formulas at the AP Exam. Studets should be able to derive it from the expressio for margi of error E i the formula for the cofidece iterval, amely, E z p(1 p) Modelig Good Aswers Oce studets have completed D8, P9, P19, or E5, show them the model aswer as a example of what is expected o the AP Exam. The aswer to D8 illustrates how to use a chart of reasoably likely sample proportios i Display 8.. Cofidece itervals are foud usig the chart ad the formula ad compared i P9. The aswer to P19 illustrates a well-writte coclusio. The aswer to E5 shows how to hadle checkig of coditios whe there is o radomizatio. A PDF file cotaiig each questio ad its model aswer is available at www.keypress.com/keyolie. Solutios Discussio D1. a. The middle 95% of a samplig distributio for a biomial proportio ˆp is cut off by the two poits p 1.96 p(1 p) 5 1.96 5(1 5) 5 5.4 So the two values are 8 ad 9. b. There are several ways to do the secod part of this questio. You could covert 145 out of 5 to the sample proportio 9 ad say that because this proportio is t i the iterval i part a, it is t reasoably likely to get 145 out of 5 whe p 5. Alteratively, usig this formula for reasoably likely umbers of successes, p 1.96 p(1 p) 5(5) 1.96 5(5)(1 5) b. The TI-83 Plus or TI-84 Plus commad radbi(,.6,1) returs the umber of successes i 1 differet samples. I Fathom, create a ew collectio ad add 1 cases (Collectio New Cases). Double-click the collectio ad defie a attribute, Success, by the formula radombiomial(,.6). View the umber of successes i each of the 1 samples with a case table. Re-collect the 1 samples by selectig Reradomize from the Collectio meu. c. To geerate a Miitab dot plot for 1 samples of size, use the meus or eter the commads MTB > radom 1 C1; SUBC > biomial.6. MTB > dotplot C1 (To eter commads, you must eable commad laguage by clickig i the sessio widow ad choosig Eable Commad Laguage from the Editor meu.) 175.9 gives a iterval from about 154.1 to 195.9. Because 145 does t fall withi this iterval, it is ot a reasoably likely evet. D. a. Usig radom digits, let 1 through 6 represet successes ad the rest represet failures. Cout the umber of successes i digits. With a TI-83 Plus or TI-84 Plus calculator, you ca use the commad radbi(,.6). The calculator will retur a sigle umber, such as 3, to idicate 3 successes out of a sample size of with p.6. 15 18 1 4 7 3 33 Sample Total ( ; p.6) Each dot i the plot gives the umber of successes i a radom sample of size take from a populatio with 6% successes. 116 Sectio 8.1 Solutios Statistics i Actio Istructor s Guide, Volume
16 14 1 1 8 6 4 A typical Fathom histogram is created by draggig the success attribute from a case table to a graph. Collectio 1 Frequecy of Success Histogram 16 18 4 6 8 3 3 34 Success The most importat cocept to poit out here (oce agai!) is that results from radom samples vary. All of these results came from radom samples of size take from a populatio with 6% successes. d. Aswers will vary accordig to the results from the simulatio. For the dot plot i part c, % of the time there were 17 successes or fewer. Also, 1% of the time there were 3 successes or more. This is as close as we ca get to.5% without goig over. So it is reasoably likely to get ay umber of heads from 17 to 3. e. Theoretically, the reasoably likely sample proportios will be betwee p 1.96 p(1 p).6 1.96.6.4 or betwee.448 ad.75, or from about 18 successes to 3 successes, which is close to the results from the simulatio i part c. The aswers for D3 7 are from the completed chart (Display 8. i the studet book, reproduced as a blacklie master at the ed of this sectio). D3. Studets should look at the horizotal lie segmet they costructed i Activity 8.1b for p.6. It stretches from proportios.448 to.75, or from about 18 successes to 3 successes. Because 7 successes out of is icluded withi this horizotal lie segmet or iterval, the the aswer is yes. Gettig 7 people of Mexica origi from a radomly chose group of