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Chap 10 answers

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

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Unformatted text preview: Introduction to Inference no.1 [3144% to sass. {b} We do not have information about the whole population; we only know about a small sam- ple. We expect our sample to give us a good estimate of the population value, but it will not be Esra-12th.-r correct. [e] The procedure used gives an estimate within 3 percentage points of the true value in 95% of all samples. 111.2 {a} The sampling distribution of i is normal with mean p. = 231] and standard deviation of‘ifii = fideEA—D m 2.1. {b} Below. [e] 2 standard deviations—Hm = 4.2. [d] Below; the confi— dence intervals drawn may vary, of course. {e} 95% {bv the 4513-95-99.? rule]. 2?}? ETEB 23TH 230 282.1 234.2 236.3 4—§—+ +——-—|—p. 1113 This is a statement about the mean score for all young men, not about individual seores. We are attempting only to estimate the center of the population distribution; the scores for individu- als are much more variable. Also, “‘95 ‘11: "' is not a probability or a proportion; it is a confidence level. 113.4 {a} The sampling distribution of E is normal 1with mean ,u and standard deviation admin: 13.415135— : snssss. {b} See the sketch on the next page. For this problem, the “numbers” below the axis would be ,u. -' {1.169'1'1, p. — l11.11314, p. — 11.115657, it, etc. {o} m = [1.11314 [2 standard deviations}. {d} 95 %. {e} See the next page. The actual confidence intervals drawn mayr vary. 161 IE: Chapter 10 p—J? p—Jl “-88 p p+lfi n+J1 F+JT 18.5 .84 i 8.18,, or .838 to .851 grams per liter. 18.8 11.28 i 8.22, or 11.81 to 12.55 years. 18.? [a] The distribution is slightly skewed to the right. {h} 224.882 i 8.829, or 223.923 to 224.831. 2239 81 2239 88288889 22 48 81 22 48 589 2241 2 18.8 i = 123.8 hur’acre, and of = 181’ V’E i 2.582 bufacre. {ch} See the table below; the intervals are x : z‘org. [8] The margin of error increases with the confidence level. lConf. _ Level 3' Interval 98911 1.845 119.8 to 128.8 bufaere 95% 1.988 118.? to 128.";I bufaere 99% 2.528 11?.1 to 138.5 bufacre 18. 9 With 11 = 88, o'—— — 182% = 1.291 bufacre. [a1959fi confidence interval. x- + 1.98817—= 121. 3 to 128. 3 hufacre {b} Smaller. with a larger sample comes more information which in turns gives less uncertainty 1“ noise“; about the value of 11.. {c} The}:r will be narrower. , 18.18 {a} 294 to 318.8. {1:} 224.? to 332.9. {c} Below—increasing confidence makes the interval wider. wo—o—p- 90% qv———————-—-—¢——-———————*-F' 9998 '-~—-——-——~ 99.9% "_1—"__|—l—'—‘—r—" 225 388 325 358 18.1 1 TIM—which is half the margin of error with n = 28. lntroductlon to Inference 163 10.12 {a} ionozos to 10.00251. {b} 22 {21.64}. 10.13 ssrssasi. 111.14 as [345?]. 10.15 {a} The computations are correct. {b} Since the numbers are based on a voluntary response, rather than an SRS, the methods of this section cannot be used—the interval does not apply to the whole population. 10.16 {a} The interval was based on a method that gives correct results 95% of the time. {1)} Since the margin of error was 2%, the true value of fl could be as low as 49%. The con— fidence interval thus contains some values of p, which give the election to Bush. {c} The proportion of voters that favor Gore is not random—either a majority favors Gore, or they don’t. Discussing probabilities about this proportion has little meaning: the “probar bilitf’ the politician asked about is either 1 or 0 [respectively]. 10.1? {a} We can be 99% confident that between 63% and 69% of all adults favor such an amend- ment. We estimate the standard deviation of the distribution of f; to be about V{0.66}{0.34)i’lfififi = 0.01161; dividing 0.03 [the margin of error] by this gives a' = 2.53, the critical value for a 99% confidence interval. 'x. H. {b} The survey excludes people without telephones {a large percentage of whom would be poor}, so this group would be underrepresented. Also, Alaska and Hawaii are not included in the sample. 