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**Unformatted text preview: **Inference for Proportions 12.1 {a} Population: the 125 residents of Tonya’s dorm; p is the proportion who like the food.
{h} p = [1.28. 12.2 {a} The population is the 2413121 students at Glen's college, and p is the proportion who believe
tuition is too high. {12:} 6 = 1.1.26. 12.3 {a} The population is the 15,6111} alumni, and £2- is the proportion who support the president’s
decision. {h} ,6 = 1.1.33. 12.4 {a} No—the population is not large enough relative to the sample. {1)} Yes—we have an SRS,
the population is 48 times as large as the sample, and the success count {38} and failure count {12}
are both greater than 111. {e} No—there were only 5 or 6 "successes” in the sample. 12.5 {a} No—npn and oil — p0] are 1eSs than 111' [they both equal 5}. [b] No—the expected number
of failures is less than 1011:1111 — p”) = 2}. {c} Yes—we have an SR5, the population is more than 16
times as large as the sample, and nit-ﬂ = 1111 — p0} = 11]. 12.6 [a] SE15 = V(ﬂ.54)(ﬂ.46};’l1]19 # 11131561, so the 95% confidence interval is [1.54 :-
[1.96]{ﬂ.[11561} = 11.51 to 115?. The margin of error is about 3%, as stated. {1)} We weren't given sample sizes for each gender. (However, students who know enough
algebra can get a good estimate of those numbers by solving the system x + y = 11119 and
131.653: + 121.43? = 5511: approximately 5113 men and 511 women.) [c] The margin of error for women alone would be greater than [1.1313 since the sample size is
smaller. 12.?F {a} The methods can he used here, since we assume we have an SR5 from a large population,
and all relevant counts are morc than 11.1. For TVs in rooms: {31 i 1166 and SE5 = V’Eassjrras43na4s 2 1101463, so the 95% confidence interval a ass : {l.96}(ﬂ.ﬂl463) i [1.631 to [1.689. For preferring Fox: .62 i 11.13 and SEP; = V(ﬂ.13§?1§1.82)f11]48 5
6.61132, so the 95% conﬁdence interval is [1.13 : {1.96)[ﬂ.1.1118?} e 11.15? to 1121.13. {b} In both cases, the margin of error for a 95% confidenCe interval ["151 cases out of 26”} was
{no more than} 5%. {e} We test H, e = as versus H, a re as. The test statistic is e = {ass—asapv egg-51 e lass,
which gives very strong evidence against 11'n [P *4: {1.1313132}; we conclude that more than half of teenagers have T‘v’s in their rooms. (Additionally, the interval from {a} does not include
{11.51.} or less.) 1|With the T1433, z = 111329 and P = 1.57"? X 113—35. 12.8 {a} 1E- = .66, and since "fl = 132 and rail — fa} = 63 are both greater than 11;}, the confidenc interval based on z can he used. The 95% Confidence interval for p is
.66 i- {1.96} V{[.66}[.34j {2111]} = .66 i 13.136565, or 6.59435 to 6.22565. 139
--———# 1 911 Chapter 1 2 11)} Yes; the 95% confidence interval contains only values that are less than 11.23, so it is
lilrely that for this particular population, p differs from 11.23 (specifically, is less than 11.23}. 12. a {a} ,5= ” =11. 1236, and sE-= V1111 212114 e 11.11413. [6] Checking conditions, in: = 15
and H111 — 31m = 69 are both at least 111. Provided that there are at least [84) {p} = 34-11 applicants in the population of interest, we are safe constructing the confidence inerval. B i 1.645 SEﬁ =
11.11193 to 11.2423. 12.111 it = {lﬁsfﬂtﬁﬂtﬂ i 355.2—use n = 356. With 113 = 11.5, SE, 5 11.11265, so the true mar-
gin of error is {1.645111111121551 = 11.11436. 12.11 {a} l1151.2-round up to 11152. {b} 11162.1—round up to 11168; 16 additional people.
12.12 4511.2—round up to 451. 12.13 {a} We do not know that the examined records came from an 5515, so we must be cautious
in drawing emphatic conclusions. Both r112 , n11 — iii) are at least 111. ﬁ = 1511131 i 1.1.3163;
= V1311 — ﬁjg’l2ll = 11.111125; the interval is I: t 1.9611 SE,-r = 11.2942 to 11.33813. {b} No: We do not know; for example, what percentage ofcyclists who were not involved in fatal accidents had
alcohol in their systems. 12.14 ﬁ= 11.115222, 2': —3. 332, and P -=: 11.1111115—very strong evidence against H], and' m favor of
H5114: m. 12.15 (a) Checking conditions: we are treating the Gallup sample as an 5115; the population of
adults is much larger than 111112435]; r111 and n11 - a} are both at least 111. The interval is 39.11% to
45.11%. [b] Since 511% falls above the 99% confidence interval, this is strong evidence against'Hﬂ:
p = 11.5 in favor of 11,1 p c: 11.5- {In fact, z = —6.25 and P is tiny.) {c} 165139.4—round up to 165911.