Hispaics is a reasoably likely evet. D4. No. The horizotal lie segmet at p goes from about 58 to.44, so a sample proportio of.6 is t a reasoably likely result for a populatio with oly 3% me. Note o D4: It may iterest your studets to kow that, o the other had, about 51% of people uder age 5 are male. [Source: www.cesus.gov] D5. As ca be see by drawig a vertical lie dowward from 34 out of or upward from a sample proportio of.85, the cofidece iterval is about 7% to 95%. (If you look closely, the vertical lie just barely misses 7%, so studets may say that the cofidece iterval is 75% to 95%.) Number of Successes i the Sample ( ) 8 16 4 3 1..9.8.7.6.4 p =.85 Proportio of Successes i the Populatio.4.6.7.8.9 1 Proportio of Successes i the Sample D6. As see from the vertical lie o the chart below, the populatios for which a sample proportio of is reasoably likely are 35% to 65%. This ca be writte as 5% 15%. (If you look closely, the vertical lie just barely misses 35% ad 65%; therefore studets may say the cofidece iterval is % to 6% ad the margi of error is 1%.) Number of Successes i the Sample ( ) 8 16 4 3 1..9.8.7.6.4 p = Proportio of Successes i the Populatio.4.6.7.8.9 1. Proportio of Successes i the Sample D7. You do t eed a cofidece iterval for ˆp because you already kow exactly what that is from our sample ad you kow that it probably would have bee differet if you had take Statistics i Actio Istructor s Guide, Volume Sectio 8.1 Solutios 117
a differet sample. What you wat is a iterval that has a good chace of capturig the true but ukow proportio of successes p i the populatio from which the sample was take. D8. a. I the chart that follows, the 95% cofidece iterval is 35% to 65%. (If you look closely, the vertical lie just barely misses 35% ad 65%, so studets may say the cofidece iterval is % to 6%.) b. Number of Successes i the Sample ( ) 8 16 4 3 1..9.8.7.6.4 p = Proportio of Successes i the Populatio.4.6.7.8.9 1. Proportio of Successes i the Sample Number of Successes i the Sample ( ) 8 16 4 3 1..9.8.7.6 p =.4 p = Proportio of Successes i the Populatio.4.6.7.8.9 1. Proportio of Successes i the Sample c. The horizotal lie segmet shows all the sample proportios ˆp that are reasoably likely for a populatio that has p 5% successes. Those sample proportios are 1.96 (1 ) 55 The edpoits of the horizotal lie segmet are the about 45 ad.655. d. The vertical lie segmet shows the populatio percetages that are i the cofidece iterval. That is, they are the populatios for which a sample proportio of is a reasoably likely result. Because it is about the same legth as the horizotal lie segmet ad is cetered at about the same poit, the edpoits of the vertical lie segmet also are about 45 ad.655. e. The 95% cofidece iterval is about 45 to.655, or about 34.5% to 65.5%. D9. If you sample from a biomial populatio with proportio of successes p ad create a samplig distributio, the reasoably likely sample proportios are those sample proportios i the middle 95% of the samplig distributio. Plausible populatio proportios refers to reasoig i the opposite directio. Suppose you have take a sample of size ad the sample cotais proportio of successes ˆp. The plausible populatio proportios are those populatio proportios for which your sample proportio, ˆp, is a reasoably likely result. D1. Yes. The sample is a simple radom sample from a biomial (success/failure) populatio. Both ˆp 4 ad (1 ˆp) 16 are at least 1. Fially, i a large city the umber of buses would be greater tha 1 times the size of the sample, which is 1(), or. D11. The quatity ˆp 4 is the umber of buses i the sample that have a safety violatio, ad the quatity (1 ˆp) 16 is the umber of buses i the sample that do ot have a safety violatio. I geeral, ˆp is the umber of successes i the sample, ad (1 ˆp) is the umber of failures. D1. No, differet umbers of eve digits i the sample give differet cofidece itervals. D13. a. If, for example, you keep ˆp fixed but double the sample