Solutions-Manual Sampling-Design-And-Analysis-2nd-Edition-Sharon-Lohr-

Solutions-Manual Sampling-Design-And-Analysis-2nd-Edition-Sharon-Lohr-

Solutions-Manual Sampling-Design-And-Analysis-2nd-Edition-Sharon-Lohr-

Solutions Manual for Sampling: Design and Analysis Second Edition by

Sharon L. Lohr

Copyright ⃝c 2009 by Sharon L. Lohr

Solutions-Manual Sampling-Design-And-Analysis-2nd-Edition-Sharon-Lohr-

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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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.Sẇ ere 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 demographics 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 nonrespondents. 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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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 98 + 102 + 154 + 133 + 190 + 175 = 142 6 (b) For each plan, we first find the sampling distribution of y¯.

2.1 (a) y¯U =

Plan 1: Sample number 1 2 3 4 5 6 7 8

P (S ) 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8

y¯£ 147.33 142.33 140.33 135.33 148.67 143.67 141.67 136.67

1 1 1 (i) E[y¯] = (147.33) + (142 .33) + · · · + (136 .67) = 142.

8 8 8 1 1 (ii) 1 V [y¯] = (147 .33 — 142) 2 + (142 .33 — 142)2 + · · · + (136 .67 — 142)2 = 18.94.

8 8 8 (iii) Bias [y¯] = E[ȳ] — y¯U = 142 — 142 = 0. (iv) Since Bias [y¯] = 0, MSE [y¯] = V [y¯] = 18.94 Plan 2: Sample number P (S ) y¯£ 1 1/4 135.33 2 1/2 143.67 3 1/4 147.33 1 1 1 (i) E[y . . .33) = 142.5. ¯] = (135 33) + (143 67) + (147 4 2 4 5

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(ii) V [y¯] =

1

(135.33 — 142.5)2 +

1

2 4 = 12.84 + 0.68 + 5.84 =

(143 .67 — 142.5)2 +

1 4

(147 .33 — 142.5)2

19.36.

(iii) Bias [y¯] = E[ȳ] — 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.

1 1 1 2.2 (a) Unit 1 appears in samples 1 and 3, so π = P (S ) + P (S + = .

3) = 1 1 8 8 4 Similarly, 1 3 5 π2 = 4 + 8 = 8 1 1 3 π3 = 8 + 4 = 8 1 3 1 5 π4 = 8 + 8 + 8 = 8 1 1 1 π5 = 8 + 8 = 4 1 1 3 5 π6 = 8 + 8 + 8 = 8 1 1 3 π7 = 4 + 8 = 8 1 1 3 1 7 π8 = + + + = .

4 8 8 8 8 Note that

Σ8

i=1 πi = 4 = n.

(b) Sample, S {1, 3, 5, 6} {2, 3, 7, 8} {1, 4, 6, 8} {2, 4, 6, 8} {4, 5, 7, 8}

P (S) 1/8 1/4 1/8 3/8 1/8

tˆ 38 42 40 42 52

Thus the sampling distribution of tˆ is: 38 40 42 52

1/8 1/8 5/8 1/8

7 2.3 No, because thick books have a higher inclusion probability than thin books.

2.4 (a) A total of ( 83) = 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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The following, then, is the sampling distribution of y¯. k 2 13 3 3 13 3 23 4 4 13 4 23 5 5 13 5 23 6 6 13 7 7 13

P (y¯ = k) 2/56 1/56 4/56 1/56 6/56 8/56 2/56 6/56 7/56 3/56 6/56 6/56 1/56 3/56

Using the sampling distribution, E [ȳ ] =

2 56

µ

1

2 3

¶ + ··· +

3 56

µ 7

1 3

¶ = 5.

The variance of y¯ for an SRS without replacement of size 3 is V [y¯] =

µ ¶2 µ ¶2 2 1 3 1 2 —5 + ··· + 7 — 5 = 1.429.

56 3 56 3

Of course, this variance could have been more easily calculated using the formula in (2.7): µ ¶ µ ¶ V [y¯] = 1 — n S2 = 1 — 3 6.8571429 = 1.429.

N 8 3 n (b) A total of 83 = 512 samples are possible when sampling with replacement.

Fortunately, we need not list all of these to find the sampling distribution of y¯. Let Xi be the value of the ith unit drawn. Since sampling is done with replacement, X1, X2, and X3 are independent; Xi (i = 1, 2, 3) has distribution k 1 2 4 7 8

P (Xi = k) 1/8 1/8 2/8 3/8 1/8

Using the independence, then, we have the following probability distribution for X̄, which serves as the sampling distribution of y¯.

9 k 1 1 13 1 23 2 2 13 2 23 3 3 13 3 23 4 41

P (y¯ = k) 1/512 3/512 3/512 7/512 12/512 6/512 21/512 33/512 15/512 47/512 48/512

k 423 5 513 523 6 613 623 7 713 723 8

P (y¯ = k) 12/512 63/512 57/512 21/512 57/512 36/512 6/512 27/512 27/512 9/512 1/512

3

The with-replacement variance of y¯ is Vwr[y¯] =

1 1 (1 — 5)2 + · · · + (8 — 5)2 = 2.

512 512

Or, using the formula with population variance (see Exercise 2.28), N

Vwr[y¯] =

1 Σ (yi — y¯U ) n i=1

2

N

=

6 = 2.

3

2.5 (a) The sampling weight is 100/30 = 3.3333. Σ (b) tˆ = i∈£ wiyi = 823.33. µ ³ ¶ n ´ s2y 2 2 ˆ ˆ (c) V( t) = N 1 — = 100 1 — 30 15.9781609 = 3728 .238, so 100 30 N n √ SE (tˆ) = 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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The data are quite skewed because 28 faculty have no publications. (b) y¯ = 1.78; s = 2.682;

r 2.682 50 1— = 0.367.

SE [y¯] = √ 807 50

(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 1Σ 3 (y — i y¯) = 28.9247040 n i∈£

so that the skewness is 28.9247040/(2.6823) = 1.499314. Note this estimate differs from SAS PROC UNIVARIATE since SAS adjusts for df using the formula skewness Σ n = (yi — ȳ) 3 /s 3 . Whichever estimate is used, however, formula (n — 1)(n — 2) i∈£ (2.23) says we need a minimum of 28 + 25(1.5)2 = 84 observations to use the central limit theorem. (d) pˆ = 28/50 = 0.56.

SE (pˆ) =

s µ (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].

11 2.07 (a) A 95% confidence interval for the proportion of entries from the South is s ¡ ¢ 175 1 — 175 175 1000 1000 ± 1.96 = [.151, .199].

1000 1000 (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 n0 ≤ N , then r zα/2

1—

s

n S √ N n

=

zα/2

S r

n0 1—

n

√ N (1 + N0 ) n0

r n0 n0 S α/2 1 + N — N √n S 0 = z α/2 z α/2S e = e = z

1+

n0 N

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 1 2 3

V (y¯) (1 — 400/4000)S2/400 = 0.00225S2 (1 — 30/300)S2/30 = 0.03S2 (1 — 3000/300,000,000)S2/3000 = 0.00033333S2

2.11 (a)

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SIMPLE PROBABILITY SAMPLES

40 0

20

frequency

60

CHAPTER 2.

10

12

14

16

18

20

age (months)

The histogram appears skewed with tail on the right. With a mildly skewed distribution, though, a sample of size 240 is large enough that the sample mean should be normally distributed. √ 2 (b) y¯ = 12.07917; s2 = 3.705003; SE [y¯] = s /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) = 57.

(0.5)2

2.12 (a) Using (2.17) and choosing the maximum possible value of (0.5)2 for S2, n0 = Then n=

(1.96)2S2 (1.96)2(0.5)2 = = 96 .04. e2 (0.1)2 n0

1 + n /N 0

=

96.04 1 + 96.04/580

= 82.4.

(b) Since sampling is with replacement, no fpc is used. An approximate 95% confidence interval for the proportion of children not overdue for vaccination is s ¡ ¢ 27 1 — 27 27 120 120 ± 1.96 = [0.15, 0.30] 120 120

13 2.13 (a) We have pˆ = .2 and µ ¶ 745 (.2)(.8) V̂ ( pˆ ) = 1 — = 0.0001557149, 2700 744 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 Sum of Weights

150 864

Class Level Information Class Variable

Levels

sex

2

Values f m

Statistics

Variable

Level

sex

f m

Mean

Std Error of Mean

95% CL for Mean

0.306667 0.693333

0.034353 0.034353

0.23878522 0.37454811 0.62545189 0.76121478

Statistics Variable

Level

sex

f m

Sum

Std Dev

95% CL for Sum

264.960000 599.040000

29.680756 29.680756

206.310434 323.609566 540.390434 657.689566

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2.15 (a) y¯ =301,953.7, s2 =118,907,450,529. s

µ ¶ s2 300 — CI : 301953.7 ± 1. 96 1 , or [264883, 339025] 3078 300 (b) y¯ = 599.06, s2 = 161795.4 CI : [556, 642] (c) y¯ = 56.593, s2 = 5292.73 CI : [48.8, 64.4] (d) y¯ = 46.823, s2 = 4398.199 CI : [39.7, 54.0] 2.16 (a) The data appear skewed with tail on right.

(b) y¯ = 5309.8, s2 = 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;

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, s2 = 3,274,784, SE [y¯] = 164.5 2.18 pˆ = 85/120 = 0.708 s 95%CI: 85/120 ± 1.96

µ

¶

= .708 ± .081, 85/120 (1 — 85/120) 120 1— 119 14938

or [0.627, 0.790].

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2.19 Assume the maximum value for the variance, with p = 0.5. Then use n0 = 1.962(0.5)2/(.04)2, n = n0/(1 + n0/N ).

City Buckeye Gilbert Gila Bend Phoenix Tempe

n0 600.25 600.25 600.25 600.25 600.25

n 535 595 446 600 598

The finite population correction only makes a difference for Buckeye and Gila Bend.

2.20 Sixty of the 70 samples yield confidence intervals, using this procedure, that include the true value t = 40. The exact confidence level is 60/70 = 0.857.

2.21 (a) A number of different arguments can be made that this method results in a simple random sample. Here is one proof, which assumes that the random number table indeed consists of independent random numbers. In the context of the problem, M = 999, N = 742, and n = 30. Of course, many students will give a more heuristic argument.

Let U1, U2, U3, . . ., be independent random variables, each with a discrete uniform distribution on {0, 1, 2, . . . , M }. Now define T1 = min{i : Ui ∈ [1, N ]} and Tk = min{i > Tk—1 : Ui ∈ [1, N ], Ui ∈ / {UT1 , . . . , UTk−1 }} for k = 2, . . . , n. Then for {x1, . . . , xn} a set of n distinct elements in {1, . . . , N }, P (S = {x1, . . . , xn}) = P ({UT1 , . . . , UTn } = {x1, . . . , xn}) P {UT1 = x1, . . . , UTn = xn} = E[P {UT1 = x1, . . . , UTn = xn | T1, T2, . . . , Tn}] µ ¶µ ¶µ ¶ µ ¶ 1 1 1 1 = N N —1 N —2 ··· N —n+1 (N — n)! .

N! Conditional on the stopping times T1, . . . , Tn, UT1 is discrete uniform on { 1, . . . , N} ; (UT2 T|1, . . . , TN , UT1 ) is discrete uniform on {1, . . . , N} — {UT1 }, and so on. Since x1, . . . , xn are arbitrary, =

P (S = {x , . . . , x }) = n!(N — n)! = 1 , ¡N¢ 1 n N! n so the procedure results in a simple random sample. (b) This procedure does not result in a simple random sample. Units starting with 5, 6, or 7 are more likely to be in the sample than units starting with 0 or 1. To see

17 this, let’s look at a simpler case: selecting one number between 1 and 74 using this procedure.

Let U1, U2, . . . be independent random variables, each with a discrete uniform distribution on {0, . . . , 9 }. Then the first random number considered in the sequence is 10U1 + U2; if that number is not between 1 and 74, then 10U2 + U3 is considered, etc. Let T = min{i : 10Ui + Ui+1 ∈ [1, 74]}.

Then for x = 10x1 + x2, x ∈ [1, 74], P (S = {x}) = P (10UT + UT +1 = x) = P (UT = x1, UT +1 = x2).

For part (a), the stopping times were irrelevant for the distribution of UT1 , . . . , UTn ; here, though, the stopping time makes a difference. One way to have T = 2 is if 10U1 + U2 = 75. In that case, you have rejected the first number solely because the second digit is too large, but that second digit becomes the first digit of the random number selected. To see this formally, note that P (S = {x}) = P (10U1 + U2 = x or {10U1 + U2 ∈ / [1, 74] and 10U2 + U3 = x} or {10U1 + U2 ∈ / [1, 74] and 10U2 + U3 ∈ / [1, 74] and 10U3 + U4 = x} or . . .) = P (U1 = x1, U2 = x2) ∞ µ t—1 \ Σ + P {Ui > 7 or [Ui = 7 and Ui+1 > 4]} t=2

i=1

¶ and Ut = x1 and Ut+1 = x2 .

Every term in the series is larger if x1 > 4 than if x1 ≤ 4. (c) This method almost works, but not quite. For the first draw, the probability that 131 (or any number in {1, . . . , 149, 170} is selected is 6/1000; the probability that 154 (or any number in {150, . . . , 169}) is selected is 5/1000. (d) This clearly does not produce an SRS, because no odd numbers can be included. (e) If class sizes are unequal, this procedure does not result in an SRS: students in smaller classes are more likely to be selected for the sample than are students in larger classes.

Consider the probability that student j in class i is chosen on the first draw.

P {select student j in class i} = P {select class i}P {select student j | class i} 1 1 = .

20 number of students in class i (f) Let’s look at the probability student j in class i is chosen for first unit in the sample. Let U1, U2, . . . be independent discrete uniform {1, . . . , 20} and let V1, V2, . . .

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be independent discrete uniform {1, . . . , 40}. Let Mi denote the number of students Σ in class i, with K = 20 i=1 Mi. Then, because all random variables are independent, P (student j in class i selected) = P (U1 = i, V2 = j) + P (U2 = i, V2 = j)P ½ +··· + P

UL+1 = i, VL+1 = j

=

∞ · 20

q=1

k=1

1 Σ Σ 1 40 — Mk 20 40 800 k=1 L=0 ¸L ∞ · 1 Σ K 1— 800 800 L=0

=

P

µ[ 20

¶

{Uq = k, Vq > Mk}

k=1

¶¸ ∞ · L µ[ 20 1 1 Σ Y {Uq = k, Vq > Mk} P 20 40 L=0

=

¶ {U1 = k, V1 > Mk}

k=1

q=1

+···

=

¾Y L

µ[ 20

1 1 800 1 — (1 — K/800)

=

¸L

1 .

K

Thus, before duplicates are eliminated, a student has probability 1/K of being selected on any given draw. The argument in part (a) may then be used to show that when duplicates are discarded, the resulting sample is an SRS. 2.22 (a) From (2.13), √ r V (y¯) n S CV(y¯) = = 1— √ .

E(ȳ) N ny¯U N Substituting pˆ for y¯, andN —1 p(1 — p) for S2, we have

s ³

s n ´ Np(1 — p) N —n1—p CV(pˆ) == 1 — = .

2 N — 1 np N (N — 1)np √ The CV for a sample of size 1 is (1 — p)/p. The sample size in (2.26) will be 2 CV2/r2. zα/2 (b) I used Excel to calculate these values.

0.001 p Fixed 4.3 Relative 4264176

0.005 21.2 849420

0.01 42.3 422576

0.05 202.8 81100

0.1 384.2 38416

0.3 896.4 9959.7

0.7 p Fixed 896.4 Relative 1829.3

0.9 384.2 474.3

0.95 202.8 224.7

0.99 42.3 43.1

0.995 21.2 21.4

0.999 4.3 4.3

0.5 1067.1 4268.4

19 2.23

P

(no missing data) =

= =

µ ¶µ ¶ 3059 19 300 0 µ3078¶ 300 (2778)(2777) . . . (2760) (3078)(3077) . . . (3060) 0.1416421.

2.24

³ n ´ S2 + c + c n.

0 1 g(n) = L(n) + C(n) = k 1 — N n kS2 dg =— + c1 dn n2 Setting the derivative equal to 0 and solving for n gives s kS2 n= . c1 The sample size, in the decision theoretic approach, should be larger if the cost of a bad estimate, k, or the variance, S2, is larger; the sample size is smaller if the cost of sampling is larger.

2.25 (a) Skewed, with tail on right. (b) y¯ = 20.15, s2 = 321.357, SE [y¯] = 1.63 2.26 In a systematic sample, the population is partitioned into k clusters, each of size n. One of these clusters is selected with probability 1/k, so πi = 1/k for each i.

But many of the samples that could be selected in an SRS cannot be selected in a systematic sample. For example, P (Z1 = 1, . . . , Zn = 1) = 0 : since every kth unit is selected, the sample cannot consist of the first n units in the population.

2.27 (a)

µ

¶µ ¶ 99,999,999 1 999 1 (you are in sample) = P µ ¶ 100,000,000 1000 99,999,999! 1000! 99,999,000! = 999! 99,999,000! 100,000,000!

1 1000 . = = 100,000 100,000,000

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CHAPTER 2.

SIMPLE PROBABILITY SAMPLES

(b)

µ P (you are not in any of the 2000 samples) =

1—

¶2000 1 = 0.9802 100,000

(c) P (you are not in any of x samples) = (1 — 1/100,000)x. Solving for x in (1 —1/100,000)x = 0.5 gives x log(.99999) = log(0.5), or x = 69314.4. Almost 70,000 samples need to be taken! This problem provides an answer to the common question, “Why haven’t I been sampled in a poll?”

2.28 (a) We can think of drawing a simple random sample with replacement as performing an experiment n independent times; on each trial, outcome i (for i∈ 1, { . . . , N )}occurs with probability pi = 1/N . This describes a multinomial experiment.

We may then use properties of the multinomial distribution to answer parts (b) and (c): E[Qi] = npi = n , N µ ¶ n 1 V [Q i] = np i(1 — pi ) = 1— , N N and Cov [Q , Q ] = —np p = —n 1 for i /= j. i j i j NN (b)

N ˆ E E[t] = n

(c)

µ ˆ = V [t] µ =

N n N n

·Σ N

¸ Qiyi =

i=1

"N Σ

¶2 V

N Σ N n n

i=1 N

yi = t.

# Qiyi

i=1

¶2 Σ N N Σ

yiyj Cov [Qij Qj]

¾ Σ ¶2½ Σ N N 2 Σ y y (—np p ) i j i j N = y npi(1 — pi) + i n i=1 j/=i i=1 ¾ Σ ¶Σ µ ¶2½ µ N N N 1 Σ yiyj 1 N n y2i — n = 1— NN N n N i=1 j=1

µ

=

½N ¾ N Σ2 y i — N y¯U2 n i=1

N Σ

=

N 2 i=1 n

(yi — y¯U )2 N

.

i=1

i=1 j/=i

21 2.29 We use induction. Clearly, S0 is an SRS of size n from a population of size n.

Now suppose Sk—1 is an SRS of size n from Uk—1 = {1, 2, . . . , n + k — 1}, where k ≥ 1. We wish to show that Sk is an SRS of size n from Uk = {1, 2, . . . , n + k}.

Since Sk—1 is an SRS, we know that P (S k—1 ) = µ

1 n+k—1 n

¶ =

n!(k — 1)! (n + k — 1)!

.

Now let Uk ∼ Uniform(0, 1), let Vk be discrete uniform (1, . . . , n), and suppose Uk and Vk are independent. Let A be a subset of size n from Uk. If A does not contain unit n + k, then A can be achieved as a sample at step k — 1 and µ n ¶ P (Sk = A) = P Sk—1 and Uk > n+k k = P (Sk—1 ) n+k n!k! . = (n + k)!

If A does contain unit n + k, then the sample at step k — 1 must contain Ak—1 = A — {n + k} plus one other unit among the k units not in Ak—1. ¶ µ n Σ P Sk—1 = Ak—1 ∪ {j} and Uk ≤ P (Sk = A) = and Vk = j n+k j∈Uk−1∩ACk−1

n!(k — 1)! n 1 (n + k — 1)! n + k n n!k! . (n + k)!

= k =

2.30 I always use this activity in my classes. Students generally get estimates of the total area that are biased upwards for the purposive sample. They think, when looking at the picture, that they don’t have enough of the big rectangles and so tend to oversample them. This is also a good activity for reviewing confidence intervals and other concepts from an introductory statistics class.

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CHAPTER 2.

SIMPLE PROBABILITY SAMPLES

Chapter 3

Stratified Sampling 3.2 (a) For each stratum, we calculate tˆh = 4y¯h Stratum 1 Sample, S1 {1, 2} {1, 3} {1, 8} {2, 3} {2, 8} {3, 8}

P (S1) 1/6 1/6 1/6 1/6 1/6 1/6

{yi, i ∈ S1} 1, 2 1, 4 1, 8 2, 4 2, 8 4, 8

tˆ1£1 6 10 18 12 20 24

P (S2) 1/6 1/6 1/6 1/6 1/6 1/6

{yi, i ∈ S2} 4, 7 4, 7 4, 7 7, 7 7, 7 7, 7

tˆ2£2 22 22 22 28 28 28

Stratum 2 Sample, S2 {4, 5} {4, 6} {4, 7} {5, 6} {5, 7} {6, 7}

(b) From Stratum 1, we have the following probability distribution for tˆ1: j 6 10 12 18 20 24

P (tˆ1 = j) 1/6 1/6 1/6 1/6 1/6 1/6

The sampling distribution for tˆ2 is:

23

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CHAPTER 3.

22 28

STRATIFIED SAMPLING

1/2 1/2

Because we sample independently in Strata 1 and 2, P (tˆ1 = j and tˆ2 = k) = P (tˆ1 = j)P (tˆ2 = k) for all possible values of j and k. Thus, j 6 6 10 10 12 12 18 18 20 20 24 24

k 22 28 22 28 22 28 22 28 22 28 22 28

P (tˆ1 = j and tˆ2 = k) 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12 1/12

j+k 28 34 32 38 34 40 40 46 42 48 46 52

So the sampling distribution of tˆstr is k 28 32 34 38 40 42 46 48 52

P (tˆstr = k) 1/12 1/12 2/12 1/12 2/12 1/12 2/12 1/12 1/12

(c) E[tˆstr] = V [tˆstr] =

Σ

kP (tˆstr = k) = 40

k Σ

(k — 40)2P (tˆstr = k) = 4713.

k

3.3 (a) y¯U = 71.83333, S2 = 86.16667. µ ¶ 6 (b) = 15.

4 (c) The 15 samples are:

25 Units in sample 1 2 3 4 1 2 3 5 1 2 3 6 1 2 4 5 1 2 4 6 1 2 5 6 1 3 4 5 1 3 4 6 1 3 5 6 1 4 5 6 2 3 4 5 2 3 4 6 2 3 5 6 2 4 5 6 3 4 5 6

(d)

3 2

¶µ

Sample Mean 69.50 69.25 66.50 72.50 69.75 69.50 75.25 72.50 72.25 75.50 73.50 70.75 70.50 73.75 76.50

µ ¶ 4 86.16667 V (y¯) = 1 — = 7.180556.

6 4

Using (2.9),

µ

y values in sample 66 59 70 83 66 59 70 82 66 59 70 71 66 59 83 82 66 59 83 71 66 59 82 71 66 70 83 82 66 70 83 71 66 70 82 71 66 83 82 71 59 70 83 82 59 70 83 71 59 70 82 71 59 83 82 71 70 83 82 71

3 2

¶ = 9.

(e) You cannot have any of the samples from (c) which contain 3 units from one of the strata. This eliminates the first 3 samples, which contain {1, 2, 3} and the three samples containing students {4, 5, 6}. The stratified samples are Units in S1 1 2 1 2 1 2 1 3 1 3 1 3 2 3 2 3 2 3

Units in S2 4 5 4 6 5 6 4 5 4 6 5 6 4 5 4 6 5 6 µ

V (ȳstr ) =

y values in sample 66 59 83 82 66 59 83 71 66 59 82 71 66 70 83 82 66 70 83 71 66 70 82 71 59 70 83 82 59 70 83 71 59 70 82 71

y¯str 72.50 69.75 69.50 75.25 72.50 72.25 73.50 70.75 70.50

¶ µ ¶2 µ ¶ µ ¶2 2 3 31 2 3 44.33333 1— + 1— = 3.14.

3 2 3 2 6 6

The variance is smaller because the extreme samples from (c) are excluded by the stratified design. The variances S2 = 31 and S2 = 44.33 are much smaller than the 1 2 population variance S2.

3.4 Here is SAS code for creating the data set:

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CHAPTER 3.

STRATIFIED SAMPLING

data acls; input stratum $ popsize returns percfem; females = round(returns*percfem/100); males = returns - females; sampwt = popsize/returns; datalines; Literature 9100 636 38 Classics 1950 451 27 Philosophy 5500 481 18 History 10850 611 19 Linguistics 2100 493 36 PoliSci 5500 575 13 Sociology 9000 588 26 ; proc print data=acls; run; data aclslist; set acls; do i = 1 to females; femind = 1; output; end; do i = 1 to males; femind = 0; output; end; /* Check whether we created the data set correctly*/ proc freq data=aclslist; tables stratum * femind; run; proc surveymeans data=aclslist mean clm sum clsum; stratum stratum; weight sampwt; var femind; run; We obtain tˆ = 10858 with SE 313. These values differ from those in Example 4.4 because of rounding.

3.5 (a) The sampled population consists of members of the organizations who would respond to the survey.

27 (b) pˆ str =

7 Σ Nh

pˆh µh=1 N ¶ µ ¶ µ ¶ 9,100 1,950 9,000 = (0.37) + (0.23) + · · · + (0.41) 44,000 44,000 44,000 = 0.334.

SE [p̂ str ] = = =

, u 7 µ ¶µ uΣ nh Nh ¶2 p̂h (1 — p̂ h ) , 1— N nh — 1 Nh h=1 √ 1.46 × 10—5 + 5.94 × 10—7 + · · · + 1.61 × 10—5 0.0079.

3.6 (a) We use Neyman allocation (= optimal allocation when costs in the strata are equal), with nh ∝ NhSh. We take Rh to be the relative standard deviation in stratum h, and let nh = 900(NhRh)/125000.

Stratum Houses Apartments Condos Sum

Nh 35,000 45,000 10,000 90,000

Rh 2 1 1

NhRh 70,000 45,000 10,000 125,000

nh 504 324 72 900

(b) Let’s suppose we take a sample of 900 observations. (Any other sample size will give the same answer.)

With proportional allocation, we sample 350 houses, 450 apartments, and 100 condominiums. If the assumptions about the variances hold, µ ¶2 µ ¶2 µ ¶2 350 (.45)(.55) 450 (.25)(.75) 100 (.03)(.97) + + Vstr [p̂str ] = 900 900 900 100 350 450 = .000215.

If these proportions hold in the population, then 35 45 10 p= (.45) + (.25) + (.03) = 0.3033 90 90 90 and, with an SRS of size 900, (0.3033)(1 — .3033) = .000235.

900 The gain in efficiency is given by Vsrs [pˆsrs] =

V str [p̂str ] .000215 = = 0.9144.

Vsrs [p̂ srs ] .000235

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CHAPTER 3.

STRATIFIED SAMPLING

For any sample size n, using the same argument as above, we have .193233 Vstr [p̂str ] = and n Vsrs [p̂srs ] =

.211322 . n

We only need 0.9144n observations, taken in a stratified sample with proportional allocation, to achieve the same variance as in an SRS with n observations.

Note: The ratio Vstr [p̂str ]/V srs [p̂ srs ] is the design effect, to be discussed further in Section 7.5.

3.7 (a) Here are summary statistics for each stratum:

average variance

Biological 3.142857 6.809524

Physical

Social

Humanities 0.4545455 0.8727273

Since we took a simple random sample in each stratum, we use (102)(3.142857) = 320.5714 to estimate the total number of publications in the biological sciences, with estimated variance µ ¶ 7 6.809524 2 102 7 (102) 1 — = 9426.327.

The following table gives estimates of the total number of publications and estimated variance of the total for each of the four strata: Stratum Biological Sciences Physical Sciences Social Sciences Humanities Total

Estimated total number of publications 320.571 652.632 267.077 80.909 1321.189

Estimated variance of total 9426.33 38982.71 14843.31 2358.43 65610.78

We estimate the total number of refereed publications for the college by adding the totals for each of the strata; as sampling was done independently in each stratum, the variance of the college total is the sum of the variances of the population stratum totals. Thus we estimate the total number of refereed papers as 1321.2, with √ standard error 65610.78 = 256.15. (b) From Exercise 2.6, using an SRS of size 50, the estimated total was tˆsrs = 1436.46, with standard error 296.2. Here, stratified sampling ensures that each division of the college is represented in the sample, and it produces an estimate with a smaller standard error than an SRS with the same number of observations.

The sample variance in Exercise 2.8 was s2 = 7.19. Only Physical Sciences had a

29 sample variance larger than 7.19; the sample variance in Humanities was only 0.87.

Observations within many strata tend to be more homogeneous than observations in the population as a whole, and the reduction in variance in the individual strata often leads to a reduced variance for the population estimate. (c)

¶

N 2hp̂ h (1 — p̂ h ) N 2 nh — 1

Nh

nh

pˆh

Nh nh pˆ µ N h 1 — Nh

Biological Sciences

102

7

1 7

.018

.0003

Physical Sciences

310

19

10 19

.202

.0019

Social Sciences

217

13

9 13

.186

.0012

Humanities

178

11

8 11

.160

.0009

Total

807

50

.567

.0043

pˆstr = 0.567 √ SE[p̂ str ] = 0.0043 = 0.066.

3.8 (a) Because the budget for interviews is $15,000, a total of 15,000/30 = 500 in-person interviews can be taken. The variances in the phone and nonphone strata are assumed similar, so proportional allocation is optimal: 450 phone households and 50 nonphone households would be selected for interview. (b) The variances in the two strata are assumed equal, so optimal allocation gives √ nh ∝ Nh/ ch.

Stratum Phone Nonphone Total

ch 10 40

Nh/N 0.9 0.1 1.0

√ Nh/(N ch) 0.284605 0.015811 0.300416

The calculations in the table imply that nphone =

0.284605 0.300416

n;

the cost constraints imply that 10nphone + 40nnon = 10nphone + 40(n — nphone) = 15,000.

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CHAPTER 3.

Solving, we have

STRATIFIED SAMPLING

nphone = 1227 nnon = 68 n = 1295.

Because of the reduced costs of telephone interviewing, more households can be selected in each stratum.

3.9 (a) Summary statistics for acres87: Region

N

n

y¯

s2

(N /N )y¯

NC NE S W Total

1054 103 220 21 1382 135 422 41 3078 300

308188.3 109009.6 212687.2 654458.7

2.943E+10 1.005E+10 5.698E+10 3.775E+11

105532.98 7791.46 95495.05 89727.61 298547.10

Σ

,

h

For acres87, y¯str =

h

h

h

h

µ

298547.1 h(Nh/N )y¯h =u Σ Nh and s2 SE(¯ ) = , ystr

)

2

N

h

Nh — nh Nh2 sh2

h

N 2 nh 30225148 2211633 76782239 156241957 265460977

Nh

¶ = 16293 .

h

nh

Of course, y¯str could also be calculated using the column of weights in the data set, as: Σ str 918927923 i £ w 3078 wiy i = y¯ = Σi ∈i∈£ = 298547.1 (b) Summary statistics for farms92: Region

N

n

y¯

s2

(N /N )y¯

NC NE S W Total

1054 103 220 21 1382 135 422 41 3078 300

750.68 528.10 578.59 602.34

128226.50 128645.90 222972.8 311508.4

257.06 37.75 259.78 82.58 637.16

h

For farms92, y¯str =

h

h

,

Σ

h

h

µ

h(Nh/N )y¯h = Σ Nhands2 u637.16

SE(¯

)=

ystr

)

,

2 h

N

h

Nh — nh Nh2 sh2 Nh

N 2 nh 131.71 28.31 300.44 128.94 589.40

¶

h

nh

= 24 28 . .

(c) Summary statistics for largef92: Region NC NE S W Total

nh

y¯h

s2h

(Nh/N )y¯h

1054 103 220 21 1382 135 422 41 3078 300

70.91 8.19 38.84 104.98

4523.34 90.16 2450.47 11328.97

24.28 0.59 17.44 14.39 56.70

Nh

Nh — nh N h2 s2h Nh N 2 nh 4.65 0.02 3.30 4.69 12.66

31 For largef92, y¯str =

Σ

h(Nh/N )y¯h = 56.70 and

, uΣ µ Nh ¶2 sh2 SE(ȳstr ) = , = 3.56.

N nh h

(d) Summary statistics for smallf92: Region

N

n

y¯

s2

(N /N )y¯

NC NE S W Total

1054 103 220 21 1382 135 422 41 3078 300

44.26 47.24 47.39 124.39

1286.43 2364.79 6205.45 100640.94

15.16 3.38 21.28 17.05 56.86

h

For smallf92, y¯str =

h

h

h

h

h

Nh — nh Nh2 sh2 Nh

N 2 nh 1.32 0.52 8.36 41.66 51.86

Σ

h(Nh/N )y¯h = 56.86 and

, Σ µ Nh ¶2 sh2 SE(ȳstr ) = ,u = 7.20.

N nh h

Here is SAS code for finding these estimates: data strattot; input region $ cards; NE 220 NC 1054 S 1382 W 422 ;

_total_;

proc surveymeans data=agstrat total = strattot mean sum clm clsum df; stratum region ; var acres87 farms92 largef92 smallf92 ; weight strwt; run; 3.10 For this problem, note that Nh, the total number of dredge tows needed to cover the stratum, must be calculated. We use Nh = 25.6 × Areah. (a) Calculate tˆh = N h ȳ h

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32 Stratum 1 2 3 4 Sum

CHAPTER 3.

Nh 5704 1270 1286 5064 13324

nh 4 6 3 5 18

Thus tˆstr = 18152 and

s2h 0.068 0.042 2.146 0.794

y¯h 0.44 1.17 3.92 1.80

SE [tˆstr] =

√

STRATIFIED SAMPLING

tˆh 2510 1486 5041 9115 18152

2

N 2h(1 — n h/N h) snhh 552718 11237 1180256 4068262 5812472

5812472 = 2411.

(b) Stratum 1 4 Sum

Nh 8260 5064 13324

nh 8 5 13

Here, tˆstr = 7229 and

s2h 0.083 0.046

y¯h 0.63 0.40

SE [tˆstr] =

tˆh 5204 2026 7229

s2

N 2h(1 — nh/N h) nhh 707176 235693 942869

√ 942869 = 971.

3.11 Note that the paper is somewhat ambiguous on how the data were collected.

The abstract says random stratified sampling was used, while on p. 224 the authors say: ‘a sampling grid covering 20% of the total area was made . . . by picking 40 numbers between one and 200 with the random number generator.” It’s possible that poststratification was really used, but for exercise purposes, let’s treat it as a stratified random sample. Also note that the original data were not available, data were generated that were consistent with summary statistics in the paper. (a) Summary statistics are in the following table: Zone 1 2 3 Total

Nh 68 84 48 200

nh 17 12 11 40

y¯h 1.765 4.417 10.545

s2h 3.316 11.538 46.073

Using (3.1), tˆstr =

Σ

N h y¯h

h

= 68(1.76) + 84(4.42) + 48(10.55) = 997.

33 From (3.3), µ ¶H s2 Nh Σ Nh2 h 1— N nh µ ¶ h=1 µ ¶ µ ¶ 17 12 11 46.073 2 3.316 2 11.538 = 1— 68 + 1— 84 + 1— 482 68 17 84 12 48 11 = 676.5 + 5815.1 + 7438.7

ˆ (t ˆ str) = V

=

13930.2,

so

SE(tˆystr) =

√

13930.2 = 118.

SAS code to calculate these quantities is given below: data seals; infile seals delimiter="," firstobs=2; input zone holes; if zone = 1 then sampwt = 68/17; if zone = 2 then sampwt = 84/12; if zone = 3 then sampwt = 48/11; run; data strattot; input zone _total_; datalines; 1 68 2 84 3 48 ; proc surveymeans data=seals total=strattot mean clm sum clsum; strata zone; weight sampwt; var holes; run; The SURVEYMEANS Procedure Data Summary Number of Strata Number of Observations Sum of Weights

3 40 200

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CHAPTER 3.

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Statistics

Variable holes

Mean

Std Error of Mean

95% CL for Mean

4.985909

0.590132

3.79018761 6.18163058

Statistics Variable holes

Sum

Std Dev

997.181818

118.026447

95% CL for Sum 758.037521 1236.32612

(b) (i) If the goal is estimating the total number of breathing holes, we should use optimal allocation. Using the values of s2h from this survey as estimates of S2h, we have: Zone 1 2 3 Total

Nh 68 84 48 200

s2h 3.316 11.538 46.073

Nhsh 123.83 285.33 325.81 734.97

Then n1 = (123.83/734.97)n = 0.17n; n2 = 0.39n; n3 = 0.44n. The high variance in zone 3 leads to a larger sample size in that zone. (ii) If the goal is to compare the density of the breathing holes in the three zones, we would like to have equal precision for y¯h in the three strata. Ignoring the fpc, that means we would like 2 2 S12 = S2 = S 3 , n1 n2 n3 which implies that nh should be proportional to S2hto achieve equal variances.

Using the sample variances s2 instead of the unknown population variances S2, this h h leads to s2 n1 = 2 12 n = 0.05n s + s + s2 1

2

n2 = 0.19n n3 = 0.76n.

3.12 We use nh = 300Nhsh/ Region Northeast North Central South West Total

Nh 220 1,054 1,382 422 3,078

3

Σ

k Nksk

Nhsh 19,238,963 181,392,707 319,918,785 265,620,742 786,171,197

nh 7 69 122 101 300

35 3.13 Answers will vary since students select different samples.

3.14 Method Role play Problem solving Simulations Empathy building Gestalt exercises

SE[p̂ str ] 0.011 0.217 0.028 0.028 0.017

pˆstr 0.96 0.82 0.45 0.45 0.11

Note that the standard errors calculated for role play and for gestalt exercises are unreliable because the formula relies on a normal approximation to pˆh: here, the sample sizes in the strata are small and p̂ h ’s are close to 0 or 1, so the accuracy of the normal approximation is questionable.

3.15 (a) An advantage is that using the same number of stores in each stratum gives the best precision for comparing strata if the within-stratum variances are the same. In addition, people may perceive that allocation as fair. A disadvantage is that estimates may lose precision relative to optimal allocation if some strata have higher variances than others. (b) y¯ =

Σ 3 Nh

str h=1

Vˆ (ȳ str ) =

N

y¯ = 3.9386. h

¶2 2 3 µ Σ Nh s

h

h=1

N

nh

= 8.85288 × 10—5 Thus, a 95% CI is

√ 3.9386 ± 1.96 8.85288 × 10—5 = [3.92, 3.96]

This is a very small CI. Remember, though, that it reflects only the sampling error.

In this case, the author was unable to reach some of the stores and in addition some of the market basket items were missing, so there was nonsampling error as well.

3.16 (a) Stratum 1 2 3 4 Total

Nh 89 61 40 47 237

nh 19 20 22 21 82

y¯h 1.74 1.75 13.27 4.10

sh2 5.43 6.83 58.78 15.59

tˆh 154.6 106.8 530.9 192.5 984.7

Vˆ (tˆh ) 1779.5 854.0 1923.7 907.2 5464.3

The estimated total number of otter holts is tˆstr = 985

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36

CHAPTER 3.

with

SE [tˆstr] =

√

STRATIFIED SAMPLING

5464 = 73.9.

Here is SAS code and output for estimating these quantities: data exer0316; infile otters delimiter=’,’ firstobs=2; input section habitat holts; if habitat = 1 then sampwt = 89/19; if habitat = 2 then sampwt = 61/20; if habitat = 3 then sampwt = 40/22; if habitat = 4 then sampwt = 47/21; ; data strattot; input habitat _total_; datalines; 1 89 2 61 3 40 4 47 ; proc surveymeans data=exer0316 total = strattot mean clm sum clsum; stratum habitat; weight sampwt; var holts; run; The SURVEYMEANS Procedure Data Summary Number of Strata Number of Observations Sum of Weights

4 82 237

Statistics

Variable holts

Mean

Std Error of Mean

95% CL for Mean

4.154912

0.311903

3.53396136 4.77586336

Statistics

37

Variable holts

Sum

Std Dev

95% CL for Sum

984.714229

73.920990

837.548842 1131.87962

3.17 (a) We form a new variable, weight= 1/samprate. Then the number of divorces in the divorce registration area is Σ (weight)h(numrecs)h = 571,185.

H

h=1

Note that this is the population value, not an estimate, because samprate= nh/Nh and (numrecs)h = nh. Thus H

H Σ Σ Nh (weight)h (numrecs)h = nh = N. h=1 nh h=1

(b) They wanted a specified precision within each state (= stratum). You can see that, except for a few states in which a census is taken, the number of records sampled is between 2400 and 6200. That gives roughly the same precision for estimates within each of those states. If the same sampling rate were used in each state, states with large population would have many more records sampled than states with small population. (c) (i) For each stratum, y¯h = pˆh =

hsblt20 + hsb20-24 . numrecs

The following spreadsheet shows calculations done to obtain tˆstr =

Σ N h pˆh h

and V̂ (t̂ str ) =

Σ h

µ ¶ nh p̂ h (1 — pˆh ) Σ N h 1— = varcont h. nh — 1 Nh 2

h

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38 state AL AK CT DE DC GA HI ID IL IA KS KY MD MA MI MO MT NE NH NY OH OR PA RI SC SD TN UT VT VA WI WY Total

CHAPTER 3. rate 0.1 1 0.5 1 1 0.1 1 0.5 1 0.5 0.5 0.2 0.2 0.2 0.1 1 1 1 1 1 0.05 0.2 0.1 1 1 1 0.1 0.5 1 1 0.2 1

nh 2460 3396 6003 2938 2525 3404 4415 2949 46986 5259 6170 3879 3104 3367 3996 24984 4125 6236 4947 67993 2465 3124 3883 3684 13835 2699 3042 4489 2426 25608 3384 3208 280983

Nh 24600 3396 12006 2938 2525 34040 4415 5898 46986 10518 12340 19395 15520 16835 39960 24984 4125 6236 4947 67993 49300 15620 38830 3684 13835 2699 30420 8978 2426 25608 16920 3208 571185

husb ≤ 24 295 371 333 238 90 440 394 380 4349 541 768 567 156 163 270 2876 432 620 458 3809 102 233 248 246 1429 93 426 591 162 2075 280 346

STRATIFIED SAMPLING

pˆh 0.11992 0.10925 0.05547 0.08101 0.03564 0.12926 0.08924 0.12886 0.09256 0.10287 0.12447 0.14617 0.05026 0.04841 0.06757 0.11511 0.10473 0.09942 0.09258 0.05602 0.04138 0.07458 0.06387 0.06678 0.10329 0.03446 0.14004 0.13166 0.06678 0.08103 0.08274 0.10786

Nhpˆh 2950 371 666 238 90 4400 394 760 4349 1082 1536 2835 780 815 2700 2876 432 620 458 3809 2040 1165 2480 246 1429 93 4260 1182 162 2075 1400 346 49039

varcont 23376 0 629 0 0 34491 0 662 0 971 1345 9685 2964 3103 22664 0 0 0 0 0 37171 4314 20900 0 0 0 32982 1027 0 0 5138 0 201422

Thus, for estimating the total number of divorces granted to men aged 24 or less, tˆstr = 49039 and

SE(tˆstr) =

√ 201,422 = 449.

