University:
Stony Brook University
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AMS 310 | Survey of Probability and Statistics
Academic year:

2029
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407

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13
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study_genius

AMS 310 Practice Final Solutions Fred Rispoli 1. Assume a population has a distribution with a mean of 100 and a standard deviation of 10. For a sample of size 50, find the following. a) P(99 < 𝑋̅ < 102) 102−100 99−100 ( 10/√50 ) - ( 10/√50) = (1.41) – (-0.71) = 0.9207 – 0.2389  .68 b) P( 𝑋̅ > 97) 97−100 1 - ( 10/√50 ) = 1 –  (-2.12) = 1- 0.0170 = 0.983 c) The 70th percentile of 𝑋̅. 𝑥̅ −100 The 70h percentile occurs when z = .52, so solve .52 = 10/√50 this gives 𝑋̅ = 100.74 2. The weight of adult bottlenose dolphins was found to follow a normal distribution with a mean of 550 pounds and a standard deviation of 50 pounds. a) What percentage of bottlenose dolphins weigh from 400 to 600 pounds? 400−550 P(400 < X < 600) = P( 50 560) = P(z > 560−550 50 = P(Z > .2) = 1 - (0.2) = 1 - .5793 = .4207 Set Y = # of dolphins that are heavier than 560. Y is a binomial rv P(Y = 5) = (59)(.4207)5 (1 − .4207)4 = 0.187 3. The survival rate of a certain type of cancer using an existing medication is known to be 40%. A pharmaceutical company claims that the survival rate of a new drug is higher. The new drug is given to 25 patients to test this claim. Let X be the number of cures out of the 25 patients. Suppose the rejection region is {X ≥ 13}. a) State the testing hypotheses. H0: p ≤ 0.40 H1: p > 0.40, claim b) Determine the type of error that can occur when the true survival rate is 35%. Find the error probability. First notice that when p = 0.35, H0 is true. So a Type 1 error may occur. 12 P(Type 1 Error) = P(X ≥ 13 given p = 0.35) = ∑25 𝑥=13 𝑏𝑖𝑛(25,0.35) = 1 - ∑𝑥=0 𝑏𝑖𝑛(25, .35) Use the normal approximation to the binomial  = np = 25(.35) = 8.75  = √𝑛𝑝(1 − 𝑝) = 2.38 z = (12.5 – 8.75)/2.38 = 1.58 P(Type 1 Error) = 1 - ∑12 𝑥=0 𝑏𝑖𝑛(25, .35) = P(Z > 1.58) = 1 - (1.58)  0.06 c) Determine the type of error that can occur when the true survival rate is 55%. Find the error probability. First note that if p = 0.55 then H0 is false. P(Type 2 Error) = P(X ≤ 12 given p = 0.55) = ∑12 𝑥=0 𝑏𝑖𝑛(25, .55) Use the normal approximation to the binomial Now  = np = 25(0.55) = 13.75 and  = √𝑛𝑝(1 − 𝑝) = 2.49 z = (12.5 – 13.75)/2.49 = -0.50 ∑12 𝑥=0 𝑏𝑖𝑛(25,0.55) = P(Z ≤ -0.50) = (-0.50)  0.31 4. Match each item in the left column with the correct item in the right column p-value = 0.02 ____c____ a. do not