 = = 50. ( ( 1 2 It follows that Sd(W = 50 = Therefore Sd ( W = Sd (W n = = Answer: D 11. A manufacturer of precision watches produced watches with variability σ 0 = 0.4 s. After changing the supplier of some of the parts, the manufacturer sampled 12 watches and found that the sample standard deviation S was 0.6. Test the null hypothesis σ = σ 0 (where σ represents the population standard deviation after the switch to the new supplier against the alternative σ > σ 0. Use S 2 as the test statistic. If the test is at 0.05 significance level, what is the endpoint of the critical region. A B C D E F G H I J Solution. The relations σ = σ 0 and σ > σ 0 are equivalent to σ 2 = σ 2 0 and σ 2 > σ 2 0. For the hypothesis test H 0 : σ 2 = σ 2 0 H a : σ 2 > σ 2 0 with n = 12 and α = 0.05, we use the test statistic S 2 with critical region: S 2 > σ2 0 n 1 χ2 α,n 1, or S 2 > ( χ2 0.05,11, or S 2 > The endpoint of the critical region is Answer: E 12. In the hypothesis test of variance presented in the preceding problem, what is the p-value? A B C D E F G H I J 0.050

5 5 Solution. We begin with the information that was not used in the preceding problem: S = 0.6. The p-value is P (S 0.6 σ = σ 0 = P ( S 2 (0.6 2 σ = σ 0 ( (n 1 S 2 = P σ 2 0 ( (n 1 S 2 = P σ 2 = P ( χ = (12 1 (0.62 σ 2 0 (12 1 (0.62 (0.4 2 σ = σ 0 The value in the last line was obtained by using software. However, the correct answer choice can be found directly from the χ 2 tables, without any interpolation at all. Answer: B 13. A container with 225 roaches was sprayed with Disencroachment, resulting in the deaths of 207 of the annoying insects. A second container with 196 roaches was sprayed with Die Roach, Die!, resulting in the deaths of 172 of the pests. Test the null hypothesis that the two sprays are equally effective against the alternative that Disencroachment is more effective. What is the p-value? A B C D E F G H I J Solution. Let p 1 be the population proportion of roaches that Disencroachment kills. Let p 2 be the population proportion of roaches that Die Roach, Die! kills. The observed sample proportions are p 1 = 207/225 and p 2 = 172/196. The observed difference is p 1 p 2 = 207/ /196 = The standard deviation of the sample statistic p 1 p 2 is given by Sd ( p 1 p 2 = p1 (1 p 1 n + p 2 (1 p 2 m (207/225 (18/225 = (172/196 (24/ The p-value is P ( p 1 p p 1 p 2 = 0 = ( p1 p 2 P Sd ( p 1 p ( p 1 p 2 = 0 = P (Z Answer: D = 1 Φ( = = After two exams in a statistics course at Midwestern University, a random sample of 8 first exam scores resulted in 16, 19, 19, 19, 19, 21, 21, 23. An independent sample of 6 second exam scores resulted in 17, 17, 18, 19, 21, 22. Let X denote the distribution of the first exam scores and Y the second. Then the sample standard deviations are S X = and S Y = Using X Y as the test statistic, test the null hypothesis that the population means satisfy µ X = µ Y against the alternative that µ X > µ Y. (Assume that the distribution of scores for each exam is normal. Do not pool the variances. Assign the degrees of freedom for the test statistic conservatively. What is the endpoint of the critical region at 5% significance level? A 0.83 B 1.07 C 1.31 D 1.55 E 1.79 F 2.03 G 2.27 H 2.51 I 2.75 J 2.99