A 95% confidence interval is 49039 ± (1.96)(449) = [48159, 49919] (ii) Similarly, for the women, tˆstr = 4600 + 664 + 1330 + · · · + 658 = 86619 and

SE(tˆstr) =

√ 33672 + 0 + 1183 + · · · 8327 + 0 = 564.

39 A 95% CI is [85513, 87725]. (d) For estimating the proportions, Pˆ =

Σ H Nh

str h=1

and ˆ (p̂str ) = V

¶2 µ H µ Σ N h

h=1

N

N

Pˆ h

¶ n h pˆ h(1 — pˆh) 1— Nh nh — 1

(i) For the men: 24600 pˆstr = 571185

µ

390 2460

¶

3396 + 571185

µ

¶ µ ¶ 647 3208 560 +··· = 0.1928.

3396 571185 3208

√ SE(p̂ st r ) = (9.06 × 10—8) + 0 + (6.54 × 10—9) + · · · + (3.37 × 10—8) + 0 = 1.068 × 10—3.

A 95% confidence interval for the proportion of men aged 40–49 at the time of the decree is 0.1928 ± 1.96(1.068 × 10—3) = [0.191, 0.195]. (ii) For the women: µ ¶ µ ¶ µ ¶ 24600 296 3396 495 3208 437 pˆstr = + + ··· + = 0.1566 571185 2460 571185 3396 571185 3208 and √ SE(p̂ st r ) = (7.19 × 10—8) + 0 + (5.80 × 10—9) + · · · + (2.83 × 10—8) + 0 = 9.75 × 10—4.

A 95% confidence interval is 0.1566 ± 1.96(9.75 × 10—4) = [0.155, 0.158].

3.18 (a)

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CHAPTER 3.

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(b) Let wh be the relative sampling weight for stratum h. Then N∝h nhwh. For each response, we may calculate ,Σ Σ y¯str = nhwhȳh nhwh; h

h

equivalently, we may define a new column ½ 1 if h = 1 or2 weighthi = 2 if h ∈ {3, 4, 5, 6} and calculate ,ΣΣ ΣΣ y¯str = weighthiyhi weighthi. h

i

h

i

Summary statistics for 1974 are: Stratum 1 2 3 4 5 6

nh 13 12 9 3 3 3

wh 1 1 2 2 2 2

Nh/N 0.213 0.197 0.295 0.098 0.098 0.098

Number of fish y¯h s2h 496.2 108528.8 101.9 10185.0 38.2 504.4 39.0 147.0 299.3 189681.3 103.7 8142.3

Weight of fish y¯h s2h 60.4 522.5 32.1 299.5 15.0 130.9 6.9 9.1 39.2 3598.4 8.2 76.4

We have, using (3.5) and ignoring the fpc when calculating the standard error, Response Number of fish Weight of fish

yˆ¯ s t r 180.5 29.0

SE(ŷ¯s t r ) 32.5 4.0

SAS code for calculating these estimates is given below.

41 data nybight; infile nybight delimiter=’,’ firstobs=2; input year stratum catchnum catchwt numspp depth temp ; select (stratum); when (1,2) relwt=1; when (3,4,5,6) relwt=2; end; if year = 1974; proc surveymeans data=nybight mean clm ; stratum stratum; var catchnum catchwt; weight relwt; run; (c) The procedure is the same as that in part (b). Summary statistics for 1975 are: Stratum 1 2 3 4 5 6

nh 14 16 15 13 3 3

Response Number of fish Weight of fish

wh 1 1 2 2 2 2

Number of fish y¯h sh2 486.9 94132.0 262.7 42234.8 119.6 9592.0 238.3 12647.2 119.7 789.3 70.7 3194.3

yˆ¯ s t r 223.9 70.6

Weight of fish y¯h s2h 127.0 3948.0 109.5 7189.8 33.5 867.8 84.1 1583.6 20.6 18.4 12.0 255.0

SE(ŷ¯s t r ) 18.5 5.7

3.19 (a) Stratum 1 2 3 4 Total

Respondents to survey 288 533 91 73 985

Respondents to breakaga, nh 232 514 86 67 899

(b) In the table, µ varconth =

nh 1— Nh

¶µ

Nh N

¶2

p̂ h (1 — p̂ h ) . nh — 1

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42 Stratum 1 2 3 4 Total

CHAPTER 3.

Nh 1374 1960 252 95 3681

nh 232 514 86 67 899

pˆh .720 .893 .872 .866

( NNh )pˆh .269 .475 .060 .022 .826

STRATIFIED SAMPLING

varcont 1.01 × 10—4 3.90 × 10—5 4.05 × 10—6 3.46 × 10—7 1.44 × 10—4

Thus, pˆstr = 0.826 √ SE (p̂ st r ) = 1.44 × 10—4 = 0.012. (c) The weights are as follows: Stratum 1 2 3 4

weight 5.922 3.813 2.930 1.418

The answer again is pˆstr = 0.826. (e)

Stratum 1 2 3 4

Employee Type Faculty Classified staff Administrative staff Academic professional

Survey Response Rate (%) 58 82 93 77

breakaga Response Rate (%) 46 79 88 71

The faculty have the lowest response rate (somehow, this did not surprise me). Stratification assumes that the nonrespondents in a stratum are similar to respondents in that stratum.

3.20 (b) This is done by dividing popsize/sampsize for each county. The first few weights for strata are: countynum 1 2 3

countyname Aitkin Anoka Beltrami

weight 1350.0 1261.4 2750.0

(c) response radon lograd

mean 4.898551 1.301306

SE 0.154362 0.028777

95% CI [4.59560775, 5.20149511] [1.24482928, 1.35778371]

(d) The total number of homes with excessive radon is estimated as 722781, with

43 standard error 28107 and 95% CI [667620, 777942]. Σ 3.22 (a) nh = 2000(NhSh/ i NiSi) Stratum 1 2 Total

Nh/N 0.4 0.6 1.0

Sh (.10)(. 90) = .3000 √ (.03)(.97) = .1706 √

ShNh/N .1200 .1024 .2224

nh 1079 921 2000

(b) For both proportional and optimal allocation, S2 = (.10)(.90) = .09 and S2 = (.03)(.97) = .0291.

1

2

Under proportional allocation n1 = 800 and n2 = 1200. .09 .0291 V (pˆ ) = (.4)2 + (.6)2 = 2.67 × 10—5. prop str 800 1200 For optimal allocation, .09 .0291 + (.6)2 = 2.47 × 10—5.

1079 921 For an SRS, p = (.4)(.10) + (.6)(.03) = .058 Vopt (pˆstr ) = (.4)2

Vsrs (pˆsrs ) =

.058(1 — .058) = 2.73 × 10—5 .

2000

3.23 (a) We take an SRS of n/H observations from each of the N/H strata, so there are a total of µ N/H ¶H · ¸H (N/H)! = (n/H)!(N/H — n/H)! n/H possible stratified samples. (b) By Stirling’s formula, µ ¶ N = n ≈

N! n!(N — n)!

µ ¶N √ N 2πN e

µ ¶N —n ³ n ´n √ √ 2π(N — n) N — n 2πn e e s NN N = 2πn(N — n) nn(N — n)N —n We use the same argument, substituting N/H for N and n/H for n in the equation above, to obtain: µ N/H ¶ s NN/H NH ≈ . n/H 2πn(N — n) nn/H (N — n)(N —n)/H

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CHAPTER 3.

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Consequently, µ

¶H

N/H n/H µ ¶ N n

"s

NN/H NH 2πn(N — n) nn/H (N — n)(N —n)/H s ≈ NN N 2πn(N — n) nn(N — n)N —n · ¸(H—1)/2 N HH/2. = 2πn(N — n)

3.24 We wish to minimize

Σ H

V [t̂str ] =

µ

h=1

#H

¶ 1 — nh N 2 Sh2 hn Nh h

subject to the constraint C = c0 +

H Σ

chnh.

h=1

Lagrange multipliers are often used for such problems. (See, for example, Thomas, G.B. and Finney, R. L. (1982). Calculus and Analytic Geometry, Fifth edition.

Reading, MA: Addison-Wesley, p. 617.)

Define ¶ µ ¶ H µ H 2 Σ Σ nh 2 Sh N f (n1, . . . , nH, λ) = 1— chnh . h n — λ C — c0 — Nh h h=1

h=1

Then

∂f

=

∂nk

2 2S —Nk 2k + λck nk

k = 1, . . . , H, and H ∂f Σ = c0 + chnh — C. ∂λ h=1

Setting the partial derivatives equal to 0 and solving gives NkSk nk = √ λc k for k = 1, . . . , H, and Σ H

which implies that

chnh = C — c0,

h=1

ΣH

√ λ=

h=1

√ chNhSh

C — c0

45 and hence that

NkSk C — c0 n = √ . k Σ H √ ck c N S h h h h=1 √ Note that we also have nk ∝ NkSk/ ck if we want to minimize the cost for a fixed variance N . Then, let g(n1, . . . , nH, λ) = c0 +

H Σ h=1

Then and

· ¶ ¸ H µ Σ nh S2h 2 Nh 1— . chnh — λ V — Nh nh h=1

∂g = ck — λNk2 Sk2 /n 2k ∂nk µ ¶ nh S H ∂g Σ 1— N 2 h2 .

V — = h ∂λ Nh nh h=1

Setting the partial derivatives equal to 0 and solving gives √ λNkSk nk = √ . ck 3.25 (a) We substitute nh,Neyman for nh in (3.4): Σµ ¶ VNeyman (t̂str ) = H 1 — nh N 2 Sh2 hn Nh h h=1 , =

H Σ

.1 — .

h=1

NhS hn

.

H

Nh

Σ

H 2Σ NLSL Sh . N 2 L=1

NLSL

h

nhNhShn

L=1

,

NhSh

H Σ

Shn = .1 — H . Σ . h=1 NLSL

H Σ

NLSL

L=1

n

L=1

=

!2 Ã H H Σ 1 Σ — NhSh NhS2h. n h=1

h=1

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CHAPTER 3.

STRATIFIED SAMPLING

(b) Vprop (tˆstr ) — V Neyman(tˆstr ) =

=

H H Σ NΣ N hS2h — NhS2h n h=1 h=1 !2 Ã H H Σ 1 Σ N S + NhS2h h h — n h=1 h=1 Ã H !2 H N Σ N S2 — 1 Σ N S h h n h=1 h h n h=1

N2 Σ N

= =

H

h

n

h=1

N "

Sh2 —

ÃH Σ Nh h=1

H Σ N h=1

" h

N

L=1

#2 H N H Σ Σ Nh 2 L = Sh — SL N Sh — 2Sh N L=1

Sh

# H H N 2 Σ Nh S2 — S Σ NL h S h n N N L h=1

But

N

!2

H N Σ L

SL + N L=1 h=1 ÃH !2 H Σ NL Σ Nh 2 S = S — , N L N h h=1

ÃH Σ N

L

L=1

N

!2 SL

L=1

proving the result. (c) When H = 2, the difference from (b) is !2 Ã 2 2 N2 Σ Nh S —Σ NL h SL n h=1 N N L=1 " µ µ ¶2 ¶2# N 2 N1 N1 N2 N1 N2 N 2 = + S2 — S1 — S2 n N S1 — N S1 — N S2 N N N " µ µ ¶2 ¶2# N 2 N1 N2 N2 N1 N2 N = S2 + S2 — 1 S1 S1 — n N N N N N N · ¸ N 2 N1 N2 N1 N2 S1 — S2 2 = n N N N + N ( ) N1 N2 = (S — S )2 .

1 2 n 3.34 (a) In the data step, define the variable one to have the value 1 for every observation. Σ Σ Then i∈£ wi1 = N . Here, i∈£ wi1 = 85174776. The standard error is zero

47 because this is a stratified sample. The weights are Nh/nh so the sum of the weights in stratum h is Nh exactly. There is no sampling variability.

Here is the code used to obtain these values: proc surveymeans data=vius mean clm sum clsum; weight tabtrucks; stratum stratum; var one miles_annl mpg; (b) The estimated total number of truck miles driven is 1.115 ×1012; the standard error is 6492344384 and a 95% CI is [1.102×1012, 1.127×1012]. (c) Because these are stratification variables, we can calculate estimates for each truck type by summing whjyhj separately for each h. We obtain: proc sort data=vius; by trucktype; proc surveymeans data=vius sum clsum; by trucktype; weight tabtrucks; stratum stratum; var miles_annl; ods output Statistics=Mystat; proc print data=Mystat; run;

Obs

VarName

VarLabel

1 2 3 4 5

MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL

Number of Miles Driven During 2002 Number of Miles Driven During 2002 Number of Miles Driven During 2002 Number of Miles Driven During 2002 Number of Miles Driven During 2002

Obs LowerCLSum 1 2 3 4 5

4.19064E11 5.32459E11 4.05032E10 3.107E10 7.12861E10

StdDev

UpperCLSum

4708839922 4408042207 395841910 348294378 518195242

4.37525E11 5.4974E11 4.2055E10 3.24353E10 7.33175E10

Sum 428294502082 541099850893 41279084490 31752656137 72301789843

(d) The estimated average mpg is 16.515427 with standard error 0.039676; a 95% CI is [16.4377, 16.5932]. These CIs are very small because the sample size is so large.

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CHAPTER 3.

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Chapter 4

Ratio and Regression Estimation 4.2 (a) We have tx = 69. ty = 83, Sx = 4.092676, Sy = 5.333333, R = 0.8112815, and B = 1.202899. (b)

49

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50 Sample Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40

CHAPTER 4. RATIO AND REGRESSION ESTIMATION Sample S {1, 2, 3} {1, 2, 4} {1, 2, 5} {1, 2, 6} {1, 2, 7} {1, 2, 8} {1, 2, 9} {1, 3, 4} {1, 3, 5} {1, 3, 6} {1, 3, 7} {1, 3, 8} {1, 3, 9} {1, 4, 5} {1, 4, 6} {1, 4, 7} {1, 4, 8} {1, 4, 9} {1, 5, 6} {1, 5, 7} {1, 5, 8} {1, 5, 9} {1, 6, 7} {1, 6, 8} {1, 6, 9} {1, 7, 8} {1, 7, 9} {1, 8, 9} {2, 3, 4} {2, 3, 5} {2, 3, 6} {2, 3, 7} {2, 3, 8} {2, 3, 9} {2, 4, 5} {2, 4, 6} {2, 4, 7} {2, 4, 8} {2, 4, 9} {2, 5, 6}

x̄ S 10.333 10.667 8.000 7.667 10.333 7.667 8.333 12.000 9.333 9.000 11.667 9.000 9.667 9.667 9.333 12.000 9.333 10.000 6.667 9.333 6.667 7.333 9.000 6.333 7.000 9.000 9.667 7.000 10.000 7.333 7.000 9.667 7.000 7.667 7.667 7.333 10.000 7.333 8.000 4.667

y¯S 10.000 11.333 8.333 6.000 11.000 8.000 7.000 13.333 10.333 8.000 13.000 10.000 9.000 11.667 9.333 14.333 11.333 10.333 6.333 11.333 8.333 7.333 9.000 6.000 5.000 11.000 10.000 7.000 12.333 9.333 7.000 12.000 9.000 8.000 10.667 8.333 13.333 10.333 9.333 5.333

B̂ 0.968 1.063 1.042 0.783 1.065 1.043 0.840 1.111 1.107 0.889 1.114 1.111 0.931 1.207 1.000 1.194 1.214 1.033 0.950 1.214 1.250 1.000 1.000 0.947 0.714 1.222 1.034 1.000 1.233 1.273 1.000 1.241 1.286 1.043 1.391 1.136 1.333 1.409 1.167 1.143

tˆSRS 90.000 102.000 75.000 54.000 99.000 72.000 63.000 120.000 93.000 72.000 117.000 90.000 81.000 105.000 84.000 129.000 102.000 93.000 57.000 102.000 75.000 66.000 81.000 54.000 45.000 99.000 90.000 63.000 111.000 84.000 63.000 108.000 81.000 72.000 96.000 75.000 120.000 93.000 84.000 48.000

tˆyr 66.774 73.313 71.875 54.000 73.452 72.000 57.960 76.667 76.393 61.333 76.886 76.667 64.241 83.276 69.000 82.417 83.786 71.300 65.550 83.786 86.250 69.000 69.000 65.368 49.286 84.333 71.379 69.000 85.100 87.818 69.000 85.655 88.714 72.000 96.000 78.409 92.000 97.227 80.500 78.857

51 Sample Number 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 Average Variance

Sample S {2, 5, 7} {2, 5, 8} {2, 5, 9} {2, 6, 7} {2, 6, 8} {2, 6, 9} {2, 7, 8} {2, 7, 9} {2, 8, 9} {3, 4, 5} {3, 4, 6} {3, 4, 7} {3, 4, 8} {3, 4, 9} {3, 5, 6} {3, 5, 7} {3, 5, 8} {3, 5, 9} {3, 6, 7} {3, 6, 8} {3, 6, 9} {3, 7, 8} {3, 7, 9} {3, 8, 9} {4, 5, 6} {4, 5, 7} {4, 5, 8} {4, 5, 9} {4, 6, 7} {4, 6, 8} {4, 6, 9} {4, 7, 8} {4, 7, 9} {4, 8, 9} {5, 6, 7} {5, 6, 8} {5, 6, 9} {5, 7, 8} {5, 7, 9} {5, 8, 9} {6, 7, 8} {6, 7, 9} {6, 8, 9} {7, 8, 9}

x̄ S 7.333 4.667 5.333 7.000 4.333 5.000 7.000 7.667 5.000 9.000 8.667 11.333 8.667 9.333 6.000 8.667 6.000 6.667 8.333 5.667 6.333 8.333 9.000 6.333 6.333 9.000 6.333 7.000 8.667 6.000 6.667 8.667 9.333 6.667 6.000 3.333 4.000 6.000 6.667 4.000 5.667 6.333 3.667 6.333

y¯S 10.333 7.333 6.333 8.000 5.000 4.000 10.000 9.000 6.000 12.667 10.333 15.333 12.333 11.333 7.333 12.333 9.333 8.333 10.000 7.000 6.000 12.000 11.000 8.000 8.667 13.667 10.667 9.667 11.333 8.333 7.333 13.333 12.333 9.333 8.333 5.333 4.333 10.333 9.333 6.333 8.000 7.000 4.000 9.000

B̂ 1.409 1.571 1.188 1.143 1.154 0.800 1.429 1.174 1.200 1.407 1.192 1.353 1.423 1.214 1.222 1.423 1.556 1.250 1.200 1.235 0.947 1.440 1.222 1.263 1.368 1.519 1.684 1.381 1.308 1.389 1.100 1.538 1.321 1.400 1.389 1.600 1.083 1.722 1.400 1.583 1.412 1.105 1.091 1.421

tˆSRS 93.000 66.000 57.000 72.000 45.000 36.000 90.000 81.000 54.000 114.000 93.000 138.000 111.000 102.000 66.000 111.000 84.000 75.000 90.000 63.000 54.000 108.000 99.000 72.000 78.000 123.000 96.000 87.000 102.000 75.000 66.000 120.000 111.000 84.000 75.000 48.000 39.000 93.000 84.000 57.000 72.000 63.000 36.000 81.000

tˆyr 97.227 108.429 81.938 78.857 79.615 55.200 98.571 81.000 82.800 97.111 82.269 93.353 98.192 83.786 84.333 98.192 107.333 86.250 82.800 85.235 65.368 99.360 84.333 87.158 94.421 104.778 116.211 95.286 90.231 95.833 75.900 106.154 91.179 96.600 95.833 110.400 74.750 118.833 96.600 109.250 97.412 76.263 75.273 98.053

7.667 3.767

9.222 6.397

1.214 0.044

83.000 518.169

83.733 208.083

(c)

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CHAPTER 4. RATIO AND REGRESSION ESTIMATION

Histogram of ratio estimate

10

Frequency

10

0

0

5

5

Frequency

15

15

20

Histogram of SRS estimate

50

100

150

SRS estimate of total

50

100

150

Ratio estimate of total

The shapes are actually quite similar; however, it appears that the histogram of the ratio estimator is a little less spread out. (d) The mean of the sampling distribution of tˆyr is 83.733; the variance is 208.083 and the bias is 83.733— 83 = 0.733. By contrast, the mean of the sampling distribution of N y¯ is 83 and its variance is 518.169. (e) From (4.6), Bias ( ŷ¯ r ) = 0.07073094.

4.3 (a) The solid line is from regression estimation; the dashed line from ratio estimation; the dashed/dotted line has equation y = 107.4.

100 120 140 160 60

80

Age

53

6

7

8

9

10

11

12

Diameter

(b, c) Method SRS, y¯ Ratio Regression

ŷ¯ 107.4 117.6 118.4

SE ( yˆ¯ ) 6.35 4.35 3.96

For ratio estimation, B̂ = 11.41946; for regression estimation, B̂ 0 = —7.808 and B̂ 1 = 12.250. Note that the sample correlation of age and diameter is 0.78, so we would expect both ratio and regression estimation to improve precision.

To calculate V̂ ( ŷ¯ r ) using (4.9), note that s2e = 321.933 so that ˆ (ȳˆ r ) = V

¶µ ¶ µ 20 10.3 2 321.933 1— = 18.96 20 1132 9.405

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CHAPTER 4. RATIO AND REGRESSION ESTIMATION

and S E ( ŷ¯ ) = 4.35. For the regression estimator, we have se2 = 319.6, so µ ¶ 20 319.6 ˆ ˆ V ( y¯ ) = 1 — = 15.7.

1132 20 From a design-based perspective, it makes little difference which estimator is used.

From a model-based perspective, though, a plot of residuals vs. predicted values exhibits a “funnel shape” indicating that the variability increases with x. Thus a model-based analysis for these data should incorporate the unequal variances. From that perspective, the ratio model might be more appropriate.

Note that the variances calculated using SAS PROC SURVEYREG are larger since they use Vˆ2 from Section 11.7.

Here is code for SAS: data trees; input treenum diam age @@; sampwt = 1132/20; datalines; 1 12.0 125 11 2 11.4 119 12 3 7.9 83 13 4 9.0 85 14 5 10.5 99 15 6 7.9 117 16 7 7.3 69 17 8 10.2 133 18 9 11.7 154 19 10 11.3 168 20 ; proc print data=trees; run;

5.7 8.0 10.3 12.0 9.2 8.5 7.0 10.7 9.3 8.2

61 80 114 147 122 106 82 88 97 99

/* proc surveymeans will estimate ratios with keyword ’ratio’ proc surveymeans data=trees total=1132 mean stderr clm sum clsum ratio ; var diam age; /* need both in var statement */ ratio ’age/diameter’ age/diam; weight sampwt; ods output Statistics=statsout Ratio=ratioout; run; /* Can get ratio estimates of totals by taking output from proc surveymeans and multiplying by N */

*/

55

data ratioout1; set ratioout; xmean = 10.3; ratiomean = ratio*xmean; semean = stderr*xmean; lowercls = lowercl*xmean; uppercls = uppercl*xmean; proc print data = ratioout1; run; /* Can also calculate ratio estimate by hand */ data treesresid; set trees; resid = age - 11.419458*diam; resid2 = resid*(10.3/ 9.405); proc univariate data=treesresid; run; proc surveyreg data=trees total=1132; model age = diam / clparm solution ; /* fits the regression model */ weight sampwt; estimate ’Mean age of trees’ intercept 1 diam 10.3; run; proc gplot data=trees; plot age*diam; run; data trees2; set trees; resid = age - (-7.8080877 + 12.2496636*diam); proc surveymeans data=trees2 total =1132; weight sampwt; var resid age diam; run; The output from proc surveyreg gives a larger standard error for the regression estimator: Analysis of Estimable Functions

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Estimate

Standard Error

t Value

Pr > |t|

118.363449

5.20417420

22.74

<.0001

Parameter Mean age of trees

4.5 There are 85 18-hole courses in the sample. For these 85 courses, the sample mean weekend greens fee is y¯d = 34.829 and the sample variance is s2d = 395.498.

Using results from Section 4.3,

r

SE[y¯ ]= d

395.498 85

= 2.16.

Here is SAS code: filename golfsrs data golfsrs; infile golfsrs delimiter="," dsd firstobs=2; /* The dsd option allows SAS to read the missing values between successive delimiters */ input RN state $ holes type $ yearblt wkday18 wkday9 wkend18 wkend9 backtee rating par cart18 cart9 caddy $ pro $; sampwt = 14938/120; if holes = 18 then holes18 = 1; else holes18=0; proc surveymeans data=golfsrs total = 14938; weight sampwt; var wkend18; domain holes18; run;

Data Summary Number of Observations Sum of Weights

120 14938

Statistics

57

Variable

N

Mean

Std Error of Mean

wkend18

85

34.828824

2.148380

Statistics Variable

95% CL for Mean

wkend18

30.5565341 39.1011129

Domain Analysis: holes18

holes18

Variable

N

Mean

Std Error of Mean

0 1

wkend18 wkend18

0 85

.

34.828824

.

2.144660

Domain Analysis: holes18 holes18

Variable

95% CL for Mean

0 1

wkend18 wkend18

. .

30.5639320 39.0937150

4.6 As you can see from the plot of weekend greens fee vs. back-tee yardage, this is not a “classical” straight-line relationship. The variability in weekend greens fee appears to increase with the back-tee yardage. Nevertheless, we can estimate the slope and intercept, with yˆ = —37.26 + 0.0113x.

(We’ll discuss standard errors in Chapter 11.) For estimating the ratio, we have y¯ 34.83 B̂ = = = 0.00545. x̄ 6392.29 Using (4.10), with s2e the sample variance of the residuals, s2 ˆ ˆ 362.578 V (B) ≈

e = —7 (85)(6292.29)2 (85)(6392.29)2 = 1.044 × 10

S E (B̂ ) = .00032.

4.7 (a) 88 courses have a golf professional. For these 88 courses, y¯d1 = 23.5983 and

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sd1 = 387.7194, so 2

V̂ (ȳd 1 ) =

387.7194 88

= 4.4059.

2 (b) For the 32 courses without a golf professional, y¯d2 = 10.6797 and sd2 = 19.146, so 19.146 = 0.5983.

V̂ (ȳd2 ) = 32

4.8 (a) (b) B̂ = y¯ / x̄ = 297897/647.7467 = 459.8975. Thus tˆyr = t x B̂ = (2087759)(459.8975) = 960,155,061.

The estimated variance of the residuals about the line y = B̂ x is s2e = 149,902,393,481.

Using (4.11), then, with farms87 as the auxiliary variable, r r 300 s2e SE [tˆyr ] = 3078 1— = 65 ,364,822.

3078 300 (c) The least squares regression equation is yˆ = 267029.8 + 47.65325x Then ŷ¯ r e g = 267029.8 + 47.65325(647.7467) = 297897.04 and

tˆyreg = 3078ŷ¯reg = 916,927,075.

The estimated variance of the residuals from the regression is se2 = 118,293,647,832, which implies from (4.19) that r r 300 s2 e = 58 065 813 SE [tˆyreg ] = 3078 1 — , , .

3078 300 (d) Clearly, for this response, it is better to use acres87 as an auxiliary variable.

The correlation of farms87 with acres92 is only 0.06; using farms87 as an auxiliary variable does not improve on the SRS estimate N y¯. The correlation of acres92 and acres87, however, exceeds 0.99. Here are the various estimates for the population total of acres92: tˆ SE [tˆ] Estimate SRS, N y¯ 916,927,110 58,169,381 Ratio, x = acres87 951,513,191 5,344,568 Ratio, x = farms87 960,155,061 65,364,822 Regression, x = farms87 916,927,075 58,065,813

59 Moral: Ratio estimation can lead to greatly increased precision, but should not be used blindly. In this case, ratio estimation with auxiliary variable farms87 had larger standard error than if no auxiliary information were used at all. The regression estimate of t is similar to N y¯, because the regression slope is small relative to the magnitude of the data. The regression slope is not significantly different from 0; as can be seen from the picture in (a), the straight-line regression model does not describe the counties with few but large farms.

4.9 We use results from Section 4.2. xi = farms92 for county i. Define Then

(a) Let yi = acres92 for county i, and

tˆy1 = N ū = 3078(161773.8) = 497,939,808 and

r

SE [tˆy1 ] = N 1 — 300 3078 (b) Now and

r

r r s2u 300 109,710,284,064 = 55 ,919 ,525. = 3078 1 — 3078 300 300

tˆy2 = 3078(136123.2) = 418, 987, 302 r r 300 53,195,371,851 = 38,938,277.

SE [tˆ ] = 3078 1 — y2

3078

300

4.10 (a)

(b) Here is code and output from SAS: filename cherries ’cherry.csv’;

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data cherry; infile cherries delimiter=’,’ firstobs=2; input diam height vol; sampwt = 2967/31; obsnum = _n_; label diam = ’diam (in) at 4.5 feet’ height = ’height of tree (feet)’ vol = ’volume of tree (cubic feet)’ sampwt = ’sampling weight’ ; /* Plot and print the data set */ proc print data = cherry; var diam height vol sampwt; proc gplot data = cherry; plot vol *diam ; proc surveymeans data = cherry total=2967 mean clm sum clsum ratio ; weight sampwt; var diam vol; ratio ’vol/diam’ vol/diam; ods output Statistics=statsout Ratio=ratioout; run; data ratioout1; set ratioout; xtotal = 41835; ratiosum = ratio*xtotal; sesum = stderr*xtotal; lowercls = lowercl*xtotal; uppercls = uppercl*xtotal; proc print data = ratioout1; run; Using this code, we obtain tˆyr = 95272.16 with 95% CI of [84098, 106,446]. (c) SAS code and output follow: proc surveyreg data=cherry total=100; model vol=diam / clparm solution; weight sampwt; estimate ’Total volume’ intercept 2967 diam 41835;

61 /* substitute N for intercept, t_x for diam */ run; Analysis of Estimable Functions

Parameter Total volume

Estimate

Standard Error

t Value

Pr > |t|

102318.860

2233.70776

45.81

<.0001

Analysis of Estimable Functions

Parameter Total volume

95% Confidence Interval 97757.0204 106880.700

Note that the estimate from regression estimation is quite a bit higher than the estimate from ratio estimation. In addition, the CI for regression estimation is narrower than the CIs for tˆyr or N y¯. This is because the regression model is a better fit to the data than the ratio model.

4.11 (a) The variable number of physicians has a skewed distribution. The first histogram excludes Cook County, Illinois (with yi = 15,153) for slightly better visibility.

The next histogram, of all 100 counties, depicts the logarithm of (number of physicians + 1) (which is still skewed).

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(b) y¯ = 297.17, sy2 =2,534,052.

Thus the estimated total number of physicians is N y¯ = (3141)(297.17) = 933,411 with

s ³ n ´ s2y SE (Ny¯) = N 1 — N n sµ ¶ 100 2,534,052 = 3141 1 — 3141 100 √ = 3141 24533.75 =

491,983.

The standard error is large compared with the estimated total number of physicians.

The extreme skewness of the data makes us suspect that N y¯ does not follow an approximate normal distribution, and that a confidence interval of the form N y¯± 1.96 SE (N y¯) would not have 95% coverage in practice. In fact, when we substitute sample quantities into (2.23), we obtain nmin = 28 + 25 (6.04)2 = 940 as the required minimum sample size for y¯ to approximately follow a normal distribution. (c) Again, we omit Cook County.

63

There appears to be increasing variance as population increases, so ratio estimation may be more appropriate. (d) Ratio estimation: y¯

B̂ =

297.17 = 0.002507. = 118531.2

x̄ ˆ t yr = B̂ t x = (.002507)(255,077,536) = 639,506.

Using (4.11),

r

³ 1 — n ´ s 2e SE (tˆyr) = sµ N n ¶ = 3141 1 100 (255077536)2 172268 — 3141 (372306374)2 100 = 87885.

N

Regression estimation: B̂ 0 = —54.23

B̂ 1 = 0.00296.

From (4.15) and (4.19), yˆ¯ r e g = Bˆ 0 + Bˆ 1 x̄ U = —54.23 + (0.00296) =

µ

255,077,536

¶

3141

186.52

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CHAPTER 4. RATIO AND REGRESSION ESTIMATION

and

r

³ n ´ s 2e 1— N n sµ ¶ 100 114644.1 = 1— 3141 100 = 33.316.

SE (ŷ¯reg ) =

Consequently, and

tˆyreg = N ŷ¯ r e g = 3141(186.52) = 585871 SE (tˆyreg) = N SE (ȳˆr e g ) = 3141(33.316) = 104645.

The standard error from proc surveyreg is smaller and equals 92535. (e) Ratio estimation and regression estimation both lead to a smaller standard error, and an estimate that is closer to the true value.

Here is SAS code for performing these analyses: data counties; infile counties firstobs=2 delimiter=","; input RN State County landarea totpop physician enroll percpub civlabor unemp farmpop numfarm farmacre fedgrant fedciv milit veterans percviet ; sampwt = 3141/100; logphys = log(physician+1); /* The following histogram is really really skewed because of Cook County */ proc univariate data=counties ; var physician; histogram / endpoints = 0 to 16000 by 500; run; data countnoCook; set counties; if physician gt 10000 then delete; proc univariate data=countnoCook; var physician ; histogram / endpoints = 0 to 4400 by 200; run; proc univariate data=counties; var logphys ;

65 histogram / endpoints = 0 to 10 by .5; run; proc surveymeans data=counties total = 3141 mean clm sum clsum; weight sampwt; var physician; run; proc gplot data=countnoCook; plot physician * totpop; run; proc surveymeans data=counties total=3141 mean clm sum clsum ratio; var physician totpop; /* need both in var statement */ ratio ’physician/totpop’ physician/totpop; weight sampwt; ods output Statistics=statsout Ratio=ratioout; run; /* Can get ratio estimates of totals by taking output from proc surveymeans and multiplying by t_x */ data ratioout1; set ratioout; xtot = 255077536; ratiotot = ratio*xtot; setot = stderr*xtot; lowercls = lowercl*xtot; uppercls = uppercl*xtot; proc print data = ratioout1; run; /* Can also calculate ratio estimate by hand */ data resid; set counties; resid = physician -0.002507*totpop; resid2 = resid*(255077536/ 372306374); /* Use g-weights in SE formula*/ proc surveymeans data=resid total=3141 mean clm sum clsum; weight sampwt; var resid resid2; run;

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proc surveyreg data=counties total=3141; model physician=totpop / clparm solution ; /* fits the regression model */ weight sampwt; estimate ’Total number of physicians’ intercept 3141 totpop 255077536; estimate ’Average number of physicians’ intercept 3141 totpop 255077536/divisor = 3141; run; 4.12 (a) The distribution appears to be skewed, but not quite as skewed as the distribution in Exercise 4.11.

(b) y¯ = 1146.87, sy2 =1,138,630 Thus the estimated total farm population, using N y¯, is Ny¯ = (3141)(1146.87) = 3,602,319 with

sµ ¶ 100 1,138,630 SE (N y¯) = 3141 1 — = 329,787.

3141 100

SAS proc surveymeans gives the same result. (c) Note that corr(farmpop,landarea) = —0.058. We would not expect ratio or regression estimation to do well here.

67

(d) Ratio estimation:

B̂ =

y¯ x̄

=

1146.87

= 1.21372.

944.92 s e = 3,297,971 2

tˆyr = B̂ t x = (1.21372)(3,536, 278) = 4,292,058 sµ ¶ 100 (3,536, 278)2 3,297,971 ) = 3141 1 ˆ — = 668 727.

SE (tyr 3141 (2967994)2 100 Note that the SE for the ratio estimate is higher than that for N y¯.

Regression estimation: B̂ 0 = 1197.35,

B̂ 1 = —.05342218, s2e = 1,134,785.

From (4.15) and (4.19), yˆ¯ r e g = B̂ 0 + B̂ 1 x̄ U = 1197.35— 0.0534(3,536,278) = 1137.2 sµ ¶ SE (ŷ¯ ) = 1 — 100 1,134,785 = 104.82. reg 3141 100 Consequently,

tˆyreg = N y¯ˆ r e g = 3141(1137.2) = 3,571,960 SE (tˆyreg) = N SE (ŷ¯r e g ) = 3141(104.82) = 329,230.

SAS proc surveyreg gives SE 350595. (e) The “true” value is ty = 3,871,583.

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Here is SAS code that may be used to compute these estimates. See Exercise 4.11 solution for reading in the data. proc univariate data=counties ; var farmpop; histogram / endpoints = 0 to 6000 by 500; run; proc gplot data=counties; plot farmpop * landarea; run; proc surveymeans data=counties total=3141 mean clm sum clsum ratio ; var farmpop landarea; /* need both in var statement */ ratio ’farmpop/landarea’ farmpop/landarea; weight sampwt; ods output Statistics=statsout Ratio=ratioout; data ratioout1; set ratioout; xtot = 3536278; ratiotot = ratio*xtot; setot = stderr*xtot; lowercls = lowercl*xtot; uppercls = uppercl*xtot; proc print data = ratioout1; run; proc surveyreg data=counties total=3141; model farmpop=landarea / clparm solution ; weight sampwt; estimate ’Total farmpop’ intercept 3141 landarea 3536278; estimate ’Average farmpop’ intercept 3141 landarea 3536278/divisor = 3141; run; 4.13 (a) As in Exercise 4.11, we omit Cook County, Illinois, from the histogram so we can see the other data points better. Cook County has 457,880 veterans. The distribution is very skewed; Cook County is an extreme outlier.

69

(b) y¯ = 12249.71, sy2 = 2,263,371,150.

N y¯ = (3141)(12249.71) = 38,476,339 sµ ¶ 100 2,263,371,150 SE (N y¯) = 3141 1 — = 14,703,478.

3141 100 (c) Again, Cook County is omitted from this plot. These data appear very close to a straight line. We would expect ratio or regression estimation to help immensely.

(d) Ratio estimation: B̂ =

y¯ x̄

12249.71 = = 0.1033 118531.2

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CHAPTER 4. RATIO AND REGRESSION ESTIMATION tˆyr = B̂ t x = (0.1033)(255,077,536) = 26,361,219.

1 Σ s2e = (y i — B̂ xi )2 = 59,771,493 99 i∈§ sµ ¶ 100 (255077536)2 59,771,493 ) = 3141 1 — = 1637046.

SE (tˆyr 3141 (372306374)2 100

Regression estimation: B̂ 0 = 1534.201,

B̂ 1 = 0.0904, s2e = 13,653,852

yˆ¯ r e g = B̂ 0 + B̂ 1 x̄ U

³ ´ = (1534.2) + (0.0904) 255,077,536 3141

= 8875.7. sµ ¶ 100 13,653,852 SE (ŷ¯reg ) = 1 — = 363.58 3141 100 tˆyreg = N yˆ¯ r e g = 27,878,564 SE (tˆyreg) = N SE (ȳˆr e g ) = 1,142,010.

The SE from proc surveyreg is 1063699.

Here is SAS code. The data are read in as in Exercise 4.11. /* The following histogram is skewed because of Cook County */ proc univariate data=counties ; var veterans; histogram / endpoints = 0 to 440000 by 10000; run; data countnoCook; set counties; if veterans gt 400000 then delete; proc univariate data=countnoCook; var veterans ; histogram / endpoints = 0 to 100000 by 5000; run; proc gplot data=countnoCook; plot veterans * totpop; run; proc surveymeans data=counties total=3141 mean clm sum clsum ratio;

71 var veterans totpop; /* need both in var statement */ ratio ’veterans/totpop’ veterans/totpop; weight sampwt; ods output Statistics=statsout Ratio=ratioout; run; data ratioout1; set ratioout; xtot = 255077536; ratiotot = ratio*xtot; setot = stderr*xtot; lowercls = lowercl*xtot; uppercls = uppercl*xtot; proc print data = ratioout1; run; /* Can also calculate ratio estimate by hand */ data resid; set counties; resid = veterans -0.002507*totpop; resid2 = resid*(255077536/ 372306374); /* Use g-weights in SE formula*/ proc surveymeans data=resid total=3141 mean clm sum clsum; weight sampwt; var resid resid2; run; proc surveyreg data=counties total=3141; model veterans=totpop / clparm solution ; /* fits the regression model */ weight sampwt; estimate ’Total number of veterans’ intercept 3141 totpop 255077536; estimate ’Average number of veterans’ intercept 3141 totpop 255077536/divisor = 3141; run; (e) Here, ty = 27,481,055.

4.15 (a) A 95% CI for the average concentration of lead is 146 127 ± 1.96√ = [101.0, 153.0].

121

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For copper, the corresponding interval is 16 35 ± 1.96 √ = [32.1, 37.85].

121 Note that we do not use an fpc here. Soil samples are collected at grid intersections, and we may assume that the amount of soil in the sample is negligible compared with that in the region. (b) Because the samples are systematically taken on grid points, we know that (Nh/N ) = (nh/n). Using (4.21), 82 31 8 = 71 + 259 + LEAD : y¯ 189 = 127 post 121 121 121 82 31 8 = 28 + 50 + COPPER : y¯ 45 = 35. post 121 121 121 Not surprisingly, these are the same numbers from the first table. The variances and confidence intervals, however, differ. From (4.22), µ ¶ µ ¶ µ ¶ 82 282 31 2322 8 792 ˆ LEAD : V (¯ypost ) = + + = 121 .8 121 121 121 121 121 121 µ ¶ µ ¶ µ ¶ 82 92 31 182 8 152 ˆ COPPER : V(¯ypost ) = + + = 1 26 . .

