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Sampling: Design and Analysis, 2nd Edition Solution Manual

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    Chapter 1 Introduction 1.1 Target population: Unclear, but presumed to be readers of Parade magazine. Sampling frame: Persons who know about the telephone survey. Sampling unit = observation unit: One call. (Although it would also be correct to consider the sampling unit to be a person. The survey is so badly done that it is difficult to tell what the units are.) As noted in Section 1.3, samples that consist only of volunteers are suspect. This is especially true of surveys in which respondents must pay to participate, as here— persons willing to pay 75 cents a call are likely to have strong opinions about the legalization of marijuana, and it is impossible to say whether pro- or anti-legalization adherents are more likely to call. This survey is utterly worthless for measuring public opinion because of its call-in format. Other potential biases, such as requiring a touch-tone telephone, or the sensitive subject matter or the ambiguity of the wording (what does “as legal as alcoholic beverages” mean?) probably make little difference because the call-in structure destroys all credibility for the survey by itself. 1.2 Target population: All mutual funds. Sampling frame: Mutual funds listed in newspaper. Sampling unit = observation unit: One listing. As funds are listed alphabetically by company, there is no reason to believe there will be any selection bias from the sampling frame. There may be undercoverage, however, if smaller or new funds are not listed in the newspaper. 1.3 Target population: Not specified, but a target population of interest would be persons who have read the book. Sampling frame: Persons who visit the website Sampling unit = observation unit: One review. 1

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    2 CHAPTER 1. INTRODUCTION The reviews are contributed by volunteers. They cannot be taken as representative of readers’ opinions. Indeed, there have been instances where authors of competing books have written negative reviews of a book, although amazon.com tries to curb such practices. 1.4 Target population: Persons eligible for jury duty in Maricopa County. Sampling frame: County residents who are registered voters or licensed drivers over 18. Sampling unit = observation unit: One resident. Selection bias occurs largely because of undercoverage and nonresponse. Eligible jurors may not appear in the sampling frame because they are not registered to vote and they do not possess an Arizona driver’s license. Addresses on either list may not be up to date. In addition, jurors fail to appear or are excused; this is nonresponse. A similar question for class discussion is whether there was selection bias in selecting which young men in the U.S ˙ were to be drafted and sent to Vietnam. 1.5 Target population: All homeless persons in study area. Sampling frame: Clinics participating in the Health Care for the Homeless project. Sampling unit: Unclear. Depending on assumptions made about the survey design, one could say either a clinic or a homeless person is the sampling unit. Observation unit: Person. Selection bias may be a serious problem for this survey. Even though the demo- graphics for HCH patients are claimed to match those of the homeless population (but do we know they match?) and the clinics are readily accessible, the patients differ in two critical ways from non-patients: (1) they needed medical treatment, and (2) they went to a clinic to get medical treatment. One does not know the likely direction of selection bias, but there is no reason to believe that the same percentages of patients and non-patients are mentally ill. 1.6 Target population: Female readers of Prevention magazine. Sampling frame: Women who see the survey in a copy of the magazine. Sampling unit = observation unit: One woman. This is a mail-in survey of volunteers, and we cannot trust any statistics from it. 1.7 Target population: All cows in region. Sampling frame: List of all farms in region. Sampling unit: One farm. Observation unit: One cow. There is no reason to anticipate selection bias in this survey. The design is a single-

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    3 stage cluster sample, discussed in Chapter 5. 1.8 Target population: Licensed boarding homes for the elderly in Washington state. Sampling frame: List of 184 licensed homes. Sampling unit = observation unit: One home. Nonresponse is the obvious problem here, with only 43 of 184 administrators or food service managers responding. It may be that the respondents are the larger homes, or that their menus have better nutrition. The problem with nonresponse, though, is that we can only conjecture the direction of the nonresponse bias. 1.13 Target population: All attendees of the 2005 JSM. Sampling population: E-mail addresses provided by the attendees of the 2005 JSM. Sampling unit: One e-mail address. It is stated that the small sample of conference registrants was selected randomly. This is good, since the ASA can control the quality better and follow up on non- respondents. It also means, since the sample is selected, that persons with strong opinions cannot flood the survey. But nonresponse is a potential problem—response is not mandatory and it might be feared that only attendees with strong opinions or a strong sense of loyalty to the ASA will respond to the survey. 1.14 Target population: All professors of education Sampling population: List of education professors Sampling unit: One professor Information about how the sample was selected was not given in the publication, but let’s assume it was a random sample. Obviously, nonresponse is a huge problem with this survey. Of the 5324 professors selected to be in the sample, only 900 were interviewed. Professors who travel during summer could of course not be contacted; also, summer is the worst time of year to try to interview professors for a survey. 1.15 Target population: All adults Sampling population: Friends and relatives of American Cancer Society volunteers Sampling unit: One person Here’s what I wrote about the survey elsewhere: “Although the sample contained Americans of diverse ages and backgrounds, and the sample may have provided valuable information for exploring factors associated with development of cancer, its validity for investigating the relationship between amount of sleep and mortality is questionable. The questions about amount of sleep and insomnia were not the focus of the original study, and the survey was not designed to obtain accurate responses to those questions. The design did not allow