size, the value of z * will eed to be multiplied by to keep the widths the same. The two itervals.6 1.96.6.4 ad.6.77.6.4 8 have the same legth, but the secod has a higher capture rate sice z * 1.96 correspods to a capture rate of 95%, but z *.77 correspods to a capture rate of about 99.4%. b. To have the same capture rate, z * must be the same. If, for example, you keep ˆp ad z * fixed but chage the sample size, the itervals will have differet widths. The two itervals.6 1.96.6.4 ad.6 1.96.6.4 8 have the same 95% capture rate, but the first iterval is loger. D14. Margis of error for samples of size 8 will be smaller because the horizotal lie segmets i a chart like Display 8. will be shorter. From Chapters 6 ad 7, studets should uderstad that the spread of the samplig distributio of the proportio of successes decreases as the sample size icreases. 118 Sectio 8.1 Solutios Statistics i Actio Istructor s Guide, Volume
D15. a. The edpoits of the horizotal lie segmets for 99% cofidece would be farther apart tha the edpoits for 95% cofidece. If you wat to cover 99% of the possible values of ˆp, you have to have a loger iterval:.576 p(1 p) rather tha 1.96 p(1 p) for each value of p. b. The margi of error would be larger for 99% cofidece tha for 95% cofidece. D16. The sample size appears i the deomiator of the fractio i the formula for the margi of error. Thus, as gets larger, that fractio, ad so the margi of error, gets smaller. D17. For E 1%, use 1.96 (1 ) 96.4 So you eed a sample size of 97, which costs 5(97), or $485. (It is customary always to roud up whe computig sample size.) For E 1%, use 1.96 (1 ).1 964 which costs 5(964), or $48,. For E %, use 1.96 (1 ).1 96, which costs 5(96,), or $4,8,. To cut the margi of error by 1 1 requires multiplyig the sample size, ad the cost, by 1. D18. E 1.96 5.45.98, or about 3% 168 D19. The term error attributable to samplig meas the same thig as samplig error or variatio due to samplig. It meas that whe we take radom samples from a give populatio, the values of ˆp do ot tur out to be the same each time ad usually are t equal to p. However, these values do ted to cluster aroud p. D. I additio to the variatio i samplig that results from takig a radom sample from a give populatio, geeral categories of sources of error iclude ot gettig a radom sample i the first place (such as volutary respose surveys where people phoe i to talk shows or write i to ewspapers) errors i codig or recordig the resposes (a poll taker misuderstads what a perso is sayig) gettig a ivalid respose from the people surveyed because they misuderstood the questio or because they did ot tell the truth about a cotroversial issue The formula for the margi of error takes ito accout oly the variatio that results from lookig at a radom sample ad ot at the etire populatio. It is geerally impossible to predict the bias that may result from the three sources of error above. Practice P1. a. The middle 95% of the samplig distributio for p.4 is bouded by the two poits p 1.96 p(1 p).4 1.96.4(1.4) 5.4 36 or 64 ad 36. b. Gettig 5 out of 5 is a sample proportio of. This is a reasoably likely evet from a populatio with p.4. P. Coditios: Radom samples are take idepedetly from a populatio whose proportio of successes is, ad the populatio is large eough that samplig with replacemet is a appropriate model. Model: Repeatedly take samples of size from a populatio with 3% successes. With radom digits, let 1 through 3 represet successes ad the rest represet failures. Cout the umber of successes i digits. With a TI-83 Plus or TI-84 Plus, use the commad radbi(,.3) ad repeatedly press Õ. Repetitio: Do this at least 1 times. Alteratively, the commad radbi(,.3,1) L1 stores the umbers of successes for 1 samples i list L1. Coclusio: Sort the 1 results i ascedig order (O the TI-83 Plus or TI-84 Plus, use the commad SortA(L1)) ad cut off the bottom.5% ad the top.5%. Those cutoff poits give the eds of the horizotal lie segmet for p. P3. The