10.1 S No: The interval refers to the mean math score, not to individual scores, which will be much more variable (indeed, if more than 95% of students score below 420, they are not doing very well}. 10.19 If we chose many samples of size 1543, then in about 95% of those samples, the percent- age found would be within :3 percentage points of the true population percentage. That is, we are using a procedure that gives results within :3% of the true percentage about 95% ofthe time. 10.20 {a} The intended population is hotel managers {perhaps specifically managers of hotels of the particular size range mentioned). However, because the sample came entirely from Chicago and Detroit, it may not do a good job of representing that larger population. There is also the problem of voluntary response. [b] The central limit theorem allows us to say that i is approximately normal for a sample of this size {n = 135] no matter what the parent distribution is. {c} 5.101 to 5.6.91. {is} one to 4.326. 10.21 The sample size for women was more than twice as large as that for men. Larger sample sizes lead to smaller margins of error {with the same confidence level) 10.22 {a} The stemplot {see next page) shows no marked deviations from normality. {b} E E 25.6? ami’hr, so the 90% confidence interval is 25.6? i 3.10 = 22.5?r to 20.27 amfhr {c} Her interval is wider: To be more confident that our interval includes the true popula- tion parameter, we must allow a larger margin of error. So the margin of error for 95% con- fidence is larger than for 90% confidence. 1 64 Chapter I D 2 1121.23 n = (m) : isms—taken = 1:94. 113.24 {a} 1.95::va = 2.352 points. {b} 1.9t'io-IV’W i 2.438 points. {c} n = {1—155}: 5 61.4?— take 11 = 62, which is under the lflflrstudent maximum. 111.25 {a} Sources of possible error mentioned in the account are: sampling error {i.e. error due to the random nature of the sampling process], Tvariations in the wording of questions [i.e. question bias}, and variations in the order of questions asked. [b] (211113.r sampling [random-chance} error is covered by the announced margin of error. The other sources of error are independent of the sampling process itself; the},r must be con- trolled by the questioner, e.g., by formulating unbiased questions that do not lead the sub- iect in a particular direction. was (a) ual 33:12. 552} 1’3 The 99% confidence interval for the mean number of years for the hotel managers is {11111, 12.5 5}. 10.2? {a} N{115, s}. {b} The actual result [see facing page} lies out toward the high tail of the curve, while 113.6 is fairl},f close to the middle. Assuming 1-1,:r is true, observing a value like 118.6 would not be surprising, but 125.?' is less iikelv, and therefore provides evidence against H]. Introduction to Inference 165 {C1 51? 1113 1118' 115 121 12?r 133 111.28 {a} N{31 ‘11., 1.518%}. {b} The lower percentage lies out in the low tail of the curve, while 311.286 is fairly close to the middle. Assuming HE is true, observing a value like 38.2% would not be surprising, but 27.6% is unlikely, and therefore presides evidence against H]. [c] Below. 26.446 22.964 29.482 31 '16 32.518 34.1136 33.5154 111.29 HQ: 11'. = 5 mm; H“: .11. 3'5 5mm. 111.311 H“: ,u. = $42,518]; H Ir: p. 3:- $42,581]. 111.31 1-1,}: p. = 511: H511 <1“. 511. 111.32 H0: 1.. = 2.6; 11,: ,u. s 213. 111.33 [a] The P—values are 11.2243 and 11.11323, respectively. [b] i = 118.6 is significant at neither level; E = 125.1r is significant at the 11.115 level, but not at the 8.111 level. 111.34 {:11 The P-values are 11.2991 and 11.11125, respectively. {b} i = 22.6 is significant at the 11.115 level. but not at the 0.111 level. 111.35 {a} i = 3‘38. {b} [t is normal because the population distribution is normal. {c} 11.81115. {d} It is significant at or = 11.115, but not at o = 11.111. This is pretty.r convincing evidence against HE. 398 1915.34 nssv 334.91 351 313115 392.11 411.115 1 56 Chapter 10 10.36 {a} Because the sample size is large {central limit theorem]. {b} 0.1004. {c} Not significant at or = 0.05. The study gives some evidence of increased compensation, but it is not very strong—it would happen 10% of the time just by chance. —l6.1'i'96 —10.'