121 121 121 121 121 121 The corresponding 95% confidence intervals are √ LEAD : 127 ± 1.96 121.8 = [105.4, 148.6] √ COPPER : 35 ± 1.96 1.26 = [32.8, 37.2].

These confidence intervals are both smaller than the CIs in part (a); this indicates that stratified sampling would increase precision in future surveys.

The following table gives the estimated coefficients of variation for the SRS and poststratified estimates: SRS Lead Copper

0.0416

Poststratified 0.0869 0.0321

4.18 As di = yi — Bxi, d¯ = y¯ — B x̄ . Then, using (A.10), V (d¯) = V (y¯ — B x̄) = V (y¯) — 2 Cov (y¯, B x̄ ) + B2V (x̄ ) µ ¶· 2 ¸ x y Sy 2 2x n — 2B RS S + B S . = 1—N n n n 4.21 From (4.6), the squaredµbias of ¶ B̂ is approximately 1 n 21 1— [BS2 — RSxSy]2 x 4 2 N n xU

73 From (4.8), µ E [(B̂ — B) ] ≈ 1 — N 2

n

¶

1 2

nx U

2 2 2 (S x Sy + B S x). y — 2BRS

The approximate MSE is thus of order 1/n, while the squared bias is of order 1/n2.

Consequently, MSE ( B̂ ) ≈ V (B̂ ).

4.22 A rigorous proof, showing that the lower order terms are negligible, is beyond the scope of this book. We give an argument for (4.6). · ¸ y¯ y¯U E [B̂ — B] = E — x̄ x̄U · µ ¶ ¸ y¯ x̄ U y¯U = E — x̄U x̄ x̄U · µ ¶ ¸ y¯ x̄ — x̄ U y¯U = E 1— — x̄U x̄U x̄ · ȳ(x̄ — x̄U ) ¸ = —E x̄ U x̄ · µ ¶¸ y¯( x̄ —2 x̄ U ) x̄ — x̄ U = E x̄ — 1 x̄U 1 = {BV (x̄) — Cov (x̄, y¯) + E[(B̂ — B)(x̄ — x̄U )2]} x̄2U 1 ≈ [BV (x̄) — Cov (x̄, y¯)]. x̄ U2 4.24 (a) From (4.5), µ ¶µ ¶ 1 tu x̄ — x̄ U y¯ — x̄ 1— 1 U1 = x̄ U tx x̄ · ¸µ ¶ U y¯ — y¯ = y¯ — ū — (1 — x̄) 1 + .

1 ty — tu x̄ — x̄ 2 U2 1 — x̄ U N — tx x̄ y¯ — y¯

and

The covariance follows because the expected value of terms involving x̄— x̄ U are small compared with the other terms. (b) Note that because xi takes on only values 0 and 1, xi 2 = xi, xiui = ui, and

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uiyi = u2. iNow look at the numerator of (4.26). ¸ ¸· N · Σ tu ty — tu (xi — x̄ U ) ui — ū U — (xi — x̄ U ) yi — y¯U — ui + ū U + tx N — tx i=1 ¸ N · ¸· Σ ty — tu tu = ui — ū U — (xi — x̄ U ) yi — ui + x tx N — tx i i=1 N · ¸ tu tu Σ u y — u2 + ty — tu tu ty — tu 2 i i = uixi — xiyi + uixi — x i tx N — tx i N — tx tx tx i=1 µ ¶ ¶µ tu ty — tu + t + x̄U — ūU — t u ty N — tx x tx ¸ N · Σ ty — tu tu ty — tu tu u + tu i = xi + 0 ui — ui — tx N — tx tx tx N — t x i=1 = =

ty — tu tu ty — tu tu — tx N — tx tx N — tx 0.

Consequently,

·µ Cov

¶ ½ ¾¸ ty — tu ū — x̄ , y¯ — ū — (1 — x̄) = 0. tx N — tx tu

4.25 We use the multivariate delta method (see for example Lehmann, 1999, p.

295). Let g(a, b) = x̄ U b a so that g(x̄, y¯) = ŷ¯r and g(x̄ U , y¯U ) = y¯U . Then, ∂g x̄ U b =— 2 , ∂a a and ∂g x̄ U = . ∂b a Thus, the asymptotic distribution of √ √ £ ¤ n[g(x̄, y¯) — g(x̄ U , y¯U )] = n ŷ¯r — y¯U is normal with mean 0 and variance µ ¶2 µ ¶µ ¶ µ ¶2 ∂g ∂g ∂g ∂g 2 S2 V = S +2 RSxSy + ∂a x¯U ,y¯U x ∂b x¯U ,y¯U y ∂a ∂b x¯U,y¯U = B2S2 + 2BSxSy + S2. x

y

75 4.26 Using (4.8) and (4.17), µ ¶ n 1 2 V (ŷ¯ r) — V (ŷ¯ reg ) ≈ 1— (Sy — 2BRS xS y + B2S2x) N n µ ¶ n 1 2 — 1— S (1 — R2) N n y µ ¶ n 1 = 1— [—2BRS S + B2S2 + R2S2] x y x y µ ¶ N n n 1 = 1— [RS — BS ]2 y x N n ≥ 0.

4.28 N Σ (yi — y¯U — B1[xi — x̄ U ]) 2 i=1

=

N —1 N

2

2

2

Σ [(yi — y¯U ) — 2B1(xi — x̄ U )(yi — y¯U ) + B1 (xi — x̄ U ) ] N —1 i=1

= S2y — 2B1RSxSy + B21S2x 2 2 2 = S2 — 2 R Sy S S + R Sy S2 y

x y

Sx 2 = S y(1 — R2).

Sx2

x

4.29 From (4.15), yˆ¯ r e g = y¯ + B̂1 (x̄U — x̄) = y¯ + B1 (x̄U — x̄ ) + (B̂ 1 — B1 )(x̄U — x̄).

Thus,

Now,

E [ŷ¯reg — y¯U ] = E[ȳ + B1 (x̄U — x̄ ) + (B̂ 1 — B1)(x̄U — x̄)] — y¯U = E [(B̂ 1 — B 1 )(x̄ U — x̄ )] = —Cov (B̂ 1 , x̄ ). Σ ˆB = i∈£ (xi — x̄ )(y i — y¯) 1 Σi∈£ (xi — x̄ ) 2

and (xi — x̄)(yi — y¯) = (xi — x̄)(yi — y¯U + y¯U — y¯) = (xi — x̄)[di + B1(xi — x̄ U ) + y¯U — y¯] = (xi — x̄)[di + B1(xi — x̄ ) + B1 (x̄ — x̄ U ) + y¯U — y¯]

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76 with Thus,

CHAPTER 4. RATIO AND REGRESSION ESTIMATION Σ Σ Σ (xi — x̄)(y i — y¯) = (xi — x̄ )d i + B1 (xi — x̄ ) 2 . i∈£

i∈£

Bˆ1 = B + 1

i∈£

Σ

U )di + (x̄ U — x̄ ) i∈£ (xi — x̄ Σ 2 i∈£ (xi — x̄)

Σ

i∈£ di

.

Let qi = di(xi — x̄ U ). Then N

Σ i=1

N

N

i=1

i=1

Σ Σ qi = (yi — y¯U )(xi — x̄ U ) — B1 (xi — x̄ U )2 = 0

by the definition of B1, so q¯U = 0. Consequently, · ¸ nq¯ + n(x̄ U — x̄ )d¯ E[(B̂ 1— B 1)(x̄ U— x̄ )] = E (x̄ U— x̄ ) (n — 1)s2x 1 2 ≈ E [q¯(x̄ U — x̄) + d̄(x̄U — x̄ ) ] S2 x 1 ¯ = {— Cov (q¯, x̄ ) + E [d (x̄ U — x̄ ) 2 ]}.

2 Sx Since Cov (q¯, x̄ ) = =

µ ¶ N 1 n 1Σ 1— (qi — q¯U )(xi — x̄ U ) N n N —1 i=1 µ ¶ N 1 n 1Σ 1— qi(xi — x̄ U ) N n N —1 i=1

and E [ d¯( x̄ U — x̄ ) 2 ] is of smaller order than Cov (q¯, x̄ ) , the approximation is shown.

4.32 From linear models theory, if Y = Xβ + ε, with E[ε] = 0 and Cov[ε] = σ2A, then the weighted least squares estimator of β is β̂ = (XT A—1X)—1XT A—1Y with V [ β̂ ] = σ2(XT A—1X)—1.

This result may be found in any linear models book (for instance, Christensen, 1996, p. 31). In our case, Y = (Y1, . . . , Yn)T , so

and

X = (x1, . . . , xn)T , and A = diag(x1, . . . , xn)

Σn ˆ =( Yi y¯ )—1 T —1 T —1 X A X X A Y = Σni=1 = β x̄ i=1 xi σ2 V [β̂ ] = σ 2(XT A—1X)—1 = Σn

i=1 xi

.

77 4.35 (a) No, because some values are 0. (b) When the entire population is sampled, tˆx = tx and tˆy = ty, so B̂ = ty/tx and t x B̂ = ty. (c) Answers will vary. (d) Letting Zi be the indicator variable for inclusion in the sample, # N "1 ty E [ b̄ — B] = E n ΣZ ib i — tx i=1 Σ N N 1Σ bixi bi — i=1 = N tx i=1 ΣN ΣN ΣN 1 ( i=1 bi)( j=1 xi) bx i=1 i i = N — tx tx "N # Σ 1 = — bixi — Nb ¯ U x̄ u tx i=1

=

(N — 1)Sbx .

— tx

(e) From linear models theory, if Y = Xβ + ε, with E[ε] = 0 and Cov[ε] = σ2A, then the weighted least squares estimator of β is β̂ = (XT A—1X)—1XT A—1Y with V [ β̂ ] = σ2(XT A—1X)—1.

Here, A = diag(x2), so i

XT A—1X =

Σ i∈£

and XT A—1Y =

Σ x i∈£

1

x 1

x =n

i 2 i xi

y =

i 2 i xi

Σ Y /x . i

i

i∈£

4.42 (a)

proc surveymeans data = vius mean sum clm clsum; weight tabtrucks; strata stratum; var miles_annl;

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CHAPTER 4. RATIO AND REGRESSION ESTIMATION

domain business; run; Business in which vehicle was most often used during 2002

Variable

For-hire tra Vehicle leas Agriculture, Mining Utilities Construction Manufacturin Wholesale tr Retail trade Information Waste manage Arts, entert Accommodatio Other servic

MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL

Mean

Std Error of Mean

56452 23306 10768 19210 15081 16714 19650 23052 17948 14927 14410 9536.588165 20461 16818

1246.317720 790.981413 415.312729 1126.305405 830.112607 381.173223 1018.600046 1184.715817 561.147469 1396.653144 726.842687 1267.485898 1300.857231 600.239019

Domain Analysis: Business in which vehicle was most often used Business in which vehicle was most often used during 2002

Variable

For-hire tra Vehicle leas Agriculture, Mining Utilities Construction Manufacturin Wholesale tr Retail trade Information

MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL

95% CL for Mean 54009.5907 58895.1519 21755.6067 24856.2511 9954.2886 11582.3131 17002.3551 21417.4684 13454.3175 16708.3561 15966.8537 17461.0515 17653.3448 21646.2535 20729.7496 25373.8316 16848.3582 19048.0543 12189.9160 17664.7915

Sum 72272793289 20024589014 24119946651 3411543277 10244675655 75906142636 15384530602 16963450921 27470445448 5622014452

79 Waste manage Arts, entert Accommodatio Other servic

MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL

Business in which vehicle was most often used during 2002 For-hire tra Vehicle leas Agriculture, Mining Utilities Construction Manufacturin Wholesale tr Retail trade Information Waste manage Arts, entert Accommodatio Other servic

Variable MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL MILES_ANNL

12985.5076 15834.7285 7052.3180 12020.8583 17911.2857 23010.6416 15641.1955 17994.1304

Std Dev 1608230919 1213307392 1354386330 360265917 942933274 2821651145 1399406209 1348917090 1422261638 923917751 901989658 353650310 677802928 2201296141

10709275945 1784083855 5816313888 35776203775

95% CL for Sum 6.91207E10 7.54249E10 1.76465E10 2.24027E10 2.14654E10 2.67745E10 2705422697 4117663856 8396528057 1.20928E10 7.03757E10 8.14366E10 1.26417E10 1.81274E10 1.43196E10 1.96073E10 2.46828E10 3.02581E10 3811137245 7432891659 8941377763 1.24772E10 1090929855 2477237856 4487821313 7144806463 3.14617E10 4.00907E10

(b) proc surveymeans data = vius mean clm ; weight tabtrucks; strata stratum; var mpg; domain transmssn; run; Domain Analysis: Type of Transmission Type of Transmission

Variable

Automatic Manual Semi-Automat Automated Ma

MPG MPG MPG MPG

Mean 16.665277 16.022122 14.846222 16.732086

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CHAPTER 4. RATIO AND REGRESSION ESTIMATION

Domain Analysis: Type of Transmission Type of Transmission

Variable

Std Error of Mean

95% CL for Mean

Automatic Manual Semi-Automat Automated Ma

MPG MPG MPG MPG

0.043659 0.097739 1.012044 1.599588

16.5797047 16.7508490 15.8305539 16.2136901 12.8626219 16.8298213 13.5969068 19.8672658

(c) proc surveymeans data = vius mean clm ; weight tabtrucks; strata stratum; var mpg; domain transmssn; run; The estimated ratio is 0.124410 with 95% CI [0.12258, 0.12624].

Chapter 5

Cluster Sampling with Equal Probabilities 5.1 If the nonresponse can be ignored, then pˆ is the ratio estimate of the proportion.

The variance estimate given in the problem, though, assumes that an SRS of voters was taken. But this was a cluster sample—the sampling unit was a residential telephone number, not an individual voter. As we expect that voters in the same household are more likely to have similar opinions, the estimated variance using simple random sampling is probably too small.

5.3 (a) This is a cluster sample because there are two levels of sampling units: the wetlands are the psus and the sites are the ssus. (b) The analysis is not appropriate. A two-sample t test assumes that all observations are independent. This is a cluster sample, however, and sites within the same wetland are expected to be more similar than sites selected at random from the population.

5.4 (a) This is a cluster sample because the primary sampling unit is the journal, and the secondary sampling unit is an article in the journal from 1988. (b) Let Mi = number of articles in journal i and ti = number of articles in journal i that use non-probability sampling designs.

From the data file,

Σ

ti = 137

i∈£

and

Σ

Mi = 148.

i∈£

81

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CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

Then, using (5.16),

Σ ti 137 = 0.926. ŷ¯r = Σ i∈£ = 148 i∈£ Mi

The estimated variance of the residuals is Σ 2 ˆ i∈£ (ti — y¯ r M i ) n—1

= 0 993221; .

using (5.18), with M £ substituted for M U , sµ ¶ 26 1 SE [ŷ¯ ] = 1— (.993221) = .034. r 1285 26(5.69)2 Here is SAS code: data journal; infile journal delimiter=’,’ firstobs=2; input numemp prob nonprob ; sampwt = 1285/26; /* weight = N/n since this is a one-stage cluster sample */ proc surveymeans data=journal total = 1285 mean clm sum clsum; weight sampwt; var numemp nonprob; ratio ’nonprob/(number of articles)’ nonprob/numemp; run; 5.5 options ls=78 nodate nocenter; data spanish; infile spanish delimiter=’,’ firstobs=2; input class score trip; sampwt = 72/10; /* weight = N/n = 72/10 since one-stage cluster sample*/ proc print data=spanish; run; proc surveymeans data=spanish total = 72 mean clm sum clsum; weight sampwt; cluster class; var trip score; run;

83

/* Construct a boxplot of the data */ proc sort data=spanish; by class; proc boxplot data=spanish; plot score * class; run; /* Since this is a one-stage cluster sample, there is no contribution to variance from subsampling. */ proc surveymeans data=spanish mean clm sum clsum; weight sampwt; cluster class; var trip score; run;

The SURVEYMEANS Procedure Data Summary Number of Clusters Number of Observations Sum of Weights

10 196 1411.2

Statistics

Variable trip score

Mean

Std Error of Mean

95% CL for Mean

0.321429 66.795918

0.079001 2.919409

0.1427164 0.5001408 60.1917561 73.4000806

Statistics Variable trip score

Sum

Std Dev

453.600000 94262

120.502946 7301.420092

95% CL for Sum 181.0034 726.197 77745.4402 110779.360

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CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

Here is a side-by-side boxplot for score.

5.6 (a) The SAS code below was used to calculate summary statistics and the ANOVA table. The output is given below. data worms; do case = 1 to 12; do can = 1 to 3; input worms @@; wt = (580/12)*(24/3); output; end; end; cards; 1 5 7 4 2 4 0 1 2 3 6 6 4 9 8 0 7 3 5 5 1 3 0 2 7 3 5 3 1 4 4 7 9 0 0 0 ; proc print data=worms; run;

85

proc glm data=worms; class case; model worms = case; mean case; run; /* SAS does not calculate the extra term for variance due to 2-stage sampling */ proc surveymeans data=worms total = 580; weight wt; cluster case; var worms; run; The GLM Procedure Class Level Information Class

Levels

case

12

Values 1 2 3 4 5 6 7 8 9 10 11 12

Number of Observations Read Number of Observations Used

36 36

Dependent Variable: worms

Source

DF

Sum of Squares

Model

11

149.6388889

13.6035354

Error

24

108.6666667

4.5277778

Corrected Total

35

258.3055556

Mean Square

F Value

Pr > F

3.00

0.0117

R-Square

Coeff Var

Root MSE

worms Mean

0.579310

58.47547

2.127858

3.638889

Level of

------------worms------------

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CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

case

N

Mean

Std Dev

1 2 3 4 5 6 7 8 9 10 11 12

3 3 3 3 3 3 3 3 3 3 3 3

4.33333333 3.33333333 1.00000000 5.00000000 7.00000000 3.33333333 3.66666667 1.66666667 5.00000000 2.66666667 6.66666667 0.00000000

3.05505046 1.15470054 1.00000000 1.73205081 2.64575131 3.51188458 2.30940108 1.52752523 2.00000000 1.52752523 2.51661148 0.00000000

From the output, we see that ŷ¯u nb = 3.639 (this works here because Mi = M for all i and mi = m for all i; in this case, (5.26) reduces to the sample mean). sµ ¶ µ ¶ 12 13.604 1 3 4.528 SE [ŷ¯ unb ] = 1— + 1— 580 (12)(3) 580 24 3 √ = 0.3701 + 0.0023 = 0.61.

Note that the second term contributes little to the standard error.

Here is the approximation from SAS, using PROC SURVEYMEANS: The SURVEYMEANS Procedure Data Summary Number of Clusters Number of Observations Sum of Weights

12 36 13920

Statistics

Variable

N

Mean

Std Error of Mean

95% CL for Mean

worms

36

3.638889

0.614716

2.28590770 4.99187008

87 5.7 We used SAS to obtain the mean and standard deviation for each city, and to plot the data.

The GLM Procedure Class Level Information Class

Levels

city

Values

6

1 2 3 4 5 6

Number of Observations Read Number of Observations Used

34 34

The GLM Procedure Dependent Variable: cases

Source

DF

Sum of Squares

Model

5

68339.9641

13667.9928

Error

28

113570.6536

4056.0948

Corrected Total

33

R-Square

Root MSE

Coeff Var

Mean Square

F Value

Pr > F

3.37

0.0166

181910.6176

cases Mean

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CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

0.375679

53.85163

63.68748

118.2647

The GLM Procedure Level of city

N

1 2 3 4 5 6

10 4 7 8 2 3

------------cases-----------Mean Std Dev 177.300000 93.250000 116.285714 104.625000 17.500000 63.000000

83.5996411 29.2389580 54.5396317 64.5930945 7.7781746 22.2710575

The SURVEYMEANS Procedure Data Summary Number of Clusters Number of Observations Sum of Weights

6 34 1267.5

Statistics

Mean

Std Error of Mean

95% CL for Mean

120.688145

19.730137

69.9702138 171.406075

Variable cases

Sum

Std Dev

95% CL for Sum

152972

56601

7473.70477 298470.742

We can also use (5.18) and (5.24) for calculating the total number of cases sold, and (5.26) and (5.28) for calculating the average number of cases sold per supermarket.

Summary quantities are given in the spreadsheet below.

City 1 2 3 4 5 6 Sum Var

Mi 52 19 37 39 8 14 169

mi 10 4 7 8 2 3 34

y¯i 177.30 93.25 116.29 104.63 17.50 63.00

si 83.60 29.24 54.54 64.59 7.78 22.27

2

mi 2 si tˆi tˆi — Miŷ¯ r (1 —M i )Mi mi 9219.6 2943.8 1526376 1771.8 -521.3 60913 4302.6 -162.9 471682 4080.4 -626.5 630534 140.0 -825.5 1452 882.0 -807.6 25461 20396.3 2716418 10952882 2138111

89 From (5.18), tˆunb = Using (5.24),

45 6

(20396.3) = 152972.

s

µ ¶ 6 10952882 45 2 ˆ SE [t unb] = 45 1 — + (2716418) 45 6 6 √ = 3,203,717,941 + 20,373,134 = 56, 781.

From (5.26) and (5.28), ŷ¯r = and SE [ŷ¯ ] = r

1

sµ

28.17 = 21.05.

20,396 169

= 120.7

¶ 6 2,138,111 1 (2716418) 1— + 45 6 6(45)

5.8 (a) The SAS code below computes estimates. data books; infile booksdat delimiter=’,’ firstobs=2; input shelf Mi number purchase replace; sampwt = (44/12)*(Mi/5); /* The crucial part for estimating the total is correctly specifying the sampling weight. */ proc print data=books; run; /* Construct the plot for part (a) */ proc boxplot data=books; plot replace * shelf; run; /*Here is the with-replacement approximation to the variance */ proc surveymeans data=books mean clm sum clsum; weight sampwt; cluster shelf; var replace; run; /* Here is the without-replacement approximation to the variance

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CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES using SAS. Note that this approximation does not include the second term in (5.21). */

proc surveymeans data=books total = 44 mean clm sum clsum; weight sampwt; cluster shelf; var replace; run;

It appears that the means and variances differ quite a bit for different shelves. (b) Quantities used in calculation are in the spreadsheet below.

Shelf Mi

mi

2 4 11 14 20 22 23 31 37 38 40 43

5 5 5 5 5 5 5 5 5 5 5 5

26 52 70 47 5 28 27 29 21 31 14 27

sum 377 var

y¯i

s2i

9.6 17.80 6.2 1.70 9.2 18.70 7.2 3.20 41.8 2666.70 29.8 748.70 51.8 702.70 61.2 353.20 50.4 147.30 36.6 600.80 54.2 595.70 6.6 4.80

2

mi si tˆi = Miy¯i (1 — M )Mi2 m (tˆi — Mi ŷ¯r )2 i i

249.6 322.4 644.0 338.4 209.0 834.4 1398.6 1774.8 1058.4 1134.6 758.8 178.2

1943.76 830.96 17017.00 1263.36 0.00 96432.56 83480.76 49165.44 9898.56 96848.96 15011.64 570.24

132696.9 819661.7 1017561.8 594901.6 8271.3 30033.9 579293.8 1188301.2 316493.1 162144.0 183399.3 210944.1

8901.2 268531.00

372463.2

5243902.9

To estimate the total replacement value of the book collection, we use the unbiased

91 estimator since M0, the total number of books, is unknown. Σ

44 tˆ i= 8901.2 = 32637.73.

12 i∈£ Σ From the spreadsheet, s2 = 268,531 and M 2(1 — mi/Mi)s2/mi = 372463.2, so N tˆunb = n t

i∈£

i

i

, u µ ¶ ´ s2 Σ N u³ mi s2i n 1 t 2 , 1— + M 1— SE (tˆunb ) = N n nN i∈£ i Mi mi sµ ¶ 12 268, 531 1 = 44 1 — 44 12 + (12)(44)(372463.2) √ = 44 16274.6 + 705.4 = 5733.52.

The standard error of tˆunb is 5733.52, which is quite large when compared with the estimated total. The estimated coefficient of variation for tˆunb is SE (tˆunb )/tˆunb =

5733.52 = 0.176.

32637.73

Here is the approximation using SAS and the with-replacement variance: Statistics

Variable

Mean

Std Error of Mean

95% CL for Mean

replace

23.610610

6.344128

9.64727746 37.5739427

Statistics Variable

Sum

Std Dev

replace

32638

6582.020830

95% CL for Sum 18150.8032 47124.6635

Note that the with-replacement variance is too large because the psu sampling fraction is large.

Here is the approximation using SAS and the without-replacement variance: Statistics

Variable

Mean

Std Error of Mean

95% CL for Mean

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CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

replace

23.610610

5.410291

11.7026400 35.5185801

Statistics Variable

Sum

Std Dev

replace

32638

5613.166224

95% CL for Sum 20283.2378 44992.2289

√ Note that 5613 = 44 16274.6; SAS calculates only the first term of the withoutreplacement variance. (c) To find the average replacement cost per book with the information given, we use the ratio estimate in (5.28): Σ tˆi 8901.2 = 23.61. ŷ¯r = Σ i∈£ = 377 i∈£ Mi In the formula for variance in (5.29), ¶ ¶ µ µ Σ 1 12 s2r mi s2i 2 ˆ ˆ , V (ȳr ) = 1 — + Mi 1 — Mi mi 44 nM 2 (12)(44)M 2 i∈£ we use M £ = 31.417, the average of the Mi in our sample. So we have Σ 5243902.93 ¯r ) 2 i∈£(tˆi — M i ŷ 2 = 476718.4455 = sr = n—1 11 and SE (ŷ¯ ) r

=

1

sµ

¶ 12 476718.4455

1— + 705.4 44 12 31.417 1 √ = 28892.0 + 705.4 = 5.476.

31.417

The estimated coefficient of variation for ŷ¯r is SE (ŷ¯r )/ ŷ¯r = 5.476/23.61 = 0.232.

5.9 (a)

93

It appears that the means and variances differ quite a bit for different shelves.

Here is SAS code and output for the purchase price of the books. filename booksdat ’books.csv’; options ls=78 nodate nocenter; data books; infile booksdat delimiter=’,’ firstobs=2; input shelf Mi number purchase replace; sampwt = (44/12)*(Mi/5); /* The crucial part for estimating the total is correctly specifying the sampling weight. */ proc print data=books; run; /* Construct the plot for part (a) */ proc boxplot data=books; plot purchase * shelf; run; /* Here is the with-replacement approx to the variance using SAS */ proc surveymeans data=books mean clm sum clsum; weight sampwt; cluster shelf; var purchase;

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CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

run; /* Here is the without-replacement approximation to the variance using SAS. Note that this approximation does not include the second term in (5.21). */ proc surveymeans data=books total = 44 mean clm sum clsum; weight sampwt; cluster shelf; var purchase; run; The SURVEYMEANS Procedure Data Summary Number of Clusters Number of Observations Sum of Weights

12 60 1382.33333

Statistics

Variable

Mean

Std Error of Mean

95% CL for Mean

purchase

12.942706

3.630539

4.95194389 20.9334672

Statistics Variable

Sum

Std Dev

purchase

17891

3854.814474

95% CL for Sum 9406.74388 26375.5228

5.10 Here is the sample ANOVA table for the books data, calculated using SAS. Source Model Error Corrected Total

DF 11 48 59

Sum of Squares 25570.98333 23445.20000 49016.18333

Mean Square 2324.635 488.442

We use (5.10) to estimate R2a: we have ^ MSW = 488.44167,

F Value 4.76

Pr > F 0.0001

95 and we (with slight bias) estimate S2 by S2 =

SSTO 49016.18333 = = 830.78, dfTO 59

so R̂ a2 = 1 — ^ MSW/Ŝ2 = 1 — 488.44/830.78 = 0.41.

The positive value of R̂ 2a indicates that books on the same shelf do tend to have more similar replacement costs.

5.11 (a) Here, N = 828 and n = 85. We have the following frequencies for ti = number of errors in claim i: Number of errors 0 1 2 3 4

Frequency 57 22 4 1 1 Σ The 85 claims have a total of i∈£ ti = 37 errors, so from (5.1), t̂ =

828 85

(37) = 360.42

and 1 Σ 2

st

=

i∈£

µ ti —

37¶2

85 · µ ¶ µ ¶2 µ ¶ 1 37 37 2 37 2 = 57 0 — + 22 1 — + 4 2— 85 85 85 84 µ ¶2 µ ¶2¸ 37 37 + 3— + 4— 85 85 1 = [10.800 + 7.02 + 9.79 + 6.58 + 12.71] 84 = 0.558263.

84

Thus the error rate, using (5.4), is ŷ¯ =

tˆ 360.42 = = 0.002025.

NM (828)(215)

From (5.6), 1 SE [yˆ¯ ] = 215

sµ ¶ 85 st2 1 = 1— (.07677) = . 000357.

828 85 215

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CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

(b) From calculations in (a), t̂ = 360.42 and, using (5.3),

sµ ¶ 85 s2 ˆ t = 63 565 SE [ t] = 828 1 — . .

828 85

(c) If the same number of errors (37) had been obtained using an SRS of 18,275 of the 178,020 fields, the error rate would be 37 = .002025 18,275 (the same as in (a)). But the estimated variance from an SRS would be µ ¶ 18,275 p̂ srs (1 — p̂ srs ) V̂ [ p̂srs ] = 1 — = 9.92 × 10—8.

178,020 18,274 The estimate varianced from (a) is (SE [ yˆ¯ ] ) 2 = 1.28 × 10—7.

Thus

Estimated variance under cluster design 1.28 × 10—7 = = 1 .29 .

Estimated variance under SRS 9.92 × 10—8

5.12 Students should plot the data similarly to Figure 5.3. Σ tˆi 85478.56 = 48.65. ŷ¯r = Σ i∈£ = 1757 i∈£ Mi s2r is the sample variance of the residuals tˆi — M i ŷ¯ r : here, Σ (tˆi — M i ŷ¯r ) 2 2 sr = i∈£ = 280 .0261.

183 Since N is unknown, (5.29) yields r 1 280.0261 SE [ŷ¯r ] = = 0.129.

9.549 184 5.13 (a) Cluster sampling is needed for this application because the household is the sampling unit. Yet the Arizona Statutes specify the statistic that must be used for estimating the error rate: it must be estimated by the sample mean. It is therefore important to make sure that a self-weighting sample is taken. In this application, a self-weighting sample will result if an SRS of n households is taken from the population of N households in the county, and if all individuals in the household are measured. It makes sense in this example to take a one-stage cluster sample.

97 (b) We know that if we were taking an SRS of persons, we would calculate n0 = (1.96)2(0.1)(0.9)/(0.03)2 = 385 and n=

n0 = 310 .

1 + n 0/N

Since we obtain at least some additional information by sampling everyone in the household, we need at most 310 households. To calculate the number of households to be sampled, we need an idea of the measure of homogeneity within the households.

We know from the equation following (5.11) that V (tˆcluster) N (M — 1) 2 =1+ Ra ˆ N —1 V (t SRS) so we can calculate a sample size by multiplying the number of persons needed for an SRS (310) by the ratio of variances, then dividing by M . The following table gives some sample sizes for different values of aR2: M 1 2 3 4 5 1 2 3 4 6 1 2 3 4 5

R2a 0.1 0.1 0.1 0.1 0.1 0.5 0.5 0.5 0.5 0.5 0.8 0.8 0.8 0.8 0.8

sample size 310 171 124 101 87 310 233 207 194 181 310 279 269 264 260

5.14 (a) Treating the proportions as means, and letting Mi and mi be the number of female students in the school and interviewed, respectively, we have the following summaries.

School M i m i smokers y¯i 1 792 25 10 0.4 2 447 15 3 0.2 3 511 20 6 0.3 4 800 40 27 0.675 Sum 2550 100 Var

s2i .010 .011 .011 .006

tiˆ tˆi — Miŷ¯ r 316.8 -24.7 89.4 -103.3 153.3 -67.0 540.0 195.1 1099.5 17943

2 2

mi Mi s i (1 — M ) mi i 243 147 139 86 614

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Then, 1099.5 2550 = 0.43 sµ ¶ 1 4 17943 1 (614) SE [ŷ¯ ] = 1— + r 637.5 29 4 4(29) √ 1 = 3867.0 + 5.3 637.5 = 0.098. ŷ¯r

=

(b) We construct a sample ANOVA table from the summary information in part (a). Note that 4 mi 4 ΣΣ Σ ssw = (yij — y¯i ) 2 = (mi — 1)s2i , i=1 j=1

i=1

and Source Between psus Within psus

df 3 96

Total

99

SS

MS

0.837

0.0087

5.15 (a) A cluster sample was used for this study because Arizona has no list of all elementary school teachers in the state. All schools would have to be contacted to construct a sampling frame of teachers, and this would be expensive. Taking a cluster sample also makes it easier to distribute surveys. It’s possible that distributing questionnaires through the schools might improve cooperation with the survey and give respondents more assurance that their data are kept confidential. (b) The means and standard deviations, after eliminating records with missing values, are in the following table:

99 School 11 12 13 15 16 18 19 20 21 22 23 24 25 28 29 30 31 32 33 34 36 38 41

Mean 33.99 36.12 34.58 36.76 36.84 35.00 34.87 36.36 35.41 35.68 35.17 31.94 31.25 31.46 29.11 35.79 34.52 35.46 26.82 27.42 36.98 37.66 36.88

Std. Dev. 1.953 4.942 0.722 0.746 1.079 0.000 0.231 2.489 3.154 4.983 3.392 0.860 0.668 2.627 2.440 1.745 1.327 1.712 0.380 0.914 2.961 1.110 0.318

There appears to be large variation among the school means. An ANOVA table for the data is shown below. school Residuals

Df 22 224

Sum of Sq 1709.187 1264.106

Mean Sq 77.69034 5.64333

F Value 13.76675

Pr(F) 0

(c) There appears to be a wider range of standard deviations for schools with higher means. (d) Calculations are in the following table.

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School Mi mi 11 33 10 12 16 13 13 22 3 15 24 24 16 27 24 18 18 2 19 16 3 20 12 8 21 19 5 22 33 13 23 31 16 24 30 9 25 23 8 28 53 17 29 50 8 30 26 22 31 25 18 32 23 16 33 21 5 34 33 7 36 25 4 38 38 10 41 30 2 Sum 628 Variance

ŷ¯r V (ŷ¯ r)

si2 y¯i 33.990 3.814 36.123 24.428 34.583 0.521 36.756 0.557 36.840 1.165 35.000 0.000 34.867 0.053 36.356 6.197 35.410 9.948 35.677 24.835 35.175 11.506 31.944 0.740 31.250 0.446 31.465 6.903 29.106 5.955 35.791 3.045 34.525 1.761 35.456 2.932 26.820 0.145 27.421 0.835 36.975 8.769 37.660 1.231 36.875 0.101

M iy¯i 1121.670 577.969 760.833 882.150 994.669 630.000 557.867 436.275 672.790 1177.338 1090.425 958.333 718.750 1667.630 1455.313 930.564 863.125 815.494 563.220 904.907 924.375 1431.080 1106.250 21241.026

2

mi si resids M i2(1 — M ) i mi 5.501 289.508 36.797 90.196 16.721 72.569 70.391 0.000 81.440 3.933 21.181 0.000 16.694 3.698 30.396 37.185 30.147 529.234 61.170 1260.867 41.903 334.393 -56.366 51.776 -59.186 19.252 -125.005 774.766 -235.852 1563.270 51.158 14.393 17.543 17.118 37.558 29.503 -147.069 9.710 -211.261 102.333 78.793 1150.953 145.795 130.978 91.551 42.525 6528.159 9349.2524

= 33.82

·µ ¶ ¸ 1 23 9349.252 6528.159 = (27.30)2 1 — 245 + (23)(245) 23 1 = [368.33 + 1.16] 745.53 = 0.50.

5.16 (a) Summary quantities for estimating y¯ and its variance are given in the table below. Here, ki denotes the number sampled in school i. We use the number of respondents in school i as mi.

101 mi

Return

y¯i

tiˆ

1 78 40 38 2 238 38 36 3 261 19 17 4 174 30 30 5 236 30 26 6 188 25 24 7 113 23 22 8 170 43 36 9 296 38 35 10 207 21 17 Sum 1961 307 281 var

19 19 13 18 12 13 15 21 23 7 160

0.5000 0.5278 0.7647 0.6000 0.4615 0.5417 0.6818 0.5833 0.6571 0.4118

39.0000 125.6111 199.5882 104.4000 108.9231 101.8333 77.0455 99.1667 194.5143 85.2353 1135.3175

School

Mi

ki

2

mi s i tˆi — Mi Mi2(1 — M ) mi i

-6.1580 -12.1786 48.4828 3.6630 -27.7087 -7.0089 11.6243 0.7455 23.1456 -34.6070

21.0811 342.3401 716.1696 207.3600 492.6675 332.8031 106.2293 158.1944 511.9485 595.3936 3484.1873

581.79702

(b) Using (5.26) and (5.28), ŷ¯r = Vˆ ( yˆ¯r ) = = =

1135.3175 = 0.5789 1961

·µ ¶ ¸ 1 10 581.797 3484.1873 1— + 46 10 (10)(46) (196.1)2 1 [45.532 + 7.574] (196.1)2 0.001381.

An approximate 95% confidence interval for the percentage of parents who returned the questionnaire is √ 0.5789 ± 1.96 0.001381 = [0.506, 0.652]. (c) If the clustering were (incorrectly!) ignored, we would have had pˆ = 160/281 = .569 with V̂ ( p̂ ) = .569(1 — .569)/280 = .000876.

5.17 (a) The following table gives summary quantities; the column y¯i gives the estimated proportion of children who had previously had measles in each school.

School Mi mi Hadmeas 1 78 40 32 2 238 38 10 3 261 19 12 4 174 30 19 5 236 30 16 6 188 25 6 7 113 23 11 8 170 43 23 9 296 38 5 10 207 21 11 Sum 1961 307 145 Var

y¯i 0.8000 0.2632 0.6316 0.6333 0.5333 0.2400 0.4783 0.5349 0.1316 0.5238

tiˆ 62.4000 62.6316 164.8421 110.2000 125.8667 45.1200 54.0435 90.9302 38.9474 108.4286 863.41

tˆi — Mi 28.0573 -42.1576 49.9262 33.5894 21.9581 -37.6546 4.2906 16.0808 -91.3787 17.2884

2

Mi2(1 — mi ) si Mi mi 12.1600 249.4572 816.4986 200.6400 417.2408 232.8944 115.3497 127.8864 235.8449 480.1837 2888.1556

1890.1486

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(b)

863.41 = 1961 = 0.4403 ·µ ¶ ¸ 1 10 1890.15 2888.16 Vˆ ( yˆ¯r ) = (196.1)2 1 — 46 + 10 (10)(46) 1 = [147.92 + 6.28] (196.1)2 = 0.004 ŷ¯r

An approximate 95% CI is

√ 0.4403 ± 1.96 0.004 = [0.316, 0.564]

5.18 With the different costs, the last two columns of Table 5.4 change. The table now becomes: Number of Stems Sampled per Site 1 2 3 4 5

ŷ¯ 1.12 1.01 0.96 0.91 0.91

S E ( yˆ¯) 0.15 0.10 0.08 0.07 0.06

Cost to Sample One Field 50 70 90 110 130

Relative Net Precision 0.15 0.14 0.13 0.12 0.12

Now the relative net precision is highest when one stem is sampled per site.

5.19 (a) Here is the sample ANOVA table from SAS PROC GLM: Dependent Variable: acres92

Source

DF

Sum of Squares

Model

41

2.0039161E13

488760018795

Error

258

1.5456926E13

59910564633

Corrected Total

299

3.5496086E13

R-Square

Coeff Var

Root MSE

acres92 Mean

0.564546

82.16474

244766.3

297897.0

Mean Square

F Value

Pr > F

8.16

<.0001

We estimate Ra2 by 1—

59910564633 = 0. 4953455 .

3.5496086 × 1013/299

103 This value is greater than 0, indicating that there is a clustering effect. (b) Using (5.32), with c1 = 15c2 and R2a= 0.5, and using the approximation M (N — 1) ≈ NM — 1, we have r 15(1 — .5) = 3.9. m̄ opt = .5 Taking m̄ = 4, we would have n = 300/4 = 75.

5.20 Answers will vary.

5.21 Answers will vary.

5.22 First note that M N Σ M Σ M N ·Σ ¸2 Σ Σ (yij — ȳU )(yik — ȳU ) = (yij — y¯U ) i=1 j=1 k=1

i=1 N

=

Σ

j=1

[M(y¯iU — y¯U)]2

i=1

= M (SSB).

Thus ΣN ΣM ICC = = = = =

i=1

ΣM

j=1

(yij — y¯U )(yik — y¯U )

k/=j

(NMΣ— 1)(M — 1)S2 N ΣM M (SSB) — (yij — y¯U )2 i=1

j=1

(NM — 1)(M — 1)S2 M (SSB) — SSTO (M — 1)SSTO M (SSTO — SSW) — SSTO (M — 1)SSTO M SSW 1— , M — 1 SSTO

proving the result.

5.23 From (5.8), ICC = 1 —

M SSW .

M — 1 SSTO

Rewriting, we have MSW

= =

1

SSTO

M —1

(1 — ICC)

N (M — 1) M NM — 1 2 S (1 — ICC).

NM

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Using Table 5.1, we know that SSW + SSB = SSTO, so from (5.8), M SSTO — (N — 1)MSB M —1 SSTO 1 M (N — 1)MSB = — + M — 1 (M — 1)(N M — 1)S2

ICC = 1 —

and MSB

=

NM — 1 M (N — 1)

2

S [1 + (M — 1)ICC] .

