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Exam 2 830am Solutions - Samoa STT 315(Secs 1-— 11 Exam#2...

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Unformatted text preview: Samoa: STT 315 (Secs 1 -— 11 ) Exam #2 - FORM A Mon Nov 2010 Gilliland (Closed book - 75 minutes) 8:30 am exam Name PID 0. You may use a calculator. All electronic devices capable of communication and/or web access must be turned off? I. In the places above, please put your name and PID. Tables are found on the last sheet. 2. Be sure to code your name, section number and PID number correctly on the answer sheet that has been provided. Sign the answer sheet. 3. On your answer sheet you must also code the letter A as the FORM of your examination in the space provided. 4. After the exam, return your answer sheet as you leave. You may keep your exam paper. 5. There are 30 questions, each worth 1.667 course points. (Total 50. 010 course points possible.) 6. Your score is determined from your answer sheet and not from what you have written on your exam paper. 1. Consider the data gathered in response to this question: What was the gross income reported by you and your husband to the [RS for the year 2009? This is an example of WWW.Mas-mwwmw (a) categorical data (:(b) numerical (1:33 fish—amrm:m.<.mzawm=mfl " 2 - 3. Below is a stem-and-Ieaf display of the ages of 51 children ranging in ages from 1 month to 38 months. 0 |112234444 s M 0 Haw ll Maxi“ 5 1 low 244 5430a «1 39 l 1 |5566899 l 2 |012344 2 I5567 3|134 3|8 H ’1'] _m.bc\~aoow\oa‘ 2. The data are (a) skewed toward small values (b Isliewed toward large values; (0) symmetric a. we.»- 3. The median age is (' (a) 11 monthsw (b) 10 months (0) 12 months (d) 26 (e) 9.5 months 51. A sample of n = 28 rats have weights that average 13 ounces and have standard deviation 1.5 ounces. The Chebyshev Inequality implies that at least how many sample rats have weights between 10.0 ounces and 16.0 ounces? yam (b) 23 (c) 27 (d) 26 (e) 24 _.L:t—-‘—.,:.~.é. g .3 l at l ‘/ a”? 0” 55-15 56" V44: 5 — 7. A MSU running back had nineteen runs of length -4, -3, —2, —2, 0_, 1, 3, 4, 5, 6, 6, 3, 9, 10, 12, 14, 15, 24, 65 yards. 5. The average run length in yards is (a) 10.6 (b) 9.0\ (c) 12.7 (d) 8.5 (e) 10.0 1°! 2.);sz 5?: 1.7.}. : QO :! l 6. The upper whisker of the box plot of these data extends to what value. (a) 65 yards (b) 15 yards (c) 16 yards (d) 22 yards (e) 24 yards 6?]:0] 4a.: 12, mass]; at: {2.4”}.5/12133535955 Largest” Value wutamu-F is “$3093 7. The 33rd percentile of these data (by the book’s method) is (a) 1.4 yards (b) 3.4 yards (0) 6. 6 K3112- .2 yards (e) none of these PaSi‘Liaulé “kW—”W ‘ t +1 \ i: {q 3533) éd’é ” I a”! 3 l+té(3‘”i)=o?|23¢fl5 r 345 Q. Which of these sample statistics is not a measure of center of the sample values? fr“ 1 (a) a? (b) 50‘h percentile (c) Q2 . (6)3 / “.4me 2. A random sample of n = 19 from a normal pepulation has mean )7 = 40.6 and standard deviation 5 = 5.6. The 98% confidence interval estimate of the population mean is 40.6 i (a) 1.99 (b) 4.74 @ (d) 2.20 (e) 5.60 f-m‘l—Lértlal 40.6 :4: 2.55:2. 3:349- w :t 3.29 w: m. A random sample of n = 100 Medicaid payments to a clinic from a large population of payments are audited. It is found to contain 40 of the payments which were in error. The 95% confidence interval estimate for the population proportion of payments in error is (a) [.3 50, .450] (b) [.234, .516] (c) [504, .496 (d) [.247, .553] (e) [.320, .480] i 1.9!: @345") its .r .