Survey Sampling Procedures

Description

Sampling procedures from the book 'Stichproben - Methoden und praktische Umsetzung mit R' by Goeran Kauermann and Helmut Kuechenhoff (2010).

Details

Index:

Author(s)

Author: Juliane Manitz <r@manitz.org>, contributions by
Mark Hempelmann <mark.hempelmann@o2online.de>,
Goeran Kauermann <gkauermann@wiwi.uni-bielefeld.de>,
Helmut Kuechenhoff <kuechenhoff@stat.uni-muenchen.de>,
Shuai Shao <shuai.shao@campus.lmu.de>,
Cornelia Oberhauser <conny.oberhauser@gmx.de>,
Nina Westerheide <nwesterheide@wiwi.uni-bielefeld.de>,
Manuel Wiesenfarth <m.wiesenfarth@uni-goettingen.de>

Maintainer: Juliane Manitz <r@manitz.org>

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

Sampling Mean Estimation

Description

The function Smean estimates the population mean out of simple samples either with or without consideration of finite population correction.

Usage

Smean(y, N = Inf, level = 0.95)

Arguments

Value

The function Smean returns a value, which is a list consisting of the components

Author(s)

Juliane Manitz

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

See Also

Sprop, sample.size.mean

Examples

data(pop)
Y <- pop$Y
Y
# Draw a random sample of size=3
set.seed(93456)
y <- sample(x = Y, size = 3)
sort(y)
# Estimation with infiniteness correction
est <- Smean(y = y, N = length(pop$Y))
est

Sampling Proportion Estimation

Description

The function Sprop estimates the proportion out of samples either with or without consideration of finite population correction. Different methods for calculating confidence intervals for example based on binomial distribution (Agresti and Coull or Clopper-Pearson) or based on hypergeometric distribution are used.

Usage

Sprop(y, m, n = length(y), N = Inf, level = 0.95)

Arguments

Details

Sprop can be called by usage of a data vector y with the observations 1 for event and 0 for failure. Moreover, it can be called by specifying the number of events m and trials n.

Value

The function Sprop returns a value, which is a list consisting of the components

Author(s)

Juliane Manitz

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

Agresti, Alan/Coull, Brent A. (1998): Approximate Is Better than 'Exact' for Interval Estimation of Binomial Proportions. The American Statistician, Vol. 52, No. 2 , pp. 119-126.

See Also

Smean, sample.size.prop

Examples

# 1) Survey in company to upgrade office climate
Sprop(m=45, n=100, N=300)
Sprop(m=2, n=100, N=300)

# 2) German opinion poll for 03/07/09 with 
# (http://www.wahlrecht.de/umfragen/politbarometer.htm)
# a) 302 of 1206 respondents who would elect SPD.
# b) 133 of 1206 respondents who would elect the Greens.
Sprop(m=302, n=1206, N=Inf)
Sprop(m=133, n=1206, N=Inf)

# 3) Rare disease of animals (sample size n=500 of N=10.000 animals, one infection)
# for 95% one sided confidence level use level=0.9
Sprop(m=1, n=500, N=10000, level=0.9)

# 4) call with data vector y
y <- c(0,0,1,0,1,0,0,0,1,1,0,0,1)
Sprop(y=y, N=200)
# is the same as
Sprop(m=5, n=13, N=200)

German Parliament Election Data

Description

Data frame with number of citizens eligible to vote and results of the elections in 2002 and 2005 for the German Bundestag, the first chamber of the German parliament.

Usage

data(election)

Format

A data frame with 299 observations (corresponding to constituencies) on the following 13 variables.

state

factor, the 16 German federal states

eligible_02

number of citizens eligible to vote in 2002

SPD_02

a numeric vector, percentage for the Social Democrats SPD in 2002

UNION_02

a numeric vector, percentage for the conservative Christian Democrats CDU/CSU in 2002

GREEN_02

a numeric vector, percentage for the Greens in 2002

FDP_02

a numeric vector, percentage for the Liberal Party FDP in 2002

LEFT_02

a numeric vector, percentage for the Left Party PDS in 2002

eligible_05

number of citizens eligible to vote in 2005

SPD_05

a numeric vector, percentage for the Social Democrats SPD in 2005

UNION_05

a numeric vector, percentage for the conservative Christian Democrats CDU/CSU in 2005