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    4 CHAPTER 1. INTRODUCTION researchers to assess whether the sample was representative of the target population of all Americans. Because of the shortcomings in the survey design, it is impossible to know whether the conclusions in Kripke et al. (2002) about sleep and mortality are valid or not.” (pp. 97–98) Lohr, S. (2008). “Coverage and sampling,” chapter 6 of International Handbook of Survey Methodology , ed. E. deLeeuw, J. Hox, D. Dillman. New York: Erlbaum, 97–112. 1.25 Students will have many different opinions on this issue. Of historical interest is this excerpt of a letter written by James Madison to Thomas Jefferson on February 14, 1790: A Bill for taking a census has passed the House of Representatives, and is with the Senate. It contained a schedule for ascertaining the component classes of the Society, a kind of information extremely requisite to the Legislator, and much wanted for the science of Political Economy. A repetition of it every ten years would hereafter afford a most curious and instructive assemblage of facts. It was thrown out by the Senate as a waste of trouble and supplying materials for idle people to make a book. Judge by this little experiment of the reception likely to be given to so great an idea as that explained in your letter of September.

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    Chapter 2 Simple Probability Samples 2.1 (a) ¯ y U = 98 + 102 + 154 + 133 + 190 + 175 6 = 142 (b) For each plan, we first find the sampling distribution of ¯ y . Plan 1: Sample number P ( S ) ¯ y S 1 1/8 147.33 2 1/8 142.33 3 1/8 140.33 4 1/8 135.33 5 1/8 148.67 6 1/8 143.67 7 1/8 141.67 8 1/8 136.67 (i) E [¯ y ] = 1 8 (147 . 33) + 1 8 (142 . 33) + · · · + 1 8 (136 . 67) = 142. (ii) V [¯ y ] = 1 8 (147 . 33 − 142) 2 + 1 8 (142 . 33 − 142) 2 + · · · + 1 8 (136 . 67 − 142) 2 = 18 . 94. (iii) Bias [¯ y ] = E [¯ y ] − ¯ y U = 142 − 142 = 0. (iv) Since Bias [¯ y ] = 0, MSE [¯ y ] = V [¯ y ] = 18 . 94 Plan 2: Sample number P ( S ) ¯ y S 1 1/4 135.33 2 1/2 143.67 3 1/4 147.33 (i) E [¯ y ] = 1 4 (135 . 33) + 1 2 (143 . 67) + 1 4 (147 . 33) = 142 . 5. 5

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    6 CHAPTER 2. SIMPLE PROBABILITY SAMPLES (ii) V [¯ y ] = 1 4 (135 . 33 − 142 . 5) 2 + 1 2 (143 . 67 − 142 . 5) 2 + 1 4 (147 . 33 − 142 . 5) 2 = 12 . 84 + 0 . 68 + 5 . 84 = 19 . 36 . (iii) Bias [¯ y ] = E [¯ y ] − ¯ y U = 142 . 5 − 142 = 0 . 5. (iv) MSE [¯ y ] = V [¯ y ] + (Bias [¯ y ]) 2 = 19 . 61. (c) Clearly, Plan 1 is better. It has smaller variance and is unbiased as well. 2.2 (a) Unit 1 appears in samples 1 and 3, so π 1 = P ( S 1 ) + P ( S 3 ) = 1 8 + 1 8 = 1 4 . Similarly, π 2 = 1 4 + 3 8 = 5 8 π 3 = 1 8 + 1 4 = 3 8 π 4 = 1 8 + 3 8 + 1 8 = 5 8 π 5 = 1 8 + 1 8 = 1 4 π 6 = 1 8 + 1 8 + 3 8 = 5 8 π 7 = 1 4 + 1 8 = 3 8 π 8 = 1 4 + 1 8 + 3 8 + 1 8 = 7 8 . Note that ∑ 8 i =1 π i = 4 = n . (b) Sample, S P ( S ) ˆ t { 1 , 3 , 5 , 6 } 1/8 38 { 2 , 3 , 7 , 8 } 1/4 42 { 1 , 4 , 6 , 8 } 1/8 40 { 2 , 4 , 6 , 8 } 3/8 42 { 4 , 5 , 7 , 8 } 1/8 52 Thus the sampling distribution of ˆ t is: k P (ˆ t = k ) 38 1/8 40 1/8 42 5/8 52 1/8