expected umber of residets without health isurace is 6 3. Reasoably likely umbers are betwee 3 6.84, or about to 4. I a radom sample of residets, you would expect betwee about 158 ad 178 residets with health isurace. Note o P4 9: Because studets are estimatig from the chart, they may ot get these umbers exactly. P4. If the coi is fair, p, so gettig aywhere from 14 to 6 heads is reasoably likely. P5. Look at the horizotal lie segmet costructed for p.65 i Display 8.. It stretches from proportios to.8 or from about to 3 successes. Withi this horizotal lie segmet, 33 successes out of is ot icluded, so the Statistics i Actio Istructor s Guide, Volume Sectio 8.1 Solutios 119
aswer is o. Whe p.65, gettig 33 out of still i school would be a rare evet. P6. from about 8 to 39 high school graduates; from about.7 to.975 P7. The vertical lie o the followig chart shows that the cofidece iterval is about 15% to %. Number of Successes i the Sample ( ) 8 16 4 3 1..9.8 p = 5.7.6.4 Proportio of Successes i the Populatio.4.6.7.8.9 1. Proportio of Successes i the Sample P8. The populatios for which a sample proportio of.45 is reasoably likely are to.6. This ca be writte as.45 5. (The vertical lie at ˆp.45 just barely misses 3%, so studets may say the cofidece iterval is 5 to.6.) P9. a. Usig the chart i Display 8., the 95% cofidece iterval is about 5% to 75% whe the sample proportio is 5, or.65. Number of Successes i the Sample ( ) 8 16 4 3 1..9.8.7.6.4 p =.65 Proportio of Successes i the Populatio.4.6.7.8.9 1. Proportio of Successes i the Sample b. The cofidece iterval has edpoits at.65(1.65 1.96.65).65 5 So the 95% cofidece iterval is about.475 to.775, or 47.5% to 77.5%. P1. a. The problem states that the sample was selected radomly from all U.S. tees aged 13 to 17. Both ˆp 6.65 39 ad (1 ˆp) 6 5 1 are at least 1. There are well over 6 1 6 tees aged 13 to 17 i the Uited States. All three coditios are met. b. The 95% cofidece iterval is.65 1.96.65 5, 6 or betwee approximately 61.% ad 68.8%. The margi of error is approximately 3.8%. c. The 9% cofidece iterval is.65 1.645.65 5, 6 or betwee approximately 61.8% ad 68.%. The margi of error is approximately 3.%. d. I terms of the chart, the horizotal lies must be loger to capture the middle 95% of the sample proportios tha the middle 9%. Therefore the (vertical) 95% cofidece iterval is loger because it will itersect more horizotal lies tha that for a 9% cofidece iterval. Also, the calculatios are idetical except for the value of z*, so the iterval with the larger value of z* is loger. P11. a. The problem states that the sample was selected radomly from all U.S. tees aged 13 to 17. Both ˆp 6.4 4 ad (1 ˆp) 6.96 576 are at least 1. There are well over 6 1 6 tees aged 13 to 17 i the Uited States. All three coditios are met. b..4 1.96.4.96, 6 or betwee approximately.4% ad 5.6% c. The iterval is shorter for 4% successes tha for 65% successes eve though both sample sizes are 6. The legth of each iterval is determied by its margi of error, which i this case differs oly i the proportios i the umerator:.65 5 75 is greater tha.4.96.384. So all else beig equal, sample proportios closer to result i loger cofidece itervals tha proportios farther from. P1. 7. If coditios are met, the method used to costruct 9% cofidece itervals captures the true populatio proportio 9% of the time. Thus, you would expect 9% of the 8 cofidece itervals, or 7 cofidece itervals, to cotai the true populatio proportio of.6. P13. Oe. Assumig that the sample size is at least so that p ad (1 p) are both at least 1, ad that more tha mice are produced i a week so that the sample size is less tha 1% of the populatio size, you would expect 95% of the itervals costructed to capture the populatio proportio of.4, meaig that 5% of the samples, or oe sample, would ot capture this proportio. P14. For samples of size 1, the horizotal lie segmets will be shorter. Thus a cofidece iterval for a 1 Sectio 8.1 Solutios Statistics i Actio Istructor s Guide, Volume