GREEN_05

a numeric vector, percentage for the Greens in 2005

FDP_05

a numeric vector, percentage for the Liberal Party FDP in 2005

LEFT_05

a numeric vector, percentage for the Left Party in 2005

Details

German Federal Elections

Half of the Members of the German Bundestag are elected directly from Germany's 299 constituencies, the other half one on the parties' land lists. Accordingly, each voter has two votes in the elections to the German Bundestag. The first vote, allowing voters to elect their local representatives to the Bundestag, decides which candidates are sent to Parliament from the constituencies. The second vote is cast for a party list. And it is this second vote that determines the relative strengths of the parties represented in the Bundestag. At least 598 Members of the German Bundestag are elected in this way. In addition to this, there are certain circumstances in which some candidates win what are known as 'overhang mandates' when the seats are being distributed.

The data set provides the percentage of second votes for each party, which determines the number of seats each party gets in parliament. These percentages are calculated by the number of votes for a party divided by number of valid votes.

Source

The data is provided by the R package flexclust.

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

Homepage of the Bundestag: http://www.bundestag.de.

Friedrich Leisch. A Toolbox for K-Centroids Cluster Analysis. Computational Statistics and Data Analysis, 51 (2), 526-544, 2006.

Examples

data(election)
summary(election)

# 1) Draw a simple sample of size n=20
n <- 20
set.seed(67396)
index <- sample(1:nrow(election), size=n)
sample1 <- election[index,]
Smean(sample1$SPD_02, N=nrow(election))
# true mean
mean(election$SPD_02)

# 2) Estimate sample size to forecast proportion of SPD in election of 2005
sample.size.prop(e=0.01, P=mean(election$SPD_02), N=Inf)

# 3) Usage of previous knowledge by model based estimation
# draw sample of size n = 20
N <- nrow(election)
set.seed(67396)
sample <- election[sort(sample(1:N, size=20)),]
# secondary information SPD in 2002
X.mean <- mean(election$SPD_02)
# forecast proportion of SPD in election of 2005
mbes(SPD_05 ~ SPD_02, data=sample, aux=X.mean, N=N, method='all')
# true value
Y.mean <- mean(election$SPD_05)
Y.mean
# Use a second predictor variable
X.mean2 <- c(mean(election$SPD_02),mean(election$GREEN_02))
# forecast proportion of SPD in election of 2005 with two predictors
mbes(SPD_05 ~ SPD_02+GREEN_02, data=sample, aux=X.mean2, N=N, method= 'regr')

Horvitz-Thompson Estimator

Description

Calculates Horvitz-Thompson estimate with different methods for variance estimation such as Yates and Grundy, Hansen-Hurwitz and Hajek.

Usage

htestimate(y, N, PI, pk, pik, method = 'yg')

Arguments

Details

For using methods 'yg' or 'ht' has to be provided matrix PI, and for 'hh' and 'ha' has to be specified vector pk of inclusion probabilities. Additionally, for Hajek method 'ha' can be specified pik. Unless, an approximate Hajek method is used.

Value

The function htestimate returns a value, which is a list consisting of the components

Author(s)

Juliane Manitz

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

See Also

pps.sampling

Examples

data(influenza)
summary(influenza)

# pps.sampling()
set.seed(108506)
pps <- pps.sampling(z=influenza$population,n=20,method='midzuno')
sample <- influenza[pps$sample,]
# htestimate()
N <- nrow(influenza)
# exact variance estimate
PI <- pps$PI
htestimate(sample$cases, N=N, PI=PI, method='yg')
htestimate(sample$cases, N=N, PI=PI, method='ht')
# approximate variance estimate
pk <- pps$pik[pps$sample]
htestimate(sample$cases, N=N, pk=pk, method='hh')
pik <- pps$pik
htestimate(sample$cases, N=N, pk=pk, pik=pik, method='ha')
# without pik just approximate calculation of Hajek method
htestimate(sample$cases, N=N, pk=pk, method='ha') 
# calculate confidence interval based on normal distribution for number of cases
est.ht <- htestimate(sample$cases, N=N, PI=PI, method='ht')
est.ht$mean*N  
lower <- est.ht$mean*N - qnorm(0.975)*N*est.ht$se
upper <- est.ht$mean*N + qnorm(0.975)*N*est.ht$se
c(lower,upper) 
# true number of influenza cases
sum(influenza$cases)

Population and Cases of Influenza for Administrative Districts of Germany

Description

The data frame influenza provides cases of influenza and inhabitants for administrative districts of Germany in 2007.