5.24 (a) From (5.24), we have that

µ ¶ N 2 ¸ · 2 µ1 — n ¶ S2 Σ 1— m 2Si t + N V (ŷ¯unb ) = ( M ) M m NM 2 N N n n µ ¶ 2 µ ¶ i=1 Σ n S m 1 N 2 = 1— S. t + 1— N nM 2 M Nnm i=1 i 1

The equation above (5.7) states that S2t = M (MSB); the definition of Si2 in Section 5.1 implies N

Σ

N

2 Si

=

M

SSW ΣΣ 1 2 = N (MSW). (yij — y¯iU ) = M —1 M —1 i=1 j=1

i=1

Thus, · ³ ¸ ´ M (MSB) ³ ´ M2 1 n N m 2 + N (MSW) V (ȳˆ unb ) = 1— N 1— N M (NM )2 n n m ³ n ´ MSB ³ m ´ MSW = 1— + 1— .

N nM M nm (b) The first equality follows directly from (5.9). Because SSTO = SSB + SSW, MSB

1 [SSTO — N (M — 1)MSW] N —1 1 = [(N M — 1)S2 — N (M — 1)S2(1 — R2)] a N —1 1 2 2 2 = [(N — 1)S + N (M — 1)S R ] a N· 1 ¸ — N (M — 1)R2 = a +1 .

S2 N —1 =

105 (c)

µ ˆ V (y¯) =

1—

n

¶

S2

·

¸

N (M — 1)R2

N

N —1

nM

µ

+1 +

a

m 1— M

¶

S2 (1 — R2a). nm

(d) The coefficient of S2R2a in (c) is µ ¶ µ ¶ n N (M — 1) m 1 (N — n)(M — 1)m — (M — m)(N — 1) 1— — 1— = N nM (N — 1) M nm Mnm(N — 1) But if N (m — 1) > nm, then (N — n)(M — 1)m — (M — m)(N — 1) = M [N (m — 1) — nm + 1] + m(n — 1) > 0, so V ( yˆ¯ ) is an increasing function of R2.a 5.25 (a) If Mi = M for all i, and mi = m for all i, then from (5.15), Σ M y¯i 1 ˆ = ŷ¯ . ŷ¯r = i∈£ = tunb unb nM NM (b) In the following, let ŷ¯ = ŷ¯r = yˆ¯ u n b Source Between clusters Within clusters Total

df

SS ΣΣ

n—1

(ȳ i — yˆ¯ ) 2

i∈£ j∈£i

n(m — 1) nm — 1

ΣΣ

(yij — ȳ i ) 2 i∈£ j∈£i Σ Σ (yij — yˆ¯ ) 2 i∈£ j∈£i

(c) Let

½ 1 if psu i in sample 0 otherwise. ΣΣ = (yij — ȳ i ) 2

Zi = S^ SW

i∈£ j∈£i · N

E[S^ SW] = E

Σ

Zi

i=1 ½ N

=E =E

Σ

i=1 ½Σ N

Σ

¸ (yij — ȳ i )

·Σ

j∈£i

ZiE

=E

¸¾ (yij — ȳ i ) | Z 2

j∈£i

¾

Zi(m — 1)E[s2i | Z] ¾

i=1

½Σ N

2

Zi(m — 1)S i 2

i=1

N nΣ = (m — 1) S2i N i=1

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Thus, E[msw] =

N 1 Σ

N i=1

S i2 = MSW.

Since Mi = M and mi = m for all i, N 1Σ 1Σ ŷ¯ unb = y¯i = Z iy¯i n n i=1

Sˆ SB =

ΣΣ

i∈£

Σ (ȳi — ŷ¯unb ) 2 = m (y¯i — ŷ¯unb ) 2

i∈£ j∈£i

i∈£

107 ·Σ ¸ 2 2 E [ S Ŝ B ] = mE (y¯ i— 2ȳ i ŷ¯u n b + ŷ¯unb) i∈£ ·Σ ¸ 2 2 ˆ = mE y¯ i — n y¯unb i∈£

µ · N ¸¶ Σ 2 2 = mE E Z i ȳ | Z1, . . . , Zn — mnE [ŷ¯unb ] i

=

=

=

=

i=1

¸

2 Zi{V (y¯i | Z) + y¯iU — mn[V (ŷ¯unb ) + y¯U ] i=1 ½µ ¾¸ ¶ 2 ·Σ N 2 Zi 1 — m S i + y¯ mE iU M m i=1 ¶ · µ ¶ ¸ N µ n S2 Σ 1 m S2i t + 1 2 1 — 1 — —mn + y¯ U N M m M2 n nN ¸ i=1 µ ¶ N m S2i + y¯2 — m 1 — n ¶ mn Σ ·µ 2 1—M m iU N M2 N St i=1 µ ¶ N mΣ 1 — m S2i — mny¯2 — U N M m i=1 Σ ¸ ¶Σ N · N µ 2 m 1 2 S 2 i m(n — 1) + mn y¯ — y¯U M i=1 m N µ 1—¶ N i=1 iU m n — 1— (MSB) M N µ ¶ N N m(n — 1) m Σ S2i mn Σ 1— + N (y¯iU — y¯U)2 N M m i=1 µ ¶ i=1 m n 1 — 1— SSB M N N —1 µ ¶N · µ ¶¸ m(n — 1) m Σ S2i m n mn 1— + 1 — N SSB — N M M (N — 1) m NM

= mE =

·Σ N

2 }

i=1

µ ¶N m(n — 1) m Σ Si2 (N — 1)mn — m(N — n) SSB = 1— + N M i=1 m NM (N — 1) µ ¶ m m = (n — 1) 1 — MSW + (n — 1) MSB. M M Thus

µ

¶ m E[msb] = 1 — MSW + MSB. M M m

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(d) From (c), µ ¶ µ ¶ M m Mm M ^ E[M SB ] = 1— MSW + MSB — — 1 MSW = MSB . m M mM m (e) From (5.22) and (5.23), µ ¶ 1 Σ ˆ tˆunb 2 ti — n — 1 i∈£ N 1 Σ ˆ 2 = (M y¯ — i M y¯ ) n—1 i∈£ 1 M2 Σ Σ = (¯yi — yˆ¯) 2 n—1 m

s2t =

i∈£ j∈£i

M2 = msb m and

Σ

i∈£ i

ΣΣ

1

s2 =

m—1 = nmsw.

(y — y¯ )2 ij

i

i∈£ j∈£i

Then, using (5.24), ¶ µ · µ ¶ ¸ 1 m M2 Σ 2 n s2t N 2 N 1— + 1— s (NM )2 N n n M m i∈£ i µ n ¶ msb 1 µ m ¶ msw 1 — 1 — + = .

N nm N M m

ˆ (y¯) ˆ = V

5.26 (a) From Exercise 5.25, msto =

1

Σ Σ( 2 yij — yˆ¯) . nm — 1 i∈£ j∈£i

109 Now, E Σ Σ (yij — ŷ¯)2 i∈£ j∈£i

^ + E[SŜB] = E[SSW] µ ¶ N nΣ m m 2 MSW + (n — 1) MSB = (m — 1) S i + (n — 1) 1 — M M N i=1 · µ ¶¸ m m = (m — 1)n + (n — 1) 1 — MSW + (n — 1) MSB M M h i m m = nm — 1 — (n — 1) MSW + (n — 1) MSB M M h i 1 m (n — 1) m = nm — 1 — (n — 1) SSW + SSB N (M — 1) N —1 M M · ¸ nm — 1 NM — 1 (n — 1)m NM — 1 m(n — 1) = NM — 1 N (M — 1) SSW + (N — 1)M nm — 1 SSB — NM (M — 1)SSW ·½ ¾ nm — 1 N —1 = 1+ SSW NM — 1 N (M — 1) ½ ¾ ¸ nm(M — 1) — (m — 1)NM — M + m + 1+ SSB (N — 1)M (nm — 1) m(n — 1) SSW — NM (M — 1) nm — 1 nm — 1 nm(M — 1) — (m — 1)NM — M + m = NM — 1 SSTO + NM — 1 SSB (N — 1)M (nm — 1) · ¸ nm — 1 N — 1 m(n — 1) + NM — 1 N (M — 1) — NM (M — 1) SSW · µ ¶ µ ¶ ¸ nm — 1 1 1 = SSTO + O SSB + O SSW .

NM — 1 n n The notation O(1/n) denotes a term that tends to 0 as n → ∞. (b) Follows from the last line of (a). (c) From Exercise 5.25, E[msw] = MSW and ³ m m´ E[msb] = MSB + 1 — MSW .

M M

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Consequently, M (N — 1) (m — 1)NM + M — m E[msb] + E[msw] m(NM — 1) m(NM — 1) ³ i M (N — 1) h m m´ = m(NM — 1) M MSB + 1 — M MSW (m — 1)NM + M — m + MSW m(NM — 1) 1 = SSB NM — 1 1 + [(N — 1)(M — m) + (m — 1)N M + M — m] MSW NM — 1 N (M — 1) 1 SSB + MSW = NM — 1 NM — 1 = S2.

E [Ŝ 2 ] =

5.27 The cost constraint implies that n = C/(c1 + c2m); substituting into (5.30), we have: (c1 + c2m)MSB MSB ³ m ´ (c1 + c2m)MSW g(m) = V (ŷ¯ unb ) = — + 1— CM NM M Cm dg c2MSB c1MSW c2MSW = — — .

Cm2 dm CM CM Setting the derivative equal to zero and solving for m, we have s c1M (MSW) . m= c2(MSB — MSW) Using Exercise 5.24, s m=

s

c1MS2(1 — R2) a

c2S2[N (M — 1)R2/(N — 1) + 1 — (1 — R2)] a a

=

c1M (N — 1)(1 — R2) c2(NM — 1)R2a

a .

5.28 This exercise does not rely on methods developed in this chapter (other than for a general knowledge of systematic sampling), but represents the type of problem a sampling practitioner might encounter. (A good sampling practitioner must be versatile.) (a) For all three cases, P (detect the contaminant) = P (distance between container and nearest grid point is less than R). We can calculate the probability by using simple geometry and trigonometry.

Case 1: R < D.

111

D

R

Since we assume that the waste container is equally likely to be anywhere in the square relative to the nearest grid point, the probability is the ratio area of shaded part area of square Case 2: D ≤ R ≤

=

πR2 4D2

.

√ 2D

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R2 − D2

D

The probability is again

area of shaded part area of square

.

The area of the shaded part is µ ¶ µ ¶µ 2 ¶ µ ¶µ 2 ¶ √ 1 √ π/2 — 2θ πR 4θ πR 2 D R2 — D2 + = D R2 — D2 + 1 — , 2 4 4 π/ π 2 where cos(θ) = D/R. The probability is thus

Case 3: R >

sµ

R D

¶2

µ —1+

4θ 1— π

¶µ

¶ πR2 .

4D2

√ 2D.

The probability of detection is 1. (b) Even though the square grid is commonly used in practice, we can increase the probability of detecting a contaminant by staggering the rows.

113

c

5.30 (a) Because the Ai’s and the εij’s are independent, ·Σ Σ ¸ ·Σ µ Σ ¶¸ ·Σ Σ ¸ VM1 bij Yij = VM1 Ai bij + VM 1 bij εij i∈£

i∈£ j∈£i

Σµ Σ

=

i∈£

½

(b) Let cij = Then

j∈£i

¶2

σA2 +

bij

j∈£i

i∈£ j∈£i

ΣΣ

bij2 σ2.

i∈£ j∈£i

bij — 1 if i ∈ S and j ∈ Si —1 otherwise.

N

Mi

ΣΣ Tˆ — T = cijyij. i=1 j=1

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CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

Using the same argument as in part (a), Mi N µΣ ¶2 N Mi Σ Σ Σ 2 2 2 VM1 [T̂ — T ] = cij σA + cij σ i=1

=

j=1

i∈£

Σ cij j∈ / £i

j∈£i

i∈£

=

cij +

Σµ Σ

+ =

i=1 j=1

Σ ·µ Σ

c2ij +

j∈£i

j∈ / £i

j=1 i ∈/ £ Mi

¶ 2 σ2 + cij

ΣΣ i∈ / £ j=1

c2ij σ 2

¾2 ¸ Σ σA2 + i∈/ £ M i2σ2A i∈£ · ¸ Σ Σ Σ 2 2 2 + i∈£ j∈£ (bij — 1) + (Mi — mi) σ + i∈/ £ Miσ

Σ

·½

Σ

¶2 ¸ ¶2 M ΣµΣ 2 cij σ2A σA +

(bij — 1) — (Mi — mi) j∈£i

·

Σ i∈£

+

Σ

i

Σ ·

Σ i∈£

b — Mi j∈£i ij

¸2 Σ σA2 + i∈/ £ M i2σ2A

¸ Σ (b 2 ij— 2b ) + M σ2 + 2 ij i j∈£i i ∈/ £ Miσ

Σ

5.32 Recall that

Tˆr =

ΣΣ

bijyij,

i∈£ j∈£

with bij = M0

Mi . Σ mi Mk k∈£

Then, from (5.36),

, Σ Σ

2 VM1[Tˆr — T ] = σA

i∈£

j∈£i

2

mi

M0Mi Σ

k∈£ Mk

à ΣΣ µ

— Mi

+

Σ

Mi2

i∈ /£

¶2

!

M0Mi M0Mi Σ —2 Σ +σ m m i k∈£ Mk i k∈£ Mk i M " i∈£ j∈£ 0 2# ¶2 Σ 2 µΣ —1 + Σ 2 Mi Mi =σA k∈£ Mk i∈£ i ∈/ £ · ¸ Σ M 2M 2 M0Mi 2 +σ + σ2 M , Σ0 i — 2Σ 0 2 mi ( k∈£ Mk ) Mk k∈£ 2

i∈£

which is minimized when

Σ

M i2/mi

i∈£

is minimized. Let g( m1, . . . , mN , λ ) =

Σ M2 i

i∈£

mi

à —λ

i—

!

Σ mi i∈£

.

+ M0

115 Then

∂g

=

Σ

m —L i

∂λ i∈£

and, for k ∈ S,

2

∂g M = — k + λ. ∂mk mk2

√ Setting the partial derivatives equal to zero, we have that Mk/mk = λ is constant for all k; that is, mk is proportional to Mk.

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CHAPTER 5. CLUSTER SAMPLING WITH EQUAL PROBABILITIES

Chapter 6

Sampling with Unequal Probabilities 6.2 (a) Instead of creating columns of cumulative Mi range as in Example 6.2, we create columns for the cumulative ψi range. Then draw 10 random numbers between 0 and 1 to select the psu’s with replacement. A table giving the cumulative ψi range for each psu follows:

117

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118 psu 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES ψi 0.000110 0.018556 0.062999 0.078216 0.075245 0.073983 0.076580 0.038981 0.040772 0.022876 0.003721 0.024917 0.040654 0.014804 0.005577 0.070784 0.069635 0.034650 0.069492 0.036590 0.033853 0.016959 0.009066 0.021795 0.059185

Cumulative ψi 0.000000 0.000110 0.000110 0.018666 0.018666 0.081665 0.081665 0.159881 0.159881 0.235126 0.235126 0.309109 0.309109 0.385689 0.385689 0.424670 0.424670 0.465442 0.465442 0.488318 0.488318 0.492039 0.492039 0.516956 0.516956 0.557610 0.557610 0.572414 0.572414 0.577991 0.577991 0.648775 0.648775 0.718410 0.718410 0.753060 0.753060 0.822552 0.822552 0.859142 0.859142 0.892995 0.892995 0.909954 0.909954 0.919020 0.919020 0.940815 0.940815 1.000000

(Note: the numbers in the “Cumulative ψi” columns were rounded to fit in the table.)

Ten random numbers I generated between 0 and 1 were: {0.46242032, 0.34980142, 0.35083063, 0.55868338, 0.62149246, 0.03779992, 0.88290415, 0.99612658, 0.02660724, 0.26350658}.

Using these ten random numbers would result in psu’s 9, 7, 7, 14, 16, 3, 21, 25, 3, and 6 being the sample. (b) Here max{ψi} = 0.078216. To use Lahiri’s method, we select two random numbers for each draw—the first is a random integer between 1 and 25, and the second is a random uniform between 0 and 0.08 (or any other number larger than max{ψi}). Thus, if our pair of random number is (20, 0.054558), we reject the pair and try again because 0.054 > ψ20 = 0.03659. If the next pair is (8, 0.028979), we include psu 8 in the sample.

6.3 Calculate tˆψ£ = ti/ψi for each sample.

119 Store A B C D

ψi 1/16 2/16 3/16 10/16

tˆψ£ 1200 600 400 120

ti 75 75 75 75

E[tˆ ] = ψ

= V [tˆ ] = ψ

=

(tˆψ£ — t)2 810,000 90,000 10,000 32,400

1

2 3 10 (1200) + (600) + (400) + (120) 16 16 16 16 300.

1 10 (810,000) + · · · + (32,400) 16 16 84,000.

6.4 Store A B C D

ψi 7/16 3/16 3/16 3/16

ti 11 20 24 245

tˆψ£ 25.14 106.67 128.00 1306.67

(tˆψ£ — t)2 75546.4 37377.8 29584.0 1013377.8

As shown in (6.3) for the general case, 7 3 3 3 E[tˆ ] = (25.14) + (106.67) + (128) + (1306.67) ψ 16 16 16 16 = 300.

Using (6.4), 7 3 3 3 V [tˆ ] = (75546.4) + (37377.8) + (29584) + (1013377.8) ψ 16 16 16 16 = 235615.2.

This is a poor sampling design. Store A, with the smallest sales, is sampled with the largest probability, while Store D is sampled with a smaller probability.

The ψi used in this exercise produce a higher variance than simple random sampling.

6.5 We use (6.5) to calculate tˆψ for each sample. So for sample (A,A), µ ¶ 1 11 11 ˆ t = 2 = = 176. ψ 2 ψA ψA For sample (A,B), · ¸ 1 11 20 1 ˆ tψ = + = [176 + 160] = 168 .

2 2 ψA ψB The results are given in the following table:

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

Sample, S (A,A) (A,B) (A,C) (A,D) (B,A) (B,B) (B,C) (B,D) (C,A) (C,B) (C,C) (C,D) (D,A) (D,B) (D,C) (D,D) Total

P (S) 1/256 2/256 3/256 10/256 2/256 4/256 6/256 20/256 3/256 6/256 9/256 30/256 10/256 20/256 30/256 100/256 1

t1/ψ1 176 176 176 176 160 160 160 160 128 128 128 128 392 392 392 392

t2/ψ2 176 160 128 392 176 160 128 392 176 160 128 392 176 160 128 392

tˆψ 176 168 152 284 168 160 144 276 152 144 128 260 284 276 260 392

P (S)(tˆ— t)2 60.06 136.13 256.69 10.00 136.13 306.25 570.38 45.00 256.69 570.38 1040.06 187.50 10.00 45.00 187.50 3306.25 7124.00

1 100 (176) + ····· + (392) = 300 256 256 1 100 V [tˆψ ] = (176 — 300)2 + · · · + (392 — 300)2 = 7124.

256 256 Of course, an easier solution is to note that (6.5) and (6.6) imply that E[tˆψ ] = t, and that V [tˆψ ] will be half of the variance found when taking a sample of one psu in Section 6.2; i.e., V [tˆψ ] = 14248/2 = 7124.

E[tˆψ] =

6.6 (a) The following table does the calculations, using (6.4) to find the variance. name

t

i

Apache 31621 Cochise 51126 Coconino 53443 Gila 28189 Graham 11430 Greenlee 3744 La Paz 15133 Mohave 80062 Navajo 47413 Pinal 81154 Santa Cruz 13036 Yavapai 81730 Yuma 74140 Sum 572221

ψ

i

0.0572 0.0969 0.0958 0.0423 0.0276 0.0070 0.0162 0.1276 0.0802 0.1480 0.0316 0.1379 0.1317 1.0000

tˆ = ti ψ (tˆ — t)2 ψ i ψ ψi 553292.1 20477223 527405.6 194693778 558108.6 19071085 667034.6 379902891 414597.1 684957758 532113.6 11318255 932417.7 2105687838 627317.4 387423351 590892.8 27974547 548502.8 83232656 412582.0 805214838 592659.0 57604012 562787.3 11723897 4789282131

tˆ

SRS

411073 664638 694759 366457 148590 48672 196729 1040806 616369 1055002 169468 1062490 963820

1

(tˆ

13

SRS

— t)2

1997590608 656992453 1155043188 3256832592 13804863397 21084888877 10845710928 16890146325 149926608 17929037997 12477690693 18489514797 11796136677 1.30534E+11

The tˆψ ’s are given in column 5, and V (tˆψ ) =4,789,282,131. (b) Using ψi = 1/13 for each i, we get the SRS estimators in column 7 with

121 V (tˆψ) =130,534,375,140.

The unequal-probability sample is more efficient because ti and Mi are highly correlated: the correlation is 0.9905. This means that the quantity ti/ψi does not vary much from sample to sample.

6.7 From (6.6),

µ ¶2 Vˆ (tˆψ) = 1 1 Σ ti — tˆψ . n n — 1 i∈R ψi

In an SRSWR, ψi = 1/N , so we have tˆψ = N t¯ and Σ N2 1 Σ ¯ 2.

Vˆ (tˆψ) = 1 1 (Nti — Nt)¯ 2 = (ti — t) n — 1 n n n — 1 i∈R i∈R 6.8 We use (6.13) and (6.14), along with the following calculations from a spreadsheet.

Academic Unit 14 23 9 14 16 6 14 19 21 11

Mi 65 25 48 65 2 62 65 62 61 41

ψi 0.0805452 0.0309789 0.0594796 0.0805452 0.0024783 0.0768278 0.0805452 0.0768278 0.0755886 0.0508055

yij 3004 2120 0010 2010 20 0225 1003 4100 2231 2 5 12 3

y¯i 1.75 1.25 0.25 0.75 1.00 2.25 1.00 1.25 2.00 5.50

tˆi 113.75 31.25 12.00 48.75 2.00 139.50 65.00 77.50 122.00 225.50

average std. dev.

tˆi/ψi 1412.25 1008.75 201.75 605.25 807.00 1815.75 807.00 1008.75 1614.00 4438.50 1371.90 1179.47

√ Thus tˆψ = 1371.90 and SE(tˆψ ) = (1/ 10)(1179.47) = 372.98.

Here is SAS code for calculating these estimates. Note that unit 14 appears 3 times; in SAS, you have to give each of these repetitions a different unit number. Otherwise, SAS will just put all of the observations in the same psu for calculations. data faculty; input unit $ Mi psi y1 y2 y3 y4; array yarray{4} y1 y2 y3 y4; sampwt = (1/( 10*psi))*(Mi/4); if unit = 16 then sampwt = (1/( 10*psi)); do i = 1 to 4; y = yarray{i};

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

if y ne . then output; end; datalines; /* Note: label the 3 unit 14s with different psu numbers */ 14a 65 0.0805452 3 0 0 4 23 25 0.0309789 2 1 2 0 9 48 0.0594796 0 0 1 0 14b 65 0.0805452 2 0 1 0 16 2 0.0024783 2 0 . .

6 62 0.0768278 0 2 2 5 14c 65 0.0805452 1 0 0 3 19 62 0.0768278 4 1 0 0 21 61 0.0755886 2 2 3 1 11 41 0.0508055 2 5 12 3 ; proc surveymeans data=faculty mean clm sum clsum; cluster unit; weight sampwt; var y; run; 6.12 (a) The correlation between ψi and ti (= number of farms in county) is 0.26.

We expect some benefit from pps sampling, but not a great deal—sampling with probability proportional to population works better for quantities highly correlated with population, such as number of physicians as in Example 6.5.

(b) As in Example 6.5, we form a new column ti/ψi. The mean of this column is

123 1,896,300, and the standard deviation of the column is 3,674,225. Thus tˆψ = 1,896,300 and

3,674,225 √ = 367,423.

100 A histogram of the ti/ψi exhibits strong skewness, however, so a confidence interval using the normal approximation may not be appropriate.

SE [tˆψ] =

Here is SAS code for producing these estimates. Note that we do not use the cluster statement in proc surveymeans since the observations are psu totals. data statepop; infile statepop delimiter=’,’ firstobs=2; input state $ county $ landarea popn physicns farmpop numfarm farmacre veterans percviet; psii = popn/255077536; wt = 1/(100*psii); /* weight = 1/(n \psi_i) */ /* Be careful when constructing your dataset that if a psu is selected multiple times, then it appears that many times in the data. Otherwise, you will get the wrong estimate.*/ proc print data=statepop; proc corr data=statepop; proc gplot data=statepop; plot numfarm*psii; run; /*Omit the ’total’ option because sampling is with replacement*/ proc surveymeans data=statepop nobs mean sum clm clsum; var numfarm; weight wt; run;

6.13 (a) Corr (population, number of veterans) = .99. We expect the unequal probability sampling to be very efficient here.

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

(b)

1 Σ ti tψˆ = = 27,914,180 100 ψi 10874532 SE [tˆψ] = √ = 1,087,453 100 Note that Allen County, Ohio appears to be an outlier among the ti/ψi: it has population 26,405 and 12,642 veterans. (c) For each county, Vietveti = veteransi ∗ percvieti/100.

We then form a column with ith entry Vietveti/ψi, and find the mean (= 8050477) and standard deviation (= 3273372) of that column. Then tˆψ = 8,050,477 SE [tˆψ ] = 327,337 Here is SAS code for calculating these estimates: data statepop; infile statepop delimiter=’,’ firstobs=2; input state $ county $ landarea popn physicns farmpop numfarm farmacre veterans percviet; vietvet = veterans*percviet/100; psii = popn/255077536; wt = 1/(100*psii); /* weight = 1/(n \psi_i) */ proc print data=statepop; run;

125 proc corr data=statepop; run; proc gplot data=statepop; plot veterans*psii; proc surveymeans data=statepop nobs mean sum clm clsum; var veterans vietvet; weight wt; run; 6.14 (a) We use (6.28) and (6.29) to calculate the variances. We have Class 4 10 1 9 14

tˆi 110.00 106.25 154.00 195.75 200.00

Vˆ (tˆi ) 16.50 185.94 733.33 2854.69 1200.00

From Table 6.7, and calculating

¶ 2 µ m si i 2 ˆ , V (tˆi) = Mi 1 — Mi mi

we have the second term in (6.28) and (6.29) is ˆ ˆ Σ V (ti) = 11355. πi i∈£

To calculate the first term of (6.28), note that if we set πii = πi, we can write Σ tˆ2 Σ Σ πik — πiπk tˆi tˆk Σ Σ πik — πiπk tˆi tˆk (1 ) i+ = .

— πi 2 πik πik πi πk πi πi πk i∈£ k∈S i∈£ i∈£ k∈£ k/=i

We obtain that the first term of (6.28) is Σ Σ πik — πiπk ˆti tˆk i∈£ k∈£

πik

= 6059.6,

πi πk

so V̂ HT (t̂ HT ) = 6059.6 + 11355 = 17414.46.

Similarly, V̂ SYG (t̂ HT ) = 6059.6 + 11355 = 66139.41.

Note how widely these values differ because of the instability of the estimators with this small sample size (n = 5). (b) Here is the pseudo-fpc calculated by SAS.

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES Statistics

Mean

Std Error of Mean

95% CL for Mean

3.450000

0.393436

2.35764732 4.54235268

Variable y

Statistics Variable y

Sum 2232.150000

Std Dev

95% CL for Sum

254.552912

1525.39781 2938.90219

This standard error is actually pretty close to the SYG SE of 257.

6.15 (a) Let J be an integer greater than max { M}i . Let U1, U2, . . ., be independent discrete uniform{1, . . . , N}random variables, and let V1, V2, . . . be independent discrete uniform{1, . . . , J}random variables. Assume that all Ui and Vj are independent. Then, on any given iteration of the procedure, P (select psu i) = P (select psu i with first pair of random numbers) +P (select psu i with second pair) +··· = P (U1 = i and V1 ≤ Mi) µ N[ ¶ +P (U2 = i, V2 ≤ Mi)P {U1 = j, V2 > Mj} j=1 k—1

+ · · · + P (Uk = i, Vk ≤ Mi)

Y

µ[ N P

L=1

¶ {UL = j, VL > Mj}

j=1

¸ N 1 Mi 1 Mi 1 Σ J — Mj = + J N J N J N j=1 · N ¸ 1 Mi 1 Σ J — Mj k—1 +··· + + ··· N J N J j=1 ¸ ∞ · N 1 Mi Σ 1 Σ J — Mj k = J N J k=0 N j=1 ·

1 Mi 1 N J 1 — Σ N (J — Mj)/(JN ) j=1 N ,Σ = Mi Mj. =

j=1

(b) Let W represent the number of pairs of random numbers that must be generated

127 to obtain the first valid psu. Since sampling is done with replacement, and hence all psu’s are selected independently, we will have that E[X] = nE[W ]. But W has a geometric distribution with success probability Σ 1 Mi p = P (U = i, V ≤ M for some i) = N .

1 1 i N J i=1

Then P (W = k) = (1 — p)k—1p and E[W ] =

1

= NJ

N ,Σ

Mi.

p

Hence, E[X] = nNJ

i=1 N ,Σ

Mi.

i=1

6.16 The random variables Q1, . . . , QN have a joint multinomial distribution with probabilities ψ1, . . . , ψN . Consequently, Σn i=1 Qi = n, E[Qi] = nψi, V [Qi] = nψi(1 — ψi), and Cov (Qi, Qk) = —nψiψk

for i /= k.

(a) Using successive conditioning, · NΣ Qi Σ tˆij ¸ E[tˆψ] = E 1 n i=1 j=1 ψi ½ · N Qi ¸¾ Σ Σ tˆij =E E 1 ψ Q1, . . . , QN n i=1 j=1 i ¯ · Σ N 1 ti ¸ =E Qi n ψi i=1 N Σ 1 ti = nψi = t. n ψi i=1

Thus tˆψ is unbiased for t. (b) To find V (tˆψ ), note that V (tˆψ ) = V [E(tˆψ | Q1, . . . , QN )] + E[V (tˆψ | Q1, . . . , QN )] Q t · NΣ · µ Σ ¸ N ¶¸ i ˆij Σ 1 1 t ¯ i =V Qi +E V n Q .

1, . . . , QN ψi ¯ n ψi i=1

i=1 j=1

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

Looking at the two terms separately, ·

1 Σ ti Qi V n ψi N

i=1

¸

· NΣ N ti 1 Σ tk ¸ = Cov 1 , Qi Qk n ψi n ψk i=1 k=1 N N 1 Σ Σ ti tk = Cov (Q i, Q k) n2 ψψ i=1 k=1 i k N µ ¶ 1 Σ ti 2 = 2 nψi(1 — ψi) n ψi i=1 ti tk N N (—nψ ψ ) 1 Σ Σ + 2 i k n i=1 k=1,k/=i ψi ψk ¶2 N µ Σ ti =1 [ψi(1 — ψi) + ψ2i] n ψi i=1N N 1 ΣΣ — ti tk n i=1 k=1 µ ti ¶2 N 1Σ = ψi —t . n ψi i=1

Qi · µ Σ ¶¸ N ·1 Σ N Qi V (tˆ ) ¸ ij Σ Σ tˆij 1 =E E V | Q , . . . , Q 1 N 2 ψ 2 i ψ n n i i=1 j=1 i=1 j=1 N ·1 Σ V i¸ =E i Q ψ2 n2 N

i=1

i

1 Σ Vi = . n i=1 ψi This equality uses the assumptions that V (tˆij ) = Vi for any j, and that the estimates tˆij are independent. Thus, ¶2 µ N N Σ 1 t 1 Σ Vi i V [tˆψ] = . ψi —t + ψi n i=1 n i=1 ψi

129 (c) To show that (6.9) is an unbiased estimator of the variance, note that E[Vˆ (tˆψ)] = E

· Σ ¸ N Qi Σ(tˆ ij/ψ i— tˆψ)2 1 n n—1 ½

i=1 j=1 N Qi Σ

Σ 1 1 = E n i=1 j=1 n — 1

·µ

tˆij ψi

¶2

tˆ — 2 ij tˆψ + ˆt2ψ ψi

¸¾

· N Qi µ ˆ ¶2 ˆ2 ¸ t ij = E 1Σ Σ 1 — tψ n i=1 j=1 n — 1 ψi n—1

Qi N ½1 ·Σ ¸¾ 1 Σ 1 µ tˆij ¶ 2 ¯ 2 E Q1, . . . , QN — [t + V (tˆψ)] n 1 ψi ¯ n n — 1 i=1 j=1 ¸ · Σ N 1 1 Qi t2i + Vi = E [t2 + V (tˆψ)] — n n — 1 ψi2 n — 1 i=1

= E

1

N 2 Σ t + Vi

1 [t2 + V (tˆ ψ)] n—1 n — 1 i=1 ψi µΣ ¶ ¶ N µΣ N 1 Vi 1 t2i 2 ˆ — V (tψ) ψi 2 — t + n—1 n—1 ψi ψi i=1 i=1

= =

i

—

N ·Σ µ ¶2 Σ ¸ N 1 Vi — 1 V (tˆψ) ψi ti — t + n—1 n—1 ψ ψi i=1 i=1 i n 1 V (tˆ ) — V (tˆ ) = V (tˆ ). ψ ψ ψ n—1 n—1

= =

6.17 It is sufficient to show that (6.22) and (6.23) are equivalent. When an SRS n(n — 1) for all i and k. So, starting of psus is selected, then πi = n/N and πik = N (N — 1) with the SYG form, µ ¶2 1 Σ Σ πiπk — πik ti tk — 2 πi πk πik i∈£ k∈S k/=i

=

=

=

1 1 π2 — π12 Σ Σ 2 π12

1

π12 1

π12

ti — tk

)2

i∈£ k∈S k/=i

1 1 π2 — π12 Σ Σ 2 π12

(

( 2+ 2 2 ) ti tk — titk

i∈£ k∈S k/=i

1 π2 — π12 Σ 1 π2 — π12 Σ Σ ( 1) 2 1 n — ti — 2 1 titk. π12 π12 π1 π12 i∈£ k∈S i∈£

k/=i

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

But in an SRS, 1 π12 — π12 ( n — 1) = π12 π12

µ

N n

¶2

n n—1 µ ¶2 ³ — n´ N — 1 (n — 1) = N N 1 — , n—1 n N N —1

which proves the result.

6.18 To show the results from stratified sampling, we treat the H strata as the N psus. Note that since all strata are subsampled, we have πi = 1 for each stratum.

Thus, (6.25) becomes H H Σ Σ tˆHT = tˆi = Niȳi. i=1

i=1

For (3.3), since all strata are sampled, either (6.26) or (6.27) gives V (tˆHT) =

H Σ V (tˆi). i=1

Result (3.4) follows from (6.29) similarly.

6.19 (a) We use the method on page 239 to calculate the probabilities.

Unit 1 2 3 4 5 6 7 8 9 πi

1 — 0.049 0.080 0.088 0.007 0.021 0.080 0.021 0.035 0.381

2 0.049 — 0.041 0.045 0.003 0.010 0.041 0.010 0.018 0.218

3 0.080 0.041 — 0.073 0.006 0.017 0.067 0.017 0.029 0.330

4 0.088 0.045 0.073 — 0.006 0.019 0.073 0.019 0.032 0.356

πjk 5 0.007 0.003 0.006 0.006 — 0.001 0.006 0.001 0.002 0.033

6 0.021 0.010 0.017 0.019 0.001 — 0.017 0.004 0.007 0.097

7 0.080 0.041 0.067 0.073 0.006 0.017 — 0.017 0.029 0.330

8 0.021 0.010 0.017 0.019 0.001 0.004 0.017 — 0.007 0.097

9 0.035 0.018 0.029 0.032 0.002 0.007 0.029 0.007 — 0.159

πi 0.381 0.218 0.330 0.356 0.033 0.097 0.330 0.097 0.159 2.000

(b) Using (6.21), V (tˆHT) = 1643. Using (6.46) (we only need the first term), VW R(tˆψ ) = 1867.

131 6.20 (a) We write ˆ ˆ Cov (tx, ty) = Cov

ÃN Σ i=1

tky tix Σ N , Z k Zi πi πk

N Σ N Σ tix tky

=

k=1

Cov (Z , Z ) i

πi πk

i=1 k=1 N Σ tix tiy

=

i=1

πi

i=1

x

N

Σ 1 — πi

y

πi

i=1

ΣN ³ N

=

i=1

n µ

+

=

=

= = =

N n

N³ n

n

+

+

πi πk

ik

—π π ) i k

(π — π π ) tix tky . ik i k πi πk i=1 k=1

n´ N

N Σ N Σ (π — π π ) tix tky ik i k πi πk i=1 k=1 k/=i

tixtiy

¶2 Σ N Σ N i=1 k=1 k/=i

·

³ n ´2¸ — tixtiy N (N — 1) N n (n — 1)

· ¸ (n — 1) n N Σ N NΣ n´ tixtiy + tixtky — 1— n N n i=1 k=1 (N — 1) N

N N ³ Σ i=1

=

tix tiy

1—

tix tiy

i=1 k=1 k/=i N Σ N Σ

(π

k/=i

(b) If the design is an SRS, Cov (tˆ , tˆ ) =

N Σ N Σ tix tky

i

i

N

Σ 1 — πi

=

k

π (1 — π ) +

πi πi

!

1—

N 1³—

n

N ´Σ

N

tixtiy —

i=1

k/=i

n ´ ΣN

N

N

1 Σ Σ tixtky tixtiy — N — 1 i=1 k=1 i=1

N

n´ N³ 1— n N

k/=i N

N ΣΣ N —n t t ix ky n i=1 k=1 N (N — 1) N

k/=i

" N Σ tixtiy —

1

N ΣΣ t t N

N — 1 i=1 k=1

ix ky +

1

N — 1 i=1 # N N ΣΣt t ix ky

" i=1 N N Σ 1 1 — tixtiy — n N N —1 N — 1 i=1 k=1 i=1 # " N tx ty Σ N2 ³ n´ 1 tixtiy — n 1— N N —1 N

N³

n´

N Σ

i=1

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

6.21 Note that

Σ

ΣM

wij yij tˆy j=1 = ΣM NM i∈£ j=1 wij

ˆ = Σ y¯

i∈£

Thus,

·µ

¶ ½ ¾¸ ty — tu (NM ) Cov û¯ — x̂¯ , ŷ¯ — û¯ — (1 — x̂¯ ) tx ¶ ½ N — tx ·µ ¾¸ t t — t y u u = Cov tˆ — tˆ , tˆ — tˆ — (NM — tˆ ) u x y u x N — tx· · ¸ tx ¸ t — t t y u = Cov tˆ , tˆ — tˆ + tˆ — u Cov tˆ , tˆ — tˆ ty — tu ˆ u y u x y u + N — t tx N — tx x tx x ½ " N2 ³ 2 N Σ tiutiy — t ty — tu tiutix = 1 n´ 1 — iu + n N N — 1 i=1 N — tx µ ¶¾¸ —tu t ix tiy — tixtiu + ty — tu t2ix N — tx tx · ty — tu 2 N³ n´ 1 + tut x —n 1 — N N — 1 tut y — tu µ N — tx ¶¸ tu — t t —t t ty — tu 2 x u + N — t tx x tx x y ½ " N2 ³ N Σ tiutiy — t2 ty — tu tiutix = 1 n´ 1 + — iu n N N — 1 i=1 N — tx µ ¶¾¸ —tu t ix tiy — tixtiu + ty — tu t2ix . tx N — tx 2

tu

(b) Now, tiy =

M Σ

yij

j=1

and tiu =

M Σ

xij yij .

j=1

So if psu i is in domain 1 then xij = 1 for every ssu in psu i, tix = M , and tiu = tiy; if psu i is in domain 2 then xij = 0 for every ssu in psu i, tix = 0, and tiu = 0. We

133 may then write the covariance as Σ½ N µ ¶¾ tu ty — tu 2 2 + ty — tu t t — t t t — t t + iu iy t ix iy ix iu iu i=1 N — tx tiutix — t x N — tx ix ½ µ ¶¾ Σ ty — tu tu ty — tu 2 = tiutiy — t2iu + tiutix — tixt iy — tix tiu + tix N — t t N — t x x x i∈domain 1 ½ µ ¶¾ Σ ty — tu tu ty — tu 2 2 + tiu tiy — tiu + tiutix — tixt iy — tix tiu + tix N — t t N — t x x x i∈domain 2 ½ µ ¶¾ Σ ty — tu tu ty — tu 2 2 = tiy tiy — tiy + t iyM — Mt iy — Mt iy + M N — tx tx N — tx i∈domain 1 Σ ½ ty — tu tu ty — tu 2 ¾ = tiy M — M tx N — tx N — tx i∈domain 1 ty — tu tx tu ty — tu 2 = tu M — N M N — tx NM tx N — tx = 0. (c) Almost any example will work, as long as some psus have units from both domains.

6.22 (a) Since E(Zi) = πi, N

E(tyˆ ) =

Σ 1

πi i=1

N

ui E(Zi ) = ΣN

Σ

N

ui =

M

Σ Σ likyk i=1 k=1

i=1

Lk

M

=

Σ yk. k=1

The last equality follows since i=1 lik = Lk. The variance given is the variance of the one-stage Horvitz-Thompson estimator. (b) Note that

N

M

M

N

Σ Zi Σ likyk Σ 1 Σ Zi ik k tyˆ = πi k=1 Lk =k=1 Lk i=1 πi l y . i=1

But the sum is over all k from 1 to M , not just the units in SB . We need to show Σ that Ni=1 Zπi lik = 0 for k /∈ SB . i

N Σ Zi

w∗ k =

π i=1 i Σ

lik

N

.

lik

i=1

But a student is in S B if and only if s/he is linked to one of the sampled units in Σ SA. In other words, k ∈ SB if and only if i∈£A lik > 0. For k /∈ SB , we must have lik = 0 for each i ∈ SA.

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

(c) Suppose Lk = 1 for all k. Then, N M N Σ Σ Zi Σ Zi tˆ = l y = y y ik k πi πi i because

ΣM k=1

i=1

i=1

k=1

likyk = yi.

(d) The values of ui are:

lik Unit i from U A

Element k from UB 1 2 1 0 1 1 0 1

1 2 3

ui 4/2 = 2 4/2 + 6/2 = 5 6/2 = 3

Here are the three SRSs from UA: tˆy Sample {1,2}

3 21 (2 + 5) = 2 2

{1,3}

3 15 (2 + 3) = 2 2

{2,3}

3 24 (5 + 3) = 2 2

Consequently,

1 E[tˆ ] = y 3

so it is unbiased. But V [tˆy] = = =

1 3 1

"µ

µ

21

+

2

15 2

+

24

¶ = 10,

2

¶2# ¶2 µ ¶2 µ 15 21 24 — 10 + — 10 + — 10 2 2 2

[0.25 + 6.25 + 4] 3 3.5.

