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Chap 9 answers

The Practice of Statistics: TI-83/89 Graphing Calculator Enhanced

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Unformatted text preview: Sampling Distributions 9.1 p. = 2.513133 is a parameter; E = 2.5009 is a statistic. 9.2 i5 = 2.2% is a statistic. 9.3 f: = 48% is a statistic; p = 52% is a parameter. 9.4 Both E1 = 335 and x2 = 239 are statistics. 9.5 {a} Since the proportion of times the toast will land butter-side clown is I15, the result of 2i] coin flips will simulate the outcomes of 213 pieces of falling toast {landing butter-side up or hut— terasicle clown]. {b} Answers will varv. {c} finswers will vary; however, it is more likely that the center of this distribution will be close to {15. and it is more likely that the shape will be close to normal. [11} Answers will vary. {e} We obtain a more accurate representation of a sampling distribution when many.r samples are taken. [9.15 [a] The results appear to be quite variable. {'3} Sampling Distributions I 55 {C} The median and mean are extremelyr close. {:1} The spread of the distribution did not seem to change. To decrease the spread, I would increase the number of trials, n. For example, use randBin tso , . 5} . 9.? {a} The scores will vary depending on the starting row. Note that the smallest possible mean is 6].?5 {from the sample 53, 62, 62, 65] and the largest is T125 {from T3, TM, SI], 82}. {bJ—[c} Answers Will vary; shown below are two 1tier-its of the sampling distribution. The first shows all possible values of the experiment {so the first rectangle is for 61.?5, the nest is for 62.1342], etc.}; the other shows values grouped from 61 to 61.?5, 62 to 62.?5, etc. {which makes the histogram less bumpy]. The tallest rectangle in the first picture is 3 units; in the second, the tallest is 23 units. 6] 5'5 69.4 T125 61 69.4 T? {:1} There are {16 X 9V2 = 45 possible samples of size 2 that can be drawn from the population. {e} The shapes and centers for the two distributions are roughly the same. However, the spread is a little larger for the distribution corresponding to n = 2. This distribution is some; what more irregular, reflecting the fact that sample means based on samples of size 2 tend to be more variable than those based on samples of size 4. 9.3 [a] Table is on the next page; histogram not shown. {b} The histogram actuallyr does not appear to have a normal shape. The sampling distribution is quite normal in appearance, but even a 156 1 Chapter 9 sample of size 100 does not necessarily show it. {c} The mean of 10 is 0.0901. The bias seems to be small. {d} The mean of the sampling distribution should be p = 0.10. {e} The mean would still be 0.10, but the spread would be smaller. in 0 Count p 11 Count p is Count 9 0.045 1 10 0.090 12 24 0.120 10 13 0.005 3 19 0.095 9 25 0.125 4. 14 0.020 2 20 0.100 2 20 0.130 1 15 0.025 5 21 0.105 5 22 0.135 2 10 0.000 11 22 0.110 0 20 0.140 2 12 0.005 12 23 0.115 2 30 0.150 1 9.9 {a} Below, left. [b] For the 22 survival times, is = 141.042 day's. [cl Means will varyr with sam- ples. {d} It would be unlikely [though not impossible] for all five E values to fall on the same side of pr. This is one implication of the unbiasedness of f: Some values will be higher and some lower than it. {But note, it is not necessarily half and half.) {e} Shown (below, right] is a stemplot for one set of 100 sample meanimatcs the sampling distribution of f . This set of means var- ied from 05 to 225 days and had mean 130.1 and standard deviation 25.9 days. The mean of the {theoretical} sampling distribution would be H; {f} Answers will vary. ' II’ 0 5 9 1209 10 1133509999 11 0000111302209 12 245500209 13 01112233000009999 14 00001111223009 l5 001223355502009 n 100 an son an an sea 15 92343 Survival time {days} E £455? 