Usage

data(influenza)

Format

A data frame with 424 observations on the following 4 variables.

id

a numeric vector

district

a factor with levels LK Aachen, LK Ahrweiler, ..., SK Zweibruecken, names of administrative districts in Germany

population

a numeric vector specifying the number of inhabitants in the specific administrative district

cases

a numeric vector specifying the number of influenza cases in the specific administrative district

Details

Data of 2007. If you want to use the population numbers in the future, be aware of local governmental reorganizations, e.g. district unions.

Source

Database SurvStat of Robert Koch-Institute. Many thanks to Hermann Claus.

References

Database of Robert Koch-Institute http://www3.rki.de/SurvStat/

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

Examples

data(influenza)
summary(influenza)

# 1) Usage of pps.sampling
set.seed(108506)
pps <- pps.sampling(z=influenza$population,n=20,method='midzuno')
pps
sample <- influenza[pps$sample,]
sample

# 2) Usage of htestimate
set.seed(108506)
pps <- pps.sampling(z=influenza$population,n=20,method='midzuno')
sample <- influenza[pps$sample,]
# htestimate()
N <- nrow(influenza)
# exact variance estimate
PI <- pps$PI
htestimate(sample$cases, N=N, PI=PI, method='ht')
htestimate(sample$cases, N=N, PI=PI, method='yg')
# approximate variance estimate
pk <- pps$pik[pps$sample]
htestimate(sample$cases, N=N, pk=pk, method='hh')
pik <- pps$pik
htestimate(sample$cases, N=N, pk=pk, pik=pik, method='ha')
# without pik just approximative calculation of Hajek method
htestimate(sample$cases, N=N, pk=pk, method='ha') 
# calculate confidence interval based on normal distribution for number of cases
est.ht <- htestimate(sample$cases, N=N, PI=PI, method='ht')
est.ht$mean*N  
lower <- est.ht$mean*N - qnorm(0.975)*N*est.ht$se
upper <- est.ht$mean*N + qnorm(0.975)*N*est.ht$se
c(lower,upper) 
# true number of influenza cases
sum(influenza$cases)

Model Based Estimation

Description

mbes is used for model based estimation of population means using auxiliary variables. Difference, ratio and regression estimates are available.

Usage

mbes(formula, data, aux, N = Inf, method = 'all', level = 0.95, ...)

Arguments

Details

The option method='simple' calculates the simple sample estimation without using the auxiliary variable. The option method='diff' calculates the difference estimate, method='ratio' the ratio estimate, and method='regr' the regression estimate which is based on the selected model. The option method='all' calculates the simple and all model based estimates. For methods 'diff', 'ratio' and 'all' the formula has to be y~x with y primary and x secondary information. For method 'regr', it is the symbolic description of the linear regression model. In this case, it can be used more than one auxiliary variable. Thus, aux has to be a vector of the same length as the number of auxiliary variables in order as specified in the formula.

Value

The function mbes returns an object, which is a list consisting of the components

Author(s)

Juliane Manitz

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

See Also

Smean, Sprop

Examples

## 1) simple suppositious example
data(pop)
# Draw a random sample of size=3
set.seed(802016)
data <- pop[sample(1:5, size=3),]
names(data) <- c('id','x','y')
# difference estimator
mbes(formula=y~x, data=data, aux=15, N=5, method='diff', level=0.95)
# ratio estimator
mbes(formula=y~x, data=data, aux=15, N=5, method='ratio', level=0.95)
# regression estimator
mbes(formula=y~x, data=data, aux=15, N=5, method='regr', level=0.95)

## 2) Bundestag election
data(election)
# draw sample of size n = 20
N <- nrow(election)
set.seed(67396)
sample <- election[sort(sample(1:N, size=20)),]
# secondary information SPD in 2002
X.mean <- mean(election$SPD_02)
# forecast proportion of SPD in election of 2005
mbes(SPD_05 ~ SPD_02, data=sample, aux=X.mean, N=N, method='all')
# true value
Y.mean <- mean(election$SPD_05)
Y.mean
# Use a second predictor variable
X.mean2 <- c(mean(election$SPD_02),mean(election$GREEN_02))
# forecast proportion of SPD in election of 2005 with two predictors
mbes(SPD_05 ~ SPD_02+GREEN_02, data=sample, aux=X.mean2, N=N, method= 'regr')