6 6 Solution. We assign the degrees of freedom to be 5: one less than the minimum of the two sample sizes. From the tables, we find that t 0.05,5 = The exdpoint of the critical region is Answer: G t 0.05,5 SX S Y , or , or After two exams in a statistics course at Midwestern University, a random sample of 7 students results in the following data for their first and second exam scores: Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Student 7 Exam 1 (X Exam 2 (Y Using X Y as the test statistic, test the null hypothesis that the population means satisfy µ X = µ Y against the alternative that µ X > µ Y. What is the p-value? A B C D E F G H I J Solution. Because values of X and Y are paired and not independent, we analyze this as a one-sample test of the difference of means. Let U denote the differences of the scores: 1, 0, 1, -1, 2, 1, 3. We calculate U = 1 and S U = The p-value is given by P ( U 1 µ X = µ Y ( U 0 = P = P / 7 µ U = / ( U µ U Sd ( U µ X = µ Y = P (t = The last line was obtained using software. The tables do not provide great accuracy. If p is a probability and t a value of t 6, then the line segment in the tp-plane joining (1.9432, 0.05 to (2.4469, is p = ( (t , ( or p = t Substituting t = results in p = Despite the slight inaccuracy, it is evident that A is the correct answer choice. Answer: A 16. The most frequently found letter in English is E. The next three most frequently occurring letters are, in order, T, A, and O. Disregarding all of the other 23 letters, the proportions of T, A, and O are estimated to be, in order, , , and In a cryptography project undertaken at Cornell University, a sample of 40,000 words resulted in the following actual counts of the letters T, A, and O : T A O Actual Count

7 7 Does the observed distribution match the estimated distribution? Answer with the p-value (using a match of the distributions as the null hypothesis. A B C D E F G H I J Solution. The total number of the three letters is 45,400. The expected counts are The test statistic is T A O Expected Number ( ( ( , or The p-value is P ( χ , or This value can eb approximated by using the given tables and interpolation. If x is a value in the χ 2 2-table and p is a value in the header, then (x, p = (5.9915, 0.05 and (x, p = (4.6052, 0.10 are two points in the table. The line segment joining the two points is or p = ( (x , ( p = x Substituting x = , we obtain p = (There is a slight inaccuracy resulting from the interpolated approximation, but the answers are spaced far enough apart that the inaccuracy has no impact. Answer: D 17. A survey of 220 voters in California (CA, 100 voters in Connecticut (CT, and 180 voters in New York (NY resulted in the following counts: CA CT NY Democrats Republicans Others For each of the three states there is a distribution of political leanings. In this problem and the two that follow, we will consider whether the three distributions of political leanings are homogeneous. You will need to calculate what all the cell entries would be under the assumption of homogenity. If that assumption were true, by how much would the expected number of surveyed New York Republicans exceed the expected number of surveyed Connecticut Democrats? A 19 B 20 C 21 D 22 E 23 F 24 G 25 H 26 I 27 J 28 Solution. The number of Democrats in the survey is 240. The number of Republicans in the survey is 200. The number of Others in the survey is 60. The proportion of Democrats is p D = 240/500 = 0.48, the proportion of Republicans is p R = 200/500 = 0.40, and the proportion of Others is p O = 60/500 = If these proportions held in each of the three states, there there would be , or 105.6, Democrats in CA, , or 48, Democrats in CT, , or 86.4, Democrats in NY, , or 88, Republicans in CA, , or 40, Republicans in CT, , or 72, Republicans in NY, , or 26.4, Others in CA, , or 12, Others in CT, and , or 21.6, Others in NY. These values are tabulated as follows:

8 8 CA CT NY Democrats Republicans Others The answer is 72-48, or 24. Answer: F 18. Refer to the data of the preceding problem. In a classical hypothesis test of homogeneity at the 10% significance level, by how much does the chi-squared test statistic fall short of the endpoint of the critical region? A 0.25 B 0.29 C 0.33 D 0.37 E 0.41 F 0.45 G 0.49 H 0.53 I 0.57 J 0.61 Solution. The number of degrees of freedom is (3 1(3 1, or 4, and χ 2 0.1,4 = The test statistic is ( ( ( ( ( ( ( ( or the answer is , or Answer: C ( , In a contemporary hypothesis test of the homogeneity of political leanings in the three states, what is the p-value? A 0.01 B 0.06 C 0.11 D 0.17 E 0.23 F 0.28 G 0.34 H 0.39 I 0.45 J 0.50 Solution. The p-value is P ( χ , which is This value can eb approximated by using the given tables and interpolation. If x is a value in the table and p is a value in the header, then (x, p = (7.7794, 0.1 and (x, p = (5.9886, 0.2 are two points in the table. The line segment joining the two points is or p = ( (x , ( p = x Substituting x = , we obtain p = (There is a slight inaccuracy resulting from the interpolated approximation, but the answers are spaced far enough apart that the inaccuracy has no impact. Answer: C 20. In a study investigating the relationship between hypertension and coronary heart disease (CHD, 600 randomly selected individuals were classified as follows:

9 9 Hypertension CHD No CHD Frequent Occasional Never If hypertension and CHD were not related, what would be the expected number of individuals in the study who suffered from CHD but were never hypertensive? A B C D E F G H I J Solution. If hypertension and CHD were not related, then the expected number in each cell would equal the product of the row total and the column total divided by the table total. Thus, the expected number of individuals in the study who suffered from CHD but were never hypertensive would be ( ( /600, or Answer: J 21. Refer to the tabulated data of the preceding problem. In a classical χ 2 test of the independence of hypertension and CHD, using the usual test statistic and significance level (NB: not 0.05, by how much does the test statistic exceed the endpoint of the critical region? A 2.32 B 2.88 C 3.44 D 4.00 E 4.56 F 5.12 G 5.68 H 6.24 I 6.80 J 7.36 Solution. The expected numbers under the null hypothesis are Hypertension CHD No CHD Frequent Occasional Never The test statistic (O E 2 /E comes to The endpoint of the critical region is χ ,(3 1(2 1, or The answer is , or Answer: H 22. The price Y of a middle-aged used car (in 100s of dollars does not seem to depend dramatically on the age X of the car (in years, other factors such as mileage and condition being approximately equal. In this problem and the three that follow, consider this data for a random sample of 6 Toyota Corolla sales: X Y Statistics for this small sample are: X = 5.5, Y = 75.5, S X = , S Y = , r = , b 0 = 104.0, and b 1 = In a classical one-sided hypothesis test of H 0 : β 1 = 0 versus the alternative H a : β 1 < 0, what is the endpoint of the critical region if b 1 is the test statistic and 0.05 is the significance level? A B C D E F G H I J Solution. With n = 6, we calculate SSE = , S e = SSE/(n 2 = , and SE (b 1 = ( 1/ n 1 Se /S X = The endpoint of the critical region is t 0.05,n 2 SE (b 1, or , or Answer: A

10 In the hypothesis test of the preceding problem, what is the p-value? A B C D E F G H I J Solution. Under the null hypothesis, b 1 /SE (b 1 has distribution t n 2. Therefore the p-value is given by ( P t n = P (t = The value in the last line was obtained by using software. If we use the tables and interpolate by finding the line joining the points (1.1896,0.15 and (1.5332,0.10 in the tp-plane, we obtain p = ( (t , ( or = p = t Substituting t = in this equation results in the p-value p = There is a small inaccuracy here, but only one answer choice, E, is at all close. The error is only and the answer choices are separated by Answer: E 24. What is the length of a 70% confidence interval for the slope of the regression line? A 5.90 B 6.66 C 7.42 D 8.18 E 8.94 F 9.70 G H I J Solution. The length of a 90% confidence interval for b 1 is 2 SE (b 1 t 0.15,4, or , or Answer: I 25. Mercedes is selling a 6 year old Toyota Corolla that is comparable to those that gave rise to the data of Problem 22. She can be 95% confident of receiving at least what amount for her car (in hundreds of dollars? A 34.9 B 36.8 C 38.7 D 40.6 E 42.5 F 44.4 G 46.3 H 48.2 I 50.1 J 52.0 Solution. Let ŷ be the sales price of Mercedes s car that is predicted by the regression line. Then SE (ŷ = SE (b 1 (6 x + (7/6 S 2 e. From the preceding work, we have SE (b 1 = and S e = Therefore, SE (ŷ = ( (7/ = Because t 0.025,4 = , it follows that the lower bound of a 95% confidence interval for the regression line prediction corresponding to x = 6 is , or Answer: B

11 11

12 12 Chi-Squared Values Right Tails.

13 Chi-Squared Values Central Hump + Right Tails. 13

14 14 Student-t Values Right Tails α = 0.45, 0.40, 0.35, 0.30, 0.25, 0.20.

15 Student-t Values Right Tails α = 0.15, 0.10, 0.05, 0.025, 0.010,