reject H0 at  = 0.10 p-value = 0.07 ____b____ b. reject H0 at  = 0.10, but not at 0.05 p-value = 0.30 ____a____ c. reject H0 at  = 0.05, but not at 0.01 p-value = 0.0006 ____d____ d. reject H0 at  = 0.01 5. a) A study of 49 bowlers showed that their average score was 186. The standard deviation of the population is 6, and the scores are normally distributed. Construct a 95% confidence interval for the mean score of all bowlers. Show all work involved. (𝑥̅ − 𝑧𝛼 2 186  1.96( 6 𝜎 √𝑛 , 𝑥̅ + 𝑧𝛼 2 𝜎 √𝑛 ) ) = 186  1.68 this gives (184.32, 187.68) as the 95% confidence interval √49 b) A study of 9 bowlers showed that their average score was 186. The standard deviation of the sample is 6, and the scores are normally distributed. Construct a 95% confidence interval for the mean score of all bowlers. Show all work involved. (𝑥̅ − 𝑡𝛼,𝑛−1 2 𝑠 √𝑛 , 𝑥̅ + 𝑡𝛼,𝑛−1 2 𝑠 √𝑛 ) df = 8 𝛼 2 = 0.025 6 186  2.306( ) = 186  4.612 this gives (181.388, 190.612) √9 6. a) Suppose that 67 murders were studied and it was found that 10 were committed by women. Construct a 90% confidence interval for the true proportion of murders in New York committed by women. Show all work involved. ( 𝑝̂ − 𝑧𝛼 √ 2 𝑝̂(1−𝑝̂) 𝑛 𝑝̂(1−𝑝̂) , 𝑝̂ + 𝑧𝛼 √ 2 (.149)(.851) 0.149  1.645 √ 67 𝑛 ) 𝑝̂ = 10 67 = 0.149 𝛼 2 = 0.05 this gives (0.077, 0.221) b) Experts claim that at most 10% of murders in New York are committed by women. Is there enough evidence at  = 0.10 to reject the claim if in a sample of 67 murders, 10 were committed by women? Use the 5 step method. H0: p ≤ 0.10 (claim) H1: p > 0.10  = 0.10 𝑝̂ = Test Statistic z = 𝑝̂−𝑝0 √𝑝0 (1−𝑝0 )/𝑛 10 67 = 0.149 Critical Value z = 1.28 Test Value z = 1.335 which is greater than 1.28, so reject H0 The data does not support the claim at  = -0.10, p-value = P(Z > 1.335) = 1 – P(Z < 1.335) = 0.091 c) Suppose that the true percentage of murders committed by women is 20%. What type of error, if any does this test involve? Explain. If p = 0.20, then H0 is false. There is no error. 7. A manufacturer claims that the standard deviation of the drying time of a certain type of paint is 18 minutes. A sample of five test panels produced a standard deviation of 21 minutes. Test the claim at  = 0.05. Use the 5 step method. 1. H0:  = 18, (claim) H1:  ≠ 18  = 0.05 df = 4 3. Test Statistic 4. Test Value 𝜒 2 = 𝜒2 = (𝑛−1)𝑠2 𝜎2 (𝑛−1)𝑠2 Critical Values: 0.4844 and 11.143 𝜎2 = (4)212 182 = 5.44 Do not reject H0 5. The data supports the claim at  = 0.05, 2(0.10) < p-value < 2(0.25) 8. A factory manager wants to determine the average time it takes to finish a certain process by workers, and he wants to test it with randomly selected