The use of p' = 11.5 is reasonable because our conﬁdence interval shows that the actual p is in the
range 11.3 to 11.2. 12.16 {a} The distribution is approximately normal with mean p. = p = 11.14 and standard devia
tion or = V111.l4}[11.36)f922 é 1111113613. {b} For testing HE: is = 11.14 versus H,: p 2* 11.14, we have is = 11.22, and the test statistic is
a = [11.22 - ﬂ.l4)f11.111136l3 i 36. We have very strong {in fact, overwhelming] evidence
that Harleys are more likely to be stolen. 12.12 {a}1-1ﬂ:p = 11.5 vs. 11:: p :n- 11.5, 2 = 1.692, P = 0.11448—reject l-lﬂ at the 5% level. fb]11.51121
to 11.2329. {c} The coffee should be presented in random order—some should get the instant cof-
fee first. and others the fresh brewed first 1213 {a} n = iii—1113111. 21111. s}— — 421a fi—use n = 4219.11): 2525\2141-11‘1—92 0.111125. 12.19 {a} Let p= Shaq s free- throw percentage during the season following his off-season training.
We wish to test H0: 1: = .533 vs. Hg is 2:3 .533. 111211 = {3911.533} = 22.1182 and 1111 — pg] =
{3911.462} = 13.213 are both greater than 111, so the onesample I test may be used. The test statistic is a = [.662 — 53312121153311.4621239} = 1.623, and the P-value = 11.114215.
There is some evidence that Shaq has, in fact, improved his free-throw percentage. {b} Type I error: concluding that Shag has improved his free-throwing when in fact he has not.
Type 11 error. concluding that Shaq has not improved his free- throwing when in fact he has. {e} We seek the power of the test when p— — 11.6. With 132— — 11.115, we reiect H” in favor of H when z :4 1.645, that is, when ,5 :4 11533 + (1 .s4sitﬁf‘sﬂ—{}= 2,1239} 11.6644. Then
PIIeiectHuwhenp=11.6}= P113 32* 11. 6644 when p=1161=P12 33- 11. 321} = 11.21153. {d1 PtType 1 error} = or = 11.115; P{Type 11 error when p .= 11.6} = 1 — 11.211513 = 11.2942. Inference for Proportions 191 12.28 This exercise is a follow-up to Activity 12.
12.21 X n 16 z P—value 14 58 .28 —.252 .2261
98 358 .28 —1.998 .8233
148 588 .28 — 2.328 .8888 Although Tonya, Frank, and Sarah all recorded the same sample proportion, 8 = .28, the P—values were all quite different. Conclude: For a given sample proportion, the larger the sample
size, the smaller the Pavalue. 12.22 [a] s, = 11 e 11.1112sne1 s, = T12 = 11.411311. {1:1} SE = e 11.115433. ss the
95% conﬁdence interval is {8.1152 — 8.4162} 111.96}[8.86438J= —8.4292 to —8.1223. As the pml}
lem notes, this interval should at least apply to the two groups about whom we have information:
skaters {with and without wrist guards} with severe enough iniuries to go to the emergency room.
{c} 45,2286 i 21.8911 did not respond. If those who did not respond were different from those who
did, then not having them represented in our sample makes our conclusions suspect. 12.25 21 = 1—1 = 11.15 and 182 = 8 2 11.5222. Then SE = W s 11.12112 so the
98% confidence interval is [8.25 — 8.5882) I {1.645}|[8.1298] = —.8.8518 to 8.3253. These methods can be used because the populations of mice are certainlyr more than 18 times as large as the sam-
ples, and the counts of successes and failures are more than 5 in both samples. 12.24 The population-to-sample ratios are large enough. and all relevant counts are at least 18.
