AMS 310 Final Exam Solutions: Practice Problems Explained

AMS 310 Practice Problems for Final Solutions Fred Rispoli 1. In a certain study you have a standard deviation of 12, a mean of 20 and a sample size of 50. What is the standard error? a) 1.70 b) 0.069 c) 0.089 d) 0.589 e) none of these a) SE = 12/sqrt(50) = 1.70 2. The standard error has been calculated as 2.6 and the sample mean is 10.00. Thus the 95% confidence interval lies between: a) 3.85 to 26 b) 4.904 to 15.096 c) 7.40 to 12.60 d) There is not enough information to work out the confidence interval b) 4.904 to 15.096 3. To calculate confidence intervals we need to make use of: a) histograms b) z-scores c) probability distributions d) none of the above c) probability distributions

4. Given a set of data, how would you calculate the 95% confidence intervals? a) To work out the 95% confidence interval you have to multiply the standard error by the standard deviation. b) To work out the 95% confidence interval you have to multiply the standard error by 1.96 c) To work out the 95% confidence interval you have to multiply the square root of the sample size by the standard deviation. d) To work out the 95% confidence interval you have to multiply the standard error by 95. e) none of the above b) To work out the 95% confidence interval you have to multiply the standard error by 1.96 5. A famous information technology company claims that their model of laser printers have mean output capacity of 400 pages. A consumer report firm wants to show the actual mean is smaller than 400. What should be the null and alternative hypotheses? a) H 0 :  ≥ 400 and H 1 :  < 400 b) H 0 :  ≤ 400 and H 1 :  > 400 c) H 0 :  ≥ 500 and H 1 :  < 500 d) H 0 :  ≤ 500 and H 1 :  > 500 a) H 0 :  ≥ 400 and H 1 :  < 400 6. In estimating a population proportion using a large sample, the estimate is 0.28, and its 95% error margin is 0.06. What is the sample size to meet the error margin? a) 216 b) 225 c) 204 d) 189 n = (0.28)(0.72)(1.96/0.06) 2 = 215.1296 round up to 216 a) 216

7. When should we use t statistics to find the confidence interval of the population mean? a) Small sample with unknown population variance b) Large sample with unknown population variance c) Large sample with known population variance d) Normal population with known population variance a) Small sample with unknown population variance 8. What distribution should we use for inferences concerning 2 population variances? a) F distribution b) t distribution c) Standard normal distribution d) Poisson distribution a) F distribution 9. The survival rate of a certain disease using a given medication is known to be 30%. A company claims that the survival rate a new drug is higher. The new drug is given to 15 patients to test for this claim. Let N be the number of cures out of the 15 patients. Suppose the rejection region is {N ≥ 8}. What is the type of error that can occur when the true survival rate is 40%? And what is the error probability? a) Type II error probability = 0.7869 b) Type I error probability = 0.0173 c) Type II error probability = 0.0173 d) Type I error probability = 0.7869 If p = .40, then H0 is false, so a Type II error can occur Error probability = P(accepting) = P(N ≤ 7) use the binomial with p = .40 = P(0) + P(1) + P(2) + … . + P(7) = .7869 a) Type II error probability = 0.7869

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10. Let X1, ..., X64 be a random sample from a Poisson distribution with a mean of 4. Find an approximate probability that the sample mean is greater than 4.2. a) 0.2119 b) 0.9772 c) 0.0228 d) 0.7881 First note that  =  = 4 and  = sqrt(4.2) and n = 64 We need to find P( 𝑋 ̅ > 4.2) = P(Z > 4.2−4 √4.2 /8 ) = P(Z > 0.7807) = 1 – P(Z < 0.7807) = 0.2119 a) 0.2119 11. 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. a) (0.077, 0.221) b) (0.064, 0.234) c) (0.037, 0.261) d) None of these ( 𝑝̂ − 𝑧 𝛼 2 √ 𝑝 ̂(1−𝑝 ̂) ? , 𝑝̂ + 𝑧 𝛼 2 √ 𝑝 ̂(1−𝑝 ̂) ? ) 𝑝̂ = 10 67 = 0.149 𝛼 2 = 0.05 0.149  1.645 √ (.149)(.851) 67 this gives a) (0.077, 0.221) 12. 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? Which of the following is correct for the appropriate hypothesis test? a) The critical value is z = 1.28 and the test value z = 1.335 so reject H 0 . b) The critical value is z = 1.96 and the test value z = 1.335 so do not reject H 0 . c) The critical value is z = 1.65 and the test value z = 1.335 so do not reject H 0 . d) The critical value is z = 1.65 and the test value z = 1.28 so do not reject H 0 .

