Chapter 10: Categorical Data10.1 a. Yes, because nˆπ = 30 > 5 and n(1 − ˆπ) = 120 > 5. Samples with n < 25 would besuspect.b. .2 ± 1.645p(.2)(.8)/150 ⇒ (0.15, 0.25) is a 90% C.I. for π.10.2 When nπ > 5 and n(1 − π) > 5.10.3 a. ˆπ = 1202/1504 = 0.8 ⇒ 95% C.I. for π : 0.8 ± 1.96p(.8)(.2)/1504 ⇒ (0.780, 0.820)b. 90% C.I. for π : 0.8 ± 1.645p(.8)(.2)/1500 ⇒ (0.783, 0.817)10.4 a. Yes, the binomial assumptions hold. The samples are indep endent, the trials are iden-tical, the probability of success remains constant, and there are two possible outcomes.b. Yes, π =13, n = 50 ⇒ nπ =503> 5 and n(1 − π) =1003> 5c. ˆπ =2154= 0.389 ⇒ 95% C.I. for π : 0.389 ± 1.96p(.389)(.611)/54 ⇒ (0.259, 0.519)The C.I. is too wide to be very informative since as an estimate of π it provides valuesfrom 26% to over 50% for π.In order to decrease the width, the sample size would need to be increased.10.5 The 95% C.I.’s are summarized here: Note that ˆπ remains essentially unchanged from %Responding because n=1230 is very large.Condition 95% C.I. on proportion having conditionSore Throat 0.30 ± 1.96p(.30)(.70)/1234 ⇒ (0.274, 0.326)Burns 0.28 ± 1.96p(.28)(.72)/1234 ⇒ (0.255, 0.305)Alcohol 0.25 ± 1.96p(.25)(.75)/1234 ⇒ (0.226, 0.274)Overweight 0.22 ± 1.96p(.22)(.78)/1234 ⇒ (0.197, 0.243)Pain 0.21 ± 1.96p(.21)(.79)/1234 ⇒ (0.187, 0.233)10.6 a. By grouping the classes into similar type, it might be possible to summarize the datamore concisely. Percentages are helpful but would not add to 100% because one adultmight use more than one of the remedies. The numerator of the percentage wouldrefer to users of an OTC remedy and the denominator to the number of patients.b. A 95% C.I. using the normal approximation requires that both nˆπ and n(1 − ˆπ)exceed 5. This condition would hold in every OTC category except Sprays/Inhalers,Anesthetic throat lozenges, Ro om vaporizers and Other products.10.7 ˆπ = 88/254 = 0.346 ⇒ 90% C.I. for π : 0.346 ±1.645p(0.346)(0.654)/254 ⇒ (0.297, 0.395)9110.8 The 95% C.I.’s are given here:Statement 95% C.I. on Prop ortionOthers Don’t Report .56 ± 1.96p(.56)(.44)/504 ⇒ (0.516, 0.604)Government is Careless .50 ± 1.96p(.50)(.50)/504 ⇒ (0.456, 0.544)Cheating can be Overlooked .46 ± 1.96p(.46)(.54)/504 ⇒ (0.416, 0.504)10.9 a. A bar chart with the responses along the horizontal axis and the percentages alongthe vertical axis would allow comparison of the responses.b. Yes, since the C.I.’s would reflect the sampling errors of the point estimators and hencebe more informative of the size of the true proportions.c. The report lists only a select few responses in the United States, ignoring the mostand least popular ones as well as almost all of the foreign figures. For those per-centages reported, it does not include the sample size and so the reader gets no ideaof the accuracy of the reported sample proportions as estimates of the populationproportions.10.10 a. A table summarizing the results is given here:Statement No YesUnderstand Radiation 70% 30%Misconceptions About Space-Rockets 40% 60%Understand How Telephone Works 80% 20%Understand Computer Software 75% 25%Understand Gross National Product 72% 28%b. Unmentioned details include a complete list of questions asked and the manner inwhich they were stated. The article also does not report how the survey was conducted.Thus, the results may be biased if the sample was not selected in a random fashion.For example, if the questionaire were given by mail, the responses would come fromonly those individuals who were able to read and write. This would bias the resultsbecause illiterate people probably understand less about technology than literate ones.Other demographic characteristics of the sample might also bias the results.10.11 a. nπo= (800)(0.096) = 76.8 > 5 and n(1 −πo) = (800)(1 −0.096) = 723.2 > 5 thus thenormal approximation would be valid.b. Ho: π ≥ 0.096 versus Ha: π < 0.096ˆπ = 35/800 = 0.04375, //z =0.04375−0.096√(0.096)(0.904)/800= −5.02 ⇒p-value = P (z < −5.02) < 0.0001 ⇒ Reject Hoandconclude there is significant evidence that π < 0.096.10.12 a. ˆπ = 562/1504 = 0.374 ⇒ 95% C.I. on π : (0.350, 0.398)Half width of C.I. is 0.02492b. Using ˆπ = 0.374, n =(1.96)2(0.374)(.626)(0.01)2= 8984.1 ⇒ n = 898510.13 ˆπ = 10/24 = 0.417 ⇒ 95% C.I. on π : (0.220, 0.614)10.14 a. ˆπAdj.=38100+34= 0.00372b. 99% C.I. on π : (0, 1 −(.005)1100) = (0, 0.0516)c. Ho: π ≥ 0.01 versus Ha: π < 0.01Because 0.01 falls in the C.I., fail to reject Hoand conclude that the data fails tosupport the company’s claim. The level of significance of the test would be α =1−.992= 0.005. The problem is that with such a small value for πo, the sample sizemust be much larger in order for the company to be able to support its claim.10.15 ˆπ1= 109/200 = 0.545 (Republicans) and ˆπ2= 86/200 = 0.43 (Democrats)z =0.545−0.43q0.545(1−0.545)200+0.43(1−0.43)200= 2.32 ⇒ p-value = 0.0102Reject Hoand conclude that a large proportion of Republicans are in favor of the incentives.10.16 Because p-value = 0.0218 < 0.05, reject Hoand conclude that the data supports thehypothesis that the rates of satisfied customers served by the two methods are different.10.17 95% C.I. on π1− π2: (0.013, 0.167)Because 0 is not contained within the C.I., Hois rejected and hence the conclusion is thesame as was in Exercise 10.16.10.18 a. 95% C.I. on π1−π2: 0.478−0.376±1.96q0.478(1−0.478)473+0.376(1−0.376)439⇒ (0.038, 0.166)b. Yes, nˆπ and n(1 − ˆπ) are greater than 5 for both samples.c. Yes, because 0 is not contained within the C.I., Hois rejected.10.19 Ho: π1= π2versus Ha: π16= π2p-value = 0.0018 < 0.05 ⇒ Reject Hoand conclude there is significant evidence that thepopulation proportions are different.10.20 a. Ho: π1= π2versus Ha: π16= π2ˆπ1= 0.32, ˆπ2= 0.20 ⇒z =0.32−0.20q0.32(1−0.32)310+0.2(1−0.2)309= 3.44 ⇒ p-value = 0.0006Reject Hoand conclude there is significant evidence in the proportion of males withnew hair growth.b. What was the amount of hair growth? What side effects were observed? Whatcharacteristics distinguished males who demonstrated hair growth from those who didnot?10.21 a. z =0.90−0.36q0.9(1−0.9)100+0.36(1−0.36)100= 9.54 ⇒ p-value = P (z > 9.54) < 0.0001Reject Hoand conclude there is significant evidence that the death rate after 30 daysis greater for Co caine group than for the Heroin
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