(e) We construct the variable u = ΣM

i k=1 likyk/Lk for each adult in the sample, where Lk = (number of other adults +1). Using weight wi = 40,000/100 = 400, we Σ calculate tˆu = i∈£ wiui = 7200 with 2 N 2 ˆ ˆ V (tu) = su = 1,900,606. n This gives a 95% CI of [4464.5, 9935.5]. Note that the without-replacement variance estimator could also be used.

The following SAS code will compute the estimates.

135 data wtshare; infile wtshare delimiter="," firstobs=2; input id child preschool numadult; yoverLk = preschool/(numadult+1); proc sort data=wtshare; by id; run; proc means data=wtshare sum noprint; by id; var yoverLk; output out=sumout sum = u; data sumout; set sumout; sampwt = 40000/100; proc surveymeans data=sumout mean sum clm clsum; weight sampwt; var u; run; 6.23 (a) We have πi’s 0.50 0.25 0.50 0.75, V (tˆHT) = 180.1147, and V (tˆψ ) = 101.4167. (b) 1 ΣΣ π iπ k 2n N

N

µ

i=1 k=1

=

N N 1 Σ Σ

2n i=1 k=1

ti tk — πi πk

¶2

·µ ti πiπk

πi

—

¶¸2 t ¶ µ tk t n — π —n k

¶µ µ ¶# t tk tk — t ¶2 — 2 ti t — πk n πi n πk — n k=1 "N µ µ N Σ N ¶#2 t ¶2 1 Σ 1Σ ti ti π π t i k = — — — πi n n πi n πi n i=1 k=1 i=1 #2 "N µ N Σ N t ¶2 1 Σ 2 1Σ ti n ψ ψ i k (ti — ψit) = — — n n nψi n N Σ N 1 Σ πiπk = 2n i=1

"µ

i=1 k=1

=

N 1Σ

n

i=1

ψi

µ ti ψi

t ¶2 + ti — πi n

¶2

µ

i=1

—t

= V (tˆψ ).

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

(c) 1 ΣΣ πiπk 2n N

ˆ HT ) = V (tˆψ) — V (t

N

µ

i=1 k=1

ti tk — πk πi

¶2

N N µ ¶2 1 Σ Σ(π π — π ) ti — tk i k ik π π —2 i k i=1 k=1 µ ¶ ³ ´ Σ Σ 1 N N π iπ ti tk 2 k— πiπk + π ik — 2 n πi πk i=1 k=1· µ ¶ ¸µ ¶ Σ 1Σ n—1 ti tk 2 N N πiπk 1 — 1 — πiπk — n 2 i=1 n πi πk

=

> =

k=1

0.

(d) If πik ≥ (n — 1)πiπk/n forµall i ¶ and k, then πik n—1 min ≥ π for all i. πk n i Consequently, µ ¶ Σ N N πik n—1Σ min ≥ πi = n — 1. πk n i=1

i=1

(e) Suppose V (tˆHT ) ≤ V (tˆψ ). Then from part (b), 0 ≤ V (tˆψ) — V (tˆHT ) µ N Σ N ¶ µ ti ¶ n—1 1Σ tk 2 πik — = π iπk — πk n 2 i=1 k/=i πi µ Σ # 1Σ " µ ¶2 N N = πik — n — 1 πiπk ¶ µ ti ¶2 ti tk tk — 2 + 2 i=1 n πi πk πi πk k/=i # ¶ "µ ¶2 ΣNΣNµ n—1 ti ti tk — = πik — n πiπk πi πi πk i=1 k/=i

=

N Σ N Σ

πik

i=1 k/=i

µ

ti πi

N

¶2

— N

N N Σ Σ π ti tk ik

i=1 k/=i

πi πk

—

µ ¶2 N n—1Σ πi(n — πi) ti n πi i=1

n —1ΣΣ + n ti tk i=1 k/=i

=

ΣN Σ µ ¶2 µ ¶2 Σ N N Σ π ti tk — (n — 1) N ti (n — 1)πi ti ik — πi π πi πi πk i i=1

+

N

i=1 k/=i N N

n —1Σ 2 n—1 ΣΣ t itk n i=1 ti + n i=1 k/=i

i=1

137 Consequently,

N

N

N

N

ΣΣ n—1 ΣΣ ti tk t it k ≥ , πik n πi πk i=1 k=1

or

N N Σ Σ

i=1 k/=i

aiktitk ≥ 0,

i=1 k=1

where aii = 1 and

n πik if i /= k. n — 1 πiπk The matrix A must be nonnegative definite, which means that all principal subma2 trices must have determinant ≥ 0. Using a 2 × 2 submatrix, we have 1 — aik ≥ 0, which gives the result. aik = 1—

6.24 (a) We have

i

πik 1 2 3 4 πk

1 0.00 0.31 0.20 0.14 0.65

k 2 0.31 0.00 0.03 0.01 0.35

3 0.20 0.03 0.00 0.31 0.54

4 0.14 0.01 0.31 0.00 0.46

πi 0.65 0.35 0.54 0.46 2.00

(b) VˆSYG (tˆHT ) tˆHT Sample, S V̂HT (t̂HT ) {1, 2} 9.56 38.10 -0.9287681 {1, 3} 5.88 -4.74 2.4710422 {1, 4} 4.93 -3.68 8.6463858 {2, 3} 7.75 -100.25 71.6674365 {2, 4} 21.41 -165.72 323.3238494 {3, 4} 3.12 3.42 -0.1793659 Σ Σ Note that i k>i π ik V̂ (t̂ HT ) = 6.74 for each method.

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

6.25 (a) P {psu’s i and j are in the sample} = P {psu i drawn first and psu j drawn second} =

+P {psu j drawn first and psu i drawn second} ai ψj aj ψi + N N 1 — ψi 1 — ψj Σ Σ ak ak k=1

=

k=1

ψi(1 — ψi)ψj

ψj(1 — ψj)ψi

+

Σ

Σ ak(1 — ψi)(1 — πi) ak(1 — ψj)(1 — πj) k=1 µ ¶ ψiψj 1 1 + N 1 — πi 1 — πj Σ ak N

N

k=1

=

k=1

(b) Using (a), P {psu i in sample} =

N Σ

πij

j=1,j/=i N

= Σ

ψiψj

ψi

k=1 ½Σ N

N j=1,j/=i Σ

=

N Σ

ψi ½

N Σ

ak ½

k=1

=

ψi N Σ

ak

ak

1 1 + 1 — πi 1 — πj

µ ψj

j=1

ak

k=1

=

µ

¶

1 1 + 1 — πj 1 — πi

¶ —

2ψi ¾ 1 — πi

N πi ¾ Σ ψj 1 + — 1 — πi 1 — πi 1 — πj j=1

ΣN

ψj 1+ 1 — πj j=1

¾

k=1

In the third step above, we used the constraint that Σ πj = n = 2, so Σ ψj = 1.

Now note that 2

ΣN

k=1 ak

N

N

j=1

j=1

N Σ ψk(1 — ψk) =2 1 — 2ψk k=1 N Σ ψk(1 — 2ψk + 1) = 1 — 2ψk k=1 N

=1+

Σ ψk k=1

1 — πk

.

139 Thus P {psu i in sample} = 2ψi = πi. (c) Using part (a), π π —π = 4ψ ψ — i j

ij

µ

¶

+ ψiψj 1 1 i j N 1 — π 1 — πj i Σ ak ¶ ¸ · µk=1 N Σ (1 — π )(1 — π ) — (1 — π + 1 — π ) 4 ψiψj ak i j j i k=1 = . ¶ µN Σ ak (1 — πi)(1 — πj) k=1

Using the hint in part (b), ¶ µΣ N ak (1 — πi)(1 — πj) — (1 — πj + 1 — πi) 4 k=1

· N Σ = 2 1+

¸ ψk (1 — πi)(1 — πj) — 2 + πi + πj 1 — πk k=1 N Σ ψk = 2(1 — π )(1 — π ) + 2 — 2π — 2π + 2π π — 2 + π + π i j i j i j i j 1 — πk k=1 N

i j Σ ψ i j i j = 2(1 — π )(1 — π ) k=1 1 —kπk — π — π + 2π π = ≥ 2(1 — πi)ψj + 2(1 — πj)ψi — πi — πj + 2πiπj = (1 — πi)πj + (1 — πj)πi — πi — πj + 2πiπj = 0.

Thus πiπj — πij ≥ 0, and the SYG estimator of the variance is guaranteed to be nonnegative. Σ 6.26 The desired probabilities of inclusion are πi = 2Mi/ 5 j=1 Mj. We calculate ψi = πi/2 and ai = ψi(1 — ψi)/(1 — πi) for each psu in the following table: psu, i 1 2 3 4 5 Total

Mi 5 4 8 5 3 25

πi 0.40 0.32 0.64 0.40 0.24 2.00

ψi 0.20 0.16 0.32 0.20 0.12 1.00

ai 0.26667 0.19765 0.60444 0.26667 0.13895 1.47437

According to Brewer’s method, P (select psu i on 1st draw) = ai

5 ,Σ

aj

j=1

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES , P (psu j on 2nd draw | psu i on 1st draw) = ψj (1 — ψi).

and Then

.26667 0.16

P {S = (1, 2)} = P {S = (2, 1)} =

1.47437 0.8 .19765 0.2

1.47437 0.84

= 0.036174, = 0.031918,

and π12 = P {S = (1, 2)} + P {S = (2, 1)} = 0.068.

Continuing in like manner, we have the following table of πij. i\j 1 2 3 4 5 Sum

1 — .068 .193 .090 .049 .400

2 .068 — .148 .068 .036 .320

3 .193 .148 — .193 .107 .640

4 .090 .068 .193 — .049 .400

5 .049 .036 .107 .049 — .240

We use (6.21) to calculate the variance of the Horvitz-Thompson estimator. ³ ´2 ti tj (π π — π ) — i j π π π t t 1 1 1 1 2 2 2 3 3 4 Sum

2 3 4 5 3 4 5 4 5 5

ij

i

j

i

j

0.068 0.193 0.090 0.049 0.148 0.068 0.036 0.193 0.107 0.049 1

0.40 0.40 0.40 0.40 0.32 0.32 0.32 0.64 0.64 0.40

0.32 0.64 0.40 0.24 0.64 0.40 0.24 0.40 0.24 0.24

20 20 20 20 25 25 25 38 38 24

25 38 24 21 38 24 21 24 21 21

i j

ij

πi

47.39 5.54 6.96 66.73 20.13 19.68 3.56 0.02 37.16 35.88 243.07

πj

Note that for this population, t = 128. To check the results, we see that ΣP (S)tˆHT £ = 128

and ΣP (S)(tˆHT £ — 128)2 = 243.07.

6.27 A sequence of simple random samples with replacement (SRSWR) is drawn until the first SRSWR in which the two psu’s are distinct. As each SRSWR in the sequence is selected independently, for the lth SRSWR in the sequence, and for

141 i /= j, P {psu’s i and j chosen in lth SRSWR} = P {psu i chosen first and psu j chosen second} +P {psu j chosen first and psu i chosen second} = 2ψiψj.

The (l + 1)st SRSWR is chosen to be the sample if each of the previous l SRSWR’s is rejected because the two psu’s are the same. Now NΣ

P (the two psu’s are the same in an SRSWR) =

ψk2,

k=1

so because SRSWR’s are drawn independently, P (reject first l SRSWR’s) =

µΣ N

2 ψk

¶L .

k=1

Thus πij = P {psu’s i and j are in the sample} =

∞ Σ

P {psu’s i and j chosen in (l + 1)st SRSWR,

L=0

and the first l SRSWR’s are rejected} µΣ N ¶L Σ (2ψiψj) = ψk2 ∞

L=0

=

k=1

2ψiψj . Σ 2 1— N k=1 ψk

Equation (6.18) implies Σ πij πi = N j=1,j/=i , µ ¶ Σ N ΣN 2 2ψ ψ i j 1— = j=1,j/=i k=1 ψk ,µ ¶ ,µ ¶ ΣN Σ 2 2 = 2ψi 1 — ψ — 2ψ 1 — N i k=1 k k=1 ψk2 ,µ ¶ ΣN 2 = 2ψi(1 — ψi) 1 — ψ . k=1

k

Note that, as (6.17) predicts, N Σ i=1

¶,µ ¶ µ N N Σ Σ 2 2 πi = 2 1 — 1— ψi ψk = 2 i=1

k=1

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

6.28 Note that, using the indicator variables, n Σ

Iki = 1 for all i,

k=1

=

x ki

P (Z = 1 | I , . . . , I i

11

Mi

Ikj Mj j=1 n

nN ) =

Σ

ΣN

j=1 Ikj Mj

n

n

=

IkiMi

k=1

and

,

ΣN

n Σ

I x ki ki k=1

N

ti Σ tα(k) ΣΣ ˆ ki Z i tRHC = = I xki . x k=1 i=1 k=1 k,α(k) We show that tˆRHC is conditionally unbiased for t given the grouping E[tˆRHC | I11, . . . , InN ] = E

·Σ n Σ N

ti

I Z

ki i

k=1 i=1 n N

xki

¸ | I11, . . . , InN

N

Σ Σ ki ti Σ kL kL I x = I xki k=1 i=1 n N

=

ΣΣ

L=1

Ikiti

k=1 i=1 N

=

Σ

ti = t.

i=1

Since E[tˆRHC | I11, . . . , InN ] = t for any random grouping of psu’s, we have that E[tˆRHC] = t.

To find the variance, note that V [tˆRHC] = E[V (tˆRHC | I11, . . . , InN )] + V [E(tˆRHC | I11, . . . , InN )].

Since E[tˆRHC |I11, . . . , InN ] = t, however, we know that V [E(tˆRHC | I11, . . . , InN )] = 0. Conditionally on the grouping, the kth term in tˆRHC estimates the total of group k using an unequal-probability sample of size one. We can thus use (6.4) within each group to find the conditional variance, noting that psu’s in different groups are selected independently. (We can obtain the same result by using the indicator

143 variables directly, but it’s messier.) Then V (tˆRHC | I11, . . . , InN ) =

n Σ N Σ

Ikixki k=1 i=1 n N

=

ΣΣ

Iki

µ t

2 ti

xki

i

xki

—

¶2 Ikjtj

j=1 N N

n

—

N Σ

ΣΣΣ I kit iI kjtj

k=1 µk=1 2 ¶ Σ NΣ NΣ N Mjt = IkiIkj i — titj .

Mi i=1

i=1 j=1

k=1 i=1 j=1

item Now to find E[V (tˆRHC| I11, . . . , InN )], we need E[Iki] and E[IkiIkj] for i /= j.

Let Nk be the number of psu’s in group k. Then E[Iki] = P {psu i in group k} =

Nk N

and, for i /= j, E[I I ] = P {psu’s i and j in group k} = Nk Nk — 1 . ki kj N N —1 Σ Thus, letting ψi = Mi/ Nj=1 Mj, V [tˆRHC] = E[V (tˆRHC | I11, . . . , InN )] µ ¶¸ ·Σ n Σ N Σ N Mjt2 = E IkiIkj i — titj Mi k=1 i=1 j=1 µ ¶ Σ Σ Σ Σ Σ Nk Nk — 1 Mjt2 n N Nk 2 n N N 2 = (t — t ) + i — titj i i N N —1 N Mi k=1 i=1 k=1 i=1 j=1,j/=i µΣ ¶µ Σ n N ¶ Nk Nk — 1 t2i 2 = — t ψi N N —1 i=1 k=1

=

µΣ n k=1

Nk Nk — 1 N N —1

¶µ NΣ

· ψi

i=1

¸2 ¶ . ti — t ψi

The second factor equals nV (tˆψ ), with V (tˆψ ) given in (6.46), assuming one stage cluster sampling.

What should N1, . . . , Nn be in order to minimize V [tˆRHC]? Note that n

Σ

n

Nk(Nk — 1) =

k=1

Σ

N 2k — N

k=1

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

is smallest when all Nk’s are equal. If N/n = L is an integer, take Nk = L for k = 1, 2, . . . , n. With this design, N µ ti ¶2 L —1 Σ ψi —t ψi N — 1 i=1 N—n V (tˆ ). ψ N —1

V [tˆRHC] = =

6.29 (a) Σ

π̃ i k =

Σ

πiπk 1 — (1 — πi)(1 — πk)/

cj j=1

k/=i

k/=i

N Σ

Σ

Σ i(1 — πi) πk — πΣ πk (1 — πk ) N c j j=1 k/=i k/=i "N # πi(1 — πi) Σ = πi(n — πi) — ΣN πk(1 — πk) — πi(1 — πi) j=1 cj k=1 = πi

= πi(n — πi) — πi(1 — πi) + π i(1 Σ — πi) N j=1 cj 2

=

πi(n — 1) +

2

π2i(1 — πi)2 . ΣN j=1 cj

(b) If an SRS is taken, πi = n/N , so Σ N cj = j=1

Σ N n ³ j=1

and

"

π̃ i k

= = = = = =

N 1

—

n´ N

³ n´ =n 1— N

# (1 — n/N)(1 — n/N) 1— ¡ ¢ N2 n 1 — Nn n h ni n—1+ N2 N · ¸ n n—1 n N N + N2 · ¸ n n—1 N —1 n—1 n + N N —1 n—1 N N2 · ¸ n n — 1 N — 1 n(N — 1) + (n — 1)N 2 N N —1 N · ¸ n n—1 N —n 1+ N N —1 (n — 1)N 2 n2

145 (c) First note that π π — π̃ i k

Then, letting B =

ΣN

ik

= π π (1 — πi)(1 — πk) . i k ΣN j=1 cj

j=1 cj,

Σ µ ¶2 N N 1Σ tk (πiπk — π̃ ik ) ti — 2 πi πk i=1 k=1 Σ µ ¶2 N N πi)(1 — πk ) ti 1Σ πiπk (1 —Σ t k = — N 2 πi πk j=1 cj i=1 k=1 ¶ µ 2 N Σ N Σ 1 ti ti t k = πiπk(1 — πi)(1 — πk) 2 2 — 2 2 ΣN πi πi πk j=1 cj

VHaj(tˆHT ) =

i=1 k=1 N Σ ΣN

µ 2 ti tk ¶ ti = ΣN πiπk(1 — πi)(1 — πk) πi2 — πi πk j=1 cj i=1 k=1 ÃN !2 N 2 Σ Σ t 1 t i πi(1 — πi) = πi(1 — πi) i — Σ N πi πi2 j=1 cj i=1 1

i=1

N Σ

ti2 1 = ci π 2 — Σ N c i j=1 j i=1 µ 2 N ¶2 Σ ti A . = ci — πi2 i=1

ÃN Σ

!2

t ci i πi i=1

6.30

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

(a) From (6.21), V (tˆHT) =

N Σ N µ ti 1Σ tk (πiπk — πik) — πk 2 i=1 k=1 πi

¶2

k/=i

Σ t µ ¶2 1 (πiπk — πik) ti — + t — tk = 2 πi n n πk i=1 k=1 N N Σ

k/=i

=

N Σ N 1Σ (π π — π ) × ik 2 i=1 k=1 i k k/=i

(µ

ti t — πi n

¶2

µ +

N Σ N Σ (πiπk — πik) =

tk

πk (µ

i=1 k=1 k/=i

—

t

¶2

µ —2

ti

¶µ

tk

t

¶)

— πk n ¶µ ¶2 ¶) t tk t ti — t — i — πi n πi n πk — n n

πi µt

—

t

n

From Theorem 6.1, we know that N Σ

πk = n

k=1

and

N Σ

πik = (n — 1)πi,

k=1

k/=i

so N ΣN µt N µ ti Σ t ¶2 t ¶2 Σ (πiπk — πik) πii — n = [πi(n — πi) — (n — 1)πi] πi — n i=1 k=1 k/=i

i=1

=

N Σ i=1

πi(1 — πi)

µ ti πi

—

t ¶2 n

.

This gives the first two terms in (6.47); the third term is the cross-product term above. (b) For an SRS, πi = n/N and πik = [n(n — 1)]/[N (N — 1)]. The first term is µ ¶ 2 Σ Nti t 2 Σ N n N N ¯ 2 St — = (ti — tU ) = N (N — 1) .

N n n n n i=1

i=1

147 The second term is

³ n ´2 N Σ i=1

µ

Nti

t — n n

N

¶2

2

= n(N — 1)

St . n

(c) Substituting πiπk(ci + ck)/2 for πik, the third term in (6.47) is µ ¶µ ¶ Σ ti t tk t N ΣN (πik — πiπk) — — π n π n i k i=1 k=1 k/=i

=

NΣ N Σ

ci + ck — 2 πiπk 2 i=1 k=1

µ

¶µ ¶ ti t tk t — — πi n πk n

k/=i

µ t ¶µ — 2 ti c + c i k t tk — π π i k = 2 πi — n πk n i=1 k=1 N Σ N Σ

=

N Σ

π i2(1 — ci)

i=1

¶ —

µ ti

t ¶2 . πi — n

Σ N

π2(ci — 1)

µ

i

i=1

ti t ¶2 — πi n

Then, from (6.47), V (tˆHT) =

N Σ

πi

µ ti

t — n πi

¶2

i=1 N Σ N Σ i=1 k=1 k/=i

µ

N Σ

πi i=1 N

+

Σ

t ti — πi n

Σ N

πi(1 — ciπi)

i=1

¶2

π2i(1 — ci)

i=1

=

i=1

(πik — πiπk)

+

≈

N µ ti t ¶2 Σ 2 — πi — π n

µ

i

ti t πi — n

¶µ

t tk πk — n

¶

µ ¶2 N Σ ti t — — π2i πi n

µ ti πi µ ti πi

i=1

— —

t ¶2 n

t ¶2 . n

—1 , then the variance approximation in (6.48) for an SRS is If ci = nn—π i

N Σ i=1

πi(1 — ciπi)

µ ti πi

—

N µ ¶ t ¶2 Σ N n—1 = 1— (ti — t¯U )2 n n N — 1 i=1 µ ¶ N (N — 1) n—1 2 = 1 — N — 1 S t. n

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CHAPTER 6. SAMPLING WITH UNEQUAL PROBABILITIES

If

´ n — 1ΣN 1 2 = ³ 2 + ci 1 — πi n k=1 πk

then N Σ

πi(1 — ciπi)

µ ti

—

πi

i=1

µ ¶ n(n — 1) N (N — 1) 1— S2t.

N—n n

t ¶2 = n

6.31 We wish to minimize µ ¶2 N N 1Σ ti 1 Σ M i2S2i ψi —t + n n ψi miψi i=1

i=1

subject to the constraint that C=E

·Σ

¸ mi = E

i∈£

·Σ N

¸ Qimi = n

i=1

N Σ

ψimi. i=1

Using Lagrange multipliers, let g(m1, . . . , mN , λ) =

N Σ M 2S2 i

i=1

∂g ∂mk ∂g ∂λ

i

miψi

µ ¶ N Σ —λ C —n ψimi . i=1

M 2S2

= — k2 k + nλψ k mk ψk N Σ =n ψ im i — C. i=1

Setting the partial derivatives equal to zero gives 1 MkSk mk = √ nλ ψk and √

√ Σ N n MiSi. λ= C i=1

Thus, the optimal allocation has mi ∝ MiSi/ψi. For comparison, a self-weighting design would have mi ∝ Mi/ψi.

6.32 Let Mi be the number of residential numbers in psu i. When you dial a number based on the method, P (reach working number in psu i on an attempt) = P (select psu i)P (get working number | select psu i) 1 Mi = .

N 100

149 Also, µ N M0 Σ Mi ¶ = 1 1 1— P (reach no one on an attempt) = — .

N 100 100N i=1

Then, P (select psu i as first in sample) = P (select psu i on first attempt) + P (reach no one on first attempt, select psu i on second attempt) + P (reach no one on first and second attempts, select psu i on third attempt) + ... µ ¶ µ ¶ 1 Mi 1 Mi M0 1 Mi M0 2 + + 1— 1— = +...

100N 100N N 100 Nµ 100 N 100 ¶j ∞ 1 Mi Σ 1 — M0 = N 100 j=0 100N =

1 Mi N 100

=

Mi M0

1 µ M0 ¶ 1— 1— 100N

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Chapter 7

Complex Surveys 7.6 Here is SAS code for solving this problem. Note that for the population, we have y¯U = 17.73, Ã2000 !2 ¶ µ Σ 2Σ000 1 354602 2 1 y i yi — /1999, = 704958 — 2000 S2 = 38.1451726 = 2000 1999 i=1 i=1 θˆ25 = 13.098684, θˆ50 = 16.302326, θˆ75 = 19.847458. data integerwt; infile integer delimiter="," firstobs=2; input stratum y; ysq = y*y; run; /* Calculate the population characteristics for comparison */ proc means data=integerwt mean var; var y; run; proc surveymeans data=integerwt mean sum percentile = (25 50 75); var y ysq; /* Without a weight statement, SAS assumes all weights are 1 */ run; proc glm data=integerwt; class stratum; model y = stratum; means stratum; run; 151

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CHAPTER 7. COMPLEX SURVEYS

/* Before selecting the sample, you need to sort the data set by stratum */ proc sort data=integerwt; by stratum; proc surveyselect data=integerwt method=srs sampsize = (50 50 20 25) out = stratsamp seed = 38572 stats; strata stratum; run; proc print data = stratsamp; run; data strattot; input stratum datalines; 1 200 2 800 3 400 4 600 ;

_total_;

proc surveymeans data=stratsamp total = strattot mean clm sum percentile = (25 50 75); strata stratum; weight SamplingWeight; var y ysq; run; /* Create a pseudo-population using the weights */ data pseudopop; set stratsamp; retain stratum y; do i = 1 to SamplingWeight; output ; end; proc means data=pseudopop mean var; var y; run; proc surveymeans data=pseudopop mean sum percentile = (25 50 75);

153 var y ysq; run; The estimates from the last two surveymeans statements are the same (not the standard errors, however).

7.7 Let y = number of species caught. y 1 3 4 5 6 7 8 9

fˆ(y) .0328 .0328 .0820 .0328 .0656 .0656 .1803 .0328

y 10 11 12 13 14 16 17 18

fˆ(y) .2295 .0491 .0820 .0328 .0164 .0328 .0164 .0164

Here is SAS code for constructing this table: data nybight; infile nybight delimiter=’,’ firstobs=2; input year stratum catchnum catchwt numspp depth temp ; select (stratum); when (1,2) relwt=1; when (3,4,5,6) relwt=2; end; if year = 1974; /*Construct empirical probability mass function and empirical cdf.*/ proc freq data=nybight; tables numspp / out = htpop_epmf outcum; weight relwt; /*SAS proc freq gives values in percents, so we divide each by 100*/ data htpop_epmf; set htpop_epmf; epmf = percent/100; ecdf = cum_pct/100; proc print data=htpop_epmf; run; 7.8 We first construct a new variable, weight, with the following values:

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154 Stratum large sm/me

CHAPTER 7. COMPLEX SURVEYS weight 245 Mi 23 mi 66 Mi 8 mi

Because there is nonresponse on the variable hrwork, for this exercise we take mi to be the number of respondents in that cluster. The weights for each teacher sampled in a school are given in the following table: dist sm/me sm/me sm/me sm/me sm/me sm/me sm/me sm/me large large large large large large large large large large large large large large large large large large large large large large large

school 1 2 3 4 6 7 8 9 11 12 13 15 16 18 19 20 21 22 23 24 25 28 29 30 31 32 33 34 36 38 41

popteach 2 6 18 12 24 17 19 28 33 16 22 24 27 18 16 12 19 33 31 30 23 53 50 26 25 23 21 33 25 38 30

mi 1 4 7 7 11 4 5 21 10 13 3 24 24 2 3 8 5 13 16 9 8 17 8 22 18 16 5 7 4 10 2

The epmf is given below, with y =hrwork.

weight 16.50000 12.37500 21.21429 14.14286 18.00000 35.06250 31.35000 11.00000 35.15217 13.11037 78.11594 10.65217 11.98370 95.86957 56.81159 15.97826 40.47826 27.04013 20.63859 35.50725 30.62500 33.20972 66.57609 12.58893 14.79469 15.31250 44.73913 50.21739 66.57609 40.47826 159.78261

155 y 20.00 26.25 26.65 27.05 27.50 27.90 28.30 29.15 30.00 30.40 30.80 31.25 32.05 32.10 32.50 32.90 33.30 33.35 33.75 34.15

fˆ(y) 0.0040 0.0274 0.0367 0.0225 0.0192 0.0125 0.0050 0.0177 0.0375 0.0359 0.0031 0.0662 0.0022 0.0031 0.0370 0.0347 0.0031 0.0152 0.0404 0.0622

y 34.55 34.60 35.00 35.40 35.85 36.20 36.25 36.65 37.05 37.10 37.50 37.90 37.95 38.35 38.75 39.15 40.00 40.85 41.65 52.50

fˆ(y) 0.0019 0.0127 0.1056 0.0243 0.0164 0.0022 0.0421 0.0664 0.0023 0.0403 0.1307 0.0079 0.0019 0.0163 0.0084 0.0152 0.0130 0.0018 0.0031 0.0020

7.10 Without weights

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CHAPTER 7. COMPLEX SURVEYS With weights

Using the weights makes a huge difference, since the counties with large numbers of veterans also have small weights.

7.13 The variable agefirst contains information on the age at first arrest. Missing values are coded as 99; for this exercise, we use the non-missing cases.

Estimated Quantity Mean Median 25th Percentile 75th Percentile

Without Weights 13.07 13 12 15

With Weights 13.04 13 12 15

Calculating these quantities in SAS is easy: simply include the weight variable in PROC UNIVARIATE. The weights change the estimates very little, largely because the survey was designed to be self-weighting.

7.14 Quantity Age ≤14 Violent offense Both parents Male Hispanic Single parent Illegal drugs

Variable age crimtype livewith sex ethnicity livewith everdrug

pˆ .1233 .4433 .2974 .9312 .1888 .5411 .8282

7.15 (a) We use the following SAS code to obtain ŷ¯ = 18.03, with 95% CI [17.48, 18.58].

157 data nhanes; infile nhanes delimiter=’,’ firstobs=2; input sdmvstra sdmvpsu wtmec2yr age ridageyr riagendr ridreth2 dmdeduc indfminc bmxwt bmxbmi bmxtri bmxwaist bmxthicr bmxarml; label age = "Age at Examination (years)" riagendr = "Gender" ridreth2 = "Race/Ethnicity" dmdeduc = "Education Level" indfminc = "Family income" bmxwt = "Weight (kg)" bmxbmi = "Body mass index" bmxtri = "Triceps skinfold (mm)" bmxwaist = "Waist circumference (cm)" bmxthicr = "Thigh circumference (cm)" bmxarml = "Upper arm length (cm)"; run; proc surveymeans data=nhanes mean clm percentile = (0 25 50 75 100); stratum sdmvstra; cluster sdmvpsu; weight wtmec2yr; var bmxtri age; run; (b) The data appear skewed.

(c) The SAS code in part (a) also gives the following.

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158 Percentile Minimum 25 50 75 Maximum

CHAPTER 7. COMPLEX SURVEYS Value 2.8 10.98 16.35 23.95 44.6

Std Error 0.177 0.324 0.425

Men: Percentile Minimum 25 50 75 Maximum

Value 2.80 9.19 12.92 18.11 42.4

Women: Percentile Minimum 25 50 75 Maximum

Value 4.00 14.92 21.94 28.36 44.6

(d) Here is SAS code for constructing the plots: data groupage; set nhanes; bmigroup = round(bmxbmi,5); trigroup = round(bmxtri,5); run; proc sort data=groupage; by bmigroup trigroup; proc means data=groupage; by bmigroup trigroup; var wtmec2yr; output out=circleage sum=sumwts; goptions reset=all; goptions colors = (black); axis3 label=(’Body Mass Index, rounded to 5’) order=(10 to 70 by 10); axis4 label=(angle=90 ’Triceps skinfold, rounded to 5’) order=(0 to 55 by 10); /* This gives the weighted circle plot */

159

proc gplot data=circleage; bubble trigroup * bmigroup= sumwts/ bsize=12 haxis = axis3 vaxis = axis4; run; /* The following draws the bubble plot with trend line */ ods graphics on; proc loess data=nhanes; model bmxtri=bmxbmi / degree = 1 select=gcv; weight wtmec2yr; ods output OutputStatistics = bmxsmooth ; run; ods graphics off; proc print data=bmxsmooth; run; proc sort data=bmxsmooth; by bmxbmi; goptions reset=all; goptions colors = (gray); axis4 label=(angle=90 ’Triceps skinfold’) order = (0 to 55 by 10); axis3 label=(’Body Mass Index’) order=(10 to 70 by 10); axis5 order=(0 to 55 by 5) major=none minor=none value=none; symbol interpol=join width=2 color = black; /* Display the trend line with the bubble plot */ data plotsmth; set bubbleage bmxsmooth; /* concatenates the data sets */ run; proc gplot data=plotsmth; bubble bmxtri*bmxbmi = sumwts/ bsize=10 haxis = axis3 vaxis = axis4; plot2 Pred*bmxbmi/haxis = axis3 vaxis = axis5; run;

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CHAPTER 7. COMPLEX SURVEYS

7.17 We define new variables that take on the value 1 if the person has been a victim of at least one violent crime and 0 otherwise, and another variable for injury.

The SAS code and output follows. data ncvs; infile ncvs delimiter = ","; input age married sex race hispanic hhinc away employ numinc violent injury medtreat medexp robbery assault pweight pstrat ppsu; if violent > 0 then isviol = 1; else isviol = 0; if injury > 0 then isinjure = 1;

161 else isinjure = 0; run; proc surveymeans data=ncvs; weight pweight; strata pstrat; cluster ppsu; var numinc isviol isinjure; run; proc surveymeans data=ncvs; weight pweight; strata pstrat; cluster ppsu; var medexp; domain isinjure; run; The SURVEYMEANS Procedure Data Summary Number of Strata Number of Clusters Number of Observations Sum of Weights

143 286 79360 226204704

Statistics

Variable numinc isviol isinjure

N 79360 79360 79360

Std Error of Mean 0.002034 0.000665 0.000316

Mean 0.070071 0.013634 0.003754

95% CL for Mean 0.06605010 0.07409164 0.01232006 0.01494718 0.00312960 0.00437754

Domain Analysis: isinjure

isinjure 0 1

Variable N medexp 79093 medexp 267

Mean 0 101.6229

Std Error of Mean 0 33.34777

95% CL for Mean 0.0000000 0.000000 35.7046182 167.541160

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CHAPTER 7. COMPLEX SURVEYS

(y) is the sum of the middle N (1 —2α) observations Σ in the population divided by N , and q1≤y≤q2 f (y) = F (q2) — F (q1) ≈ 1 — 2α.

Consequently, sum of middle N (1 — 2α) observations in the population . y¯U α = N (1 — 2α) To estimate the trimmed mean, substitute fˆ, qˆ1, and qˆ2 for f , q1, and q2. q1≤y≤q2 yf

7.21 As stated in Section 7.1, the yi’s are the measurements on observation units.

If unit i is in stratum h, then wi = Nh/nh. To express this formally, let ½ 1 if unit i is in stratum h xhi = 0 otherwise.

Then we can write

ΣN w = H hx i nh hi h=1

and

Σ

Σ

yiwi ˆ( ) = i∈£ Σ y yf y wi i∈£

H Σ Σ yi (Nh/nh)xhi h=1 = i∈£ H ΣΣ (Nh/nh)xhi i∈£ h=1 H

Σ

Nh

= h=1 H Σ

Σ

(xhiyi/nh)

i∈£

Nh h=1

Σ (xhi/nh) i∈£

H Σ N h ȳ h

= h=1 H Σ

Nh

h=1

=

H Σ Nh h=1

N

y¯h.

7.22 For an SRS, wi = N/n for all i and Σ i∈£:yi=y fˆ(y) = Σ N n i∈£

N n .

163 Thus,

Σ

y2fˆ(y) = Σ yi , n y i∈£ Σ Σ yi yfˆ(y) = = y¯, n y

and

2

i∈£

·Σ ½Σ ¸2¾ N 2 ˆ ˆ yf (y) y f (y) — N —1 y y ½Σ 2 ¾ N yi = — y¯2 N —1 n i∈£ N Σ (yi — y¯)2 = N —1 n i∈£ N n—1 2 = s.

N —1 n

Sˆ2 =

If n < N , Sˆ2 is smaller than s2 (although they will be close if n is large).

7.23 We need to show that the inclusion probability is the same for every unit in S2. Let Zi = 1 if i ∈ S and 0 otherwise, and let Di = 1 if i ∈ S2 and 0 otherwise.

We have P (Zi = 1) = πi and P (Di = 1 | Zi = 1) ∝ 1/πi.

P (i ∈ S2) = P (Zi = 1, Di = 1) = P (Di = 1 | Zi = 1)P (Zi = 1) 1 π i = 1. ∝ πi 7.24 A rare disease affects only a few children in the population. Even if all cases belong to the same cluster, a disease with estimated incidence of 2.1 per 1,000 is unlikely to affect all children in that cluster.

7.25 (a) Inner-city areas are sampled at twice the rate of non-inner-city areas. Thus the selection probability for a household not in the inner city is one-half the selection probability for a household in the inner city. The relative weight for a non-inner-city household, then, is 2. (b) Let π represent the probability that a household in the inner city is selected.

Then, for 1-person inner city households, P (person selected | household selected) P (household selected) = 1 × π.

For k-person inner-city households, 1 P (person selected | household selected) P (household selected) = π. k Thus the relative weight for a person in an inner-city household is the number of adults in the household. The relative weight for a person in a non-inner-city household is 2 × (number of adults in household).

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CHAPTER 7. COMPLEX SURVEYS

The table of relative weights is: Number of adults 1 2 3 4 5

Inner city 1 2 3 4 5

Non-inner city 2 4 6 8 10

Chapter 8

Nonresponse 8.1 (a) Oversampling the low-income families is a form of substitution. One advantage of substitution is that the number of low-income families in the sample is larger. The main drawback, however, is that the low-income families that respond may differ from those that do not respond. For example, mothers who work outside the home may be less likely to breast feed and less likely to respond to the survey. (b) The difference between percentage of mothers with one child indicates that the weighting does not completely adjust for the nonresponse. (c) Weights were used to try to adjust for nonresponse in this survey. We can never know whether the adjustment is successful, however, unless we have some data from the nonrespondents. The response rate for the survey decreased from 54% in 1984 to 46% in 1989. It might have been better for the survey researchers to concentrate on increasing the response rate and obtaining accurate responses instead of tripling the sample size.

Because the survey was poststratified using ethnic background, age, and education, the weighted counts must agree with census figures for those variables. A possible additional variable to use for poststratification would be number of children.

8.2 (a) The respondents report a total of Σ

yi = (66)(32) + (58)(41) + (26)(54) = 5894

hours of TV, with Σ 2 yi = [65(15)2 + 66(32)2] + [57(19)2 + 58(41)2] + [25(25)2 + 26(54)2] = 291725.

Then, for the respondents,

s2

5894

= 39.3 150 291725 — (150)(39.3)2 = = 403.6 149

y¯ =

165

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CHAPTER 8.

and

sµ SE(y¯) =

1—

150 2000

¶

NONRESPONSE

403.6 = 1.58.

150

Note that this is technically a ratio estimate, since the number of respondents (here, 150) would vary if a different sample were taken. We are estimating the average hours of TV watched in the domain of respondents. (b) GPA Group 3.00–4.00 2.00–2.99 Below 2.00 Total X2

=

Respondents 66 58 26 150

Non respondents 9 14 27 50

Σ (observed — expected)2 expected cells

=

Total 75 72 53 200

[66 — (.75)(75)]2

+

[9 — (.25)(75)]2

+ ··· +

[27 — (.25)(53)]2

(.75)(75) (.25)(75) = 1.69 + 5.07 + 0.30 + 0.89 + 4.76 + 14.27 = 26.97

(.25)(53)

Comparing the test statistic to a χ2 distribution with 2 df, the p-value is 1.4× 10—6.

This is strong evidence against the null hypothesis that the three groups have the same response rates.

The hypothesis test indicates that the nonresponse is not MCAR, because response rates appear to be related to GPA. We do not know whether the nonresponse is MAR, or whether is it nonignorable. (c) i ΣΣ

3

SSB

=

n

(y¯i — y¯)2 = 9303.1

i=1 j=1

MSW = s2 = 403.6.

The ANOVA table is as follows: Source Between groups Within groups Total, about mean

df 2 147 149

SS 9303.1 59323.0 68626.1

MS 4651.5 403.6

F 11.5

p-value 0.0002

Both the nonresponse rate and the TV viewing appear to be related to GPA, so it would be a reasonable variable to consider for weighting class adjustment or poststratification. (d) The initial weight for each person in the sample is 2000/200=10. After increasing the weights for the respondents in each class to adjust for the nonrespondents, the

167 weight for each respondent with GPA ≥ 3 is initial weight ×

Sample Size 75 72 53 200

sum of weights for sample sum of weights for respondents

Number of Respondents (nR) 66 58 26 150

Weight for each Respondent (w) 11.36364 12.41379 20.38462

75(10) 66(10) = 11.36364. = 10 ×

y¯ 32 41 54

wn R ȳ 24000 29520 28620 82140

wnR 750 720 530 2000

Then tˆwc = 82140 and y¯wc = 82140/2000 = 41.07.

The weighting class adjustment leads to a higher estimate of average viewing time, because the GPA group with the highest TV viewing also has the most nonresponse. (e) The poststratified weight for each respondent with GPA ≥ 3 is wpost = initial weight ×

population count sum of respondent weights

= 10 ×

700 (10)(66)

= 10.60606.

Here, nR denotes number of respondents. nR

66 58 26 150

Population Count 700 800 500 2000

wpost 10.60606 13.79310 19.2307

y¯ 32 41 54

wpost ȳ n R 22400 32800 27000 82200

wpostnR 700 800 500 2000

The last column is calculated to check the weights constructed—the sum of the poststratified weights in each poststratum should equal the population count for that poststratum. tˆpost = 82140 and 82140 = 41.07. yˆpost = 2000 8.6 (a) For this exercise, we classify the missing data in the “Other/Unknown” category. Typically, raking would be used in situations in which the classification variables were known (and known to be accurate) for all respondents.

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CHAPTER 8.