£1 = 8.5895, {11 = 8.3261. and the interval is —8.8288 to 8.14264. 12.25 To test 151,: e, = 11, versus 11,: e, :s 21,, we find ,5, = gas a 111111155915, 15;. = 31,—, e 1111114545 and s = ,J 3:1,. e 8.888586. Then SE = V511 411111111 + “11 e 111111211114, se .2 = 1.11, _ {ELISE % 8.98. This gives P = 8.1655, so we don't have enough evidenCe to conclude that the
death rates are different. 12.26 [a] Diagram below. {b} To test Ho: 11. = p, versUs H“: p. as p2, we ﬁnd 8. = % é 8.1249, 112 = 1155;, e 11.111515, and ,1. = $5121, 2 111111113. Then SE = V211 —ﬁ][ﬁ + Elm) 2
8.818892, so 2 = (in. — {12:12 815] % 2.23. This gives P = 8.8864, which is very,r strong evidence that
the proportions are different. {c} With the same hypotheses {for the pro ortions of deaths], this time we ﬁnd 11,: 1181 = 1111112122 = 23;, 2 11.1121, and 11: 133 1 1,, e 11.11125. Then SE
é 8.818949, so 2 i —8.16; this gives P = 8.8228—virtuall:-,r no evidence of a difference. Group 1 Treatment 1 / 1649 patients Aspirin \
Random Observe health assignment for 2 years
Grou 2 Treatment 2 /
p ——- Aspirin 121: 1658 patients (1'.de damn}: 12.22 For home computer access, we test H”: p, = 112 versus H“: 11, as p2 and find 151 —- % = 8.6565, 2, = 111%- : 11.15122, and ,1: = H 2 11151511. Then SE = V1111 — £11.41 + gigs) i
8.84394, so 2 = [81 - 11212813 i L8]. This givEs P =. 8.3124—little evidence that the proportions are
different. With the same hypotheses for the proportions with PC access at work, we ﬁnd 1 92 l(Chapter 1 2 in. = iii a (1.714534. rs = his a asses, and s = H a 0.6019. Then se 2 0.044202, so
z i 3.90. This gives P s: 0.0004, so we have very strong evidence ofa difference (specifically, that
higher-income blacks have more access at work}. 12.23 {a} For eventual contact, we test HO: p] = 112 versus P1,: p, *4: p2 and find 151 = 75% = 0.50, q s, = % e asses, and s = dist—551 = asses. Then se = van — amt, + $11: assess,
so 1 = {$1 — 61),! SE i — 1.515. This gives P = 0.0254—fair1y strong evidence that the pro—
portions and different. {b} With the same hypotheses for the proportions completing the survey, we ﬁnd
rs. = 95, = £1.33, a, = a 2 cases, and s = fist—120. = 0.42711. Then ss 2 assess, so
z i —~2.20. This gives P = 0.0113, so we have strong evidence of a difference. {c} For survey completion, the standard error for a confidence interval is SE= WW 5 0.05536, so, for example, a 95% confidence interval for the differ-
ence p. - p2 is {0.33 — 0.4605} : {1.96}{0.05536} = —0.2300 to #00220. Although the dif-
fcrence for “eventual contact” was not significant, we might examine the difference any
way: The confidence interval standard error is SE 2 0.05634, so a 95% conﬁdence interval for the difference p. — 1:2 is [0.53 — 0.6023] 2: {l.96}{0.05634} = —0.21?? to 0.0031. This
indicates that the differences could he very small, but they might he substantially large, and so have the potential to reduce nonresponse rates in surveys. At the very least, fur-
ther study is warranted. 12.251 {a} H0: £3, = 152 vs. Ha: p1 ::- 113; the populations are much larger than the samples, and 1? {the
smallest count] is greater than 5. ~
{b} ,5 = 0.0632, z = 2.926, and P = 0.0012—the difference is statistically significant. {c} Neither the subjects not the researchers who had contact with them knew which sub- jects were getting which drug—if anyone had known, they might have confounded the nuts
. come by letting their expectations or biases affect the results. 12.30 {a} HE: p1 = {:2 vs. H“: p] at .452; F = 0.0020—tbe difference is statistically significant. [b] 0.1 122
to 0-3919. 12.31 Hg: 11. = 12-2 vs. HE: pl at p3; P = 0.9305v—insufﬁcient evidence to reject Hg. 12.32 {a} Hg: in] = p2 val-1,11151L :2- 13:52:13II = 0.0335—reject ffﬂjat the 5'36 level}. [b] —0.0053 to 0.2336. 12.33 The population—to-sample ratio is large enough, and the smallest count is 10——twice as big
as it needs to be to allow the z procedures. Fatal heart attacks: 2 = —2.6'?, P = 0.0026. Non-fatal
heart attacks: 1 = —4.50, P sf. 0.000005. Strokes: z = 1.43, P = 0.1525. The proportions for both
kinds of heart attacks were significantly different; the stroke proportions were not. 12.34 {a} —9.91% to —4.0036. Since 0 is not in this interval, we would reject H“: p, = p2 at the 1% level. {b} —0.?044 to —0.4056. Since 0 is not in this interval, we would reject H“: psi = #2 at the 1'16