workers. He wants to be able to assert with 95% confidence that the mean of his sample is off by at most 1 minute. If the population standard deviation is 3.2 minutes, how large of a sample must he take? n=( 𝑧𝛼⁄ 𝜎 2 2 𝐸 (1.96)(3.2) 2 ] 1 ) =[ = 39.3 round up to 40 9. A cell phone store has sold 150 phones of brand A and had 14 of them returned as defective. Additionally it has sold 125 phones of Brand B and had 15 of them returned as defective. Is there statistical evidence that Brand A has a smaller chance of being returned than Brand B? Use  = 0.05 to test the claim. 14 15 𝑝 ̂𝐴 = 0.093 = 150 𝑝̂𝐵 = 0.120 = 125 1. H0: pA ≥ pB H1: pA < pB (claim) 𝑥+𝑦 29 𝑝̂ = 𝑚+𝑛 = 275 = 0.1055 2.  = 0.05 3. Test Statistic 𝑍= 𝑝̂1 −𝑝̂2 1 1 𝑚 𝑛 √𝑝̂(1−𝑝̂)( + ) Critical Value Z = -1.645 0.093−0.120 4. Z = √(0.1055)(0.8945)( 1 1 + ) 150 125 = -0.726, do not reject H0 5. The data does not support the claim at  = 0.05, p-value = (-0.726) = 0.234 10. To test whether a college course is working a pre- and post-test is arranged for the students. The results are given below. Compare the scores with a t-test as indicated. Use  = 0.05 to test the claim. a) Assuming the scores are randomly selected from the two groups. Assume equal variances. b) Assume that the scores are pairs of scores for ten students. Test Pre-test Post-test a) 77 88 56 74 64 83 H0: pre = post Scores 60 57 68 58 53 50 72 67 62 64 65 74 66 60 H1: pre ≠ post  = 0.05 df = 10 + 10 – 2 = 18 Test statistic t = ̅̅̅ (𝑥 −𝑦̅)− Δ 1 1 𝑚 𝑛 𝑠𝑝 √ + , where 𝑠𝑝2 = (𝑚−1)𝑠12 +(𝑛−1)𝑠22 𝑚+𝑛−2 Critical Values: t = 2.101 𝑠𝑝2 = 9.672 Test Value t = -1.25 Do not reject H0, there is no significant difference in scores at  = 0.05, (2)(.10) < p-value < (2)(.25) 0.2 < p < 0.5 Not required but included just as a check: P-Value = 0.228 b) Here we develop the test as a paired data t test Test Pre-test Post-test Difference ̂ = -5.40 𝐷 77 88 -11 56 74 -18 64 83 -19 Scores 60 57 68 58 -8 -1 53 50 3 72 67 5 62 64 -2 65 74 -9 66 60 6 sd = 9.03 n = 10 ̂=0 1. H0: 𝐷 ̂≠0 H1: 𝐷 2.  = 0.05 df = 9 3. Test Statistic: 𝑇 = 𝑆 ̅ 𝐷 𝑑 /√ 𝑛 Critical Values, use df = 9 and to get 2.262 4. Test Value = -1.89 Do not reject H0 5. There is no significant difference in scores at  = 0.05, 0.025(2) < p-value < 0.05(2) 11. A study has been done to test the effect of classical music on the brain. Twenty sixth-grade students were randomly divided into two groups of 10 students, and the same math test was given to these students. The first group of students listened to classical music for 20 minutes right before taking the test, and the other group took the test without listening to the music. The scores are given