?E64 —5.3932 0 5.3932 10.2864 16.1296 10.3? Comparing men’s and women’s earnings for our sample, we observe a difference so large that it would only occur in 3.3% of all samples if men and women actually eamed the same amount. Based on this, we conclude that men earn more. _ While there is almost certainly some difference between earnings of black and white students in our sample, it is relatively small—if blacks and whites actually earn the same amount, we would still observe a difference as big as what we saw almost half {42.6%} of the time. 10.30 [a] Hui p. = 11.5 vs. 1:11,: p. a5 11.5. [b] a = 0.362. {c} P = 05214. This is reasonable variation when the null hypothesis is true, so we do not reject H“. 10.351 {a} HE: ,u = 300 vs. H“: p. s: 300. {b} .1 = —0.?893. {c} P = 0.2150—this is reasonable variation when the null hypothesis is true. so we do not reject Ha. 10.40 {a} z = —2.200. {1:} Yes, because lzl :a- 1.960. {c} No, because Isl 1: 2.526. {d} lal lies between 2.054 and 2.326. The P—value falls between 0.02 and 0.04. 10.41 {a} The command rand|[100]l —> LI generates 100 random numbers in the interval {0,1} and stores them in list L]. Here's a histogram of our simulation, and the 1-variab1e statistics {your results will be slightly different}: The z test statistic is z = —.516, and the P—value is 0.6058. We fail to reject flu. E-Teat u#.5 Z—Teet In?“ Data m E=.635??93639 x=.4351 n=1EIEi DE :5 KICI Introduction to Inference 16? There is no evidence to suggest that the mean of the random numbers generated is differ- ent from 11.5. 111.42 {a} Yes, because 2 1: 1.645. {'13} Yes, because z :5 2.326. {c} The value of 5 lies between 2.526 and 2.526. The P-value thus lies between 11.11115 and 11.111. 111.45 {a} 99.66 to 1116.41. {b} Because 1115 falls in this 91196 confidence interval. we cannot rciect Hail-'1 = 1115 in favor of H”: ,u. 515 11:15. 16.44 [a] With 5 2 165.64, our 55% confidence interval for p. is 165.64 i [1.966] (my?) e 165.64 a 5.26, or 11111.56 to 111.1219 points. {b} Our hypotheses are H”: Fr 2 11111 versus H“: pt. 515 11111. Since 11111 does not fall in the 9556 confidence interval, we reiect HD at the 5'16 level in favor of the two~sided alternative. {6} Since the scores all came from a single school, the sample may not be representative of the school district as a whole. 111.45 {:1} NW, {1.11559Hbelow}. {b} E = {1.27 lies out in the tail of the curve, while 11.119 is fairly close to the middle. Assuming 1-1,;I is true, observing a value like 11.119 would not be surprising, but 11.2? is unlikely, and therefore provides evidence against H“. {c} P = 11.4214 [the shaded region below}. {d} P = {1.1111'3. 41.541211? 41.22626 —11.1153-9 11 11.11539 11.22626 11.341117r 111.46 HE: is. = 12511 v5.11“: p. 5: 12511. 111.47 Hg: ,6. =13 95.1.1": p. {13. 111.48 Hypotheses: Hui III. = —11.545 vs. H“: is ‘2:- —11.545. Test statistic: z = 1.952. vaalue: P = 11.11252. We conclude that the mean freezing point really is higher, and thus the supplier is appar- ently adding water. 111.49 {a} No, because lzl c: 1.961}. 1b} No, because lel 5: 1.645. 111.511 P = 11.1292. Although this sample showed some difference in market share between pio- neers with patents or trade secrets and those without, the difference was small enough that it could have arisen merely by chance. The observed difference would occur in about 1596 of all samples even if there were no difference between the two types of pioneer companies. 16.51 Significance at the 1‘16 level means that the P-value for the test is less than {1.111. So, it must also be less than 11.115. A result that is significant at the 596 level, by contrast, may or may not be significant at the 196 level. w- 111.52 The explanation is not correct; either H9 is true {in which case the “probability” that 111,,I is true equals 1} or H0 is false {in which case this “probability” is 11]. “Statistically significant at the 1 68 Chapter 1 ID or = H.135 level” means that if H{, is true, we have observed outcomes that occur less than 5% of the time. “3.53 {a} Reject jf-fu if 2 “L:- l.ti45. [b]: Reject HQ, if lrl 1?:- l.