Ph.D. Master’s Other/Unknown Industry Academia Government Other/Unknown

Population 10235 7071 1303 5397 6327 2047 4838

Respondents 3036 1640 325 1809 2221 880 91

NONRESPONSE

Response Rate (%) 30 23 25 34 35 43 19

These response rates are pretty dismal. The nonresponse does not appear to be MCAR, as it differs by degree and by type of employment. I doubt that it is MAR—I think that more information than is known from this survey would be needed to predict the nonresponse. (b) The cell counts from the sample are: PhD non-PhD

Industry 798 1011 1809

Academia 1787 434 2221

Other 451 520 971

3036 1965 5001

Other 1678.2 1934.9 3613.1

11297.1 7311.9 18609.0

The initial sum of weights for each cell are: PhD non-PhD

Industry 2969.4 3762.0 6731.4

Academia 6649.5 1614.9 8264.5

After adjusting for the population row counts (10235 for Ph.D. and 8374 for nonPh.D.) the new table is: PhD non-PhD

Industry 2690.2 4308.5 6998.7

Academia 6024.4 1849.5 7873.9

Other 1520.4 2216.0 3736.4

10235 8374 18609

Raking to the population column totals (Industry, 5397; Academia, 6327; Other, 6885) gives: PhD non-PhD

Industry 2074.6 3322.4 5397.0

Academia 4840.8 1486.2 6327.0

Other 2801.6 4083.4 6885.0

9717.0 8892.0 18609.0

As you can see, the previous two tables are still far apart. After iterating, the final table of the weight sums is:

169

PhD non-PhD

Industry 2239.2 3157.8 5397.0

Academia 4980.6 1346.4 6327.0

Other 3015.2 3869.8 6885.0

10235.0 8374.0 18609.0

The raking has dramatically increased the weights in the “Other” employment category.

To calculate the proportions using the raking weights, create a new variable weight.

For respondents with PhD’s who work in industry, weight = 2239.2/798 = 2.806.

For the question, “Should the ASA develop some sort of certification?” the estimated percentages are:

No response Yes Possibly No opinion Unlikely No

Without Weights 0.2 26.4 22.3 5.4 6.7 39.0

With Raking Weights 0.3 25.8 22.3 5.4 6.9 39.3

(c) I think such a conclusion is questionable because of the very high nonresponse rate. This survey is closer to a self-selected opinion poll than to a probability sample.

8.7 Discipline Literature Classics Philosophy History Linguistics Political Science Sociology

Response Rate (%) 69.5 71.2 73.1 71.5 73.9 69.0 71.4

Female Members (%) 38 27 18 19 36 13 26

The model implicitly adopted in Example 4.3 was that nonrespondents within each stratum were similar to respondents in that stratum.

We can use a χ2 test to examine whether the nonresponse rate varies among strata.

The observed counts are given in the following table, with expected counts in parentheses:

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Literature Classics Philosophy History Linguistics Political Science Sociology

Respondent 636 (651.6) 451 (450.8) 481 (468.6) 611 (608.9) 493 (475.0) 575 (593.2) 588 (586.8) 3835

Nonrespondent 279 (263.4) 182 (182.2) 177 (189.4) 244 (246.1) 174 (192.0) 258 (239.8) 236 (237.2) 1550

NONRESPONSE

915 633 658 855 667 833 824 5385

The Pearson test statistic is X2 =

Σ (observed —expected)2 expected

= 6. 8

cells

Comparing the test statistic to a χ26distribution, we calculate p-value 0.34. There is no evidence that the response rates differ among strata.

The estimated correlation coefficient of the response rate and the percent female members is 0.19. Performing a hypothesis test for association (Pearson correlation, Spearman correlation, or Kendall’s τ ) gives p-value > .10. There is no evidence that the response rate is associated with the percentage of members who are female.

8.12 (a) The overall response rate, using the file teachmi.dat, was 310/754=0.41. (b) As with many nonresponse problems, it’s easy to think of plausible reasons why the nonresponse bias might go either direction. The teachers who work many hours may be working so hard they are less likely to return the survey, or they may be more conscientious and thus more likely to return it. (c) The means and variances from the file teachnr.dat (ignoring missing values) are responses y¯ s2 V̂ ( ȳ )

hrwork 26 36.46 2.61 0.10

size 25 24.92 25.74 1.03

preprmin 26 160.19 3436.96 132.19

assist 26 152.31 49314.46 1896.71

The corresponding estimates from teachers.dat, the original cluster sample, are: ŷ¯r ˆ V (yˆ¯r )

hrwork 33.82 0.50

size 26.93 0.57

preprmin 168.74 70.57

assist 52.00 228.96

8.14 (a) We are more likely to delete an observation if the value of xi is small.

Since xi and yi are positively correlated, we expect the mean of y to be too big. (b) The population mean of acres92 is y¯U = 308582.

171 8.16 We use the approximations from Chapter 3 to obtain: , , N Σ ZiRiwiyi ,, ,, ΣN ( ZiRiwi — φi ) i=1 E [ y¯ˆ R ] = E 1 — i=1 N N Σ Σ ,, ZiRiwi φi , , , i=1

Σ

i=1

N

≈

φiyi

i=1 N

Σ i=1

≈

1

φi N Σ

N φ̄U i=1

φ iy i.

Thus the bias is N 1 Σ (φ — φ̄ )y i U i N φ¯ U i=1 1 ΣN = (φ — φ̄ )(y — y¯ ) i U i U N φ̄U i=1 1 Cov (φi, yi) ≈ ¯ φU

Bias [ŷ¯ ] ≈ R

8.17 The argument is similar to the previous exercise. If the classes are sufficiently large, then E [1/φ̃ c ] ≈ 1/φ̄ c .

8.19 V ( yˆ¯ w c )

"

# N N n1 1 Σ n2 1 Σ ZiRixiyi + ZiRi(1 — xi)yi = V n n1R i=1 n n2R i=1 #) ( " N N ¯ n2 1 Σ n1 1 Σ = E V ZRxy + ZiRi(1 — xi)yi¯Z1, . . . , ZN n n2R n n1R i=1 i i i i i=1 #) ( " N N Σ n2 1 Σ ¯ n 1 1 ZiRixiyi + +V E ZiRi(1 — xi)yi¯Z1, . . . , ZN n n2R n n1R i=1 i=1 ( " ΣN ΣN #) i i(1 — xi)yi ¯ Z ,...,Z = E V n1 i=1 ZiRixiyi + n2 i=1 Z R ¯ 1 n ΣN N n ΣN i=1 ZiRixi i=1 ZiRi(1 — xi) ( " ΣN ΣN #) i i(1 — xi)yi ¯ Z ,...,Z . +V E n1 i=1 ZiRixiyi + n2 i=1 Z R ΣN ¯ 1 N n ΣN n i=1 ZiRixi i=1 ZiRi(1 — xi)

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We use the ratio approximations from Chapter 4 to find the approximate expected values and variances. " Σ # n1 N ZiRixiyi ¯ i=1 Z ,...,Z E 1 N ¯ n ΣN i=1 ZiRixi à ! # " ΣN ΣN n1 ZiRixiyi Zi[Ri — φ1]xi ¯ i=1 i=1 1— Z ,...,Z = E ΣN ΣN N ¯ 1 n φ1 i=1 Zixi i=1 ZiRixi # "N ΣN Z [R — φ ]x ¯ N Σ Σ 1 i i 1 i Z ,...,Z 1 N = E Zi R i x i y i — ZiRixiyi i=1 Σ N ¯ nφ1 Z R x i=1 i i i i=1 i=1 1Σ 1 Σ N N ≈ Zxy — Z V (R )x y i i i i i i i 2 n (nφ1) i=1

i=1 N

≈

1Σ n i=1

Z ix i y i .

Consequently, (

"

=

³ n ´ S2y 1— N n ,

ΣN ΣN (1 — xi)yi ¯ #) n1 i=1 ZiRixiyi Z i Ri Z , . . . , Z n 2 i=1 + V E n ΣN ZiRixi N n ΣN — xi) ¯ 1 i=1 i=1 ZiRi(1 ) ( N N 1Σ 1Σ Zi(1 — xi)yi Z i xi y i + ≈ V n n i=1 i=1

the variance that would be obtained if there were no nonresponse. For the other term, " Σ # n1 N ZiRixiyi ¯ Z ,...,Z i=1 V Σ N 1 N n i=1 ZiRixi ¯ " ! # Ã ΣN N 1 Σ ¯ Z [R — φ ]x i i 1 i = V Z ,...,Z ZiRixiyi 1 — i=1 ΣN 1 N ¯ nφ1 i=1 Z R x i i i i=1 1 Σ Z iV (R i)x iy2i (nφ1)2 N

≈

i=1

≈

N

φ1(1 — φ1) Σ Zix iy2i. (nφ1)2 i=1

173 Thus, (

"

Σ ΣN #) i i(1 — xi)yi ¯ n1 Ni=1 ZiRixiyi Z ,...,Z n2 i=1 Z R + E V ¯ 1 N n ΣN n ΣN i=1 ZiRixi i=1 ZiRi(1 — xi) ( ) N N Σ φ1(1 — φ1) Σ φ (1 — φ ) 2 2 Zi(1 — xi)y2 i Zixiy2i + ≈ E 2 2 ( nφ1) (nφ2 ) N

=

i=1

N

i=1

φ1(1 — φ1) Σ 2 φ2(1 — φ2) Σ xi yi + (1 — xi)y2i .

2 nφ21 nφ 2 i=1 i=1

8.20 (a) Respondents are divided into 5 classes on the basis of the number of nights the respondent was home during the 4 nights preceding the survey call.

The sampling weight wi for respondent i is then multiplied by 5/(ki + 1). The respondents with k = 0 were only home on one of the five nights and are assigned to represent their share of the population plus the share of four persons in the sample who were called on one of their “unavailable” nights. The respondents most likely to be home have k = 4; it is presumed that all persons in the sample who were home every night were reached, so their weights are unchanged. (b) This method of weighting is based on the premise that the most accessible persons will tend to be overrepresented in the survey data. The method is easy to use, theoretically appealing, and can be used in conjunction with callbacks. But it still misses people who were not at home on any of the five nights, or who refused to participate in the survey. Since in many surveys done over the telephone, nonresponse is due in large part to refusals, the HPS method may not be helpful in dealing with all nonresponse. Values of k may also be in error, because people may err when recalling how many evenings they were home.

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NONRESPONSE

Chapter 9

Variance Estimation in Complex Surveys 9.1 All of the methods discussed in this chapter would be appropriate. Note that the replication methods might slightly overestimate the variance because sampling is done without replacement, but since the sampling fractions are fairly small we expect the overestimation to be small.

9.2 We calculate y¯ = 8.23333333 and s2 = 15.978, so s2/30 = 0.5326.

For jackknife replicate j, the jackknife weight is wj(j) = 0 for observation j and wi(j) = (30/29)wi = (30/29)(100/30) = 3.44828 for i/= j. Using the jackknife weights, we find y¯(1) = 8.2413, . . . , y¯(30) = 8.20690, so, by (9.8), Σ30 2 ˆ (y¯) = 29 [y¯ VJK (j) — y¯] = 0.5326054.

30 j=1

9.3 Here is the empirical cdf F̂ (y): Obs 1 2 3 4 5 6 7 8 9 10 11 12

y

COUNT

PERCENT

CUM_FREQ

CUM_PCT

2 3 4 5 6 7 8 9 10 12 14 15

3.3333 10.0000 3.3333 6.6667 16.6667 10.0000 10.0000 6.6667 10.0000 6.6667 6.6667 6.6667

3.3333 10.0000 3.3333 6.6667 16.6667 10.0000 10.0000 6.6667 10.0000 6.6667 6.6667 6.6667

3.333 13.333 16.667 23.333 40.000 50.000 60.000 66.667 76.667 83.333 90.000 96.667

3.333 13.333 16.667 23.333 40.000 50.000 60.000 66.667 76.667 83.333 90.000 96.667

epmf

ecdf

0.03333 0.10000 0.03333 0.06667 0.16667 0.10000 0.10000 0.06667 0.10000 0.06667 0.06667 0.06667

0.03333 0.13333 0.16667 0.23333 0.40000 0.50000 0.60000 0.66667 0.76667 0.83333 0.90000 0.96667

175

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CHAPTER 9. VARIANCE ESTIMATION IN COMPLEX SURVEYS 17

3.3333

3.3333

100.000

100.000

0.03333

1.00000

Note that F̂ (7) = .5, so the median is θˆ0.5 = 7. No interpolation is needed.

As in Example 9.12, F̂ (θ 1/2 ) is the sample proportion of observations that take on value at most θ1/2, so µ ¶ ³ n ´ 0.25862069 30 0.25862069 ˆ ˆ ˆ V [ F ( θ1/2 )] = 1 — = 1— = 0.006034483.

100 30 N n This is a small sample, so we use the t29 critical value of 2.045 to calculate q 2.045 V̂ [F̂ (θ̂ 1/2)] = 0.1588596.

The lower confidence bound is F̂ — 1(.5 — 0.1588596) = Fˆ — 1(0.3411404) and the upper confidence bound for the median is Fˆ — 1(.5 + 0.1588596) = F̂ — 1(0.6588596).

Interpolating, we have that the lower confidence bound is 5+

0.34114 — 0.23333 (6 —5) = 5 .6 0.4 — 0.23333

and the upper confidence bound is 8+

0.6588596 — 0.6 (9 —8) = 8 .8.

0.666667 — 0.6

Thus an approximate 95% CI is [5.6, 8.8].

SAS code below gives approximately the same interval: data srs30; input y @@; wt = 100/30; datalines; 8 5 2 6 6 3 8 7 10 14 3 4 17 10 ;

6 10 7 15 6 14 12 7

9 15 8 12

3 9

5

6

/* We use two methods. First, the "hand" calculations" */ /* Find the empirical cdf */ proc freq data=srs30; tables y / out = htpop_epmf outcum; weight wt; run;

177

data htpop_epmf; set htpop_epmf; epmf = percent/100; ecdf = cum_pct/100; run; proc print data=htpop_epmf; run; /* Find the variance of \hat{F}(median) */ data calcvar; set srs30; ui = 0; if y le 7 then ui = 1; ei = ui - .5; proc univariate data=calcvar; var ei; run; /* Calculate the stratified variance for the total of variable ei */ proc surveymeans data=calcvar total = 100 sum stderr; weight wt; var ei; run; /* Method 2: Use sas directly to find the CI */ proc surveymeans data=srs30 total=100 percentile=(25 50 75) nonsymcl; weight wt; var y; run; Quantiles Variable

Percentile

Estimate

Std Error

95% Confidence Limits

y

25% Q1 50% Median 75% Q3

5.100000 7.000000 9.833333

0.770164 0.791213 1.057332

2.65604792 5.8063712 5.64673564 8.8831609 7.16875624 11.4937313

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9.5 θˆ1 θˆ2 θˆ3 θˆ4 θˆ5 θˆ6 θˆ7

(a) Age 0.12447 0.09528 0.08202 0.21562 0.21660 0.07321 0.02402

(b) Violent 0.52179 0.43358 0.36733 0.37370 0.42893 0.48006 0.51201

(c) Bothpar 0.29016 0.31309 0.34417 0.25465 0.30181 0.30514 0.27299

(d) Male 0.90160 0.84929 0.99319 0.96096 0.91314 0.96786 0.96558

(e) Hispanic 0.30106 0.20751 0.17876 0.08532 0.14912 0.15752 0.25170

(f) Sinpar 0.55691 0.52381 0.51068 0.55352 0.54480 0.55350 0.54490

(g) Drugs 0.90072 0.84265 0.82960 0.80869 0.74491 0.82232 0.84977

θˆ θ˜ ˆ V 1(θˆ) Vˆ2(θˆ)

0.12330 0.11875 0.00076 0.00076

0.44325 0.44534 0.00055 0.00055

0.29743 0.29743 0.00012 0.00012

0.93119 0.93594 0.00036 0.00036

0.18877 0.19014 0.00072 0.00072

0.54108 0.54116 0.00004 0.00004

0.82821 0.82838 0.00032 0.00032

9.6 From Exercise 3.4, B̂ = 11.41946, ŷ¯r = t x B̂ = 10.3B̂ = 117.6, and SE ( y¯ˆ r ) = 3.98. Using the jackknife, we have B̂ ( · ) = 11.41937, y¯ˆ r ( · ) = 117.6, and SE ( yˆ¯ r ) = √ 10.3 0.1836 = 4.41. The jackknife standard error is larger, partly because it does not include the fpc.

9.7 We use

10 Σ ˆ (ŷ¯ ) = n — 1 VJK ( yˆ¯r(j) — ŷ¯r )2. r n j=1

The ŷ¯r ( j ) ’ s for returnf and hadmeas are given in the following table: School, j 1 2 3 4 5 6 7 8 9 10 For returnf,

returnf, ŷ¯ r ( j ) 0.5822185 0.5860165 0.5504290 0.5768984 0.5950112 0.5829014 0.5726580 0.5785320 0.5650470 0.5986785

hadmeas, ŷ¯ r ( j ) 0.4253903 0.4647582 0.4109223 0.4214941 0.4275614 0.4615285 0.4379689 0.4313120 0.4951728 0.4304341

9 Σ 10 Vˆ (ŷ¯ ) = ( yˆ¯ — 0.5789482)2 = 0.00160 JK r r(j) 10 j=1

For hadmeas, ˆ (ŷ¯r ) = 9 VJK 10

10 Σ j=1

(ŷ¯ r(j) — 0.4402907)2 = 0.00526

179 9.8 We have B̂ ( · ) = .9865651 and V̂ J K (B̂ ) = 3.707 × 10—5. With the fpc, the linearization variance estimate is V̂ L ( B̂ ) = 3.071 × 10—5; the linearization variance q 300 estimate if we ignore the fpc is 3.071 × 10—5/ 1 — 3078 = 3.232 × 10—5.

9.9 The median weekday greens fee for nine holes is θˆ = 12. For the SRS of size 120, (.5)(.5) = 0.0021.

V [Fˆ (θ 0.5 )] = 120 An approximate 95% confidence interval for the median is therefore √ √ [Fˆ — 1(.5 — 1.96 .0021), Fˆ — 1(.5 + 1.96 .0021)] = [Fˆ — 1(.4105), Fˆ — 1(.5895)].

We have the following values for the empirical distribution function: y ˆ F (y)

10.25 .3917

10.8 .4000

11 .4167

11.5 .4333

y F̂ (y)

13 .5250

14 .5417

15 .5833

16 .6000

12 .5167

Interpolating, ˆ—1( 4105) = 10 8 + .4105 — .4 (11 10 8) = 10 9 F . .

— . . .4167 — .4 and

.5895 — .5833 ˆ—1 = 15 + (16 —15) = 15.4.

F ( 5895) . .6 — .5833 Thus, an approximate 95% confidence interval for the median is [10.9, 15.4].

Note: If we apply the bootstrap to these data, we get 1 Σ ˆ∗ θr = 12.86 1000 1000 r=1

with standard error 1.39. This leads to a 95% CI of [10.1, 15.6] for the median. '' 9.13 remainder term µ is 2 ¶ ∫ x(a) Since h (t) = —2t,∫the x x a2 ( ) ''( ) = 2 ( ) = 2 2 + = ( )2 x — t h t dt — x — t dt — x — — ax — x—a .

2 2 a a

Thus, h(pˆ) = h(p) + h' (p)(p̂ — p) — (p̂ — p)2 = p(1 — p) + (1 — 2p)(p̂ — p) — (p̂ — p)2. (b) The remainder term is likely to be smaller than the other terms because it has (p̂ — p)2 in it. This will be small if pˆ is close to p.

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(c) To find the exact variance, we need to find V (pˆ—p̂ 2 ), which involves the fourth moments. For an SRSWR, X = npˆ ~Bin(n, p), so we can find the moments using the moment generating function of the Binomial: MX(t) = (pet + q)n So, E(X) = M ' (t) X

E(X2)

¯t=0

= n(pet + q)n—1pet

¯t=0

= np

=

M '' (t)

=

[n(n — 1)(pet + q)n—2(pet)2 + n(pet + q)n—1pet] ¯

X

¯t=0 t=0

= n(n — 1)p2 + np = n2p2 + np(1 — p) E(X3) = MX'''(t) ¯

= np(1 — 3p + 3np + 2p2 — 3np2 + n2p2)

t=0

E(X4) = np(1 — 7p + 7np + 12p2 — 18np2 + 6n2p2 — 6p3 + 11np3 — 6n2p3 + n3p3) Then, V [pˆ(1 — pˆ)] = V (pˆ) + V (p̂2 ) — 2Cov (pˆ, p̂ 2 ) £ ¤ = E [p̂ 2] — p 2+ E [ p̂ 4] — [E (p̂2 )]2 — 2E pˆ3 + 2pE (p̂2 ) p(1 — p) = n p + 3 (1 — 7p + 7np + 12p2 — 18np2 + 6n2p2 — 6p3 + 11np3 — 6n2p3 + n3p3) n· ¸2 — p2 + p(1 — p) n · ¸ p p(1 — p) 2 2 2 2 2 —2 (1 — 3p + 3np + 2p — 3np + n p ) + 2p p + n2 n p(1 — p) = (1 — 4p + 4p2) n 1 1 + 2 (—2p + 14p2 — 22p3 + 12p4) + 3(p — 7p2 + 12p3 — 6p4) n n Note that the first term is (1 — 2p)2V (pˆ)/n, and the other terms are (constant)/n2 and (constant)/n3. The remainder terms become small relative to the first term when n is large. You can see why statisticians use the linearization method so frequently: even for this simple example, the exact calculations of the variance are nasty.

181 Note that with an SRS without replacement, the result is much more complicated.

Results from the following paper may be used to find the moments.

Finucan, H. M., Galbraith, R. F., and Stone, M. (1974). Moments Without Tears in Simple Random Sampling from a Finite Population Biometrika, 61, 151–154.

9.14 (a) Write B1 = h(txy, tx, ty, tx2 , N ), where h(a, b, c, d, e) =

a — bc/e d — b2/e

=

ea — bc ed — b2

.

The partial derivatives, evaluated at the population quantities, are: ∂h = ∂a

e ed — b2

∂h c ea — bc = — + 2b 2 ∂b ed — b (ed — b2)2 c 2bB1 = — + ed — b2 ed — b2 e ·c b ¸ = — — B — bB 1 1 e ed — b2 e e e (B — B x̄ ) 0 1 U = — ed — b2 e ∂h b =— x̄ = — 2 ed — b2 U ∂c ed — b ∂h e B1 = — ∂d ed — b2 ∂h a d(ea — bc) = — 2 ∂e ed — b (ed — b2)2 a dB1 = — 2 ed — b ed — b2 e B0 x̄ U . = ed — b2

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The last equality follows from the normal equations. Then, by linearization, B̂1 — B1 ∂h ∂h ∂h ≈ (tˆxy — txy ) + (tˆx — tx) + (tˆy — ty) ∂a ∂b ∂c ∂h ∂h + (txˆ2 — tx2) + ( Nˆ — N ) ∂d ∂e £ N = tˆ — txy — (B0 — B1x̄ U)(tˆx — tx) Ntx2 — (tx)2 xy i ˆ ˆ —x̄ (t — t ) — B (t 2 — t 2 ) + B x̄ ( N̂ — N ) U y y 1 x 0 U x " # Σ © } N 2 = w x y — (B — B x̄ )x — x̄ y — B x + B x̄ i i i 0 1 U i U i 1 i 0 U Ntx2 — (tx)2 i∈£ N [txy — tx(B0 — B 1 x̄ U ) — x̄ U ty — B1tx2 + B0N x̄ U ] — 2 Ntx2 — (tx) Σ i i N = Ntx2 — (tx)2 w (y — B0 — B1 xi ) (xi — x̄U ). i∈£

9.15 (a) Write S =

N Σ

1

2

,

N — 1 i=1 yi — N

(b) Substituting, we have ˆ2 = S

N 1Σ

µ ¶ 1 t2 = t1 — t3 t3 — 1

2

yj

j=1

1

µ

tˆ2

¶

ˆ t1 — 2 tˆ3

tˆ3 — 1

(c) We need to find the partial derivatives: 1 t3 — 1

∂h ∂t1

=

∂h ∂t2 ∂h ∂t3

= —2

t2 µ t3(t3 — 1) 1 = — t (t3 — 1)2

1—

t2 t3

¶ +

1 t2 t3 — 1 t23

Then, by linearization, Sˆ2 — S2

≈

∂h ∂t1

(tˆ — t ) + 1

1

∂h ∂t2

(tˆ — t ) + 2

2

∂h

(tˆ — t )

∂t3

3

3

183 Let qi = = =

∂h ∂t1

yi2 +

∂h

∂h yi ∂t2 + ∂t3

µ ¶ 1 t2 1 t2 1 t2 y2i — 2 yi — 1 t3 — 1 µ t3(t3 — 1) (t3 — 1)2 t — t3 + t3 — 1 t23 µ ¶ ¶ 1 t2 t2 t2 1 y2i — 2 y i — 1— 2 t + (t3 — 1) t3 — 1 t t t 3

3

3

9.16 (a) Write R = h(t1, . . . , t6), where d — ab/f fd — ab (a, b, c, d, e, f ) = = . √ √ h (c — a2/f )(e — b2/f ) (fc — a2)(fe — b2) The partial derivatives, evaluated at the population quantities, are: µ ¶ ∂h 1 a(fd — ab) —b + = √ ∂a fc — a2 (fc — a2)(fe — b2) —ty txR = N (N — 1)S S + N (N — 1)S2 x y x µ ¶ ∂h 1 b(fd — ab) = √ —a + fe — b2 ∂b (fc — a2)(fe — b2) —ty tyR = + N (N — 1)SxSy N (N — 1)S2y µ ¶ ∂h 1 f (fd — ab) = —√ ∂c fc — a2 2 (fc — a2)(fe — b2) R 1 = — 2 (N — 1)S2x ∂h f √ 2 ∂d = (fc — a )(fe — b2) µ ¶ ∂h 1 f (fd — ab) = —√ ∂e fe — b2 2 (fc — a2)(fe — b2) R 1 = — 2 (N — 1)S2y µ ¶ ∂h d fd — ab e c — √ = √ + ∂f (fc — a2)(fe — b2) 2 (fc — a2)(fe — b2) fe — b2 fc — a2 µ ¶ ty2 txy R tx2 = — + 2 N (N — 1)SxSy 2 N (N — 1)S y N (N — 1)S2x

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CHAPTER 9. VARIANCE ESTIMATION IN COMPLEX SURVEYS

Then, by linearization, R̂ — R ≈

=

=

∂h ˆ ∂h ˆ ∂h ˆ (tx — tx ) + (t y — ty ) + (t x2 — tx2 ) ∂a ∂b ∂c ∂h ˆ ∂h ˆ ∂h ˆ + (txy — txy·µ )+ (t y2 — t y2) + ( N — N) ¸ ∂d 1 ∂e t RS ¶ ∂f tyRSx x y ˆ ˆ —t + (t — t ) + (—t + )(t — t ) y x x x y y Sx Sy N (N — 1)SxSy R 1 1 — (tˆ — t ) + (tˆ — t ) 2 2 xy xy x x 2 2 (N — 1)Sx (N — 1)SxSy R 1 — (tˆ — t ) y2 2 2 (N — 1)Sy y 2 · µ ¶¸ ty2 txy R tx2 + N (N — 1)S — + ·µ ¶ — 1)S2y 2 N (N Nµ(N — 1)S2x ¶(N̂ — N ) xSy 1 txRSy tyRSx —t + (tˆ — t ) + —t + (tˆ — t ) y x x x y y Sx Sy N (N — 1)SxSy NRSy ˆ NRS x — (t — t ) + N (tˆ — t ) — (tˆ — t ) x2 x2 y2 y2 xy xy 2S¸y ½2Sx µ ¶¾ R ty2 Sx tx2 Sy + txy — + ( Nˆ — N ) 2 Sy Sx

This is somewhat easier to do in matrix terms. Let ·µ ¶ µ ¶ x̄ U RSy y¯U RSx RSy RSx δ = —y¯U + , —x̄ U + ,— ,— , 1, 2Sx 2Sy Sx Sy µ ¶¸T ty2 Sx tx2 Sy txy R — + , N 2N Sy Sx then

ˆ Cov (R) ≈

1

T

δ [(N — 1)S xS y]2

ˆ Cov (t)δ.

9.17 Write the function as h(a1, . . . , aL, b1, . . . , bL). Then ∂h ¯ ∂aL ¯t1,...,tL,N1,...,NL and

∂h ¯ ∂bL ¯t1,...,tL,N1,...,NL

=1

=—

tL .

NL

Consequently, L

L

L=1

L=1

Σ Σ tL h(t1ˆ , . . . , tLˆ , N̂1 , . . . , N̂L ) ≈ t + (tˆ L— t L) — ( Nˆ L — N L) NL

185 and

·Σ ¶¸ L µ t ˆ ˆ post) ≈ V V (t tˆL — L NL .

NL L=1

9.18 From (9.5),

R

1 1 Σ ˆ Vˆ2(θˆ) = (θ r— θˆ)2.

RR—1 r=1

Without loss of generality, let y¯U = 0. We know that y¯ =

ΣR r=1

ȳ r /R.

Suppose the random groups are independent. Then y¯1, . . . , y¯R are independent and identically distributed random variables with E [ȳ r ] = 0, S2 = κ2 (¯y1 ), m E[ȳr4 ] = κ4 (ȳ1 ).

V [ȳ r ] = E[ȳ2r ] =

We have E"

#

2 1 ( 1) Σ(y¯r — ȳ) R R — r=1

R

=

£ 2 ¤ R 2 ( 1 1) Σ E y¯ — (ȳ) r R R — r=1 R

= = = =

Σ 1 [V (ȳr ) — V (y¯)] R(R — 1) r=1 Σ R · 2 S S2 ¸ 1 — n R(R — 1) r=1 m ¸ R · Σ 1 S2 S2 R — R(R — 1) n n r=1 S2 . n

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CHAPTER 9. VARIANCE ESTIMATION IN COMPLEX SURVEYS

Also, E

( R Σ

)2 (y¯r — y¯)2

r=1

= E

( R Σ

)2 y¯r2 — Ry¯2

r=1 # Σ " R Σ R 2 2 2 R 2 2 4 = E Σ ȳ ȳ — 2Ry¯ y¯ + R ȳ r=1 s=1

= EΣ

r s

µ Σ R

R

r=1 s=1

r=1

r

¶ 2 1 ΣΣΣΣ y¯2 y¯2 — ȳ 2 y¯2 + y¯jy¯ky¯ry¯s r s R2 R r s r s j

k

µ ¶ R µ ¶ R R 2 1 Σ 4 2 3 ΣΣ = E 1— + y¯ + 1 — + 2 y¯r2y¯s2 R R2 r=1 r R R r=1 s/=r ¶ µ ¶ µ 3 2 1 2 + 2 R(R — 1)κ 22(ȳ 1) = 1 — + R2 Rκ (y¯ 4 1) + 1 — R R R Consequently, h i E V̂ 2 (θˆ) 2

¶

"

Σ 1 (y¯r — y¯)2 1) R(R — R

# =

r=1

=

" CV

µ ¶ ¸ 2 3 2 Rκ (y¯ ) + 1 — + R(R — 1)κ (ȳ )

R2(R — 1)2

=

V

1

2 + 1— 4 1 R R2 1 1 κ (y¯ ) + (R2 — 2R + 3)κ 2 (ȳ ) 4 1 2 1 R3 R3(R — 1) 1

=

and

·µ

1

R(R — 1) r=1

(y¯r — y¯)2

1

µ 2 ¶2 S 1 R2 — 2R + 3 2 κ (ȳ ) + κ (ȳ ) — 4 1 1 2 3 n R R 3 (R — 1) µ 2 ¶2 µ 2 ¶2 2 1 R — 2R + 3 S S κ (ȳ ) + — 4 1 R3 R3(R — 1) m Rm

= =

2

R2

R2 — 2R + 3 13 κ 4 (ȳ 1 ) + 3 R R (R — 1)

#

Σ R

2

R

· 1 κ 4 (ȳ1 )m 2 R S4

µ 2 ¶2 µ 2 ¶2 S S — m Rm

µ S2 ¶2 ¸ Rm R—3 — .

R—1

187 We now need to find κ 4 (ȳ 1 ) = E [ȳ 4r] to finish the problem. A complete argument giving the fourth moment for an SRSWR is given by Hansen, M. H., Hurwitz, W. N., and Madow, W. G. (1953). Sample Survey Methods and Theory, Volume 2. New York: Wiley, pp. 99-100.

They note that y¯4 =

r

1 Σ m4

y4 + 4

Σ

i/=j

i∈£r

Σ

κ (y¯ ) = E [ȳ 4 ] = 4

1

r

y2y2

i j

i/=j

yi2 yjyk +

i/=j/=k

so that

Σ

i j

i

+6

y3y + 3

Σ yiyjykyL i/=j/=k/=L

ΣN m—1 4 (y — y¯ )4 + 3 S. i U m3 m3(N — 1) 1

i=1

This results in

#

" CV 2

Σ(y¯r — ȳ)2 1 ( 1) R R — r=1 R

= =

· ¸ 1 κ 4 (ȳ1 )m 2 R —3 — R S4 R —1 · ¸ 1 κ m—1 R—3 +3 — .

R m m3 R—1

The number of groups, R, has more impact on the CV than the group size m : the random group estimator of the variance is unstable if R is small.

9.19 First note that y¯ (α ) — y¯ str

r

str

ΣN Σ N y + yh2 = H h y (α ) — H h h1 N 2 N h r h=1 ¶ Σ h=1 Σ H N µα + 1 H N y +y α 1 h rh h h1 h2 = yh2 — yh1 — rh N N 2 2 2 h=1 h=1 H Σ Nh yh1 — yh2 = αrh .

2 N h=1

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CHAPTER 9. VARIANCE ESTIMATION IN COMPLEX SURVEYS

Then ΣR ˆ (y¯str ) = 1 [y¯ str (α r) — y¯str ]2 VBRR R r=1 R ·Σ H Σ Nh y — yh2 ¸2 = 1 αrh h1 R N 2 r=1 R Hh=1H

yh1 — yh2 Nl y — yl2 1 Σ Σ Σ Nh α rl l1 αrh 2 2 N R r=1 h=1 l=1 N 2 µ ¶ 2 Σ Σ (y — y ) 1 R H N h h1 h2 2 = αrh R N 4

=

r=1Hh=1 H

=

R

1 Σ Σ Nh yh1 — yh2 Nl yl1 — yl2 Σ + αrhαrl 2 2 N R h=1 l=1,l/=h N r=1 ¶ H µ Σ N 2 (y — y )2 h

h1

h=1

h2

4

N

= Vˆstr (ȳ str ).

The last step holds because

ΣR

= h. r=1 αrhαrl = 0 for l /

9.20 As noted in the text, V̂str (ȳstr ) =

H µ ¶ Σ Nh 2 (yh1 — yh2)2

Also,

4

N

h=1

.

Σ H αrh Nh (y — y ) + y¯ θˆ(α ) = y¯ (α ) = h2 r str r str 2 N h1 h=1

so

H

θˆ(αr ) — θˆ(—α r )= and, using the property

ΣR r=1

R

1 Σ ˆ [θ (α )r— θˆ(—αr)]2 4R r=1

Σ αrh

h=1

Nh (y — yh2 ) N h1

αrhαrk = 0 for k /= h, R

H

H

=

1 ΣΣΣ N N α rhα rk h k (yh1 — yh2 )(y k1 — yk2 ) N N 4R r=1 h=1 k=1

=

1 ΣΣ 4R R

H

r=1 h=1

=

µ

Nh N

H 1 Σ µ Nh ¶2

4

h=1

N

¶2

(yh1 — yh2)2

(yh1 — yh2)2 = V̂ str (ȳ str ).

189 Similarly, 1 ΣR ˆ ˆ2 ˆ ˆ2 {[θ (α ) — r θ ] + [θ (—α r) — θ ] } 2R r=1

= = =

H R ½· Σ 1 Σ α

¸2

·Σ H

rh Nh

+ 2 N (yh1 — yh2) µ ¶2 R H 2 Nh 1 Σ Σ αrh (yh1 — yh2)2 2 2R N 2R r=1

h=1

r=1 h=1 H 1 Σ µ Nh ¶2

4

N

h=1

h=1

—α 2

¸2¾

N rh

h

N

(yh1 — yh2)

(yh1 — yh2)2.

9.21 Note that tˆ(αr) =

·

H Σ

Nh h=1 H

=

αrh (y — yh2) + y¯h 2 h1

¸

Σ Nharh — y ) + tˆ h2 2 (yh1 h=1

and

[tˆ(α )]2 =

H Σ H Σ NhNkαrhαrk

r h=1 k=1 H

(y h1

4

— y )(y h2

k1

—y ) k2

Σ Nhαrh +2tˆ (yh1 — yh2) + tˆ2.

2 h=1

Thus, tˆ(αr) — tˆ(—αr) =

H

Σ

Nhαrh(yh1 — yh2),

h=1

Σ [tˆ(αr)]2 — [tˆ(—αr)]2 = 2tˆ Nhαrh(yh1 — yh2), H

h=1

and

Σ θˆ(αr) — θˆ(—αr) = (2atˆ+ b) Nhαrh(yh1 — yh2).

H

h=1

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CHAPTER 9. VARIANCE ESTIMATION IN COMPLEX SURVEYS

Consequently, using the balanced property

ΣR

= h, we have r=1 αrhαrk = 0 for k /

1 ΣR ˆ [θ (α )r— θˆ(—αr)]2 4R r=1

= =

R

H

H

ΣΣΣ 1 ˆ+ b)2 NhNk αrh αrk (yh1 — yh2 )(yk1 — yk2) (2at 4R r=1 h=1 k=1 H Σ 1 2 (2atˆ+ b) N 2(y — y )2. h2 h h1 4 h=1

Using linearization,

h(tˆ) ≈ h(t) + (2at + b)(tˆ— t),

so

VL(θˆ) = (2at — b)2V (tˆ)

and

H

1Σ

VLˆ (θˆ) = (2atˆ— b) N h2(y h1 — yh2 )2, 4 h=1 Σ which is the same as 1 R [θˆ(αr) — θˆ(—αr)]2.

2

4R

r=1

9.23 We can write ˆ = (w, y, x , . . . , x ) = tpost g 1 L

Σ Σ NL wjxLjyj L j∈£ Σ . wx j Lj

L=1

j∈£

Then, zi =

∂g(w, y, x1, . . . , xL) ∂wi

,

Σ NLxLi wjxLjyj , , , Nx y L , , , L Li i — , j∈£ Σ Σw x 2 Σ = L=1 j Lj , , j∈£ wjxL j , , j∈£ , , ) ( L Σ N Lx Liyi NLxLitˆyL = — Nˆ 2 N̂L L L=1 L µ ¶ Σ NL tˆ yi — yL . x Li = N̂L N̂ L L=1

191 Thus, V̂ (t̂post ) =

à ! Σ Vˆ wizi . i∈£

Note that this variance estimator differs from the one in Exercise 9.17, although they are asymptotically equivalent.

9.24 From Chapter 5, V (tˆ) ≈ N 2

M MSB

n N 2M NM — 1 2 = S [1 + (M — 1)ICC] n M (N — 1) NM NM ≈ p(1 — p)[1 + (M — 1)ICC] n M

Consequently, the relative variance v can be written as β0 + β1/t, where β0 = — 1 [1 + (M — 1)ICC] + 1 N [1 + (M — 1)ICC]. nM

nt

9.25 (a) From (9.2),

"½ ¾2# t 1 y ˆ ≈ E ˆ x — tx) + (t ˆ y — ty) V [B] — 2 (t tx tx "½ ¾2# ty2 1 ˆ 1 ˆ = 2E — (tx — tx) + (ty — ty) tx tx ty · µ ¶¸ ty2 V (txˆ ) V (tyˆ tˆx , tˆy = + 2 — 2Cov t2x ty tx ty t2x = =

· ¸ ¡ ¢ t2y V (tˆx) V (tˆy B ˆ — V tx 2 t2x + t2y tx ty t2x · ¸ t2y V (tyˆ V (tˆx) — 2 t2y tx t2x

Using the fitted model from (9.13), V̂ (tˆx ) txˆ2 and

= a + b/tˆ x

Vˆ (tˆy) = a + b/tˆ y tˆ2 y

Consequently, substituting estimators for the population quantitites, · ¸ b b 2 ˆ V [ B̂ ] = B̂ a+ —a— , tˆy tˆx

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192

CHAPTER 9. VARIANCE ESTIMATION IN COMPLEX SURVEYS

which gives the result. (b) When B is a proportion, · V̂ [B̂] = B̂

2

¸ · ¸ b b b b bB̂ (1 — B̂ ) 2 — = B̂ — = . tˆy tˆx tˆx tˆx t̂x B̂

Chapter 10

Categorical Data Analysis in Complex Surveys 10.1 Many data sets used for chi-square tests in introductory statistics books use dependent data. See Alf and Lohr (2007) for a review of how books ignore clustering in the data.

10.3 (a) Observed and expected (in parentheses) proportions are given in the following table: Abuse No

.7542 (.7109)

.1017 (.1451)

.0763 (.1196)

.0678 (.0244)

Symptom

(b)

2

X G2

· ¸ (.7542 — .7109)2 (.0678 — .0244)2 = 118 + ···+ .7109 .0244 = 12.8 · µ ¶ µ ¶¸ .7542 .0678 = 2(118) .7542 ln + · · · + .0678 ln .7109 .0244 = 10.3.

Both p-values are less than .002.

Because the expected count in the Yes-Yes cell is small, we also perform Fisher’s exact test, which gives p-value .0016.

10.4 (a) This is a test of independence. A sample of students is taken, and each student classified based on instructors and grade. (b) X2 = 34.8. Comparing this to a χ2 3distribution, we see that the p-value is 193

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194CHAPTER 10. CATEGORICAL DATA ANALYSISIN COMPLEX SURVEYS less than 0.0001. A similar conclusion follows from the likelihood ratio test, with G2 = 34.5. (c) Students are probably not independent–most likely, a cluster sample of students was taken, with the Math II classes as the clusters. The p-values in part (b) are thus lower than they should be.

10.5 The following table gives the value of θˆ for the 7 random groups: Random Group 1 2 3 4 5 6 7 Average std. dev.

θˆ 0.0132 0.0147 0.0252 -0.0224 0.0073 -0.0057 0.0135 0.0065 0.0158

√ Using the random group method, the standard error of θˆ is 0.0158/ 7 = 0:0060, so the test statistic is θˆ2 = 0 79 . .