## 3) money sample
data(money)
mu.X <-  mean(money$X)
x <- money$X[which(!is.na(money$y))]
y <- na.omit(money$y)
# estimation
mbes(y~x, aux=mu.X, N=13, method='all')

## 4) model based two-phase sampling with mbes() 
id <- 1:1000
x <- rep(c(1,0,1,0),times=c(10,90,70,830))
y <- rep(c(1,0,NA),times=c(15,85,900))
phase <- rep(c(2,1), times=c(100,900))
data <- data.frame(id,x,y,phase)
# mean of x out of first phase
mean.x <- mean(data$x)
mean.x
N1 <- length(data$x) 
# calculation of estimation for y 
est.y <- mbes(y~x, data=data, aux=mean.x, N=N1, method='ratio')
est.y
# correction of standard error with uncertaincy in first phase
v.y <- var(data$y, na.rm=TRUE)
se.y <- sqrt(est.y$ratio$se^2 + v.y/N1)
se.y
# corrected confidence interval
lower <- est.y$ratio$mean - qnorm(0.975)*se.y
upper <- est.y$ratio$mean + qnorm(0.975)*se.y
c(lower, upper)

Money Data Frame

Description

Data provides guesses and true values for students wallet money.

Usage

data(money)

Format

A data frame with 13 observations (corresponding to the students) on the following 3 variables.

id

a numeric vector of identification number

X

a numeric vector of secondary information, guesses of money in the wallet

y

a numeric vector of primary information, counted money in the wallet. NA means subject was not included into the sample.

Details

In a lesson an experiment was made, in which the students were asked to guess the current amount of money in their wallet. A simple sample of these students was drawn, who counted the money in their wallet exactly. Using this secondary information, model based estimation of the population mean is possible.

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

Examples

data(money)
print(money)

# Usage of mbes()
mu.X <-  mean(money$X)
x <- money$X[which(!is.na(money$y))]
y <- na.omit(money$y)
# estimation
mbes(y~x, aux=mu.X, N=13, method='all')

Small Suppositious Sampling Example

Description

pop is a suppositious data frame for a small population with 5 elements. It is used for simple illustration of survey sampling estimators.

Usage

data(pop)

Format

A data frame with 5 observations on the following 3 variables.

id

a numeric vector of individual identification values

X

a numeric vector of first characteristic

Y

a numeric vector of second characteristic

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

Examples

data(pop)
print(pop)

## 1) Usage of Smean()
data(pop)
Y <- pop$Y
Y
# Draw a random sample pop size=3
set.seed(93456)
y <- sample(x = Y, size = 3)
sort(y)
# Estimation with infiniteness correction
est <- Smean(y = y, N = length(pop$Y))
est

## 2) Usage of mbes()
data(pop)
# Draw a random sample of size=3
set.seed(802016)
data <- pop[sample(1:5, size=3),]
names(data) <- c('id','x','y')
# difference estimator
mbes(formula=y~x, data=data, aux=15, N=5, method='diff', level=0.95)
# ratio estimator
mbes(formula=y~x, data=data, aux=15, N=5, method='ratio', level=0.95)
# regression estimator
mbes(formula=y~x, data=data, aux=15, N=5, method='regr', level=0.95)

Sampling with Probabilities Proportional to Size

Description

The function provides sample techniques with sampling probabilities which are proportional to the size of a quantity z.

Usage

pps.sampling(z, n, id = 1:N, method = 'sampford', return.PI = FALSE)

Arguments

Details

The different methods vary in their run time. Therefore, method='sampford' is stopped if N > 200 or if n/N < 0.3. method='tille' is stopped if N > 500. In case of large populations use method='midzuno' or method='madow'.

Value

The function pps.sampling returns a value, which is a list consisting of the components

Author(s)

Juliane Manitz

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

See Also

htestimate

Examples

## 1) simple suppositious example
data <- data.frame(id = 1:7, z = c(1.8, 2 ,3.2 ,2.9 ,1.5 ,2.0 ,2.2))
# Usage of pps.sampling for Sampford method
set.seed(178209)
pps.sample_sampford <- pps.sampling(z=data$z, n=2, method='sampford', return.PI=FALSE)
pps.sample_sampford
# sampling elements
id.sample <- pps.sample_sampford$sample
id.sample
# other methods
set.seed(178209)
pps.sample_tille <- pps.sampling(z=data$z, n=2, method='tille')
pps.sample_tille
set.seed(178209)
pps.sample_midzuno <- pps.sampling(z=data$z, n=2, method='midzuno')
pps.sample_midzuno
set.seed(178209)
pps.sample_madow <- pps.sampling(z=data$z, n=2, method='madow')
pps.sample_madow

## 2) influenza
data(influenza)
summary(influenza)

set.seed(108506)
pps <- pps.sampling(z=influenza$population,n=20,method='midzuno')
pps
sample <- influenza[pps$sample,]
sample

Sample Size Calculation for Mean Estimation

Description

The function sample.size.mean returns the sample size needed for mean estimations either with or without consideration of finite population correction.