19 223 20 21 22 5 Frequency 9.10 {a} Large bias and large variability. {b} Small bias and small variability. {:2} Small bias. large vari- ‘abilitv. {:21} Large bias, small variability. 9.11 [:11 Since the smallest number of total tax returns [i.e., the smallest population} is still more than 100 times the sample size. the variability will be [approximately] the same for all states. [b] Yes, it will change—the sample talren from Wyoming will be about thesame size, but the sam— ple in. e.g., California will be oonsiderably larger, and therefore the variability will decrease. 9.12 E = 04.5 is a statistic; ,u. = 03 is a parameter. 9.13 f) = 4.5% = .045 is a statistic. 9.14 {3] Use digits 0 and 1 [or any other 2 of the 10 digits} to represent the presence of egg mass es. Reading the first 10 digits from line 110, for example, gives YNNHN WWW—2 square yards with egg masses, 0 without—so B = 0.2. Sampling Distributions 15? {h} The stemplot might look like the one below {which is close to the sampling distribution of ,5}. [c] The mean would he {1 = [1.2. {d} '14. [LID Eli] [Li] 55555 [3.1 nuance [L1 5555 [3.2 CID [1.2 5 5.15 {a} a = 5.5. or = 1.555. n=.1i 5555 {h} This is equivalent to rolling a pair of fair, sixrsidcd dice. {c} SR5 of size 2 5 Lu 1 aims L5 altars 2 t1l312L4 15 11A4l3$2L5 5 a11514,352Ls 55 a23,5a45,5as 4 a5h554,s 45 a4,5as 5 a5,5 5 asF s . . . .5: {d} Histogram below. The center is identical to that of the population distribution. The shape 1: 585% is normal {symmetric and hell-shaped} rather than uniform. WM that - of the population distribution; the probability of observing a value at some distance fromdthc center remains constant for the population distribution but decreases with increasing dis- tance for the histogram corresponding to n = 2. \J'l - - wan ;.=.'_:Wuflj No.0 2???? 158 Chapter 9 9.16 Answers will vary. 6 sample histogram is shown below. 1|While the center of the distribution remains the same, the spread is smaller than that of the histogram in Exercise 9.15. max-{1.5! 333 3: has. 9.1? Assuming that the poll’s sample size was less than 289,999—1996 of the population of New Jersey—the variability would be practically the same for either population. [The sample size for this poll would have been considerably less than 289,999.} 9.18 {a} The digits 1 to 41 are assigned to adults who say that they have watched Survivor 11. The pro gram outputs a proportion of “Yes"r answers. For {b}, {c}, {d}, and {e}, answers will vary; however. as the sample size increases from 5 to 25 to 199, the variability of the sample proportions should decrease. v.19 (ally. = o = or; o- = 1.; 9 = V—mn—mlml] = 9.9144. {b} The population {all 1.1.8. adults} is clearly at least 19 times as large as the sample {the 1912 surveyed adults}. {c} up = [1912} {a} = 298.4 a is; n{1 — p} = [1912] {a} = was a is. [d] PH: E .62} = HE E — 9.25} = 9.9186 —this is a fairly unusual result if 2996 of the pop; ulation actually drinks the cereal milk. [a] Multiply the sample size by 4; we would need to sample {191214} = 4948 adults. {f} It would probably be higher, since teenagers {and children in general} have a greater ten- dency to drink the cereal milk. ' 9.29 {a} ,u. = p = 9.4, o' = V[9.4}[9.51 + 1'28 = 9.9116. [b] The population {U.S. adults} is con- siderably larger than 19 times the sample size. {c} up = 214, of] — p} = 1921—both are much big- ger than 19. {d} 1319.32 at”. f} s: 9.43} = Pi—2.586 a: Z a: 2.586) = 9.9994. Over 9996 of all samples should give 8 within i396 of the true population proportion. 