%. {c} For tests at a fixed significance level {a}, we reject H0 when we observe values of our statistic that are so extreme {far from the mean, or other “center” of the sampling distribution} that they would rarely occur when Hg is true. (Specifically, they occur with probability no greater than e.) For a two-sided alternative, we split the rejection region—this set of extreme values—into two pieces, while with a one-sided altema- tive, all the extreme values are in one piece, which is twice as large [in area} as either of me two pieces used for the two-sided test. 113.54 {a} Test flu: is = '3 mg vs. H“: ,u. at F mg; since 5' is not in the interval {1.9 to 6.5 mg], we have evidence against Hg. {b} No, since 5 is in the interval. 10.55 {a} H“: ,u. = 450, Ha: n L? 450. [h] z = 2.46. {c} P = H.001 This represents rather strong evi- dence against the null hypothesis; if 1% were true, then the observed value of E would occur in less than 1 it of all samples. We therefore reject Hg. There is strong evidence against the claim that the mean math score for all California seniors is no more than 455]. 111156 {a} In Example 1t'JLl4, H“: p. = NS, Ha: ,u. -=: NS, and cr = so. Specifying a r test and enter- ing 2?? for the sample mean, the THEE screens that specify the information and present the results of the test are shown below. We specified “Calculate.” Z—Teat Inet=Data nn=2?5 =63 =2?2 =34B =¢un hen aloula 9 Draw E 1: I1 p. I: The results, 2 = — 1.45 and P-value = .UTEE, agree with the results in Example 10.14. {c} llfl: nu = Sill}, II,: p. a: SIDE], and o- = 3. Entering the data into list L1 and specifying a z test, here are the T133 screens that specify the information and present the results of the test. Again, we specified “Calculate." E-Test u:#un ens Calcula e Draw The test statistic is z = -.?tl9, and the P—value is 0.215. There is insufficient evidence to conclude that the mean contents of cola bottles is less than Elli]. Introduction to Inference 169 113.5? {a} a = 1.64; not significant at 5% level [P = 0.35135}. {h} z = 1-65; significant at 5% level {P = 0.0495}. 13.53 {a} P = 3.5321. 331:- = 3.1214. (3)12 = 3.3314. 13.53 22 = 133452.24 3;. 533.23. r1 = 1,333: 435.35 to 433.15. in. = 13,333: 425.42 to 433.53. 10.1% bio—the percentage was based on a voluntary response sample, and so cannot be assumed to be a fair representation of the population. Such a poll is likely to draw a higher-than—actual pro- portion of people with a strong opinion, especially a strong negative opinion. 13.31 {a} No—in a sample of size 5131], we expect to see about 5 people who have a “P—value” of Lllll or less. These four might have ESP, or they may simply be among the "lucky” ones we expect to see. [b} The researcher should repeat the procedure on these four to see if they again perform well. 10.62 1°11 test of Significance answers question {b} Hindi We might conclude that customers prefer design A, but perhaps not “strongly."r Because the sample size is so large, this statistically significant difference may not be of any practical importance. 111164 {a} illl5. {b} Out of T? tests, we can expect to see about 3 or 4 {3:85, to be precise) significant tests at the 5% level. 113.155 {a} The P-values indicate that the results observed would happen very rarely if H, were true; specifically, if drivers with radar detectors were the same as those without, the_"before radar” speeds would be this different less than 1% of the time and “at radar” speeds would be this different less than [till it of the time. Since these results are so unlikely when HE, is true, we conclude that H] is not tnie. [b] In order for a 24mph difference to be statistically significant {with small sample sizes}, there must be little variation in the speeds; that is, those with radar detectors must all be driving at speeds very close to Ft] mph, whereas those without detectors are all traveling very close to 63 mph. ,-';.