sample of size 1 will be shorter because a vertical lie will cross fewer of the shorter horizotal lie segmets. Agai, remid studets that the spread of the samplig distributio of the proportio of successes decreases as the sample size icreases. P15. Use z* 1.8 because z 1.8 cuts off the outer 1% o either ed of a ormal distributio. The margi of error would be smaller for 8% cofidece tha for 95% cofidece because 1.8 ˆp(1 ˆp) is smaller tha 1.96 ˆp(1 ˆp). I terms of the chart, the lie segmet that covers the middle 95% of all possible sample proportios has to be loger tha the lie segmet that covers oly the middle 8%. P16. You should have used a sample size 9 times as big, or 9. The margi of error is give by the formula z* ˆp(1 ˆp) For this to be 1_ 3 as large as it was before, you must solve for the ew sample size, m, i the equatio 1 3 z * ˆp(1 ˆp) z* ˆp(1 ˆp) m which will give the result m 9. P17. Because you have o estimate for p, use p. a. For E % with 95% cofidece, use (z*) p(1 p) E 1.96 (1 ). 1 b. For E 1% with 99% cofidece, use (z*) p(1 p) E.576 (1 ).1 16,589.44 16,59 c. For E % with 9% cofidece, use (z*) p(1 p) E 1.645 (1 ).5 7,65 or 7,61 P18. Gallup should use a sample size of 879. (As usual, roud up.) (z*) p(1 p) E 1. 96 9.71.3 878.87 P19. a. There is o idicatio that the sample is radomly selected. The populatio of iterest is studets aged 1 to 17 who have Iteret access, ad 971 of these were surveyed. Both ˆp 971.76 737.96 ad (1 ˆp) 971 4 33.4 are at least 1. The populatio is about millio, much more tha 1 971, so the populatio is more tha 1 times the sample size. The secod two coditios are met, but you do t kow about the first. b. The 95% cofidece iterval is about 73.3% to 78.7%: ˆp z* ˆp(1 ˆp).76 1.96.76 4 971.76.7 c. the proportio of all studets aged 1 to 17 with Iteret access who would say they go olie to get ews or iformatio about curret evets d. If you could ask all studets aged 1 to 17 with Iteret access whether they go olie to get ews or iformatio about curret evets, you are 95% cofidet that the proportio who would say yes would be somewhere i the iterval 73.3% to 79.7%. e. Suppose you could take 1 radom samples from this populatio ad costruct the 1 resultig cofidece itervals. You would expect that the true proportio of all studets aged 1 to 17 with Iteret access who would say they go olie to get ews or iformatio about curret evets would be i 95 of these itervals. Exercises E1. a. No, a sample proportio of.9 is t a reasoably likely result for a populatio with 75% successes i the populatio. The horizotal lie segmet for p.75 does ot cotai a sample proportio of.9. b. Because 48% is close to 5%, you ca estimate pretty well by usig the lie segmet for the populatio proportio of 5%. The largest is about.65 ad the smallest about 5. c. about 3% to 55% Statistics i Actio Istructor s Guide, Volume Sectio 8.1 Solutios 11
E. a. Yes, the horizotal segmet for p. goes from about 5 to 5, so a sample proportio of is i this iterval ad is a reasoably likely result for a populatio with % successes. b. Usig the horizotal lie segmet for p 5 to estimate p 6, the largest reasoably likely sample proportio is about. c. Drawig a vertical lie dowward from 8 out of or upward from a sample proportio of ˆp 8.7 itersects horizotal lie segmets for populatio proportios of about 5 to.85. The cofidece iterval is approximately 55% to 85%. E3. a. About.5 to 5. (The lie for barely misses the sample proportio 5, so studets might give a iterval from.5 to.) b. Usig the formula gives 5 1.96 5(1 5) 5 11, or.39 to 61. c. They are similar but the iterval i part b is a bit off. There are several reasos for the slight differece i the cofidece itervals from parts a ad b. The chart would give a reasoably accurate cofidece iterval but it does t iclude all possible values of p. Also, ˆp 5 6 1, so the ormal approximatio to the biomial used i part b is ot reliable. (The cofidece iterval for 5 usig the exact