9.21 For n = 399: o' = 9.92828 and P = 9.2198. For n = 1299: o' = 9.91414 and P = 9.9669. For n = 4899: or = 9.9929? and P = 1 [approximately]. Larger sample sizes give more accurate results {the sample proportions are more likely to be close to the true proportion]. 9.22 {a} The distribution is approximately normal with mean ,u = p = 9.14 and standard deviation . a = W“ .; t = V—Ja—W = 9.9155. {b} ease. or more Harley owners is unlikely; Hi:- 2:- 43.213} = P{Z “a 3.82] a: 9.9992. There is a fairly good chance of finding at least 1596 Harley owners; 1‘18 :2- 9.15312: P{Z :5 9.64] = 9.2611. 9.23 {a} 9.86 [8696]. [b] We use the normal approximation {Rule of Thumb 2 is just satisfied— n|[1 — p} = 19}. The standard deviation is 9.93, and PH: E 9.86} = HE E —1.33} = 9.9918. (Note: The exact probability is 9.1239.) {c} Even when the claim is correct, there will be some variation in sample proportions. In particular. in about 1996 of samples we can expect to observe 86 or fewer orders shipped on time. 9.24 The calculation for Exercise 9.22 should be more accurate. This calculation is based on a larg- er sample size {599. as opposed to the 199 of Exercise 9.23}. Rule of Thumb 2 is easily satisfied in Exercise 9.22 but just barely satisfied in Exercise 9.23. Sampling Distributlons 159 9.23 1a} .11- = p = 11.15, or = 1211115111135} + 15411 = 11.111191. 1b} The population 11.1.3. adults} is considerably larger than 111 times the sample size 115411}. 1e] no = 231, 1111 - p} = 13119—both are much bigger than 11]. 1d} 19111.13 *5. fi s: 11.1?) = P1—2.19l3 4: Z sf. 2.193} = 11.9222. [e] To achieve or = .4113, we need a sample nine times as large-about 13,8611. 9.26 For n = 21111: or = 11.112525, and the probability is P = 11.571114. For n = 80-11: or = 11.111262 and P = 11.114358. For n = 321111: a = 11.11631 and P = 11.9934. Larger sample sizes give more accurate results 1the sample proportions are more likely to be close to the true proportion}. 9.2? as 4: .sss) = P(2 a 1”“ -" ) = P12 :_= .8165} = .2921. {The exact answer is .213.) 9.23 {a} p. = 11.52, or = 11.112234. 1b} up and i111 — p} are 2611 and 2411 respectively. F16 a 11.511] = P12 2 —11.3951]| = 11.8159. 9.29 {a} P113 5 11.211} = P12 5 —l.155]| = 11.1241. [b11011] 5 11.211) = P12 5 —l.826} = 11.11339. {e} The test must contain 41111 questions. 1d} The answer is the same for Laura. 9.311 1a) up = 11511113} = 4.5—this fails Rule of Thumb 2. 1b} The population size 1316) is not at least 111 times as large as the sample size 1311}—this fails Rule of Thumb 1. 1e} P1X 5 3} = binomcdfllfi, .3, .31 = .2969. 9.31 13111 = "3.51111, 1r 2 assess/s = 11.62B%.1b}P1}1 a 5%] =sz a 11.3269]=.3219.1eiP1§ “3—” 5911] = P12 2 11.23} = 11.2321. {(1} F12 s: 11} = P12 s: [1.311) = 11.6179. Approximately 62 91:1 of all five- stoek portfolios lost money. 9.32 1a1P1X a 21] = P12 3: 11.41168} = 113421.1an = 18.6, o' = 3.919611 = 11.6344. This result is independent of distribution shape. 1e} P1 i 2 21} :5 P12 2 2.87164} = 9.1111211. 9.33 1a} ofV'3 é 5.71135 mg.1b}So1ye 911% = 31V? 2 131159 11 = 11-1 91 12‘ The average of sev- eral measurements is more likely than a single measurement to be close to the mean. 9. 34 1a} 11 we choose many samples, the average of the i-yalues from these samples will be elose to p.. 1I.e., '2 is “correct on the average” in many samples.) 1b} The larger sample will giye more information, and therefore more precise results; dust is, it is more likely to be close to the population truth. also, :7: for a larger sample is less affect- ed by outliers. 