below. Group with music: 91, 77, 58, 89, 83, 78, 74, 81, 91 and 88 𝑥̅ = 81 s1 = 10.1 Group without music: 81, 65, 69, 69, 67, 61, 67, 87, 64, 81 𝑦̅ = 71.1 s2 = 8.7 Assume equal variance for performing the following calculations. a) Construct a 95% confidence interval for the mean difference. b) Test if the students who listened to classical music before the test scored higher than the other group of students using  = 0.05. c) Conduct the same test as in part (b) using  = 0.01. First note that both 𝑚 = 𝑛 = 10 are small. 𝑥̅ = 81, 𝑠1 = 10.1; and 𝑦̅= 71.1, 𝑠2 = 8.7 (a) (𝑠𝑝)2 = (𝑚 − 1)𝑠12 + (𝑛−1) 𝑠22 /(𝑚+𝑛−2) = (9(10.1)2+9(8.7)2 ) /(10 + 10 – 2) = 88.85 Since 𝑡𝛼/2, 𝑚+𝑛−2 = 𝑡0.025,18 = 2.101, 𝑥̅ − 𝑦̅ ± 𝑡𝛼/2,+𝑛−2 𝑠𝑝 SQRT(1/𝑚+1/𝑛) = 9.9 ± 8.86. Thus, a 95% confidence interval for 𝜇1 − 𝜇2 is (1.04, 18.76). (b) 𝐻0: 𝜇1 = 𝜇2, H1: 𝜇 1 > 𝜇2 𝛼 = 0.05 The test statistic is 𝑇 = (𝑋̅ − 𝑌̅) /(𝑆𝑝SQRT(1/𝑚+1/𝑛)) The rejection region is 𝑡 ≥ 𝑡𝛼,+𝑛−2 = 𝑡0.05,18 = 1.734. 𝑡 = (81−71.1)/(SQRT(88.85)SQRT(1/10+1/10)) = 2.349 > 1.734 Reject 𝐻0. 𝑝-value = (𝑇 > 2.349) = 0.015 (c) Do not reject at 𝛼 = 0.01 because the 𝑝-value of 0.015 is larger than 0.01. 12. A survey on computers requiring repairs within two years was conducted. 21 out of 200 computers from company A, and 37 out of 200 computers from company B required repairs. Do these data show that computers from company A are more reliable than computers from company B? Test using  = 0.01. Sample Proportions 𝑝 ̂1 = 0.105 1. Hypotheses: 𝑝 ̂2 = 0.185 H0: p1 = p2 Pooled proportion: 0.145 H1: p1 < p2 (claim) 2.  = 0.01 3. Test Statistic: 𝑍= 𝑝̂1 −𝑝̂2 1 1 𝑚 𝑛 Rejection Region: -2.33 √𝑝̂(1−𝑝̂)( + ) 4. Value of Test Statistic: z = -2.27 Fail to reject H0 5. The data does not support the claim that computers from A are more reliable at  = 0.01. P-value: 0.0116 13. In animal cancer research the potential of human drugs and other substances are studied. Four hundred ppm of benzidine dihydrochloride is given to each of the male and female mice from a certain strain. In one of these experiments, tumors were found in 54 out of 484 male mice, and 127 out of 429 female mice. a) Find a 99% confidence interval for the difference between the tumor rate of male mice and female mice. b) Test if the tumor rate of female mice is higher than the tumor rate of male mice. Use  = 0.01 to test the claim. (a) 𝑚 = 484, 𝑥 = 54, 𝑛 = 429, 𝑦 = 127. 𝑥 54 𝑦 127 Thus, 𝑝̂1 = 𝑚 = 484 = 0.112 and 𝑝̂ 2 = 𝑛 = 429 = 0.296. ̂ (1 − 𝑝 ̂2 ) 𝑝̂1 (1 − 𝑝̂1 ) 𝑝 𝑝̂1 − 𝑝̂2 ± 𝑧𝛼 √ + 2 𝑚 𝑛 2 (0.112)(0.888) (0.296)(0.704) = 0.112 − 0.296 ± 𝑧0.005 √ + 484 429 = −0.184 ± 2.58 ∙ 0.026 = −0.184 ± 0.068 Therefore, a 99% confidence interval for 𝑝1 − 𝑝2 is (−0.252, −0.116). (b) 1. 