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    7 2.3 No, because thick books have a higher inclusion probability than thin books. 2.4 (a) A total of ( 8 3 ) = 56 samples are possible, each with probability of selection 1 56 . The R function samplist below will (inefficiently!) generate each of the 56 samples. To find the sampling distribution of ¯ y , I used the commands samplist <- function(popn,sampsize){ popvals <- 1:length(popn) temp <- comblist(popvals,sampsize) matrix(popn[t(temp)],nrow=nrow(temp),byrow=T) } comblist <- function(popvals, sampsize) { popsize <- length(popvals) if(sampsize > popsize) stop("sample size cannot exceed population size") nvals <- popsize - sampsize + 1 nrows <- prod((popsize - sampsize + 1):popsize)/prod(1:sampsize) ncols <- sampsize yy <- matrix(nrow = nrows, ncol = ncols) if(sampsize == 1) {yy <- popvals} else { nvals <- popsize - sampsize + 1 nrows <- prod(nvals:popsize)/prod(1:sampsize) ncols <- sampsize yy <- matrix(nrow = nrows, ncol = ncols) rep1 <- rep(1, nvals) if(nvals > 1) { for(i in 2:nvals) rep1[i] <- (rep1[i - 1] * (sampsize + i - 2))/(i - 1) } rep1 <- rev(rep1) yy[, 1] <- rep(popvals[1:nvals], rep1) for(i in 1:nvals) { yy[yy[, 1] == popvals[i], 2:ncols] <- Recall( popvals[(i + 1):popsize], sampsize - 1) } } yy } temp1 <-samplist(c(1,2,4,4,7,7,7,8),3) temp2 <-apply(temp1, 1, mean) table(temp 2)

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    8 CHAPTER 2. SIMPLE PROBABILITY SAMPLES The following, then, is the sampling distribution of ¯ y . k P (¯ y = k ) 2 1 3 2/56 3 1/56 3 1 3 4/56 3 2 3 1/56 4 6/56 4 1 3 8/56 4 2 3 2/56 5 6/56 5 1 3 7/56 5 2 3 3/56 6 6/56 6 1 3 6/56 7 1/56 7 1 3 3/56 Using the sampling distribution, E [¯ y ] = 2 56 ( 2 1 3 ) + · · · + 3 56 ( 7 1 3 ) = 5 . The variance of ¯ y for an SRS without replacement of size 3 is V [¯ y ] = 2 56 ( 2 1 3 − 5 ) 2 + · · · + 3 56 ( 7 1 3 − 5 ) 2 = 1 . 429 . Of course, this variance could have been more easily calculated using the formula in (2.7): V [¯ y ] = ( 1 − n N ) S 2 n = ( 1 − 3 8 ) 6 . 8571429 3 = 1 . 429 . (b) A total of 8 3 = 512 samples are possible when sampling with replacement. Fortunately, we need not list all of these to find the sampling distribution of ¯ y . Let X i be the value of the i th unit drawn. Since sampling is done with replacement, X 1 , X 2 , and X 3 are independent; X i ( i = 1 , 2 , 3) has distribution k P ( X i = k ) 1 1/8 2 1/8 4 2/8 7 3/8 8 1/8 Using the independence, then, we have the following probability distribution for ¯ X , which serves as the sampling distribution of ¯ y .