‘Mé 11 — 12. The Michigan Bureau of Elections believes that about proportion p = .70 of the signatures in a large petition drive are valid. 11. Using the value p = .70 for planning purposes, what is the minimum sample size required so that the 90% z-interval estimate of p has a margin of error of i .02? (a) 1,083 (b) 385 ((3.724 (g) 1,421 i (e) 897 2. . — | I ('3 “3- 'Wm ”"3; “54—14297 1‘02 12. Suppose that the population size is N = 1,000. Using the value p = .70 for planning purposes, what is the minimum sample size required so that the 90% z—interval estimate of p has a margin of error of + .02? (a) 444 (b) 685 (c) 721 (d) 144 (.6) 533 W a c - 1 + W3 1 Q. The mean u of a population of size N = 400 is the parameter of interest. The population standard deviation is known, 6 = 5. A simple random sample of size n— — 100 is drawn; the sample average is 2?- ~— 29.0. Determine the 95% upper estimate for a. Do not forget the fpcf= 1/(1\1‘— n)/(N— 1) )in the calculation of S D (x) (a) 30.12 (1))” 29.71 :1 (c) 29. 35 (d) 29.55 (e) 29.96 71.21% “#3 11?: fl. Consider the confidence interval estimate 2? i 2,1,20' 1’ J; of a population mean ,u. Which case has the narrowest interval? . H4 (a) 95% confidence, n= 100 (b) 95% confidence, n= 400 (c) 99% confidence n= 1001;: (d) 90% confidence, n= 400 "— "4.1..- ,5, m I‘VE. _“,,I._._.m--.--.-- _1_§. The Food and Drug Administration (FDA) needs to estimate the average content of an additive in a given food product. A random sample of 121 portions of the product has average 8. 9 units and standard deviation 0. 66 units. The 95% confidence interval for the average number of units of additive 1n the population of portions of this foo mprpduct has endpoints 8. 9 + (a) 0.19 (b) 1.60 (c) 0.04 \‘ 24334.1 34-41440 W p E2! “€032.45“ _1_§_. A random sample of 200 accounts is selected from a population of N = 600 accounts and audited. Forty of the sample accounts are found to have errors. Give a 90% confidence interval for the population proportion of accounts with errors. .w'” ”W (a) .20 4 .060 (@ij 033%) (c) .20 4 .047 (d) .20 4 .073 (e) .20 i .086 Argo”. 490—194: agave) :. + l: ., so .lDi-LMS 670-! @4335”— .20... .038 17 - 18. The mean usable space ,u for utility poles in Michigan is to be estimated with a 99% confidence interval X i Za/ZJ/WfT—l where e = 15 ft. A simple random sample of size n = 100 poles is selected and the average usable space on the sample poles is found to be 14.5 fl. 17. The 99% confidence interval estimate for ,u is 14.5 ft i (a) 3.1;": (b) 2.41% (c) 5.7ft (d)1.2fi @ .2) /%5f2.57éL5; :- [9.5’4- 39 11733 '- 18. What sample size will reduce the margin of error by 50%? / \ (a) 144 (b) 200 (’1' (c) 400 (d) 266 (e) 346 -_.._____ @Mflfwfle n xeesa lee $1: 51199 «.9154 “6496* Me by 559% 19 - 20. A company that conducts surveys of current jobs for executives wants to estimate the average salary of an executive at a given level to within $1,000 with 95% confidence with the z-interval. From previous surveys it is known that the standard deviation of executive salaries is $12,000. 19. What is the minimum required sample size based on the book’s formula for sample size? (a) 231 (b) 144 (c) 312 ("31) 5549”,.) (e) 614 Mb: 1.4170?” 