9.35 E has approximately a 1911.6, 11.11849} distribution; the probability is P12 :3 4.?l}—essenfially 11. 9.36 E {the mean return} has approximately a 191991., 4.12496} distribution; P1 E 3::- 15911} = F12 :2: 1.432] = fléigffi; P12 s: 596} = P12 sf. —11.9583} = 11.16911. 9.3? {a} 1111123, 11.114619]. 11)] P12 3:- 21.651—essentially 11. 9.33 1a} Mean: 411.125, standard deviation: 11.11111; normality is not needed. 1b} No: We eannot com- plete p [E 3:- 49.12?) based on a sample of size 4 because the sample size must be larger to justify use of the central limit theorem if the distribution type is unknown. ass 1a}P1X 4:. 295} = 912 a —1}= 611%. [sins 4: 2115:}: P12 e —2.44ss3 = anon. 9.411 [a] N1556116,451111y’VE} = N1556611, 1591}. {s} P12 4: —2.sn; = 11.11222. 9.41 1a] N122, 11.1941}. 16} as a 2} .. as e —1.oso4i = 11.1513. {e} as 4: 1,9) has a 4.425?) = oniss. 9.42 ,1 — 1.645m1v’r—i = 121513. -— - -r-a_.-n- -i-IH-ar'nl.‘n=. .umAmumfltm 160 Chapter 5 9.43 {a} p = 88% = .88 is a parameter; {3 = 713% = 8.?3 is a statistic. {bl}; = p = 8.88, or = V1“; ”1' = V—la—E'Mi' = 0.0001. 101140 a .03}: 13-12 a 1.5120}: 00040. There is a 1000 {one in ten} chance that an observation of :5 greater than or equal to the observed value of .58 will be seen. 9.44 [a], {b} Answers will vary. In one simulation, we obtained a total of 51 simulated values of ,8 less than or equal to 8.85, a percentage of 18%. {c} 0 has an approximately normal distribution. ,0. = 0 = 0.0.0 = “4—1.1“ = 04-81808 = 004501011010 :4 0.0510 H2 5 -1.0011= 0150010115 result is reasonably close to the 18% obtained in our simulation. {e} For n = 1888. ,8 is again approximately normal, 01111 a = 0 = 0.0. 0 = V81 .j ”l = Vat—‘8'” = 0.0145. 10.30 s 0.05; 4: HE E 41.45} = 8.8883. 1'": is less variable for larger sample sizes. so the probability of seeing a value of I: less than or equal to 8.85 decreases. 0.45 PIPE :4 20.4} = 2142 :4 02.5%] =4 21012 :0 1.004) = 2(0.1502) = .2004. _ 9.445 {a} 18 has anapproiiimately normal distribution with p. = p = 8.4T, o- = Vial .: M = V—ms—i' 8'] = 0.0150. {b} The middle 95% of all sample results will fall within 20' 0= 8.8512 of the mean 8.4T, that is, in the interval 8.4388 to 8.5812. {01-10110 4: 0.45144 P{Z 4: —1.200} = 0.0000. 9.4? The mean loss from fire. by definition. is the long-ten'n average of many observations of the random variable X = fire loss. The behavior of X is much less predictable if only a small number of observations are made. If only 12 policies were sold, then the company would have no protec- tion against the large expense that would be incurred if one of the 12 policyholders happened to lose his or her home. If thousands of policies were sold. then the average fire loss for these poliv cies would be far more likely to be close to a, and the company’s profit would not be endangered by the few large fire-loss payments that it would have to make. 0.40 Pf‘i :- 200} 2 13(2 :0 3%) a. 1-42 :0 0.55}‘= 0.0004. 0.40 {a} P[Z :0 Mir-SE]. = 0(2 :01): 0.50044. [15} Mean: 100; standard deviation: 100040. {c} P[Z :0 8005040) = 102 :4- 2.50201 = 0.00401. {01 The answer to {a} 00010 be quite different: {b} would be the same {it does not depend on normality at all]. The answer we gave for [c] would still be fairly reliable because of the central limit theorem. 9.58 {a} No—a count assumes only whole-number values, so it cannot be normally distributed. {b} 1011.5. 0.02055}. {0] PE :0 %) = r42 :0 1.2500} = 0.10500. 9.51 p. + 2.334153% = 1.4825. . 0.52 {a} 00 = (25000110141) = 5525. [b] 10X 2 0500) = 142 a wgfigfl) = 142 a —0.4545} = 0.0052. 0.53 1:13.520 4: 5 a: “—53; = 1401.002 4: z a: 2.5001 = 0.05300. ...
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