biomial distributio is.5% to 3%.) E4. a. about 65% to 9% b..8 1.96.8(1.8).8 4, or 67.6% to 9.4% c. They are similar. There are several reasos for the slight differece i the cofidece itervals from parts a ad b. The chart would give a reasoably accurate cofidece iterval but it does t iclude all possible values of p. Also, (1 ˆp) (1.8) 8 1, so the ormal approximatio to the biomial used i part b is ot reliable. E5. a. This is a multistage samplig pla with o apparet radomizatio. The Epilepsy Foudatio selected affiliates (assumig they have more tha that), each of which selected schools, presumably i their local area. The surveys were passed out to studets i these schools. You probably should cosider this a atiowide coveiece sample. The lack of radomizatio makes it impossible to draw reliable coclusios from the sample. Note: The data beig weighted by age ad regio meas that, for example, if % of the tees i their sample were 14-year-olds from the South, but 4% of the tees atiowide are 14-year-olds from the South, they would double-cout the resposes of each 14-year-old from the South i the sample. b. No, there is o idicatio of a radom sample. However, ˆp 19,441 1 9,914.91 ad (1 ˆp) 19,441.49 9,56.9 are both greater tha 1, ad there are more tha 19,441 1 194,441 tees i the Uited States, so the other two coditios have bee met. c. yes E z* ˆp(1 ˆp) 1.96 1(1 1) 19,441.7.1 d. the proportio of all U.S. tees who kow that epilepsy is t cotagious e. Suppose you could take 1 radom samples from this populatio ad costruct the 1 resultig cofidece itervals. You d expect that the actual proportio of all U.S. tees who kow that epilepsy is t cotagious would be i 95 of these itervals. E6. a. You should woder whether the sample was selected radomly from all tees ad adults i the Uited States to see if coditios for a cofidece iterval are met. b. As stated i part a, the simple radom sample coditio may or may ot have bee met. Several sample proportios are listed. The most extreme (farthest from ) proportio listed for all tees is 9%. Sice ˆp 5.9 45 ad (1 ˆp) 5.91 455 are both at least 1, ad there are more tha 5 1 5 tees i the Uited States, the other two coditios for a cofidece iterval for each of the proportios of all tees are met. For the statemet that 4% of girls picked egieerig as the field that most iterested them, you ca t determie whether ˆp is greater tha 1 because you do t kow how may of the sampled tees were girls. However, the populatio is clearly more tha 1 times the sample size. For adults, ˆp ad (1 ˆp) are both at least 1 because the sample size is bigger tha for tees ad all the proportios are larger tha 9%. Also, there are more tha 1 13 adults i the Uited States. c. The largest sample proportio give for the tees was 33%, so the margi of error for tees would be at most 1.96 3.67 5.41. Givig the margi of error as 4% is slightly uderstatig it. For adults, the largest sample proportio give was 45%, the margi of error would be 1.96.45 5 13.34, which is very close to 3%. d. You are 95% cofidet that the proportio of all tees i the Uited States who thik 1 Sectio 8.1 Solutios Statistics i Actio Istructor s Guide, Volume
egieerig is a attractive career choice is betwee ad 8. e. Suppose you could take 1 radom samples from this populatio ad costruct the 1 resultig cofidece itervals. You d expect that the actual proportio of all U.S. tees who thik egieerig is a attractive career choice would be i 95 of these itervals. E7. The problem states that the sample was radomly selected. Both ˆp 549 85.48 ad (1 ˆp) 549.48 63.5 are both more tha 1. There are more tha 1 549 5,49 teeagers aged 13 to 17 i the Uited States, assumig this is the populatio from which the sample was take. All three coditios have bee met. The 9% cofidece iterval is 1.645.48 549.35, or about.485 to 55. You are 9% cofidet that if all tees aged 13 to 17 were asked, betwee 48.5% ad 55.5% of them would respod that it is appropriate for parets to istall a computer program limitig what tees ca access o the Iteret. E8. There is o idicatio of how the sample was selected, so the simple radom sample coditio may ot have bee met. However, ˆp 885. 