𝐻0 : 𝑝1 = 𝑝2, 𝐻1 : 𝑝1 < 𝑝2 2. 𝛼 = 0.01 3. The test statistic is 𝑍= 𝑝̂ = 𝑝 ̂ 1 − 𝑝̂ 2 . √𝑝̂ (1 − 𝑝̂ ) ( 1 + 1) 𝑚 𝑛 𝑥+𝑦 54 + 127 181 = = = 0.198 𝑚 + 𝑛 484 + 429 913 The rejection region is 𝑍 ≤ −𝑧𝛼 = −𝑧0.01 = −2.33. 4. 𝑧= 0.112 − 0.296 √(0.198)(0.802) ( 1 + 1 ) 484 429 = −6.96 < −2.33 Reject 𝐻0 . 5. There is enough evidence that the tumor rate of female mice is higher than that of male mice. 𝑝-value= 𝑃(|𝑍| > 4.56) = 2Փ(−4.56) ≈ 0 14. The foot size of each of 16 men was measured, resulting in the sample means of 27.32 cm. Assume that the distribution of foot size is normal with  = 1.2 cm. a) Test if the population mean of men's foot sizes is 28.0 cm using  = 0.01. b) If  = 0.01 is used, what is the probability of a type II error when the population mean is 27.0 cm? c) Find the sample size required to ensure that the type II error probability is (27) = 0.1 when  = 0.01. (a) 𝐻0 : 𝜇 = 28 versus 𝐻1 : 𝜇 ≠ 28 𝛼 = 0.01 The test statistic is 𝑍= 𝑋̅ − 𝜇0 𝜎/√𝑛 and the rejection region is |𝑧| > zα/2 = z0.005 = 2.58. 𝑥̅ − 𝜇0 27.32 − 28 𝑧= 𝜎 = = −2.27 1.2 √𝑛 √16 Since |𝑧| = 2.27 < 2.58, do not reject 𝐻0 . (b) 𝛽(27) = Փ (𝑧.005 + = Փ (2.58 + 𝜇0 − 𝜇 ′ 𝜎⁄√𝑛 ) − Փ (−𝑧.005 + 𝜇0 − 𝜇 ′ 𝜎⁄√𝑛 ) 28 − 27 28 − 27 ) − Փ (−2.58 + ) = Փ(5.91) − Փ(0.75) 1.2⁄√16 1.2⁄√16 = 1 − 0.7734 = 0.2266 (c) 2 2 2 σ(zα/2 + zβ ) σ(z0.005 + z0.1 ) 1.2(2.58 + 1.28) 𝑛=( ) = ( ) = ( ) = 21.46 𝜇0 − 𝜇 ′ 𝜇0 − 𝜇 ′ 28 − 27 The required sample size is 22. 15. The random variables X and Y have the following joint probability distribution. X f(x,y) 1 2 3 Y 1 .1 .2 .15 2 .05 .05 .05 3 .04 .15 .10 4 .01 .04 .06 a) Find 𝑃(𝑋 + 𝑌 ≤ 5). b) Find the marginal probability distributions 𝑓1 (𝑥) and 𝑓2 (𝑦). c) Find 𝑃(𝑋 < 2|𝑌 = 3). d) Are X and Y independent? Thoroughly explain your answer. e) Find E(X) and Var(X) f) Find the correlation coefficient of X and Y. a) The X,Y combinations that sum to 5 or less are in red below. X f(x,y) Y 1 2 3 1 .1 .2 .15 2 .05 .05 .05 3 .04 .15 .10 4 .01 .04 .06 𝑃(𝑋 + 𝑌 ≤ 5) = .1 + .05 + .04 + .01 + .2 + .05 + .15 + .15 + .05 = .80 b) X f1(x) 1 .45 2 .15 3 .29 Y f2(y) 1 .2 2 .44 3 .36 4 .11 c) 𝑃(𝑋 < 2|𝑌 = 3) = 0.15/0.36 d) Not independent f(1,1) = .1 ≠ f1(1) f2(1) = (.45)(.2) = .09 e) E(X) = 1(.45) + 2(.15) + 3(.29) + 4(.11) = .45 + .30 + .87 + .44 = 2.06 X2 f1(x) 1 .45 4 .15 9 .29 16 .11 E(X2) = 1(.45) + 4(.15) + 9(.29) + 16(.11) = .45 + .60 + 2.61 + 1.76 = 5.42 Var(X) = E(X2) - E(X)2 = 5.42 – 2.062 = 1.176 X = SQRT(1.176) = 1.084 f) Use 𝝆 = 𝑪𝒐𝒗(𝑿,𝒀) 𝝈𝑿 𝝈𝒀 E(Y) = 1(.2) + 2(.44) + 3(.36) = 2.16 Y2 f2(y) 1 .2 4 .44 9 .36 E(Y2) = 1(.2) + 4(.44) + 9(.36) = 5.20 Var(Y) = E(Y2) - E(Y)2 = 5.20 – 2.162 = 0.5344 Y = SQRT(0.5344) = 0.731 𝐶𝑜𝑣(𝑋, 𝑌) = 𝐸(𝑋𝑌) − 𝐸(𝑋)𝐸(𝑌) Thus we need to know E(XY), the procedure is the following: P(XY=1)=P(X=1,Y=1)=0.1 P(XY=2)=P(X=1,Y=2)+ P(X=2,Y=1)=0.2+0.05=0.25 P(XY=3)=P(X=1,Y=3)+ P(X=3,Y=1)=0.04+0.15=0.19 P(XY=4)=P(X=4,Y=1)+ P(X=2,Y=2)=0.01+0.05=0.06 P(XY=6)=P(X=2,Y=3)+ P(X=3,Y=2)=0.05+0.15=0.2 P(XY=8)=P(X=4,Y=2)=0.04 P(XY=9)=P(X=3,Y=3)=0.1 P(XY=12)=P(X=4,Y=3)=0.06 E(XY)=1(0.1)+2(0.25)+3(0.19)+4(0.06)+6(0.2)+8(0.04)+9(0.1)+12(0.06) = 4.55 𝐸(𝑋𝑌) − 𝐸(𝑋)𝐸(𝑌) 4.55 − 1.98 ∗ 2.06 𝜌= = = 0.1266 𝜎𝑥 𝜎𝑦 1.084 ∗ 0.731 16. Let X and Y be independent random variables representing the lifetime (in 100 hours) of type A and B lightbulbs, respectively. Both variables have exponential distributions, and the mean of X is 2 and the mean of Y is 3. a) Find the joint pdf f(x,y) of X and Y. b) Find the conditional pdf f2(yx) c) Given the probability that a type A bulb fails at 200 hours, find the probability that a type B bulb fails lasts longer than 300 hours. a) By independence, 𝑓(𝑥,𝑦)=𝑓1(𝑥)𝑓2(𝑦)={(1/6)𝑒−𝑥/2−𝑦/3, 𝑥 ≥ 0, 𝑦 ≥ 0 0, otherwise b) By independence, 𝑓2(𝑦|𝑥)=𝑓2(𝑦)=1/3𝑒−𝑦/3, 𝑦≥0 c) By independence, 𝑃(Y>3|X =2)=𝑃(Y>3)=1−𝐹2(3)= 1 – (1- 𝑒−3/3 ) = 𝑒 −1 = 0.3679 17. Three random variables X, Y and Z are independent. Each has a binomial distribution with success probability 0.3 and 5 trials. a) Find the joint probability distribution function f(x,y,z). b) Find the probability P(Z > Y > X > 1). a) X, Y and Z are all binomial(5, 0.3), so f(x,y,z) = ((𝑥5)0. 3𝑥 . 75−𝑥 ) ((𝑦5 ) 0.3𝑦 . 75−𝑦 ) ((5𝑧)0. 3𝑧 . 75−𝑧 ) b) Add the probabilities for (x,y,z) in (2,3,4), (2,3,5), (2,4,5), (3,4,5) 18. The joint pdf of X and Y is given as f(x,y) = 2(𝑥+2𝑦) 3 , 0 < x < a, 0 < y < 1, where a is a constant. a) Find the value of a. b) Using the value of a obtained in part(a), determine the marginal pdf of X and Y. c) Determine the conditional pdf 𝑓1 (𝑥|𝑦) d) Find P(X ≤ 1/2|Y = 1/2) (a) The marginal pdf of 𝑋 is ∞ 1 1 2(𝑥 + 2𝑦) 2(𝑥𝑦 + 𝑦 2 ) 2(𝑥 + 1) (𝑥) 𝑓1 = ∫ 𝑓(𝑥, 𝑦)𝑑𝑦 = ∫ 𝑑𝑦 = | = ,0 < 𝑥 < 𝑎 3 3 3 −∞ 0 𝑦=0 ∞ 𝑎 𝑎 2(𝑥 + 1) 𝑥 2 + 2𝑥 𝑎2 + 2𝑎 1 = ∫ 𝑓1 (𝑥)𝑑𝑥 = ∫ 𝑑𝑥 = | = ⇒ 𝑎2 + 2𝑎 = 3 3 3 3 −∞ 0 0 ⇒ (𝑎 − 1)(𝑎 + 3) = 0 The value of 𝑎 must be 1 because 𝑎 > 0. (b) From part (a), 𝑓1 (𝑥) = ∞ 2(𝑥 + 1) , 3 0<𝑥<1 1 1 2(𝑥 + 2𝑦) 𝑥 2 + 4𝑥𝑦 4𝑦 + 1 𝑓2 (𝑦) = ∫ 𝑓(𝑥, 𝑦)𝑑𝑥 = ∫ 𝑑𝑥 = | = ,0 < 𝑥 < 1 3 3 3 −∞ 0 𝑥=0 (c) 𝑓1 (𝑥|𝑦) = 𝑓(𝑥, 𝑦) 2(𝑥 + 1)/3 2(𝑥 + 1) = = , 𝑓2 (𝑦) (4𝑦 + 1)/3 4𝑦 + 1 0 < 𝑥 < 1, 0 < 𝑦 < 1 (d) 𝑓1 (𝑥|𝑦 = 1/2) = 2(𝑥 + 1) 2(𝑥 + 1) = , 1 3 4 (2) + 1 1/2 1/2 𝑃(𝑋 ≤ 1/2|𝑌 = 1/2) = ∫ 𝑓1 (𝑥|𝑦 = 1/2)𝑑𝑥 = ∫ 0 0 = 1 2 1 (2) + 2 (2) 3 = 5 12 0<𝑥<1 1/2 2(𝑥 + 1) 𝑥 2 + 2𝑥 𝑑𝑥 = | 3 3 0

Practice Final Solutions Winter 2019