level. ' 12.35 {a} ﬁ = %‘% : 0.44m and SE,3 = weamﬁhssseeylﬁf 31 2 access, so the 95% con-
fidencc interval is asses : [13610004365] = 11.4315 to 0.4486. [bi s e. sees, as =e 1-131 = 0.3200, and se .. eases; the 95% Confidehee interval is {cease
— cases} : {1.06}{0.00‘32}, or 0.102 to hiss. {c} In a cluster of cars, where one driver's behavior might affect the others, we do not have
independence—one of the important properties of a random sample. 12.36 {a} We must make sure that we draw the samples from a representative cross-section of all
Illinois high-school freshmen and seniors. A multistager'cluster samming procedure could be Inference for Proportions 193 used in order to ensure that all geographical areas of the state are equally likely to contain
sampled individuals. {b} {1 = £12925, and since ab = 34 and n(l — ﬁr) = 1645 are both greater than 19, the con-
fidence interval based on r can be used. The 95% confidence interval for p is
.ﬂZEIZS I [l.96}V{{.02i129j{.92925],31529j = .92925 : [109624, or 9.1211351 to (1.02699. {e} Letting p1 = the proportion of freshmen who have used steroids and £3 = the propor-
tion of seniors who have used steroids, we test HUT p1 = :12 vs. Ha: to] is £12. From the data, ﬁll = .92925,ﬁ2 = 31111352, and f1 = .Gl995.njf1, n.(l —ﬁ}, 113?}, and oylil — 35] are all greater
than 5, so a normal approximation can be used. Test statistic is — {1121123 — .ﬂ1’25"”;VTE—ﬂl'amﬂﬁrﬁﬂlﬂoﬁﬂ 1213s 5: 11.33s2. site] the P-valuc = 9.59. There is no reason to reject H”; the difference between {1], p, is not significant. 12.33 [a] s e c.14ss ssti SE = x/{D—JositosE—iwns e o.o2323, se the 353s eeniitiettee inter-
val is 111436 e [l.9ﬁ}|flll]2923} a 11321313 to 1124:1333. [b] it = (ﬂattenssiiossmi : 3113.2_ttse T1 = 3114. {We sheuiti not use if = 11.3 here
since we have evidence that the true value of p is not in the range 11.3 to {1.2.} {c} Aside from the 45% nooresponse rate, the sample comes from a limited area in Indiana,
focuses on only one kind of business, and leaves out ”any businesses not in the Yellow Pages
[there might be a few of these; perhaps they are more likely to fail}. It is more realistic to
believe that this describes businesses that match the above profile; it might generalize to food-and-drink establishments eISewhere, but probably not‘to hardware stores and other
types of business. 12.38 H}: p, = 132 vs. 1-1,: p, 4E 11,; P = 9.6931—insufﬁcient evidence to reject Hg. 12.333 [13}me = n.1415,s, = 111551213 = {135981. {b}: = 2.12, F' = 11113311. {e} FIDITI [a]: 43.11355 to (1.1559. From {b}: {1.1391223 to 9.949936. The larger samples make the margin of error [and thus the
length of the conﬁdence interval} smaller. —-.—.-n—.--n—-._p——..n—_ u. .|.. 12.49I For testing Hg: ,1: = 12" 3 versus H“: p :5 12’ 3, we have fit i 0.3236, and the test statistic is z = (£13186 ~ 1&va e 2.32. This gives F = i1.ﬂi133——very sttettg evidence that these
than one-third of this group never use condoms. 12.4] {a} [1.2465 to {1.3359—since II} is not in this interval, we would reject HD: p1 = p, at the 1%
level [in fact, P is practically 9}. [b] No: i = —ﬁ.8653, which gives a P—value close to {1.4. 12.42 No—the data is not based on an SR5, and thus the z procedures are not reliable in this case.
In particular, a voluntary response sample is typically biased. l2.43 f.- = 321% i 9.1351, and SE; = "tx’ﬁl —ﬁlf316 2 (1.996981, so the 99% confidence interval
is [1.135] t {2.5261(90960811 = [1.1194 to 9.1398. 12.44 To test 11,: it, = e, vs. 11,: e, e: :32, we ﬁnd ti, = 1%: 13.163931 = % 2 11.3531, and the peeled value ,3 = H : 11.23333. Then 511 = V’sti '— 133% + 11-5) e ELD3966, so
I = if), — ﬁzijE i —4.32. This gives a tiny P-value {2.2 X 19—7}, so we conclude that bnpropion
increases the success rate. - 12.45 {a} s, = egg; 2 13.3911. [13} ,3 = 1% 2 11.3893, e, i U.ﬂl33,and s = [5 — payo}; e —2=3.1, so
P i l] [regardless of whether H, is {1 =1.” jog or £1 4-": on}. This is very strong evidence against He; we conclﬂglgutlgat Mexican ﬁmericaos are underrepresented on juries. {c} I), = %3 = 9.3892, while
_ — 33 _._ 3 — 1m — {3.2931}. Then iii 9.2911 {the value of en from part (all, s,, = 13.9133, and
r & —29.2-—ar1d again, we have a tiny P-value and reject Hg. - ...

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