e) None of these a) The critical value is z = 1.28 and the test value z = 1.335 so reject H 0 . 13. 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? a) 40 b) 69 c) 28 d) None of these n = ( ? 𝛼 2 ⁄ 𝜎 ? ) 2 = [ (1.96)(3.2) 1 ] 2 = 39.3 round up to 40 a) 40 14. Match each item in the left column with the correct item in the right column p-value = 0.02 ________ a. do not reject H 0 at  = 0.1 p-value = 0.07 ________ b. reject H 0 at  = 0.1, but not at 0.05 p-value = 0.3 ________ c. reject H 0 at  = 0.05, but not at 0.01 p-value = 0.0006 ________ d. reject H 0 at  = 0.01 p-value = 0.02 ____ c ____ p-value = 0.07 ____ b ____ p-value = 0.30 ____ a ____ p-value = 0.0006 ____ d_ ___ 15. 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.

a) Computers from company A are more reliable than computers from company B using  = 0.01. b) Computers from company A are not more reliable than computers from company B using  = 0.01. c) There is not enough information to determine which is more reliable. Sample Proportions 𝑝 1 ̂ = 0.105 𝑝 2 ̂ = 0.185 Pooled proportion: 0.145 Hypotheses: H 0 : p 1 = p 2 H 1 : p 1 < p 2 (claim) Test Statistic: 𝑍 = 𝑝̂ 1 −𝑝 ̂ 2 √𝑝̂(1−𝑝̂)( 1 ? + 1 ? ) Rejection Region: -2.33 Value of Test Statistic: z = -2.27 Fail to reject H 0 b) Computers from company A are not more reliable than computers from company B using  = 0.01. 16. Assume a population has a distribution with a mean of 100 and a standard deviation of 10. For a sample of size 50, P(99 < 𝑋 ̅ < 102) rounded to 2 decimal places and the 70 th percentile of 𝑋 ̅ is given by: a) P(99 < 𝑋 ̅ < 102) ≈ 0.68 and the 70 th percentile is 100.74 b) P(99 < 𝑋 ̅ < 102) ≈ 0.92 and the 70 th percentile is 100.74 c) P(99 < 𝑋 ̅ < 102) ≈ 0.68 and the 70 th percentile is 105.2 d) Neither of these  ( 102−100 10/√50 ) -  ( 99−100 10/√50 ) =  (1.41) –  (-0.71) = 0.9207 – 0.2389  0.68 The 70h percentile occurs when z = .52, so solve .52 = ?̅−100 10/√50 this gives 𝑋 ̅ = 100.74 a) P(99 < 𝑋 ̅ < 102) ≈ 0.68 and the 70 th percentile is 100.74 17. 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 9 bowlers. a) (181.388, 190.612) b) (182.08, 189.92)

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c) (180.84, 191.96) d) Neither of these (𝑥̅ − ? 𝛼 2 ,?−1 𝑠 √? , 𝑥̅ + ? 𝛼 2 ,?−1 𝑠 √? ) df = 8 𝛼 2 = 0.025 186  2.306( 6 √9 ) = 186  4.612 this gives (181.388, 190.612) a) (181.388, 190.612) 18. 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}. Determine the type of error that can occur when the true survival rate is 35% is ______, and the error probability is _______. First notice that when p = 0.35, H 0 is true. So a Type 1 error may occur. 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 - ∑ 𝑏𝑖𝑛(25, .35) 12 ?=0 = P(Z > 1.58) = 1 -  (1.58)  0.06 19. 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. Assume the scores are randomly selected from the two groups, and assume equal variances. Test Scores Pre-test 77 56 64 60 57 53 72 62 65 66 Post-test 88 74 83 68 58 50 67 64 74 60 H 0 :  pre =  post H 1 :  pre ≠  post  = 0.05 df = 10 + 10 – 2 = 18 Test statistic t = (? ̅̅̅ −? ̅)− Δ 𝑠 𝑝 √ 1 ? + 1 ? , where ? 𝑝 2 = (?−1)𝑠 1 2 +(?−1)𝑠 2 2 ?+?−2 Critical Values: t =  2.101 Test Value t = -1.25 ? 𝑝 2 = 9.67 2 Do not reject H 0 , 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 20. 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. Assume that the scores are pairs of scores for ten students. Test Scores Pre-test 77 56 64 60 57 53 72 62 65 66 Post-test 88 74 83 68 58 50 67 64 74 60 Test Scores Pre-test 77 56 64 60 57 53 72 62 65 66 Post-test 88 74 83 68 58 50 67 64 74 60 Difference -11 -18 -19 -8 -1 3 5 -2 -9 6 𝐷 ̂ = -5.40 s d = 9.03 n = 10 H 0 : 𝐷 ̂ = 0 H 1 : 𝐷 ̂ ≠ 0 df = 9 Test Statistic : 𝑇 = ? ̅ 𝑆 𝑑 /√? Critical Values, use df = 9 and  to get  2.262 Test Value = -1.89 Do not reject H 0 There is no significant difference in scores at  = 0.05, 0.025(2) < p-value < 0.05(2)

AMS 310 Practice Problems for Final Exam Solutions

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