V (θˆ) Since our estimate of the variance from the random group method has only 6 df, we compare the test statistic to an F (1, 6) distribution rather than to a χ21 distribution, obtaining a p-value of 0.4.

10.6 (a) The contingency table (for complete data) is as follows:

Faculty Classified staff Administrative staff Academic professional

Break again?

No Yes 65 167 55 459 11 75 9 58 140 759

232 514 86 67 899

X2 = 37.3; comparing to a χ2 distribution gives ρ-value < .0001. We can use the p

3

χ2 test for homogeneity because we assume product-multinomial sampling. (Class is the stratification variable.) (b) Using the weights (with the respondents who answer both questions), we estimate the probabilities as

195

No

Work No Yes 0.0832 0.0859

0.1691

0.6496 0.7328

0.8309 1.0000

Breakaga Yes

0.1813 0.2672

To estimate the proportion in the Yes-Yes cell, I used: sum of weights of persons answering yes to both questions pˆyy = . sum of weights of respondents to both questions Other answers are possible, depending on how you want to treat the nonresponse. (c) The odds ratio, calculated using the table in part (b), is 0.0832/0.0859 = 0.27065.

0.6496/0.1813 (Or, you could get 1/.27065 = 3.695.)

The estimated proportions ignoring the weights are

No

Work No Yes 0.0850 0.0671

0.1521

0.6969 0.7819

0.8479 1.0000

breakaga Yes

0.1510 0.2181

Without weights the odds ratio is 0.0850/0.0671 = 0.27448 0.6969/0.1510 (or, 1/.27448 = 3.643).

Weights appear to make little difference in the value of the odds ratio. (d) θˆ = (.0832)(.1813) — (.6496)(.0859) = —0.04068. (e) Using linearization, define qi = p̂22 y11i + p̂11 y12i — p̂12 y21i — p̂21 y22i where yjki is an indicator variable for membership in class (j, k). We then estimate V (q¯str) using the usual methods for stratified samples. Using the summary statistics, ´ Faculty C.S. A.S. A.P. Total

1374 1960 252 95 3681

228 514 86 66 894

—.117 —.059 —.061 —.076

0.0792 0.0111 0.0207 0.0349

4.04 × 10—5 4.52 × 10—6 7.42 × 10—7 1.08 × 10—7 4.58 × 10—5

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196CHAPTER 10. CATEGORICAL DATA ANALYSISIN COMPLEX SURVEYS Thus V̂ L (θ̂ ) = 4.58 × 10—5 and 2

XW =

0.00165 θˆ2 = = 36.2.

Vˆ L(θˆ) 4.58 × 10—5

We reject the null hypothesis with p-value < 0.0001.

10.7 Answers will vary, depending on how the categories for zprep are formed.

10.8 (a) Under the null hypothesis of independence the expected proportions are:

Smoking Status

Recommended .241 .020 .186

Current Occasional Never

Fitness Level Minimum Acceptable .140 .011 .108

Unacceptable .159 .013 .123

Using (10.2) and (10.3), r

X

2

=n

c

Σ Σ (pˆij — pˆi + p̂+j )2 i=1 j=1 r Σ c Σ

G2 = 2n

= 18.25 pˆi + pˆ+ j µ ¶ pˆij pˆij ln p̂i+ p̂+j = 18.25

i=1 j=1

Comparing each statistic to a χ24distribution gives p-value= .001. (b) Using (10.9), E[X2] ≈ E[G2] ≈ 6.84 4X2 (c) 2 2 2 XF = G F = = 10.7, with p-value = .03 (comparing to a χ4 distribution).

6.84 10.9 (a) Under the null hypothesis of independence, the expected proportions are: Males 0.076 0.018 0.064 0.103 0.279

Decision-making managers Advisor-managers Supervisors Semi-autonomous workers Workers

Females 0.065 0.016 0.054 0.087 0.238

Using (10.2) and (10.3), r

X

2

=n

c

Σ Σ (pˆij — p̂i+ p̂+j )2 i=1 j=1 r Σ c Σ

G2 = 2n

= 55.1 pˆi + pˆ+ j µ ¶ pˆij pˆij ln p̂i+ p̂+j = 56.6

i=1 j=1

Comparing each statistic to a χ2 distribution with (2 — 1)(5 — 1) = 4 df gives “p-values” that are less than 1 × 10—9.

197 (b) Using (10.9), we have E[X2] ≈ E[G2] ≈ 4.45 (c) df = number of psu’s — number of strata = 34 (d) XF2 = 2

GF

=

4X2 = .899X2 = 49.5 4.45 2 4G 4.45

= 50.8

The p-values for these statistics are still small, less than 1.0 × 10—9. (e) The p-value for X2s is 2.6 × 10—8, still very small.

10.11 Here is SAS code and output: options ovp nocenter ls=85; filename nhanes ’C:\nhanes.csv’; data nhanes; infile nhanes delimiter=’,’ firstobs=2; input sdmvstra sdmvpsu wtmec2yr age ridageyr riagendr ridreth2 dmdeduc indfminc bmxwt bmxbmi bmxtri bmxwaist bmxthicr bmxarml; bmiclass = .; if 0 > bmxbmi and bmxbmi < 25 then bmiclass = 1; else if bmxbmi >= 25 and bmxbmi < 30 then bmiclass = 2; else if bmxbmi >= 30 then bmiclass = 3; if age < 30 then ageclass = 1; else if age >= 30 then ageclass = 2; label age = "Age at Examination (years)" riagendr = "Gender" ridreth2 = "Race/Ethnicity" dmdeduc = "Education Level" indfminc = "Family income" bmxwt = "Weight (kg)" bmxbmi = "Body mass index" bmxtri = "Triceps skinfold (mm)" bmxwaist = "Waist circumference (cm)" bmxthicr = "Thigh circumference (cm)" bmxarml = "Upper arm length (cm)"; run; proc surveyfreq data=nhanes ; stratum sdmvstra; cluster sdmvpsu;

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198CHAPTER 10. CATEGORICAL DATA ANALYSISIN COMPLEX SURVEYS weight wtmec2yr; tables bmiclass*ageclass/chisq deff; run; Table of bmiclass by ageclass

bmiclass

ageclass

Frequency

Weighted Frequency

Std Dev of Wgt Freq

Percent

Std Err of Percent

Design Effect

1

1 2

881 75

9566761 2686500

716532 495324

6.0302 1.6934

0.3131 0.2434

0.8572 1.7639

Total

956

12253261

1083245

7.7236

0.3951

1.0855

1 2

788 1324

19494074 57089320

1615408 4300350

12.2878 35.9853

0.6689 1.2334

2.0573 3.2727

Total

2112

76583394

5452762

48.2731

1.2969

3.3381

1 2

627 1262

15269676 54539886

1538135 4631512

9.6250 34.3783

0.6261 1.2349

2.2338 3.3499

Total

1889

69809562

5816243

44.0033

1.4066

3.9794

1 2

2296 2661

44330511 114315706

3336974 8538003

27.9430 72.0570

1.0011 1.0011

2.4666 2.4666

Total

4957

158646217

11335366

100.000

2

3

Total

Frequency Missing = 4686 Rao-Scott Chi-Square Test Pearson Chi-Square Design Correction

525.1560 1.6164

Rao-Scott Chi-Square DF Pr > ChiSq

324.8848 2 <.0001

F Value Num DF Den DF Pr > F

162.4424 2 30 <.0001

There is strong evidence of an association.

10.12 Here is SAS code and output: data ncvs; infile ncvs delimiter = "," firstobs=2; input age married sex race hispanic hhinc away employ numinc

199 violent injury medtreat medexp robbery assault pweight pstrat ppsu; if violent > 0 then isviol = 1; else if violent = 0 then isviol = 0; run; proc surveyfreq data=ncvs; stratum pstrat ; cluster ppsu; weight pweight; tables isviol*sex/chisq; run;

Table of isviol by sex

isviol

sex

Frequency

Weighted Frequency

Std Dev of Wgt Freq

Percent

Std Err of Percent

0

0 1

36120 42161

107752330 115149882

1969996 1906207

47.6830 50.9566

0.1620 0.1638

Total

78281

222902212

3813397

98.6396

0.0663

0 1

558 441

1746669 1327436

101028 81463

0.7729 0.5874

0.0445 0.0341

Total

999

3074105

155677

1.3604

0.0663

0 1

36678 42602

109498999 116477318

1985935 1930795

48.4560 51.5440

0.1597 0.1597

Total

79280

225976317

3853581

100.000

1

Total

Frequency Missing = 80

Rao-Scott Chi-Square Test Pearson Chi-Square Design Correction

30.6160 1.0466

Rao-Scott Chi-Square DF Pr > ChiSq

29.2529 1 <.0001

F Value Num DF Den DF Pr > F

29.2529 1 143 <.0001

There is strong evidence that males are more likely to be victims of violent crime

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200CHAPTER 10. CATEGORICAL DATA ANALYSISIN COMPLEX SURVEYS than females.

10.13 This test statistic does not in general give correct p-values for data from a complex survey. It ensures that the sum of the “observed” counts is n but does not adjust for stratification or clustering.

To see this, note that for the data in Example 10.4, the proposed test statistic is the same as X2 because all weights are equal. But in that example X2/2, not X2, has a null χ21 distribution because of the clustering.

10.14 (a) For the Wald test, θ = p11p22 — p12p21 and

θˆ = p̂11 p̂22 — p̂12p̂21 .

Then, using Taylor linearization, θˆ ≈ θ + p22 (p̂11 — p11) + p11 (p̂22 — p22) — p12 (p̂21 — p21) — p21 (p̂12 — p12) and VL(θˆ) = V [p22 p̂11 + p11 p̂ 22 — p12 p̂ 21 — p21 p̂12 ] = p2 V (p̂ 11 ) + p2 V (p̂ 22 ) + p2 V (p̂ 21 ) + p2 V (p̂ 12 ) 22

11

12

21

+2p11p22 Cov (p̂ 1 1 , p̂ 2 2 ) — 2p22p12 Cov (p̂ 1 1 , p̂ 2 1 ) —2p22p21 Cov (p̂ 11 , p̂ 1 2 ) — 2p11p12 Cov (p̂ 22 , p̂ 2 1 ) —2p11p21 Cov (p̂ 22 , p̂ 1 2 ) + 2p12p21 Cov (p̂ 21 , p̂ 12 ).

To estimate VL(θˆ), define yjki =

½

1 if unit i in cell (j, k) 0 otherwise

for j, k ∈ {1, 2} and let qi = p̂22 y11i + p̂11 y12i — p̂12 y21i — p̂21 y22i .

Then

V̂ L (θ̂) = V (qˆ¯).

(b) For multinomial sampling, V (θˆ) = p2 p11(1 — p11) + · · · + p2 p12(1 — p12) 21 L 22 n n p p p p 11 22 12 21 2 p2 p11 p2 p2 22 —2 +2 + · · · — 2 12 21 n n n 1 2 2 2 2 = {—4p11 p22 — 4p12 p21 + 8p11 p22 p12 p21 n +p11p22(p11 + p22) + p12p21(p12 + p21)}.

201 Under H0 : p11p22 = p12p21,

1 1 1 V (θˆ) = p p = p p = p p p p L 11 22 12 21 1+ +1 2+ +2 n n n and 1 1 p +p p +p 1 1 1 11 22 21 12 + + + = + = . p11 p12 p21 p22 p11p22 p12p21 p11p22 Thus, if H0 is true, µ ¶ 1 1 1 1 1 = n p11 + p12 + p21 + p22 VL(θˆ) µ ¶ 1 1 1 1 = n + + + . p1+p+1 p1+p+2 p2+p+1 p2+p+2 Also note that, for any j, k ∈ {1, 2}, θˆ2 = (p̂11 p̂ 22 — p̂12 p̂21 )2 = (p̂jk — p̂j+ p̂+k )2 .

Thus, estimating VL(θˆ) under H0, 2

2

2

Σ Σ (p̂ j k — p̂ j + p̂ + k ) X2W = n = X2p. p̂j+ p̂+k j=1 k=1

10.15 (a) We can rewrite θ = log(p11) + log(p22) — log(p12) — log(p21).

Then, using Taylor linearization,

and

pˆ11 — p11 pˆ22 — p22 pˆ12 — p12 pˆ21 — p21 θˆ ≈ θ + + — — p11 p22 p12 p21 µ ¶ pˆ11 pˆ22 pˆ12 pˆ21 V (θˆ) = V + — — L p p p p21 22 12 2 211 ΣΣ 1 = V (pˆjk) 2 p jk j=1 k=1 2 2 + Cov (pˆ , pˆ ) — Cov (pˆ , pˆ ) 11 22 11 12 p11p22 p11p12 2 2 — p11p21 Cov (p̂11 , p̂ 21 ) — p 12p22 Cov (p̂ 22 , p̂ 12 ) 2 — 2 Cov (p̂22 , p̂21 ) + p p p p Cov (p̂ 12 , p̂ 21 ).

22 21

12 21

(b) Under multinomial sampling, 2

2

Σ Σ pjk(1 — pjk) 4 VL(θˆ) = +n np 2jk j=1 k=1 µ ¶ 1 1 1 1 1 = + + + p12 p21 p22 n p11

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202CHAPTER 10. CATEGORICAL DATA ANALYSISIN COMPLEX SURVEYS µ ¶ 1 1 1 1 1 ˆ ˆ VL (θ ) = + + + n pˆ11 pˆ12 pˆ21 pˆ22 1 1 1 1 = . + + + x11 x12 x21 x22 This is the estimated variance given in Section 10.1.1. and

10.17 In a multinomial sample, all design effects are 1. From (10.9), under H0 r

E[X2] =

ΣΣ

c

r

c

Σ

Σ

i=1

j=1

(1 — pij) — (1 — pi+) — (1 — p+j)

i=1 j=1

= rc — 1 — (r — 1) — (c — 1) = (r — 1)(c — 1).

10.18 (a) Write

YT CY = YT Σ—1/2Σ1/2CΣ1/2Σ—1/2Y.

Since C is symmetric and positive definite, so is Σ1/2CΣ1/2 and we can write Σ1/2CΣ1/2 = PΛPT for an orthogonal matrix P and diagonal matrix Λ, where each diagonal entry of Λ is positive. Let U = PT Σ—1/2Y; then Y T CY = U TΛU =

k Σ

λiU i2.

i=1

Since Y ∼ N (0, Σ), U ∼ N (0, PT Σ—1/2ΣΣ—1/2P) = N (0, I), so Wi = U 2 ∼ χ2 i

and the Wi’s are independent.

1

(b) Using a central limit theorem for survey sampling, we know that V(θ̂ ) —1/2 (θ̂ — θ) has asymptotic N (0, I) distribution under H0 : θ = 0. Using part (a), then, T

T

θ̂ A—1 θ̂ = θ̂ V(θ̂) — 1/2V(θ̂)1/2A—1 V(θ̂)1/2 V(θ̂)—1/2 θ̂ has the same asymptotic distribution asΣ λiWi, where the λi’s are the eigenvalues of V(θ̂)1/2 A—1 V(θ̂)1/2 . (c) T

Σ

T

Σ

E[θ̂ A—1 θ̂ ] ≈

V [θ̂ A—1 θ̂ ] ≈ 2

λi , λi,

10.19 This sample is self-weighting, so the estimated cell probabilities are Male Female

S .3028 .2254 .5282

N .1056 .3662 .4718

.4084 .5916 1.0000

203 The variances under the (incorrect) assumption of multinomial sampling are: Male Female

S .001487 .001229 .001755

N .000665 .001634 .001755

.001701 .001701

We use the information in Table 10.1 to find the estimated variances for each cell and margin using the cluster sample. For the schizophrenic males, define tSMi = number of schizophrenic males in psu i, and similarly for the other cells. Then we use equations (5.4)–(5.6) to estimate the mean and variance for each cell. We have the following frequency data: Freq.

0 1 2 yˆ¯ Vˆ ( yˆ¯ )

SM 41 17 13 .3028 .0022

SF 45 20 6 .2254 .0015

NM 58 11 2 .1056 .0008

NF 34 22 15 .3662 .0022

M 30 24 17 .4085 .0022

F 17 24 30 .5915 .0022

S 24 19 28 .5282 .0026

N 28 19 24 .4718 .0026

Thus the estimated variances using the clustering are Male Female

S .002161 .001488 .002604

N .000796 .002209 .002604

.002244 .002244

and the estimated design effects are Male Female

S 1.453 1.210 1.484

N 1.197 1.352 1.484

1.319 1.319

Using equation (10.9), E[X2] is estimated by Σ2 Σ 2 Σ2 Σ2 (1 — p̂ij )dij — (1 — p̂i+ )dRi — (1 — p̂+j )dcj = 1.075. i=1 j=1

i=1

Then 2

XF =

j=1

17.89 X2 = = 16.7 1.07 1.07

with p-value < 0.0001.

10.20 Both statistics are very large. We obtain the Rao-Scott χ2 statistic is 2721, while the Wald test statistic is 4838. There is strong evidence that the variables are associated.

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204CHAPTER 10. CATEGORICAL DATA ANALYSISIN COMPLEX SURVEYS

Chapter 11

Regression with Complex Survey Data 11.3 The average score for the students planning a trip is y¯1 = 77.158730 and the average score for the students not planning a trip is y¯2 = 61.887218. Using SAS PROC SURVEYREG, we get y¯1 y¯ —2 = 15.27 with 95% CI [7.6247634, 22.9182608].

Since 0 is not in the CI, there is evidence that the domain means differ.

11.4 (a) From SAS, the fitted regression line for the truncated data set is data anthrop; infile anthrop firstobs=2 delimiter=","; input finger height ; one = 1; run; proc sort data=anthrop; by height; data anthrop1; /* Keep the lowest 2000 values in the data set */ set anthrop; if _N_ <= 2000; run; proc reg data = anthrop1; model height = finger; output out=regpred pred=predicted residual=resid; run; goptions reset=all; goptions colors = (black); 205

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CHAPTER 11. REGRESSION WITH COMPLEX SURVEY DATA

axis1 label=(’Left Middle Finger Length (cm)’) order = (10 to 13.5 by .5); axis2 label=(angle=90 ’Height (inches)’) order=(55 to 75 by 5); axis3 order=(55 to 75 by 5) major=none minor=none value=none; symbol interpol=join width=2 color = black; proc sort data=regpred; by finger height; run; proc means data=regpred noprint; by finger height; var one predicted resid; output out=circlepred sum=sumn sumpred sumresid mean = meanone meanpred meanresid; run; proc gplot data=circlepred; bubble height*finger=sumn/haxis=axis1 vaxis=axis2; plot2 meanpred*finger/haxis=axis1 vaxis =axis3 ; run;

Parameter Estimates

Variable

DF

Parameter Estimate

Standard Error

t Value

Pr > |t|

Intercept finger

1 1

43.78957 1.78861

0.80553 0.07094

54.36 25.21

<.0001 <.0001

The line is much flatter than one the on Figure 11.4.

207

(b) We use exactly the same code as before, except now we sort the data by finger instead of by height. proc sort data=anthrop; by finger; data anthrop2; set anthrop; if _N_ <= 2000; run; proc reg data = anthrop2; model height = finger; output out=regpred pred=predicted residual=resid; run; Parameter Estimates

Variable

DF

Parameter Estimate

Standard Error

t Value

Pr > |t|

Intercept finger

1 1

31.19794 2.96393

1.33050 0.11821

23.45 25.07

<.0001 <.0001

These values of the slope and intercept are quite close to the values given in Figure 11.4. The standard errors are larger, however, reflecting the lesser number of data points and the reduced spread of the x’s.

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CHAPTER 11. REGRESSION WITH COMPLEX SURVEY DATA

(c) Regression uses conditional expectation, conditional on x. Thus, if the model holds for all the observations, you should get unbiased estimates of the parameters if you take a subset of the data using x to define the mechanism.

11.5 We obtain B̂ 0 = 14.2725 and B̂ 1 = 0.08138. Using equation (11.8), with Σ Σ qi = (yi — Bˆ 0 — B̂ 1 x i )(xi — x̂¯) and x̂¯ = wixi/ wi = 180.541, we have ³Σ Σ ¸2 ˆ (ˆ ) Vˆ w q ´, ·Σ w x2 — ( wixi)2 = VL B1 Σ i i i i wi Σ ¶ wi qi µ Σ Σ = Vˆ Σ wix2 — ( wixi)2/ wi i = 0.000261 SE L (B̂ 1 ) = .016.

Here is output from SAS: data nybight; infile nybight delimiter="," firstobs=2; input year stratum catchnum catchwt numspp depth temp ; if stratum = 1 or stratum = 2 then relwt = 1; else if (stratum ge 3 and stratum le 6) then relwt = 2; if year = 1974; run; proc surveyreg data=nybight; weight relwt; stratum stratum; model catchwt = catchnum; run;

209

Estimated Regression Coefficients

Parameter

Estimate

Standard Error

t Value

Pr > |t|

Intercept catchnum

14.2725002 0.0813836

2.13426397 0.01612252

6.69 5.05

<.0001 <.0001

11.7 Using the weights as in Exercise 11.4, the estimated regression coefficients are B̂ 0 = 7.569 and B̂ 1 = 0.0778. From equation (11.8), V̂ L (B̂ 1 ) = 0.068, (alternatively, V̂ J K (B̂ 1 ) = 0.070). The slope is not significantly different from 0. Here is SAS output: Estimated Regression Coefficients

Parameter

Estimate

Standard Error

t Value

Pr > |t|

95% Confidence Interval

Intercept temp

7.56852711 0.07780553

4.31739366 0.26477065

1.75 0.29

0.0879 0.7705

-1.1793434 16.3163976 -0.4586708 0.6142818

11.10 (a)

(b)

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CHAPTER 11. REGRESSION WITH COMPLEX SURVEY DATA

(c) Here is output from SAS. Estimated Regression Coefficients

Parameter

Estimate

Standard Error

t Value

Pr > |t|

95% Confidence Interval

Intercept purchase

9.61708567 1.08119004

3.56639563 0.12136848

2.70 8.91

0.0208 <.0001

1.76750181 17.4666695 0.81405982 1.3483203

NOTE: The denominator degrees of freedom for the t tests is 11.

We use (number of psus) — (number of strata) = 12 — 1 = 11 df.

11.14 Here is code and output from SAS: data nhanes; infile nhanes delimiter=’,’ firstobs=2; input sdmvstra sdmvpsu wtmec2yr age ridageyr riagendr ridreth2 dmdeduc indfminc bmxwt bmxbmi bmxtri bmxwaist bmxthicr bmxarml; if riagendr = 1 then x = 0; /* x=0 is male*/ if riagendr = 2 then x = 1; /* x=1 is female */ if age ge 15 then over15 = 1; else if age lt 15 then over15 = 0; else over15=.; one = 1; label age = "Age at Examination (years)"

211 agesq = "Age^2" riagendr = "Gender" ridreth2 = "Race/Ethnicity" dmdeduc = "Education Level" indfminc = "Family income" bmxwt = "Weight (kg)" bmxbmi = "Body mass index" bmxtri = "Triceps skinfold (mm)" bmxwaist = "Waist circumference (cm)" bmxthicr = "Thigh circumference (cm)" bmxarml = "Upper arm length (cm)"; run; proc surveyreg data=nhanes; stratum sdmvstra; cluster sdmvpsu; weight wtmec2yr; model bmxtri=bmxbmi /clparm; output out=quad pred=quadpred residual=quadres; ods output ParameterEstimates = quadcoefs; run; /* Do a weighted bubble plot of data with the regression line. */ goptions reset=all; goptions colors = (gray); axis4 label=(angle=90 ’Triceps Skinfold’) order=(0 to 50 by 10); axis3 label=(’Body Mass Index’) order = (10 to 70 by 10); axis5 order=(0 to 50 by 10) major=none minor=none value=none; symbol interpol=join width=2 color = black; proc sort data=quad; by bmxtri bmxbmi; proc means data=quad noprint; by bmxtri bmxbmi; var wtmec2yr quadpred quadres; output out=quadplot sum=sumwts sumquad sumres mean=meanwt meanquad meanres; run; proc gplot data=quadplot; bubble bmxtri*bmxbmi= sumwts/bsize=10 haxis = axis3 vaxis=axis4; plot2 meanquad*bmxbmi/ vaxis=axis5; run;

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CHAPTER 11. REGRESSION WITH COMPLEX SURVEY DATA

/* Plot residuals vs predicted values */ goptions reset=all; goptions colors = (gray); axis3 label=(’Predicted Values’) order = (15 to 30 by 5); axis4 label=(angle=90 ’Residuals’) order=(-20 to 40 by 10); axis5 order=(10 to 70 by 10) major=none minor=none value=none; proc gplot data=quadplot; bubble meanres*meanquad= sumwts; run;

Estimated Regression Coefficients

Parameter

Estimate

Standard Error

t Value

Pr > |t|

95% Confidence Interval

Intercept bmxbmi

-3.4248758 0.8581404

0.47196343 0.02265388

-7.26 37.88

<.0001 <.0001

-4.4308420 -2.4189096 0.8098548 0.9064260

NOTE: The denominator degrees of freedom for the t tests is 15.

R2 = 0.38. Note the pattern in the residuals vs. predicted values plot. You may want to use a model with log transformations instead.

11.15 Estimated Regression Coefficients

213

Parameter Intercept bmxthicr

Estimate

Standard Error

5.02260509 1.38143897 1.68332900 0.02540473

t Value Pr > |t| 3.64 66.26

0.0024 <.0001

95% Confidence Interval 2.07813762 7.96707255 1.62918010 1.73747790

NOTE: The denominator degrees of freedom for the t tests is 15.

R2 = 0.57.

11.16 data ncvs; infile ncvs delimiter = ","; input age married sex race hispanic hhinc away employ numinc violent injury medtreat medexp robbery assault pweight pstrat ppsu; agesq = age*age; if violent ge 1 then isviol = 1; else if violent = 0 then isviol = 0; run; proc surveylogistic data=ncvs; stratum pstrat; cluster ppsu; weight pweight; model isviol (event=’1’)= age sex ; run;

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CHAPTER 11. REGRESSION WITH COMPLEX SURVEY DATA

proc surveylogistic data=ncvs; stratum pstrat; cluster ppsu; weight pweight; model isviol (event=’1’)= age agesq sex ; run; Analysis of Maximum Likelihood Estimates

Parameter

DF

Estimate

Standard Error

Wald Chi-Square

Pr > ChiSq

Intercept age sex

1 1 1

-2.6743 -0.0418 -0.2925

0.0891 0.00211 0.0637

900.7557 392.9618 21.0673

<.0001 <.0001 <.0001

From this model, younger people and males are more likely to have at least one violent victimization. A quadratic term in age is not significant.

11.17 From (11.4), N Σ

B1 =

xiyi —

i=1

Σ N i=1

ÃN Σ

!Ã N Σ yi

xi i=1

à N x2 — Σ i

!

, N

i=1

!2 xi

/N

i=1

N1 ȳ1U — N1 ȳU N1 — N12/N y¯1U — (N1 ȳ1U + N2 ȳ2U )/N = 1 — N1/N = y¯1U — y¯2U . =

From (11.5), ty — B1tx N = y¯U — B 1 x̄ U N 1 ȳ 1U + N 2 ȳ 2U N = — 1 (y¯ — y¯ ) 2U N N 1U = y¯2U .

B0 =

11.18 (a) (b) βˆ0 = —4.096; βˆ1 = 6.049 (c) We estimate the model-based variance using the regression software: VˆM (β̂ 1 ) =

215 0.541.

Σ n (xi — x̄)2 (y i — βˆ0 — β̂ 1 xi )2 ˆ ˆ Σ VL( β)1 = (n — 1)[ (xi — x̄) 2 ] 2 = 0.685

VˆL is larger, as we would expect since the plot exhibits unequal variances.

11.19 From (11.10), for straight-line regression, Ã !—1 Σ Σ T wixix i wixiyi B̂ = i∈£

i∈£

with x = [1 x ]T . Here, i

i

Σ i∈£ wi Σ wixi

Σ wixixT = i

i∈£

Σ i∈£

i∈£

and

wixi wix2i

i∈£

Σ wiyi Σi∈£ w xi yi i

Σ x = wi iyi i∈£

so

Σ

,

i∈£

B̂

à Σ

=

1

!Ã Σ

!

i∈£

Σ

wix

i

Σ

Σwxy — i i i

B̂0 =

i∈£

Ã

!Ã ! Ã ! , Σwx Σwy Σw i i

wix2i — Σ

i∈£

wiyi — i

Σ

=

i

i∈£

i∈£

! —1 " Σ i∈£

wi

i∈£

i∈£

!2

,

wixi

i

i

i∈£

Ã Σ i∈£

Σ

! wi

Σ wixi

Σ

wi à Σ

Σi∈£ wixiyi i∈£

i∈£

wix2

wiyi

i∈£

Σ

Σ

Σ w i xi

à Σ

i∈£

i∈£

and

Σ

i∈£ Σ wi

wixi

i∈£

Thus,

!2 wixi

i∈£

—

wix2i

i∈£

—

B̂ 1 =

—

2

i∈£ wi

à Σ

i∈£

wixiyi i∈£ 2

wi xi

wiyi — B̂ 1

Σ

—

Σ

i∈£

2

wi x i i∈£ # wi xi .

i∈£

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CHAPTER 11. REGRESSION WITH COMPLEX SURVEY DATA

11.20 In matrix terms, , Σ T ˆ B= wj xj xj

—1

Σ wj xjyj.

j∈£

j∈£

Then, using the matrix identity in the hint, zi =

∂ B̂ ∂wi

—1 , Σ Σ ∂ wjxjyj wjxjxjT ∂wi j∈£ j∈£

=

+

, Σ

—1

∂ Σ

wjxjxjT

j∈£ —1

, Σ

w j xjx Tj

= —

, Σ

j∈£

+ j∈£

=

, Σ

xixT i

,∂w Σ

i j∈£

wjxjyj —1

T wjxjx j j∈£

Σ

wjxjyj

j∈£

—1

w j xjx Tj —1

wjxjxjT

³

xiyi T

ˆ + xiyi —xixi B

´

j∈£

=

, Σ

—1

wjxjxTj

³ ´ xi yi — xTi B ˆ .

j∈£

Then the estimated variance is à ! Σ ˆ ˆ V (B̂ ) = V wizi i∈£

=

, Σ

—1

wjxjxjT

Ã Σ ˆV

j∈£

wiqi

!, Σ

1 T wjxjxj

i∈£ j∈£

11.21 First express the estimator as a function of the weights: tˆyGREG = tˆy + (tx — t̂ x ) T B̂, with B̂ =

Ã Σ i∈£

wi

1

T 2 xix i

σi

!—1

1 Σ wi 2 xiyi. σi i∈£

.

217 Thus, tˆyGREG =

Σ

à wiyi + tx —

i∈£

Σ

!T wixi B̂

i∈£

Using the same argument as in Exercise 11.20, , —1 ∂ B̂ Σ ∂ Σ 1 1 T = wj 2 xjyj w j 2 xj xj ∂wi j∈£ σj ∂wi j∈£ σj , Σ

—1

∂ Σ 1 1 T + wj 2 xjxj wj 2 xjyj i σj σj ∂w j∈£ , j∈£ , Σ Σ —1 = — wj 1 xjxT wj 1 xjxT 1 xi xT i j j 2 2 σj σi σj2 j∈£ j∈£ , —1 Σ 1 1 xiyi wj 2 xjxTj + σi2 σj = =

zi =

—1

Σ wj 1 xjyj σj2 j∈£

j∈£

, Σ

—1

T 1 wj 2 xjxj j σ j∈£

, Σ

µ

—1

1 T wj 2 xjxj σ j j∈£

1 T ˆ 1 — 2 xixi B + 2 xiyi σi σi

¶

T ´ ³ 1 ˆ 2 xi yi — xi B σi

∂tˆyGREG ∂wi

T B̂ ¡ — ¢ ∂ = yi — xi B̂ + tx t̂x ∂w i —1 , ³ ´ 1 Σ 1 x xT T ¡ x — t̂ x¢T x y — x B̂ i j j i i iT w σ2 j i = y — x B̂ + t σi2 j T

j∈£

= gi(yi — x B̂ ) . iT

11.26 The OLS estimator of β is β̂ = (XT X)—1XT Y. Thus,

CovM (β̂ ) = (XT X)—1XT Cov (Y)[(XT X)—1XT ]T = (XT X)—1XT ΣX(XT X)—1.

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CHAPTER 11. REGRESSION WITH COMPLEX SURVEY DATA

For a straight-line regression model,

Σ

n XT X = (X

T

X)—1

=

Σ

x2i

1 Σ( xi — ¯) x2 ·

and

Σ

xi

xi

XT ΣX =

Σ 2 x i/n —x̄ —x̄

1

¸ Σ 2 Σ σi xiσ2i Σ Σ x2σ2 . xiσ2i i i

Thus, in straight-line regression, · 1Σ 2 ¸· Σ 2 ¸· 1 Σ 2 Σ 1 x —x̄ σ xiσ2 x Cov ( ) = i i i i n n Σ Σ Σ x2σ2 M β̂ xiσi2 —x̄ 1 —x̄ [ (x i — x̄)2 ]2 i i V (βˆ ) = M

1

=

—x̄ 1

³

´ hΣ i2 Σ Σ Σ x̄ 2 σ2 — 2 x¯ x σ2 + x2σ2 / (x — x̄ ) 2 i i i i i i Σ 2 2 (xi — x̄ ) σ i. Σ [ (xi — x̄) 2 ]2

To see the relation to Section 11.2.1, let Qi = Yi — β0 — β1xi.

Then VM (Qi) = σ2; isince observations are independent under the model, ·Σ ¸ Σ VM (xi — x̄ )Q i = (xi — x̄) 2 σ 2 . i

i

i

If we take wi = 1 for all i, then Equation (11.8) provides an estimate of VM (β̂ 1 ).

11.29 For this model (see Exercise 11.19) Σ 1 XT£ W£ Σ—1 £ X£ = σ2

i∈£ Σ

wi

Σ Σ

wixi i∈£

and σ12

wix2i

i∈£

Σ

T 1 X £W £Σ £ y£ =

wixi

i∈£

wiyi Σi∈£ = wxy i i i

1 tˆy . ˆ t 2 σ xy

i∈£

Using (11.20), ˆ B=

·

N̂ tˆ x tˆx tˆx2

¸—1 ·

tˆ y tˆxy

¸ =

1 N̂ t̂ x 2 — (tˆx)2

·

tˆ 2tˆ — tˆ tˆ x

y

x xy

—tˆxtˆy + N̂ t̂ xy

¸

¸

219 and tˆyGREG = tˆy + [N — N̂ , tx — tˆx ] B̂ 1 = tˆ + [(N — N̂ ) ( tˆ tˆ — tˆ tˆ ) + (t — tˆ )(—tˆ tˆ + N̂t̂ )]. y x xy x x x y xy x2 y N̂t̂x2 — (tˆx)2 If y = x, tˆ xGREG

= tˆ + x

= tx.

1 N̂t̂x2 — (tˆx)2

[0 + (t — tˆ )(—tˆ2 + N̂ tˆ )] x

x

x

x2

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CHAPTER 11. REGRESSION WITH COMPLEX SURVEY DATA

Chapter 12

Two-Phase Sampling 12.1 We use (12.4) to estimate the total, and (12.7) to estimate its variance. We obtain 100,000 Σ nh h ˆ(2) = 1000 tstr mh r = 48310. cells

From (12.7), we estimate s

2(2)

h

= p̂

(2)

h

(1 — p̂(2) )/(mh — 1) and obtain h

¶ h 2(2) h H µ ³ ´ Σ n h — 1 — m — 1 n sh (2) ˆ ˆ N — 1 n t = N (N — 1) V str n—1 mh h=1 µ ¶Σ H 2 N n nh + 1— (y¯(2) — ŷ¯ ( 2 ) ) 2 str N n h n—1 h=1 1000002 = 100000(99999)0.000912108 + 0 109282588 .

999 = 10214904. ˆ ) = 3196.

Thus SE(tstr (2)

Note that we can do a rough check using SAS PROC SURVEYMEANS, which will capture the variability due to the phase II sample. data exer1201; input strat nh mh diabcount ; nondiab = mh - diabcount ; datalines; 1 241 96 86 2 113 45 17 3 174 35 29 4 472 47 8 ;

221

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CHAPTER 12. TWO-PHASE SAMPLING

data exer1201; set exer1201; do i = 1 to diabcount; sampwt = nh/mh*(100000/1000); diab = 1; output; end; do i = 1 to nondiab; sampwt = nh/mh*(100000/1000); diab = 0; output; end; proc surveymeans data=exer1201 mean clm sum clsum; stratum strat; weight sampwt; var diab; run; This code gives ˆt = 48310 with SE 3059.0785. We can then add the second term in (12.7) to obtain ³ ´ 1000002 2 ˆV ˆ(2) tstr = 3059.0785 + 0.109282588 = 10451881.

999 This is a little bit larger than the estimate obtained above because we did not incorporate fpcs.

12.3 Using the population and sample sizes of N = 2130, n(1) = 201, and n(2) = 12, the phase I weight is wi(1) = 2130/201 = 10.597 for every phase I unit. For the units in phase II, the phase II weight is wi(2) = 201/12 = 16.75. We have the following information and summary statistics from shorebirds.dat: Σ ˆ = t(1) wi(1)xi = 44284.93. x i∈£(1)

tˆx(2) =

Σ

w(1) wi(2)xi = 34790, i

i∈£(2)

and tˆy = 43842.5. Using (12.9), (2)

tˆ(2) 43842.5 tˆ(2) = tˆ(1) y = 44284.93 = 55808. yr x (2) 34790 tˆx

223 We estimate the variance using (12.11): we have s2 = 115.3561, s2 = 7.453911, and Ã

!

Ã

y

!

e

n(1) s2y n(2) s2e 2 2 N N 1 + ) = — 1 — Vˆ ( ˆt(2) yr n(1) n(2) N n(1) µ ¶ µ ¶ 201 115.3561 12 7.453911 2 2 = (2130) 1 — + (2130) 1 — 2130 201 201 12 = 2358067 + 2649890 = 5007958, so the standard error is 2238.

Note that the paper by Bart and Earnst has some inconsistencies so it is possible that the data set constructed for this problem does not reflect the distribution of shorebirds in the region.

12.4 (a) The phase 2 weight is 1049/60 =17.48 for stratum 1, 237/48 = 4.9375 for stratum 2, and 272/142 =1.915 for stratum 3. (b) We use (12.5) to estimate (2) yˆ¯str =

H Σ nh

h=1

n

y¯h(2) = 0.3030 + 0.1078 + 0.1426 = 0.5534.

Then, using (12.8) (which we may use since the fpc is negligible), H

Σ nh — 1 nh s

2(2)

H

1 Σ nh h (2) 2) 2 Vˆ (ŷ¯str ) ≈ + (y¯(2)h — yˆ¯ (str ) n — 1 n mh n—1 n h=1 h=1 = 0.00186941 + 9.924 × 10—05 + 3.201 × 10—05 1 + (0.007192 + 0.003654 + 0.012126) 1558 = 0.002015 so the standard error is 0.045. Note that the second term adds little to the variability since the phase I sample size is large.

12.5 (a) We use the final weights for the phase 2 sample to calculate the proportions in the table, and use (12.8) to find the standard error for each (given in parentheses behind the proportion).

Case?

Proportion (SE) Male Gender Female

No 0.2265 (0.0397)

Yes 0.1496 (0.0312)

0.2164 (0.0399) 0.4430 (0.0449)

0.4075 (0.0426) 0.5570 (0.0449)

0.3761 (0.0444) 0.6239 (0.0444)

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CHAPTER 12. TWO-PHASE SAMPLING

(b) We can calculate the Rao-Scott correction for a test statistic based on a sample of size n = 1558. Then, (10.2) gives r

2

X =n

c

Σ Σ (pˆij — p̂ i + p̂ +j 2) i=1 j=1

pˆi + pˆ+ j

= (1558)0.062035726 = 96.65

To find the design effect, we divide the estimated variance from part (a) by the variance that would have been obtained if an SRS of size 1558 had been selected, namely pˆ(1 — pˆ)/1558. We obtain the following table: Case?

No Yes 13.987 11.929

Design effect Male Gender Female

14.661 12.715

11.708 12.715

13.072 13.072

Using (10.9), r

E[X ] ≈ 2

ΣΣ

c

(1 — pij)dij —

i=1 j=1

Σ

r

c

(1 — pi+)d i — R

i=1

Σ

(1 — p+j)dCj = 13.601.

j=1

Then, from Section 10.2.2, X2/E[X2] approximately follows a χ2 1distribution if the null hypothesis is true. We calculate X2/E[X2] = 96.65/13.601 = 7.1, so the p-value is approximately 0.008. There is evidence against the null hypothesis of independence. Σ 2 Σ ij—pˆi+pˆ+j ) Note that we could equally well have calculated X2 as m r i=1 cj=1 (pˆpˆ i+pˆ+j and calculated the variance under an SRS as pˆ(1 — pˆ)/m; this gives the same result.

Since the phase I sample size is relatively large, and the second term in (12.8) is small relative to the first term, we can use SAS PROC SURVEYFREQ to obtain an approximate check on our results. data exer1205; input strat nh mh gender $ case count ; datalines; 1 1049 60 m 0 16 1 1049 60 m 1 8 1 1049 60 f 0 17 1 1049 60 f 1 19 2 237 48 m 0 9 2 237 48 m 1 8 2 237 48 f 0 5 2 237 48 f 1 26

225 3 272 142 m 0 15 3 272 142 m 1 28 3 272 142 f 0 8 3 272 142 f 1 91 ; data exer1205; set exer1205; do i = 1 to count; sampwt = nh/mh; output; end; proc surveyfreq data=exer1205; stratum strat; weight sampwt; tables gender*case / chisq deff; run; This code treats the phase I sample as a population, so it underestimates the variance slightly. But since in this case n is large, the results are very close. SAS calculates the Rao-Scott chi-square statistic as 7.01, and p-value as 0.008.