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    9 k P (¯ y = k ) k P (¯ y = k ) 1 1/512 4 2 3 12/512 1 1 3 3/512 5 63/512 1 2 3 3/512 5 1 3 57/512 2 7/512 5 2 3 21/512 2 1 3 12/512 6 57/512 2 2 3 6/512 6 1 3 36/512 3 21/512 6 2 3 6/512 3 1 3 33/512 7 27/512 3 2 3 15/512 7 1 3 27/512 4 47/512 7 2 3 9/512 4 1 3 48/512 8 1/512 The with-replacement variance of ¯ y is V wr [¯ y ] = 1 512 (1 − 5) 2 + · · · + 1 512 (8 − 5) 2 = 2 . Or, using the formula with population variance (see Exercise 2.28), V wr [¯ y ] = 1 n N ∑ i =1 ( y i − ¯ y U ) 2 N = 6 3 = 2 . 2.5 (a) The sampling weight is 100/30 = 3.3333. (b) ˆ t = ∑ i ∈S w i y i = 823 . 33 . (c) ˆ V (ˆ t ) = N 2 ( 1 − n N ) s 2 y n = 100 2 ( 1 − 30 100 ) 15 . 9781609 30 = 3728 . 238, so SE (ˆ t ) = √ 3728 . 238 = 61 . 0593 and a 95% CI for t is 823 . 33 ± (2 . 045230)(61 . 0593) = 823 . 33 ± 124 . 8803 = [698 . 45 , 948 . 21] . The fpc is (1 − 30 / 100) = . 7, so it reduces the width of the CI. 2.6 (a)

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    10 CHAPTER 2. SIMPLE PROBABILITY SAMPLES The data are quite skewed because 28 faculty have no publications. (b) ¯ y = 1 . 78; s = 2 . 682; SE [¯ y ] = 2 . 682 √ 50 √ 1 − 50 807 = 0 . 367 . (c) No; a sample of size 50 is probably not large enough for ¯ y to be normally distributed, because of the skewness of the original data. The sample skewness of the data is (from SAS) 1.593. This can be calculated by hand, finding 1 n ∑ i ∈S ( y i − ¯ y ) 3 = 28 . 9247040 so that the skewness is 28 . 9247040 / (2 . 682 3 ) = 1 . 499314. Note this estimate differs from SAS PROC UNIVARIATE since SAS adjusts for df using the formula skewness = n ( n − 1)( n − 2) ∑ i ∈S ( y i − ¯ y ) 3 /s 3 . Whichever estimate is used, however, formula (2.23) says we need a minimum of 28 + 25(1 . 5) 2 = 84 observations to use the central limit theorem. (d) ˆ p = 28 / 50 = 0 . 56. SE (ˆ p ) = √ (0 . 56)(0 . 44) 49 ( 1 − 50 807 ) = 0 . 0687 . A 95% confidence interval is 0 . 56 ± 1 . 96(0 . 0687) = [0 . 425 , 0 . 695] .

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    11 2.07 (a) A 95% confidence interval for the proportion of entries from the South is 175 1000 ± 1 . 96 √ 175 1000 ( 1 − 175 1000 ) 1000 = [ . 151 , . 199] . (b) As 0.309 is not in the confidence interval, there is evidence that the percentages differ. 2.08 Answers will vary. 2.09 If n 0 ≤ N , then z α/ 2 √ 1 − n N S √ n = z α/ 2 √ 1 − n 0 N (1 + n 0 N ) S √ n 0 √ 1 + n 0 N = z α/ 2 √ 1 + n 0 N − n 0 N S √ n 0 = z α/ 2 S z α/ 2 S e = e 2.10 Design 3 gives the most precision because its sample size is largest, even though it is a small fraction of the population. Here are the variances of ¯ y for the three samples: Sample Number V (¯ y ) 1 (1 − 400 / 4000) S 2 / 400 = 0 . 00225 S 2 2 (1 − 30 / 300) S 2 / 30 = 0 . 03 S 2 3 (1 − 3000 / 300 , 000 , 000) S 2 / 3000 = 0 . 00033333 S 2 2.11 (a)