354 ad (1 ˆp) 885.6 531 are both at least 1 ad there are more tha 885 1 885 adults atiowide, so the other two coditios are met. The 9% cofidece iterval is ˆp z * ˆp(1 ˆp). 1.645.4.6 885.4.7, or about 37.3% to 4.7%. Of course, you ca have cofidece i this iterval oly if the samplig was doe radomly. However, if you kew this, you could say that you are 95% cofidet that if all adults atiowide were asked this questio, betwee 37.3% ad 4.7% of them would respod that they were very worried that popular culture is lowerig the moral stadards i this coutry. E9. No, Display 8. applies oly to samples of size. E1. Yes, you ca use Display 8. to estimate the 95% cofidece iterval whe the sample proportio is.5 ad get a cofidece iterval from to. Although ˆp.5 is ot at least 1, the horizotal lie segmets i Display 8. for the populatio proportios of,.5,, 5, were made usig the biomial probability distributio formula that was leared i Sectio 6., ot by usig a ormal approximatio. E11. You are told that you ca assume the sample was radomly selected. Both ˆp.81(1) 81 ad (1 ˆp) 9(1) 19 are at least 1. Fially, the umber of adults i the Uited States is greater tha 1(1). Coditios are met. You are 95% cofidet that if you were to ask all adults from the geeral public whether they thought TV cotributed to a declie i family values, the proportio would be betwee.786 ad.834. The computatios follow: ˆp z * ˆp(1 ˆp).81 1.96.81(1.81) 1.81 43 E1. No, because the retur rate was oly about 9.4%, it is ulikely that the group who retured the surveys were a radom sample of Hollywood leaders. I geeral, those who feel strogly about issues ted to retur surveys. E13. a. The sample size appears i the deomiator of the fractio i the formula for the margi of error. Thus, as gets larger, that fractio gets smaller, ad so the width of the cofidece iterval gets smaller. b. The width of the cofidece iterval icreases. To have more cofidece that the iterval captures the true populatio percetage p, the iterval must be loger. This is see i the formula, as z * must be larger to have a larger probability of havig ˆp i the iterval aroud p. E14. The symbol p is used for the proportio of successes i the populatio from which we are drawig a sample. This is the ukow parameter the value that we are tryig to estimate. The symbol ˆp is used for the proportio of successes i a sample draw from the populatio with proportio of successes p. The value of ˆp varies from sample to sample. Whe costructig a cofidece iterval, the value of ˆp from the sample is at the ceter of the cofidece iterval ad so is always i it. The value of p may or may ot be i the cofidece iterval. E15. a. For a 95% cofidece iterval, the margi of error is 1.96 ˆp(1 ˆp) 1.96.48 549.4 The margi of error is actually closer to 4% tha 5%, but they may have reported 5% to avoid uderestimatig the error. b. For a margi of error of 3%, Gallup would eed a sample size of 168 because (z*) p(1 p) E 1.96 ().3 167.11 Statistics i Actio Istructor s Guide, Volume Sectio 8.1 Solutios 13
E16. Icrease the sample size by a factor of 16. Because quadruplig the sample size cuts the margi of error i half, quadruplig it agai cuts it by oe-fourth. To see this algebraically, suppose that a sample of size gives a margi of error E. The to get a error of E_, 4 the ew sample size would have to be z p(1p) (E 4) 16 z p(1 p) 16 E17. A, D, F, ad H are the correct iterpretatios. E18. D E19. a. For 95% cofidece, the margi of error would be 1.96 ˆp(1 ˆp) 1.96.67 3.13 5 or about 1.3%. b. We must solve for. 1.96.67 3. 1.96.67 3. 1.96.67 3. 