Usage

sample.size.mean(e, S, N = Inf, level = 0.95)

Arguments

Value

The function sample.size.mean returns a value, which is a list consisting of the components

Author(s)

Juliane Manitz

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

See Also

Smean, sample.size.prop

Examples

# sample size for precision e=4
sample.size.mean(e=4,S=10,N=300)
# sample size for precision e=1
sample.size.mean(e=1,S=10,N=300)

Sample Size Calculation for Proportion Estimation

Description

The function sample.size.prop returns the sample size needed for proportion estimation either with or without consideration of finite population correction.

Usage

sample.size.prop(e, P = 0.5, N = Inf, level = 0.95)

Arguments

Details

For meaningful calculation, precision e should be chosen smaller than 0.5, because the domain of P is between values 0 and 1. Furthermore, precision e should be smaller than proportion P, respectively (1-P).

Value

The function sample.size.prop returns a value, which is a list consisting of the components

Author(s)

Juliane Manitz

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

See Also

Sprop, sample.size.mean

Examples

## 1) examples with different precisions
# precision 1% for election forecast of SPD in 2005
sample.size.prop(e=0.01, P=0.5, N=Inf)
data(election)
sample.size.prop(e=0.01, P=mean(election$SPD_02), N=Inf)
# precision 5% for questionnaire
sample.size.prop(e=0.05, P=0.5, N=300)
sample.size.prop(e=0.05, P=0.5, N=Inf)
# precision 10%
sample.size.prop(e=0.1, P=0.5, N=300)
sample.size.prop(e=0.1, P=0.5, N=1000)

## 2) tables in the book
# table 2.2
P_vector <- c(0.2, 0.3, 0.4, 0.5)
N_vector <- c(10, 100, 1000, 10000)
results <- matrix(NA, ncol=4, nrow=4)
for (i in 1:length(P_vector)){
  for (j in 1:length(N_vector)){
    x <- try(sample.size.prop(e=0.1, P=P_vector[i], N=N_vector[j]))
    if (class(x)=='try-error') {results[i,j] <- NA}
    else {results[i,j] <- x$n}
  }
}
dimnames(results) <- list(paste('P=',P_vector, sep=''), paste('N=',N_vector, sep=''))
results
# table 2.3
P_vector <- c(0.5, 0.1)
e_vector <- c(0.1, 0.05, 0.03, 0.02, 0.01)
results <- matrix(NA, ncol=2, nrow=5)
for (i in 1:length(e_vector)){
  for (j in 1:length(P_vector)){
    x <- try(sample.size.prop(e=e_vector[i], P=P_vector[j], N=Inf))
    if (class(x)=='try-error') {results[i,j] <- NA}
    else {results[i,j] <- x$n}
  }
}
dimnames(results) <- list(paste('e=',e_vector, sep=''), paste('P=',P_vector, sep=''))
results

Stratified Sample Mean Estimation

Description

The function stratamean estimates the population mean out of stratified samples either with or without consideration of finite population correction.

Usage

stratamean(y, h, Nh, wh, level = 0.95, eae = FALSE)

Arguments

Details

If the absolute stratum sizes Nh are given, the variances are calculated with finite population correction. Otherwise, if the stratum weights wh are given, the variances are calculated without finite population correction.

Value

The function stratamean returns a value, which is a list consisting of the components

Author(s)

Shuai Shao and Juliane Manitz

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

See Also

Smean, Sprop

Examples

# random data
testy <- rnorm(100)
testh <- c(rep("male",40), rep("female",60))
stratamean(testy, testh, wh=c(0.5, 0.5))
stratamean(testy, testh, wh=c(0.5, 0.5), eae=TRUE)

# tax data
data(tax)
summary(tax)

nh <- as.vector(table(tax$Class))
wh <- nh/sum(nh)
stratamean(y=tax$diff, h=as.vector(tax$Class), wh=wh, eae=TRUE)

Sample Size Calculation for Stratified Sampling

Description

The function stratasamp calculates the sample size for each stratum depending on type of allocation.