-.- ' | 13.33 {a}Hfl: the patient is ill {or “the patient should see a doctor”); H“: the patient is healthy [or “the l patient should not see a doctor”). A Type I error means a false negative—clearing a patient ' il. who should be referred to a doctor. A Type 11 error is a false positive—sending a healthy :4 patient to the doctor. 5] [b] One might Tivish to lower the probability of a false negative so that most ill patients are E treated. [in the other hand, if money is an issue, or there is concern about sending too many patients to see the doctor, lowering the probability of false positives might be :- desirable. . ‘l' 13.32 {:5} 3.31334 H, if z 4: —2.523. {b} 3.31 [the significance 1344111. {3} We accept H, a a s 223.135, 3:" so when p. = 223, arm II error} = 135 2 223.135} Hg,- 1 a W] = 3.4344. 3'5, fill." U 113.68 {a} 43.511]. {b} 13.184]. {c} flllflli. 5i 11169 {a} Type I error: concluding that the local mean income exceeds $45,flflfl when in fact it s does not [and therefore, Opening your restaurant in a locale that will not support it]. ! Type [I error. concluding that the local mean income does not exceed $45,0Ufl when in |3= | fact it does {and therefore, deciding not to open your restaurant in a locale that would support it}. I re Chapter I D {b} Type I error is more serious. If you opened your restaurant in an inappropriate area, then you would sustain a financial loss before you recognized the mistake. If you failed to open your restaurant in an appropriate area, then you would miss out on an opportunity to earn a profit there, but you would not necessarily lose money [e.g., if you chose another appro priate location in its place}. {c} Hg: ya = 45,131], H“: p. 333- 45,1130. {d} or = [1.1.11 would be most appropriate, because it would minimize your probability of com- mitting a Type I error. [e] In order to reject HIJ at level or = [1.1211, we must have .2 a 2.326, or E 2: 45,366 + [2.32615flflflr’ m] = 46645. The sample mean would have to be at least $46,645. a} P{faii to reiect H, when a. = when} = Pry e 46645 when a = 4-1606} = P{Z e mi = we «r. —e.s} = sass. When ,u = 4111131], the probability of committing a Type II error is .3635, or about 31%. 11].?[1 {a} Reject Hfl if i 2 11.52112. {13} 0.9666. 11].?‘1 z 2 2.326 is equivalent to E a 431} + 2.326 {lllflfmj i 46121.4, so the power is Pfl'eject Ha when p. = 461]] = PE} 2 461.131 when p. = 46E} =' 112 :- tigfi) = Hz a 6.6894) = 6.4644. This is quite a bit less than the “311% power” standard; this test is not very sensitive to a lflrpoint increase in the mean score. 1U.7"2 [a] Reject HEl if i £ 291985, so the power against p. = 299 is 3.2331 {b} The power against is. = 295 is 1.1.9926. {e} The power against is. —-— 291] would be greater—it is further from pin [Still], so it is easier to distinguish from the null hypothesis. less {a} asses. {in} 0.9543. 16.24 {a} Reiect if E a 2'99}? or E 5 2511.23. {b} Power: (1.85632. {c} 1 - [1.85632 = 13.14363. WIFE {a} We reiect H,J if E 2 131.46 or E E 124.54. Power: (1.9246. [b] Power: [1.9246 [same as (all. IDyer @696 of the time, this test will detect a difference of 6 [in either the positive or nega- tive direction]. {c} The power would be higher—it is easier to detect greater differences than smaller ones. 16.?6 A test having low power may do a good job of not incorrectly rejecting the null hypothesis, but it is likely to accept HIJ even when some alternative is correct, simply because it is difficult to distinguish between flEl and "nearby” alternatives. 1&2? {a} The test reiects HE, when Isl a 2.526. The test statistic is _ c.3404 — ass _ _ z _ seesaw“; _ 299' We therefore reiect 1-1,, {most emphatically: the P—value of the test is approximately 6 x lfl‘lfl. {b} The test reiects 1-1,, when either I a 2.526 {that is, i a {1.36 + {2.52...
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