1,344.44 You would eed at least 1,345 rus to get a margi of error less tha.. Note o E19b: It is difficult for some studets to see the basic simplicity of part b. This problem is similar to problems o the AP Exam. E. a. Coditios: The give proportios of defective items are correct, ad the defects o a lie ad betwee lies occur idepedetly of each other. Model: Choose oe digit to represet each selected item. For the first assembly lie, let represet a defective item ad 1 9 represet a acceptable item. For the secod lie, let or 1 represet a defective item ad 9 represet a acceptable item. Select te digits to represet the first assembly lie, followed by te digits to represet the secod lie. Record the total umber of defective items represeted i the set of digits. This questio does ot ask you to complete the simulatio so the Repetitio ad Coclusio portios are ot eeded. b. From the 1 rus, the estimate of the probability of seeig five or more defectives E is 1. For a margi of error of.1 ad 95% cofidece, the total umber of rus eeded is p(1 p) (z*) E 1.96 1.89.1 4748.4 You eed 4749 1, or 4649, additioal rus. E1. a. Because x(1 x) is i the form p(1 p), which we are tryig to maximize. Usig y ad x allows you to graph the fuctio o a graphig calculator ad makes the algebra seem more familiar. The domai of x is restricted because a probability ca be at most 1 ad must be at least. b. The graph of the quadratic y x x is a parabola that opes dow as show here. y x c. The maximum y-value occurs at the vertex. The vertex for a parabola y ax bx c occurs at x (a) b. Here, a 1 ad b 1, so the vertex 1 occurs at x (1) 1_. The value of y at x 1_ is y 1_ 1 1_ 1_ 4. If you graph this fuctio o a graphig calculator, you ca also locate the coordiates of the vertex usig the maximum fuctio. To use this fuctio, graph the equatio, press ND [CALC], select 4:MAXIMUM, select a left boud, right boud, ad guess, ad press Õ. E. The method used i this sectio is based o the idea that the middle 95% of sample proportios will be withi about 1.96 stadard errors of the populatio proportio. That 1.96 comes from a ormal model, so this idea depeds o the distributio of the sample proportio beig approximately ormal. As you leared i Chapter 7, larger samples give the distributio of the sample proportio a more ormal shape, ad whe the sample is large eough so that p ad (1 p) are both at least 1, the the ormal approximatio to the shape of the samplig distributio gives reasoably accurate results. Suppose, for example, a sample of size 5 has 3 successes, so that ˆp is.6. Usig the formula of this sectio, the 95% cofidece iterval would be (.6, 6). Obviously, these are ot all plausible values for the populatio proportio of successes, as this parameter caot be egative. 1 14 Sectio 8.1 Solutios Statistics i Actio Istructor s Guide, Volume
E3. a. The distributio of the values of ˆp has mea p ad ca be approximated by a ormal distributio with mea p provided both p ad (1 p) are at least 1. I a ormal distributio, 95% of all values lie withi 1.96 stadard errors of the mea. b. If 95% of all values lie withi 1.96 stadard errors of p, the there is a 95% chace that the proportio ˆp from a radom sample lies withi 1.96 stadard errors of p. Basically, parts a ad b are sayig the same thig. c. If ˆp is withi 1.96 stadard errors of p, the p is withi 1.96 stadard errors of ˆp. d. This formula says i algebra exactly what part c said i words: There is a 95% chace that p is withi 1.96 stadard errors of ˆp. Symbolically, the argumet may be writte this way. From the Cetral Limit Theorem you kow that whe is large eough, P p 1.96 p(1 p) ˆp p 1.96 p(1 p) Solvig for p i the iequality, this is equivalet to p(1 p) p ˆp 1.96 p(1 p).95 P ˆp 1.96 This is almost the formula for a cofidece iterval, except that you eed to use p i the formula for the stadard error ad you do t kow it. So estimate p with ˆp. E4. Results will vary. For a 98% cofidece iterval, use z *.33. Suppose, for example, that 16 of the studets carry backpacks. The, if you were to check all studets o the campus, you would be 98% cofidet that the proportio carryig backpacks would be i the cofidece iterval (, 8)..95 Statistics i Actio Istructor s Guide, Volume Sectio 8.1 Solutios 15