Usage

stratasamp(n, Nh, Sh = NULL, Ch = NULL, type = 'prop')

Arguments

Value

The function stratasamp returns a matrix, which lists the strata and the sizes of observation depending on type of allocation.

Author(s)

Shuai Shao and Juliane Manitz

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

See Also

stratamean, stratasize, sample.size.mean

Examples

#random proportional stratified sample
stratasamp(n=500, Nh=c(5234,2586,649,157))
stratasamp(n=500, Nh=c(5234,2586,649,157), Sh=c(251,1165,8035,24725), type='opt')

Sample Size Determination for Stratified Sampling

Description

The function stratasize determinates the total size of stratified samples depending on type of allocation and determinated by specified precision.

Usage

stratasize(e, Nh, Sh, level = 0.95, type = 'prop')

Arguments

Value

The function stratasize returns a value, which is a list consisting of the components

Author(s)

Shuai Shao

References

Kauermann, Goeran/Kuechenhoff, Helmut (2011): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

See Also

stratasamp, stratamean

Examples

#random proportional stratified sample
stratasize(e=0.1, Nh=c(100000,300000,600000), Sh=c(1,2,3))

#random optimal stratified sample
stratasize(e=0.1, Nh=c(100000,300000,600000), Sh=c(1,2,3), type="opt")

Sub-sample Mean Estimation

Description

The function submean estimates the population mean out of sub-samples (two-stage samples) either with or without consideration of finite population correction in both stages.

Usage

submean(y, PSU, N, M, Nl, m.weight, n.weight, method = 'simple', level = 0.95)

Arguments

Details

If the absolute sizes M and Nl are given, the variances are calculated with finite population correction. Otherwise, if the weights m.weight and n.weight are given, the variances are calculated without finite population correction.

Value

The function submean returns a value, which is a list consisting of the components

Author(s)

Shuai Shao and Juliane Manitz

References

Kauermann, Goeran/Kuechenhoff, Helmut (2011): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

See Also

Smean, stratamean

Examples

y <- c(23,33,24,25,72,74,71,37,42)
psu <- as.factor(c(1,1,1,1,2,2,2,3,3))
# with finite population correction
submean(y, PSU=psu, N=700, M=23, Nl=c(100,50,75), method='ratio')
# without finite population correction
submean(y, PSU=psu, N=700, m.weight=3/23, n.weight=c(4/100,3/50,2/75), method='ratio')

# Chinese wage data
data(wage)
summary(wage)
submean(wage$Wage,PSU=wage$Region, N=990, M=33, Nl=rep(30,14))

Hypothetical Tax Refund Data Frame

Description

Simulated tax refund data frame including the estimated and actual refund value

Usage

data(tax)

Format

A data frame with 9083 observations on the following 5 variables.

id

a numeric vector indicating the tax payer

estRefund

a numeric vector representing the estimated value of tax refund by the tax payer

actRefund

a numeric vector representing the actual tax refund calculated by the financial authority

diff

difference between estimated and acture tax refund

Class

a factor with levels 1, 2, 3, and 4 indicating the strata

Source

Due to data protection this is a simulated data set reflecting the real data.

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

Examples

data(tax)
summary(tax)

# illustration of stratamean
nh <- as.vector(table(tax$Class))
wh <- nh/sum(nh)
stratamean(y=tax$diff, h=as.vector(tax$Class), wh=wh, eae=TRUE)

Chinese wage data

Description

A data frame with hypothetical Chinese wages differenciated by region and industrial sector.

Usage

data(wage)

Format

A data frame with 231 observations on the following 3 variables.

Region

factor, Chinese regions with 14 levels.

Sector

factor, industrial sector with 30 levels.

Wage

a numeric vector, average wage in the region and sector measured in Chinese yuan.

Details

The dataset is hypothetical. Its structure imitates the data in the Chinese Statistical Yearbook. The values are simulated corresponding to the distribution of the real data which are not publicly accessible.

References

Kauermann, Goeran/Kuechenhoff, Helmut (2010): Stichproben. Methoden und praktische Umsetzung mit R. Springer.

Examples

# Chinese wage data
data(wage)
summary(wage)
submean(wage$Wage,PSU=wage$Region, N=990, M=33, Nl=rep(30,14))