1213001: 553.3L 1 em?" 20. What is the minimum required sample size if the population of current jobs is size 1,200? (a) 431 ,r’fbrs'ff‘émfi (c) 212 (d) 243 (e) 196 ”16 m: aha-1 H "W a. A simple random sample n = 150 parts is selected from a bin containing 2,000 parts. The sample contains 12 defective parts so that the sample proportion is f3 = .08. The 95% z—confidence interval estimate for the proportion of parts in the bin that are defective is 0.08 i 0.04. Carefully consider the following statements and then answer the question. (ii) It IS probably true that 8% of the parts in the bin are defective .. (iii) We do not know the percentage of parts in the bin that are defective but [email protected] that it is between 4% and 12%. (iv) For about 95% of all simple random samples of size 150, the interval estimate captures the unknown p. Therefore, we can be about 95% confident that the interval estimate 0. 08 i 0. 04 contains p. Which of the above statements are correct statements? (a) (iii) only (b) (ii) only (c) (i) and (iii) only \Ei) (iv) only p‘1(e) (i) only Aflfiificmmu' .w ur- " 22 - 24. Carryout an 0. = 0.05 level test of H0: 11 = 65 v. H]: p. at 65 for a large population. Use the t-test and the sample statistics 11 =16, fobs = 71.7, S = 20. ‘ 22. The observed value of the test statistic is tabs = (a) .10 (b) 1.96 (c) .05 @3b (6) 2.58 .. .eéS” .4651 W? 3': (131}! erg/J7; 23. The p-value of tabs is closest to (You may use the z—table to compute this if you wish.) (b) .01 (c) .001 (d) .10 (e) .05 aéale‘é: 1122.12.10 = g—fall 5° aloobihe (a) .201 Frm 72611.? 3 ) we 24. What decision is made at level (I. = 0.05? (a) Reject H0: 11. = 65 in favor of H1: 11 ¢ 65 (x: (b) Retain Ho gnaw p—wiue :1' .10 >1 . 05”.“: Eve'i‘dm H‘D l—eads to the decision to Retain —.H0 Then a W”; W HT“ m“ "-~._, (a) Type I error has been made (b) Type 11 error has been mad\ (0) correct decision has been made H1 wWae we 5% a net-21.14,! 26 — 27. Consider testing H): p S .50 v. H1: p > .50 for a population of size N = 1,000. Suppose that a random sample of size n = 400 is drawn from a population and the sample has x055 *—“ 204 successes, that is, 33 = 1:13;; 2“ . 26. The value of the test statistic 20,” is (Do not forget the fiocf.) (a) 1.645 (b) 0.52 .. : (c) 1.96 (d) 1.47 (e) 1.03 51 -' 5b bé I“ j . two new ‘ _. area 27. What is the p-value of 20...? (a) 0.70 (b) 0.20 (c) 0.30 7 (d) 0.10 (e) 0.05 E. A company must decide whether the percentage of defectives in a large shipment is 20% or less than 20%. It decides to use a random sample of n = 20 parts to test H0: p = .20 v. Hg: p < .20. Let X denote the number of defectives in the sample. Suppose that xm = 3 so that p = 3% = .15. Use the Binomial to determine the p-value of x055 = 3. I”: ..,.m.._..._,l_fi_“' _,.r"""(';1) 0.411, .5 (b) 0.323 (c) 0.104 (d) 0.589 (e) 0.05 . “f1.“ .7 gem @mmai 80(535 X .51!“ :10 :10 If} "‘ ’1’ “12:01.4 1 2. A company must decide whether the percentage of defectives in a large shipment is 20% or less than 20%. It decides to use a random sample of n = 100 parts to test H0: p = .20 v. H1: p < .20. LetX denote the number of defectives in the sample. Suppose that x053 = 15 so that 33 = if?) = .15. Use the Binomial or normal approximation to determine the p- value of x01”. = 15, equivalently, the p—value of [6 = .15. . "”""~>-~.... (a) about .03 (b) about .09 (0) about .12 3 (d) about .24 (e) about .18 a; F Fm TI’ b\noMCoQ'P(lGO).ZJ\5 5 I.” 3(23‘5’ lie! “42.0 .. “hag Named cappwx =' 2': ammo.) "' ii“??? W275. 4.75) z“ . 105 e 10. Which of these statements is correct? (a) The p-value of a test statistic is the probability that the null hypothesis H0 is true. (b) The p-value of a test statistic is the probability that the alternative hypothesis H1 is true. The p-value of a test statistic is a kind of “credibility rating” for H0 in light of the evidence. d) A large p-value leads to Rejection of H0. ...
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