12.9 We estimate Wh by nh/n, and estimate S2 by s2(2). Then, using (12.17), we h

have Stratum Yes No Not available Total

Ŵ h 0.3658 0.3895 0.2447 1.0000

Sˆ2h 0.1995 0.1313 0.2437

h

Ŵ h Sˆh 2 0.0730 0.0511 0.0596 0.1837

νn 0.40 0.32 0.44

We estimate S2 using H

H

h=1

h=1

Σ Σ (n — 1)Sˆ2 = (nh — 1)Sˆh2 + nh (p̂h — p̂) 2 , which gives Sˆ2 = 0.2468.

This allocation takes many more observations in the “Yes” and “No” strata than did the allocation that was used. Proportional allocation would have ν1 = ν2 = ν3.

12.10 From property 5 in Section A.4, V (tyˆ(2)) = V (tyˆ(1)) + E(V [tˆy(2) | Z]).

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CHAPTER 12. TWO-PHASE SAMPLING

Because the phase I sample is an SRS, V (tˆ(1)) = N

à 2

y

1 — n(1) N

!

S2y . n(1)

Because the phase II sample is also an SRS,

, P (Di = 1 | Zi = 1) = n(2) n(1), n(2)( n(2) — 1) for j /= i, n(1)(n(1) — 1) N wi(1) = (1) , n

P (DiDj = 1 | ZiZj = 1) =

and

(2)

wi =

n(1) Zi. n(2)

In addition, P (Zi = 1) = and

n(1) N

(1) (1) P (ZiZj = 1) = n ( n — 1) .

N (N — 1)

Thus, using (12.1) to write tˆy , (2)

V [tyˆ(2) | Z] ·Σ N ¸ N n(1) ZiDi (1) (2) yi | Z = V n n i=1 ¸ · ¶ N µ N Σ Σ N 2 Z Z D D y y | Z i k i k i k = — [tyˆ(1)]2 E n(2) i=1 k=1 ¶ N ¶2 Σ N µ N µ (2) n(2)(n(2) — 1) N N 2Σ Σ 2n = — [tˆ(1)]2 Z Z y y Z y + i j i j i i n(2) n(2) n(1) n(1)(n(1) — 1) y i=1

i=1 j=1,j/=i

227 E(V [tˆ(2) | Z]) = y

µ

N n(2)

¶2 Σ N

n(1) 2 n(2) y N i n(1)

i=1

µ

¶2 Σ N N (1) — 1) yiyj n(1) Σ (1) — n(1) N(n (N — 1) N n (2)(n (n(2) — 1) + (2) i=1 j=1,j/=i n —V [tˆ(1)] — (E[tˆ(1)])2 N

y

y (2)

N

N

N Σ 2 N (n — 1) Σ Σ = n(2) yi + n(2)(N — 1) yi yj i=1 j=1,j/=i

i=1

=

µ [tˆy(1)] — t2y —V 2 1 n(2) ¶ S2y [ˆ(1)] N — — V ty .

N n(2)

Thus,

µ ˆ(2)

V [t

]=N

2

y

1 — n(2) N

¶ S2

y .

n(2)

12.11 Conditioning on the phase I units, · 2(2) ¸ ¶ sh ¯ mh — 1 n h E — ¯Z n—1 N —1 n mh h=1 ¸ ¶ HΣ · (2) ¯ 2 µ (2) + N 1— n E nh (y¯ — yˆ¯ )2 Z . str ¯ n h n—1 N

ˆ (t ˆ(2))|Z] = N (N — 1) E[V str

H µ Σ n h— 1

h=1

" 2(2) # 2(1) sh sh ¯ ¯Z = E mh mh

Now

and Σ H

hn E

h=1

= E

(2)

(y¯ n h "H Σn h

h

=

i #

2) 2 (y¯h(2) )2 — ( ŷ¯ (str ) ¯Z

n H ¶ Σ nh · sh2(1) µ (1) ¸ 1 — mh + (y¯h )2 n mh nh h=1 ¸2 ¶ 2(1) µ ·Σ H nh (1) ¶ H Σ 2µ m s nh 1— h h — — y¯ n n h nh mh h=1 h=1 " µ # Σ 1 ¶ 2(1) ¶µ H Σ nh 1 — nh (n — 1)s2(1) — H (nh — 1)s 2(1) mh s h ; + 1— y h n n nh mh n h=1 h=1

=

ˆ(2) ¯ — y¯ )2 Z str ¯

h=1

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CHAPTER 12. TWO-PHASE SAMPLING

the last equality follows by applying the sums of squares identity µ ¶2 Σ Σ Σ H H H nk (1) 2(1) 2(1) (1) (n — 1)s = (nh — 1)s + n h ȳ — y¯ y h h n k h=1

h=1

k=1

to the phase I sample.

Plugging in to the first equation, we have µ ¶ 2 N n (2) 2(1) ˆ ˆ E[V (tstr )|Z] = 1— sy n N µ ¶ 2(1) ½ H Σs N — 1 nh — 1 mh — 1 nh h +N 2 — N n—1 N —1 n mh h=1 µ ¶· µ ¶µ ¶ ¸¾ 1 n nh nh mh mh(nh — 1) + 1— 1— 1— — n—1 N n n nh n µ ¶ ¶2µ ¶ H 2(1) µ Σ N2 n 2(1) m sh nh sy + N 2 = 1— 1— h n nh mh n N h=1

(The last equality follows after a lot of algebra.) Since E[s2(1)] = S2, the unbiasedy

ness is shown.

y

12.12 (a) Equation (A.9) implies these results. (b) From the solution to Exercise 12.10, (2) V (tˆ(2)) = V [tˆ(1)] + E[V (tˆ | Z)], y µ yr d ¶ (1) 2 (1) S2y , V [tˆ ] = N 1 — n y N n(1)

and (2) E [V (tˆd | Z)]

= = = =

µ ¶ n(2) S2d (1) N 1— — V [tˆd ] N (2) µ ¶n µ ¶ n(2) S2 n(1) S2 2 2 N 1— 1— d —N d N n(2) N n(1) ¸ · (2) N — n(1) 2 N —n NSd — n(1) µ n(2) ¶ N 2 1 — n(2) S2d . n(1) n(2) 2

(c) Follows because s2 and s2 estimate S2 and S2, respectively. y

e

y

d

229 12.13 Using the hint, 2

Sy

N

=

1 Σ ( i — y¯U)2 y N — 1 i=1

=

Σ 1 (yi — Bxi + Bxi — Bx̄U )2 N — 1 i=1

N

N

= =

Σ£ 1 U i i i U ¤ (y i — Bx i)2 + B2(x i — x̄ )2 + 2(y — Bx )B(x — x̄ ) N —1 i=1 S2 + B2S2 + 2BSxd. d

x

Then,

à 2 (1) V (tˆ(2)) ≈ N 1 — n yr N Ã

Ã

!

S2y + N 2 1 — n(2) n(1) n(1)

!

S2d n(2)

! Ã !

(2) n S2d 1 — n(1) 2BSxd + B2S2x + S2d N 2 — (1) + 1 (1) n n N n(2) Ã Ã ! !

S2d . = N 2 1 — n(1) 2BSxd + B2S2x N 2 1 — n(2) + N N n(1) n(2) =

N2

12.14 (a) (1) zi

=

ˆ(2) ∂tyr

∂wi(1) tˆ(2) y = xi(2) ˆ tx

and (2)

zi

Thus,

, V̂DR (t̂ (2) ) =

Vˆ

ˆ(2) ∂tyr ∂wi " # yi ˆ(2) (1) = tˆx — xi t y (2) (2) (2) tˆx tˆx tˆx i ˆ ˆ tˆx(1) h (2) (2) = y — x t i i xB (2) ˆ tx =

Σ wi(1)zi(1) +

yr

i∈£(1)

, =

Vˆ

wiz (2) i

i∈£(2)

Σ w x Bˆ i i

i∈£(1)

Σ

(1)

(2)

Σ + i∈£(2)

i ˆ(2) ˆ(2) yi — xitx B

tˆx(1) h wi

tˆx(2)

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CHAPTER 12. TWO-PHASE SAMPLING

12.16 Note that ˆ ˆ E[VHT (t(1) y ) | Z]

Σ =

(1) (1) Σ π(1) yi ik — πi πk

E

(1) (2) πik π ik

i∈£(2) k∈£(2)

Σ

= E

Σ

(1)

DiDk

yk (1) (1) πi πk

(1) (1)

πik — πi πk (1) (2) πik π ik

i∈£(1) k∈£(1)

|Z

yi yk |Z π(1) π(1) i

Σ

k

(1) Σ πik — πi(1)π(1) yi yk | Z k (1) (1) (1) π π i πk ik i∈£(1) k∈£(1) (1) ˆ(1) = V̂ (t )

= E

HT

y

12.17 (a) Since mh = νhnh and νh is known, E[mh] = νhE[nh] ¸ ·Σ N Z x = νhE i ih i=1

= νh

N Σ n i=1

N

xih

= nνhWh. Thus, E[C] = cn + n

H Σ

chνhWh.

h=1

(b) Using the constraint, set E[C]

n= c+

H Σ

.

chνhWh

h=1

Then Σ H

V ( yˆ¯ ( 2 ) ) = S2

c+

chνhWh

h=1

str

E[C]

—1 N

H

c+ +

Σ

µ ¶ chνhWh H 1 Σ 2 WhS h —1 νh E[C]

h=1

h=1

231 and ∂V (yˆ¯ ( 2 ) ) str

∂νk

· ¶¸ H µ1 H ¶ WkS 2 µ Σ Σ S2 1 2 k = .

WhS h chνhWh ckWk+ ckWk c+ 1 — 2 E[C] νk E[C] νh — h=1

h=1

Setting the derivatives equal to 0, we have S c kν kW k + c kν kW k 2

H Σ

2

WhS h

¶ ¶ W S2µ H Σ k k c+ chνhWh = 0 —1 — νk νh h=1

µ1

h=1

for k = 1, . . . , H. Thus µ Σ Σ · ¶ ¶¸ Σ H µ1 H H WkS k2 — 2 2 c+ —1 chνhWh 0 = ckνkWk S + WhS h h=1 k=1 h=1 νh νk k=1 · ¸ H H H Σ Σ Σ WkSk2 2 2 S — W S h — = ckνkWk c h νk k=1 h=1 k=1 Σ H

and

ΣHW h S2 h=1

νh

h =

¶Σ H H 1µ Σ 2 S2 — WhS h ckνkWk. c k=1

h=1

Substituting into (∗), we have Σ Σ · ¸ ¶Σ H H H 1µ 2 2 2 2 W S + h 0 = ckνkWk S — S — WhS h chνhWh h h=1 h=1 c h=1 ¶ µ H WkS k2 Σ chνhWh c+ — νk h=1 Σ Σ µ ¶ H H ¶ µ 2 2 1 chνhWh = ckνkWk S — h=1 WhS h c c + h=1 Ã !

H Σ WkS2k c+ chνhWh , — νk h=1 which implies that

and, consequently,

¶ µ H ckν k2 2 Σ 2 2 WhS S — h = S k, c h=1

,

cS2

. νk = H u Σ , c (S2 — W S2) k

k

h

h=1

h

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CHAPTER 12. TWO-PHASE SAMPLING

(c) To meet the expected cost constraint with the optimal allocation, set E[C]

n=

H Σ

c+

,

chνh∗Wh

h=1

with

, u cS2h νh∗ = .

H u Σ u 2 2 , ch(S — WjS j) j=1

12.18 Let A = S2y—

H Σ

WhS2h. Then, from, (12.17),

h=1

, u νh,opt = u , u , ch and

s c(1)S2h 2

S

—

C∗

(1)

n opt = c(1) +

H Σ h=1

chWhνh,opt

H Σ

= WjS2

c(1)Sh2 chA

j

j=1

C∗

=

H

c(1) +

Σ h=1

√ WhSh ch

r

c(1) A

.

233 Then, Vopt (ŷ¯str ) = (2)

= = = =

=

=

¶ H µ S2y 1 Σ 1 2 + — WhS h νh,opt — 1 N n(1) n(1) opt opt h=1 ÃH ! r H Σ S2y Sy2 1 Σ √ A WhS2h + (1) — — WhSh ch (1) (1) c N h=1 nopt nopt h=1 r ! Ã H S2y Sy2 Σ √ A 1 2 + (1) + A — Sy — WhSh ch (1) c(1) N nopt nopt h=1 ! Ã H Sy2 √ r A 1 Σ WhSh ch +A — (1) c(1) N nopt h=1 ! Σ Ã !Ã r r H H A Sy2 Σ 1 √ √ (1) (1) + c + A c c c W S W S h h h h — h h h=1 c(1) C∗ A N h=1 ÃH !2 H Σ Σ 1 √ √ √ (1) c A c + W W ch ∗ h hSh hSh C h=1 h=1 Σ # √ H Sy2 (1) (1) +Ac + c A WhSh √c — h=1 h N Sy2

1 C∗

, 2 H Sy2 Σ √ u √ 2 Wh S h — .

Wh Sh ch + c(1) , S2 — N h=1 h=1

H Σ

12.19 The easiest way to solve this optimization problem is to use Lagrange multipliers. Using the variance in (12.10), the function we wish to minimize is µ ¶ µ ¶ h i 1 1 1 1 2 2 (1) (1) (2) (2) g(n(1), n(2), λ) = — S + — S — λ C — c n — c n . y d n(1) n(2) n(1) N Setting the partial derivatives with respect to n(1), n(2), and λ equal to 0, we have S2 ∂g Sd2 =— y + + λc(1) = 0, £ ¤ £ (1)¤2 ∂n(1) n(1) 2 n S2d ∂g = — £ ¤2 + λc (2) = 0, ∂n(2) n(2) and

h i ∂g = — C — c(1)n(1) — c(2)n(2) = 0. ∂λ Consequently, using the first two equations, we have h i2 S2 — S2 d n(1) = y λc(1)

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and

h

Taking the ratios gives

2 i2 n(2) = S d . λc(2)

à (2) ! 2 c(1)S2d n = (2) c (S2 — S2) , n(1) y d

which proves the result.

12.20 (a) These results follow directly from the contingency table. For example, µ ¶ N1 = C21 = C21 C2+ = 1 — C22 C2+ = (1 — S )p. p1 2 C2+ N C2+ N N N N2 C C C p2 = 22 = 22 2+ = S 2p.

N N C2+ N The other results are shown similarly. (b) From (12.19), 2 (2) Sh Vopt (p̂str ) ≈ Σ + W h VSRS (pˆ) Sy h=1

s

s c(1) c(2)

Sy2 —

Σ2

2

2 h=1 WhSh

Sy2

.

Using the results from part (a), Σ s s 2 µ N1 ¶2 p1(1 — p1) µ N2 ¶2 p2(1 — p2) Sh Wh = + h=1 Sy N p(1 — p) N p(1 — p) s s (1 — S2)p(1 — p)S1 pS2(1 — p)(1 — S1) = + p(1 — p) p(1 — p) √ √ = (1 — S2)S1 + S2(1 — S1).

For the second term, note that N

N

Σ Σ (xi — x̄ U )(yi — y¯U ) = xi(yi — p) = C22 — NpW2, i=1

i=1

N

Σ and

N

(xi — x̄ U ) = 2

Σ

N

(xi — W2) = 2

Σ

i=1

i=1

N

N

N

i=1

i=1

i=1

xi — NW 22 = NW1W2,

i=1

Σ Σ Σ (yi — y¯U )2 = (yi — p)2 = yi — Np2 = Np(1 — p),

Consequently, SyR =

C22 — NpW2 p(S2 — W2) √ = √ .

N W1W2 W1W2

235 S2 —

Σ 2

WhS2

y

= p(1 — p) — W1p1(1 — p1) — W2p2(1 — p2)

h h=1

= W1p21 + W2p22— p2 (1 — S2)2p2 S2p2 = + 2 —p2 W1 W2 2 £ ¤ p = W2 (1 — S2)2 + W 1S22— W 1W 2 W1W2 £ ¤ p2 = W 2 — 2W S + S2 1 2 2 2 Wp1W 2 2 = [ 2 S2 — W2 ] W1W2 = Sy2R2. (c) The following calculations were done in Excel.

S1 0.5 0.6 0.7 0.8 0.9 0.95

Cost Ratio 0.01 0.1 1.02 1.06 1.02 1.15 0.93 1.15 0.76 1.04 0.48 0.78 0.28 0.54

0.0001 1.00 0.97 0.85 0.65 0.37 0.20

0.5 1.15 1.42 1.61 1.68 1.53 1.23

1 1.21 1.64 2.01 2.25 2.25 1.92

12.21 Suppose that Sis a subset of m units from U , and supposeS has mh units from stratum h, for h = 1, . . . , H. Let Zi = 1 if unit i is selected to be in the final sample and 0 otherwise; similarly, let Fi = 1 if unit i is selected to be in the stratified sample and 0 otherwise, and let Di = 1 if unit i is selected to be in the subsample chosen from the stratified sample and 0 otherwise. Then the probability that S is chosen to be the sample is P (S) = P (Zi = 1, i ∈ S, and Zi = 0, i /∈ S) We can write P (S) as P (S) = P (Fi = 1, i ∈ S)P (Di = 1, i ∈ S | F1, . . . , FN ).

Then, P (Fi = 1, i ∈ S) µ ¶µ ¶µ ¶µ ¶ µ ¶µ ¶ m1 N1 — m1 m2 N2 — m2 . . . mH NH — mH m1 n1 — m1 m2 n2 — m2 nH — mH mH = µ ¶µ ¶ µ NH ¶ N1 N2 . . . n1 n2 nH total number of stratified samples containing S = number of possible stratified samples

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Also, P (Di = 1, i ∈ S | F) = P (Mh = mh, h = 1, . . . , H)P (Di = 1, i ∈ S|M = m) µ ¶ µ NH ¶ N m11 . . . m 1 H = ¶ µ ¶ µ ¶µ µ ¶. n1 n2 N . . . nH m m1 m2 mH Note that for each h, µ ¶µ ¶ N h — mh Nh nh — mh mh µ ¶µ ¶ Nh nh nh mh (Nh — mh)!

Nh! nh!(Nh — nh)! mh!(nh — mh)! = (nh — mh)!(Nh — nh)! mh!(Nh — mh)!

Nh! nh! = 1.

Consequently, P (S) = P (Fi = 1, i ∈ S)P (Di = 1, i ∈ S | F1, . . . , FN ) µ ¶µ ¶ —m Nhh Y N h n m 1 h — mhh H = µ N ¶ h=1 µ Nh ¶ µ nh ¶ nh mh m 1 = µ ¶, N m so this procedure results in an SRS of size m.

Chapter 13

Estimating Population Size 13.1 Students may answer this in several different ways. The maximum likelihood estimate is n1n2 (500)(300) N̂ = = = 1250 m 120 with 95% CI (using likelihood ratio method) of [1116, 1422]. A bootstrap CI is [1103, 1456]. xmat <- cbind(c(1,1,0),c(1,0,1)) y <- c(120,380,180) captureci(xmat,y) bootout <- captureboot(converty(y,xmat[,1],xmat[,2]), crossprod(xmat[,1],y),nboot=999,nfunc=nmle) 13.2 (a) The maximum likelihood estimate is N̂ =

n1n2 (7)(12) = = 21. m 4

Chapman’s estimate is (n1 + 1)(n2 + 1) (8)(13) —1= — 1 = 19.8.

5 m+1 Because of the small sample sizes, we do not wish to employ a confidence interval that requires N̂ or Ñ to be approximately normally distributed. Using the R function captureci gives N̂ = 21 with approximate 95% confidence interval [15.3, 47.2].

Alternatively, we could use the bootstrap to find an approximate confidence interval for Ñ . (Theoretically, we could also use the bootstrap to find a confidence interval for N̂ as well; in this data set, however, m∗ for resamples can be 0, so we only use the procedure with Chapman’s estimator) The bootstrap gives a 95% confidence interval [12, 51] for N , using Chapman’s estimator.

Ñ =

Here is the code from R: 237

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CHAPTER 13. ESTIMATING POPULATION SIZE

xmat <- cbind(c(1,1,0),c(1,0,1)) y <- c(4,3,8) captureci(xmat,y) bootout <- captureboot(converty(y,xmat[,1],xmat[,2]), crossprod(xmat[,1],y),nboot=999,nfunc=nchapman) (b) N̂ = 27.6 with approximate 95% confidence interval [24.1, 37.7]. (c) You are assuming that the two samples are independent. This means that fishers are equally likely to be captured on each occasion.

13.3 We obtain N̂ = 65.4 with 95% CI [60.9, 74.4].

Here is the code from R: xmat <- cbind(c(1,1,0),c(1,0,1)) y <- c(33,15,12) captureci(xmat,y) 13.4 (a) We treat the radio transmitter bears and feces sample bears as the two samples to obtain N̂ = 483.8 with 95% CI [413.7, 599.0]. xmat <- cbind(c(1,1,0),c(1,0,1)) y <- c(36,311-36,20) captureci(xmat,y) (b)N̂ = 486.5 with 95% CI [392.0, 646.4]. xmat <- cbind(c(1,1,0),c(1,0,1)) y <- c(28,239-28,57-28) captureci(xmat,y) (c)N̂ = 450 with 95% CI [427, 480]. xmat <- cbind(c(1,1,0),c(1,0,1)) y <- c(165, 311-165,239-165) captureci(xmat,y) 13.5 The model with all two-factor interactions has G2 = 3.3, with 4 df. Comparing to a χ2 4distribution gives a p-value 0.502. No simpler model appears to fit the data.

Using this model and function captureci in R, we estimate 3645 persons in the missing cell, with approximate 95% confidence interval [2804, 4725].

239 13.6 (a)N̂ = 336, with 95% CI [288, 408]. Ñ = 333 with 95% CI [273, 428]. The linearization-based standard errors are r n2n2(n2 — m) ˆ 1 SE (N) = = 37 m3 s

and ˜) = SE ( N

(n1 + 1)(n2 + 1)(n1 — m)(n2 — m) = 29. (m + 1)2(m + 2)

xmat <- cbind(c(1,1,0),c(1,0,1)) y <- c(49,73,86) captureci(xmat,y) bootout <- captureboot(converty(y,xmat[,1],xmat[,2]), crossprod(xmat[,1],y),nboot=999,nfunc=nchapman) (b) The following SAS code may be used to obtain estimates for the models. data hep; input elist dlist tlist count; datalines; 0 0 1 63 0 1 0 55 0 1 1 18 1 0 0 69 1 0 1 17 1 1 0 21 1 1 1 28 ; proc means data=hep sum; var count; run; proc print data=hep; run; proc catmod data=hep; weight count; model elist*dlist*tlist = _response_ /pred=freq ml=nr; loglin elist dlist tlist; /* Model of independent factors */ run;

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proc genmod data=hep; CLASS elist dlist tlist / param=effect; MODEL count = elist dlist tlist / dist=poisson link=log type3; run; proc catmod data=hep; weight count; model elist*dlist*tlist = _response_ /pred=freq ml; loglin elist dlist tlist elist*dlist; /* Model with elist and dlist dependent */ run; proc catmod data=hep; weight count; model elist*dlist*tlist = _response_ /pred=freq ml; loglin elist dlist tlist tlist*dlist; /* Model with tlist and dlist dependent */ run; proc catmod data=hep; weight count; model elist*dlist*tlist = _response_ /pred=freq ml; loglin elist dlist tlist elist*tlist; /* Model with elist and tlist dependent */ run; proc catmod data=hep; weight count; model elist*dlist*tlist = _response_ /pred=freq; loglin elist|dlist|tlist@2; /* Model with all 2-way intrxns */ run; 13.7 (a) The assumption of independence of the two sources is probably met, at least approximately. The registry is from state and local health departments, while BDMP is from hospital data. Presumably, the health departments do not use hospital newborn discharge information when compiling their statistics. However, there might be a problem if congenital rubella syndrome is misclassified in both data sets, for instance, if both sources tend to miss cases.

We do not know how easily records were matched, but the paper said matching was not a problem.

The assumption of simple random sampling is probably not met. The BDMP was from a sample of hospitals, giving a cluster sample of records. In addition, selection of the hospitals for the BDMP was not random—hospitals were self-selected. It is

241 unclear how much the absence of simple random sampling in this source affects the results. (b) Year 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985

Ñ 244.33 95 48 79.5 44.5 114 41.67 30.5 62.33 159 31.5 4 35 3 3 1

The sum of these estimates is 996.3333. (c) Using the aggregated data, the total number of cases of congenital rubella syndrome between 1970 and 1985 is estimated to be Ñ = (263 + 1)(93 + 1)/(19 + 1) — 1 = 1239.8.

Equation (12.8) results in Vˆ ( Ñ ) = 53343, and in an approximate 95% confidence √ interval 1240 ± 1.96 (53343) = [787, 1693]. Using the bootstrap function gives a 95% confidence interval of [855, 1908].

The aggregated estimate should be more reliable—each yearly estimate is based on only a few cases and has large variability. (d) Many methods could be used to assess whether the incidence of congenital rubella syndrome has changed, using these data. You could use a change-point test or divide the data into two groups and test whether the incidence is the same in both. The plot below shows the relation between year and log(Ñ ), estimated in the table in part(b). The decrease after 1980 is apparent.

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3 0

1

2

log(N)

4

5

CHAPTER 13. ESTIMATING POPULATION SIZE

1970

1975

1980

1985

Year

13.8 Model Independence 1*2 1*3 2*3

G2 11.1 2.8 10.7 9.4

df 3 2 2 2

p-value 0.011 0.250 0.005 0.00

The model with interaction between sample 1 and sample 2 appears to fit well.

Using that model, we estimate N̂ = 2378 with approximate 95% confidence interval [2142, 2664].

13.9 A positive interaction between presence in sample 1 and presence in sample 2 (as there is) suggests that some fish are “trap-happy”—they are susceptible to repeated trapping. An interaction between presence in sample 1 and presence in sample 3 might mean that the fin clipping makes it easier or harder to catch the fish with the net.

13.10 (a) The maximum likelihood estimate is N̂ = 73.1 and Chapman’s estimate is Ñ = 70.6. A 95% confidence interval for N , using N̂ and the function captureci, is [55.4, 124.1]. Another approximate 95% confidence interval for N , using Chapman’s estimate and the bootstrap, is [52.7, 127.8].

243 13.12 For the data in Example 13.1, pˆ = for p is

r 0.2 ± 1.96

20 100

(0.2)(0.8)

100 The confidence limits L(pˆ) and U (pˆ) satisfy

=

1

and a 95% confidence interval 5

= [0.12, 0.28].

P {L(pˆ) ≤ p ≤ U (pˆ)} = 0.95.

Because N = n1/p, we can write P {n1/U (pˆ) ≤ N ≤ n 1 /L(p̂)} = 0.95.

Thus, a 95% confidence interval for N is [n1/U (pˆ), n 1 /L(p̂)]; for these data, the interval is [718, 1645]. The interval is comparable to those from the inverted chisquare tests and bootstrap; like them, it is not symmetric.

13.13 Note that

µ

¶ N — 1 — n1 n2 — m (N — 1|n , n ) = L 1 2 µ ¶ N —1 n2 µ ¶µ ¶ n1 N — n1 N — n1 — (n2 — m) m n2 — m N — n1 µ ¶ = N N — n2 N n2 N — n1 — n2 + m N = L(N |n , n ) .

1 2 N — n1 N — n2 n1 m

¶µ

Thus, if N > n1 and N > n2, L(N ) ≥ L(N — 1)

iff iff

N — n1 — n2 + m N — n1 mN ≤ n1n2.

N N — n2

≤ 1

Take N̂ to be the integer part of n1n2/m. Then for any integer k ≤ N̂ , mk ≤ m

n1n2 = n 1n 2 , m

so L(k) ≥ L(k — 1) for k ≥ N̂ . Similarly, for k > N̂ (k integer), n1n2 mk > m >n n , 1 2 m so L(k) < L(k — 1) for k > N̂ . Thus N̂ is the maximum likelihood estimator of N .

13.14 (a) Setting the derivative equal to zero, we have m ( N̂ — n1) = (n2 — m)n1,

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CHAPTER 13. ESTIMATING POPULATION SIZE

or N̂ =

n2n1 . m

Note that the second derivative is d2 log L(N ) = dN 2 =

(n2 — m)n1(2N — n1) m — 2 N N 2(N — n1)2 2 mN — 2Nn1n2 + n12n2 , N 2(N — n1)2

which is negative when evaluated at N̂ . (b) Noting that E[m] = n1n2/N , the Fisher information is · 2 ¸ ∂ I(N ) = —Em log L(N/m, n1, n2) ∂N 2 · ¸ mN 2 — 2Nn1n2 + n 21 n2 = —Em N 2(N — n1)2 Nn1n2 — 2Nn1n2 + n12n2 — = ___ N 2 (N — n1) 2 n1n2 = .

N 2(N — n1) Consequently, the asymptotic variance of N̂ is V (N̂ ) =

1 N 2(N — n1) = .

I(N ) n1n2

13.15 Substituting C — n1 for n2 in the variance, we have g(n1) = V (N̂ ) =

N 2(N — n1) . n1(C — n1)

Taking the derivative, dg dn1

N2 N 2(N — n1)(C — 2n1) — n1(C — n1) n12(C — n1)2 2 N — 2 [n1(C — n1) + (N — n1)(C — 2n1)]. n1 (C — n1 )2

= — =

Equating the derivative to 0 gives n21 — 2Nn1 + NC = 0, or n1 = Since n1 ≤ N , we take

2N ±

n1 = N —

√

4N 2 — 4NC .

2

√

N (N — C)

245 and n2 = C — N +

√ N (N — C)

13.16 (a) X is hypergeometric; µ ¶µ ¶ n1 N — n1 m n2 — m P (X = m) = . µ ¶ N n2 (b) In the following, we let q = n2 + 1 · ¸ (n1 + 1)(n2 + 1) E [Ñ ] = E —1 X +1 µn ¶µ N — n ¶ 1 1 n2 m n2 — m (n1 + 1)(n2 + 1) Σ = 1 µN ¶ m+1 m=0 n2 µ ¶µ ¶ n1 + 1 N + 1 — (n1 + 1) n2 = Σ (N + 1) m + 1 µnN2 ++11— ¶ (m + 1) m=0

µ q

=

(N + 1)

Σ k=1

= (N + 1)

Σq k=0

¶µ

—1

n2 + 1

¶ n1 + 1 N — n1 k q—k —1 µ ¶ N +1 q µ ¶µ ¶ µ ¶ n1 + 1 N — n1 N — n1 k q—k q ¶ µ ¶ — µ — 1.

N +1 N +1 q q

The first term inside the brackets is Σ P (Y = k) for Y a hypergeometric random k ¡ ¢ 1 variable; it thus equals 1. If n2 ≥ N — n1, then q > N — n1 and N —n = 0. Hence, q if n2 ≥ N — n1, E [Ñ ] = N .

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Chapter 14

Rare Populations and Small Area Estimation Σ riyi 14.2 (a) Note that y¯1 = i∈£ Σ

, so using properties of ratio estimation [see equa-

1

ri

i∈£1

tion (4.10)],

N1

2

Σ 1 1 S 2 V (y¯1 ) ≈ (r yi i— r i y¯ (M1 — 1)S12 ≈ 1 .

1U ) = 2 2 n1(N1 — 1)p1 i=1 n1(N1 — 1)p1 n1p1 Consequently, V ( yˆ¯ d ) = A2V (ȳ 1 ) + (1 — A)2V (ȳ 2 ) ≈ A2S12 (1 — A)2S2 2 . n p + n p 1 1

2 2

(b) With these assumptions, we have A =

N 1p 1 Np

n2p2 = f2p2N2, and V ( yˆ¯d ) ≈ = =

, 1—A =

N2p2 Np

, n p = kf p N , 1 1

2 1

1

¶ A2 (1 — A)2 + kN1p1 N2p2 "µ # ¶2 µ ¶2 N1p1 1 N2p2 1 2 S1 + Np kf2N1p1 Np f2N1p1 ¶ µ N1p1 N p .

S21 + 2 2 2 (Np) f2 k S21 f2

µ

The constraint on the sample size is n = n1 + n2 = kf2N1 + f2N2; solving for f2, we have n f2 = .

N1k + N2 247

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248 CHAPTER 14.

RARE POPULATIONS AND SMALL AREA ESTIMATION

Consequently, we wish to minimize ¶ N1p1 + N2p2 k ¶ µ N1p1 N p , S21 + 2 2 (N1k + N2 ) (Np)2n k 2

S V (ŷ¯d ) ≈ (Np)12f 2 =

µ

or, equivalently, to minimize

µ

g(k) = (N1 k + N 2) Setting the derivative

N1p1 k

¶ + N 2p 2 .

dg

= N N p — N1N2p1 1 2 2 dk k2

to 0 gives k2 = p1/p2.

14.3 (a) The estimator is unbiased because each component is unbiased for its respective population quantity. The variance formula follows because the random variables for inclusion in sample A are independent of the random variables for inclusion in sample B. (b) We take the derivative of the variance with respect to θ. £ A¤ £ ¤ d © £ ˆA¤ V ta + θ2V tˆab + 2θCov tˆaA, tˆAab dθ £ ¤ £ ¤ £ ¤} +(1 — θ)2V tˆBab + V tˆBb + 2(1 — θ)Cov tˆBb , tˆBab £ ¤ £ ¤ £ ¤ £ ¤ B = 2θV tˆAab + 2Cov tˆA,atˆA ab — 2(1 — θ)V tˆBab — 2Cov tˆbB, tˆab

d V (tˆy,θ ) = dθ

Setting the derivative equal to 0 and solving gives the optimal value of θ.

14.7 (a) We write θ˜d(a) — θd = a(xT β + vd + ed) + (1 — a)xT β — (xT β + vd). d

Then

d

d

E[θ˜d(a) — θd] = E[a(vd + ed) — vd] = 0.

(b) © } V [θ˜d(a) — θd] = E [a(vd + ed) — vd]2 = (a — 1)2σv2 + a2ψd since ed and vd are independent. d [(a — 1)2σv2 + a2ψ d] = 2(a — 1)σv2 + 2aψ d; da

249 setting this equal to 0 and solving for a gives a = σ2/(σ2 + ψd) = αd. The minimum v

variance achieved is V [θ˜d(αd) — θd] = (αd — 1)2σ2 + α2ψd µ = = =

v

d

¶2

µ

v

σv2 σv + σv2 + ψd ψdσ4 v + (σv2 + ψd)2 (σv2 + ψd)2 ψdσ2(ψd + σ2) ψd σv2 + ψd ψd2σv2 v

(σ2v + ψd)2 = αdψd.

2

¶2 ψd

v

14.8 Here is SAS code for construction the population and samples:

options ls=78 nodate nocenter; data domainpop; do strat = 1 to 20; do psu = 1 to 4; do j = 1 to 3; y = strat; dom = 1; output; end; end; do psu = 5 to 8; do j = 1 to 3; y = strat; dom = 2; output; end; end; end; proc sort data=domainpop; by strat psu; proc print data=domainpop; run; /* Select SRS of 2 psus from each stratum */ data psuid; do strat = 1 to 20; do psu = 1 to 8;

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250 CHAPTER 14.

RARE POPULATIONS AND SMALL AREA ESTIMATION

output; end; end; proc surveyselect data=psuid out=psusamp1 sampsize=2 seed=425; strata strat; proc sort data=psusamp1; by strat psu; proc print data=psusamp1; run; /* Merge back with data */ data samp1 ; merge psusamp1 (in=Insample) domainpop ; /* When a data set contributes an observation for the current BY group, the IN= value is 1. */ by strat psu; if Insample ; /*delete obsns not in sample */ run; proc print data=samp1; run; /* Here is the correct analysis */ proc surveymeans data=samp1 nobs mean sum clm clsum; stratum strat; cluster psu; var y; weight SamplingWeight; domain dom; run; /*Now do incorrect analysis by deleting observations not in domain*/ data samp1d1; set samp1; if dom = 1; proc surveymeans data=samp1d1 nobs mean sum clm clsum; stratum strat; cluster psu; var y;

251 weight SamplingWeight; run;

data samp1d2; set samp1; if dom = 2; proc surveymeans data=samp1d2 nobs mean sum clm clsum; stratum strat; cluster psu; var y; weight SamplingWeight; run;

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RARE POPULATIONS AND SMALL AREA ESTIMATION

Chapter 15

Survey Quality 15.2 This is a stratified sample, so we use formulas from stratified sampling to find φ̂ and Vˆ (φ̂ ) .

Nh

nh

yes

φˆh

Nh ˆ φh N

Nh — nh N 2 s2 h h Nh N 2 nh

8972 1548 860 11380

900 150 80 1130

123 27 27 177

0.1367 0.1800 0.3375

0.1077 0.0245 0.0255 0.1577

7.34 × 10−5 1.66 × 10−5 1.47 × 10−5 1.05 × 10−4

Stratum Undergrads Graduates Professional Total

Thus φ̂ = 0.1577 with Vˆ ( φ̂ ) = 1.05× 10—4. The probability P that a person is asked the sensitive question is the probability that a red ball is drawn from the box, 30/50. Also, pI = P (white ball drawn | red ball not drawn) = 4/20.

Thus, using (12.10), pˆS = and

φ̂ — (1 — P )pI 0.1577 — (1 — .6)(.2) = = 0.130 P .6 —4

1.05 × 10 ˆ V (p̂S ) = (0.6)2

= 2.91 × 10

4

so the standard error is 0.17.

15.3 (a) P (“1”) = P (“1” | sensitive)ps + P (“1” | not sensitive)(1 — ps) = θ1ps + θ2(1 — ps) (b) Let pˆ be the proportion of respondents who report “1.” Let pˆs =

p̂ — θ2 . θ1 — θ2 253

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CHAPTER 15. SURVEY QUALITY

(We must have θ1 /= θ2.) (c) If an SRS is taken, V (p̂ s ) = =

1

V (pˆ) (θ1 — θ2)2 1 p(1 — p) .

2 (θ1 — θ2) n — 1

Appendix A: Probability Concepts Used in Sampling A.1

A.2

µ ¶µ ¶ 53 30 (10)(435) 4350 2 (match exactly 3 numbers) = = = µ ¶ P 324,632 35 324,632 5 P (match at least 1 number) = 1 — P µ (¶mµa t c h¶ n o numbers) 5 30 0 5 =1 — µ ¶ 35 5 142,506 182,126 =1— = .

324,632 324,632 µ ¶µ ¶ 3 5 5 0 4 (no 7s) = = µ ¶ P 8 70 4 µ ¶µ ¶ 3 5 1 3 30 (exactly one 7) = = µ ¶ P 8 70 4 µ ¶µ ¶ 3 5 2 2 30 (exactly two 7s) = = . µ ¶ P 8 70 4

A.3 Property 1: Let Y = g(x). Then P (Y = y) =

Σ

P (X = x)

x:g(x)=y

255

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CHAPTER 15. SURVEY QUALITY

so, using (A.3), Σ

E[Y ] =

yP (Y = y)

y

Σ

=

Σ

P (X = x)

y y

x:g(x)=y

Σ Σ

=

g(x)P (X = x)

y x:g(x)=y

Σ

=

g(x)P (X = x).

x

Property 2: Using Property 1, let g(x) = aX + b. Then Σ E[aX + b] = (ax + b)P (X = x) Σ

x

= a

xP (X = x) + b

Σ

x

=

P (X = x)

x

aE[X] + b.

Property 3: If X and Y are independent, then P (X = x, Y = y) = P (X = x)P (Y = y) for all x and y. Then ΣΣ E[XY ] = xyP (X = x, Y = y) x

=

y

ΣΣ x

=

·Σ

xyP (X = x)P (Y = y)

y

xP (X = x)

¸· Σ

x

¸ yP (Y = y)

y

=

(EX)(EY ).

=

E[(X — EX)(Y — EY )]

=

E[XY — Y (EX) — X(EY ) — (EX)(EY )]

=

E[XY ] — (EX)(EY ).

Property 4: Cov [X, Y ]

257 Property 5: Using Property 4, ·Σ ¸ n m Σ Cov aiXi + bi, cjYj + dj i=1

j=1

¸ ·Σ n Σ m = E (aiXi + bi)(cjYj + dj) i=1 j=1 ·Σ n

¸ ·Σ ¸ m (aiXi + bi) E (cjYj + dj)

—E

i=1

j=1

ΣΣ [aicjE(XiYj) + aidjEXi + bicjEYj + bidj] n

=

m

i=1 j=1 n

m ΣΣ — [aiE(Xi) + bi][cjE(Yj) + dj] i=1 j=1 m

ΣΣ n

=

aicj[E(XiYj) — (EXi)(EYj)]

i=1 j=1 n m

ΣΣ

=

aicjCov [Xi, Yj].

i=1 j=1

Property 6: Using Property 4, V [X] = Cov (X, X) = E[X2] — (EX)2.

Property 7: Using Property 5, V [X + Y ]

= Cov [X + Y, X + Y ] = Cov [X, X] + Cov [Y, X] + Cov [X, Y ] + Cov [Y, Y ] = V [X] + V [Y ] + 2Cov [X, Y ].

(It follows from the definition of Cov that Cov [X, Y ] = Cov [Y, X].)

Property 8: From Property 7, ¸ · X ¸ · X ¸ · Y Y V = V +V √ + √ √ √ V (Y ) V (X) V (X) · V (Y ) ¸ X Y +2Cov √ , √ V (Y ) V (X) = 2 + 2Corr [X, Y ].

Since the variance on the left must be nonnegative, we have 2 + 2 Corr [X, Y ] ≥ 0, which implies Corr [X, Y ] ≥ —1.

Similarly, the relation 0

≤

V

· √

X V (X)

Y

—√ V (Y )

¸ = 2 — 2 Corr [X, Y ]

Solutions-Manual Sampling-Design-And-Analysis-2nd-Edition-Sharon-Lohr-

258

CHAPTER 15. SURVEY QUALITY

implies that Corr [X, Y ] ≤ 1.

A.4 Note that Z2 = Zi, so E[Z2] = E[Zi] = n/N . Thus, i

i

V [Zi] = E[Z2i] — [E(Zi)]2 n µ n ¶2 = — N N n(N — n) = .

N2 Cov [Zi, Zj] = E[ZiZj] — (EZi)(EZj) µ ¶2 n(n — 1) n = — N (N — 1) N n(n — 1)N — n2(N — 1) = N 2(N — 1) n(N — n) =— 2 .

N (N — 1) A.5

Cov [ x̄ , y¯] Corr [¯ x,y¯] = √ V [x̄]V1[y¯] n (1 — )RS S x y r N n = 1 n 2 1 n 2 [ (1 — )Sx][ (1 — )Sy ] n N n N = R.