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    12 CHAPTER 2. SIMPLE PROBABILITY SAMPLES 10 12 14 16 18 20 0 20 40 60 age (months) frequency The histogram appears skewed with tail on the right. With a mildly skewed distri- bution, though, a sample of size 240 is large enough that the sample mean should be normally distributed. (b) ¯ y = 12 . 07917; s 2 = 3 . 705003; SE [¯ y ] = √ s 2 /n = 0 . 12425. (Since we do not know the population size, we ignore the fpc, at the risk of a slightly-too-large standard error.) A 95% confidence interval is 12 . 08 ± 1 . 96(0 . 12425) = [11 . 84 , 12 . 32] . (c) n = (1 . 96) 2 (3 . 705) (0 . 5) 2 = 57. 2.12 (a) Using (2.17) and choosing the maximum possible value of (0 . 5) 2 for S 2 , n 0 = (1 . 96) 2 S 2 e 2 = (1 . 96) 2 (0 . 5) 2 (0 . 1) 2 = 96 . 04 . Then n = n 0 1 + n 0 /N = 96 . 04 1 + 96 . 04 / 580 = 82 . 4 . (b) Since sampling is with replacement, no fpc is used. An approximate 95% confi- dence interval for the proportion of children not overdue for vaccination is 27 120 ± 1 . 96 √ 27 120 ( 1 − 27 120 ) 120 = [0 . 15 , 0 . 30]

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    13 2.13 (a) We have ˆ p = . 2 and ˆ V (ˆ p ) = ( 1 − 745 2700 ) ( . 2)( . 8) 744 = 0 . 0001557149 , so an approximate 95% CI is 0 . 2 ± 1 . 96 √ 0 . 0001557149 = [ . 176 , . 224] . (b) The above analysis is valid only if the respondents are a random sample of the selected sample. If respondents differ from the nonrespondents—for example, if the nonrespondents are more likely to have been bullied—then the entire CI may be biased. 2.14 Here is SAS output: The SURVEYMEANS Procedure Data Summary Number of Observations 150 Sum of Weights 864 Class Level Information Class Variable Levels Values sex 2 f m Statistics Std Error Variable Level Mean of Mean 95% CL for Mean __________________________________________________________________ sex f 0.306667 0.034353 0.23878522 0.37454811 m 0.693333 0.034353 0.62545189 0.76121478 Statistics Variable Level Sum Std Dev 95% CL for Sum __________________________________________________________________ sex f 264.960000 29.680756 206.310434 323.609566 m 599.040000 29.680756 540.390434 657.689566 __________________________________________________________________

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    14 CHAPTER 2. SIMPLE PROBABILITY SAMPLES 2.15 (a) ¯ y =301,953.7, s 2 =118,907,450,529. CI : 301953 . 7 ± 1 . 96 √ s 2 300 ( 1 − 300 3078 ) , or [264883 , 339025] (b) ¯ y = 599 . 06, s 2 = 161795 . 4 CI : [556 , 642] (c) ¯ y = 56 . 593, s 2 = 5292 . 73 CI : [48 . 8 , 64 . 4] (d) ¯ y = 46 . 823, s 2 = 4398 . 199 CI : [39 . 7 , 54 . 0] 2.16 (a) The data appear skewed with tail on right. (b) ¯ y = 5309 . 8, s 2 = 3 , 274 , 784, SE [¯ y ] = 164 . 5 Here is SAS code for problems 2.16 and 2.17: filename golfsrs ’C:\golfsrs.csv’; options ls=78 nodate nocenter; data golfsrs; infile golfsrs delimiter="," dsd firstobs=2; /* The dsd option allows SAS to read the missing values between successive delimiters */ sampwt = 14938/120;

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    15 input RN state $ holes type $ yearblt wkday18 wkday9 wkend18 wkend9 backtee rating par cart18 cart9 caddy $ pro $ ; /* Make sure the data were read in correctly */ proc print data=golfsrs; run; proc univariate data= golfsrs; var wkday9 backtee; histogram wkday9 /endpoints = 0 to 110 by 10; histogram backtee /endpoints = 0 to 8000 by 500; run; proc surveymeans data=golfsrs total = 14938; weight sampwt; var wkday9 backtee; run; 2.17 (a) The data appear skewed with tail on left. (b) ¯ y = 5309 . 8, s 2 = 3 , 274 , 784, SE [¯ y ] = 164 . 5 2.18 ˆ p = 85 / 120 = 0 . 708 95%CI: 85 / 120 ± 1 . 96 √ 85 / 120 (1 − 85 / 120) 119 ( 1 − 120 14938 ) = . 708 ± . 081 , or [0.627, 0.790].

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