Nothing
inst/scripts/EPA.2009/Chapter22Examples.R
In EnvStats: Package for Environmental Statistics, Including US EPA Guidance
#####
# File: Script - EPA09, Chapter 22 Examples.R
#
# Purpose: Reproduce Examples in Chapter 22 of the EPA Guidance Document
#
# USEPA. (2009). "Statistical Analysis of Ground Water
# Monitoring Data at RCRA Facilities, Unified Guidance."
# EPA 530-R-09-007.
# Office of Resource Conservation and Recovery,
# Program Information and Implementation Division.
# March 2009.
#
# Errata are listed in:
# USEPA. (2010). "Errata Sheet - March 2009 Unified Guidance."
# EPA 530/R-09-007a.
# Office of Resource Conservation and Recovery,
# Program Information and Implementation Division.
# August 9, 2010.
#
# Author: Steven P. Millard
#
# Last
# Updated: October 4, 2012
#####
#NOTE: Unless otherwise noted, differences between what is shown in the
# EPA Guidance Document and results shown here are due to the
# EPA Guidance Document rounding intermediate results prior to
# the final result.
#######################################################################################
library(EnvStats)
# Example 22-1, pp. 22-6 to 22-8
#-------------------------------
EPA.09.Ex.22.1.VC.df
# Year Quarter Period Well VC.ppb
#1 1 1 Background GW-1 6.3
#2 1 2 Background GW-1 9.5
#3 1 3 Background GW-1 8.1
#4 1 4 Background GW-1 11.9
#5 2 1 Compliance GW-1 7.3
#6 2 2 Compliance GW-1 11.2
#7 2 3 Compliance GW-1 6.0
#8 2 4 Compliance GW-1 7.5
#9 3 1 Compliance GW-1 8.4
#10 3 2 Compliance GW-1 6.4
#11 3 3 Compliance GW-1 8.9
#12 3 4 Compliance GW-1 4.9
#13 4 1 Compliance GW-1 9.6
#14 4 2 Compliance GW-1 9.7
#15 4 3 Compliance GW-1 8.7
#16 4 4 Compliance GW-1 8.7
#17 1 1 Background GW-2 5.9
#18 1 2 Background GW-2 3.0
#19 1 3 Background GW-2 8.8
#20 1 4 Background GW-2 12.0
#21 2 1 Compliance GW-2 11.2
#22 2 2 Compliance GW-2 8.6
#23 2 3 Compliance GW-2 12.6
#24 2 4 Compliance GW-2 7.2
#25 3 1 Compliance GW-2 13.8
#26 3 2 Compliance GW-2 5.6
#27 3 3 Compliance GW-2 11.0
#28 3 4 Compliance GW-2 9.8
#29 4 1 Compliance GW-2 6.3
#30 4 2 Compliance GW-2 10.4
#31 4 3 Compliance GW-2 7.5
#32 4 4 Compliance GW-2 9.7
# Step 1: Summary Statistics and Time Series Plots
#--------------------------------------------------
# Summary Statistics by Well
#---------------------------
summaryStats(VC.ppb ~ Well, data = EPA.09.Ex.22.1.VC.df,
combine.groups = TRUE, digits = 1, stats.in.rows = TRUE)
# GW-1 GW-2 Combined
#N 16 16 32
#Mean 8.3 9 8.6
#SD 1.9 2.9 2.4
#Median 8.6 9.2 8.7
#Min 4.9 3 3
#Max 11.9 13.8 13.8
# Summary Statistics by Period and Well
#--------------------------------------
VC <- EPA.09.Ex.22.1.VC.df$VC.ppb
Period <- EPA.09.Ex.22.1.VC.df$Period
Well <- EPA.09.Ex.22.1.VC.df$Well
summaryStats(VC, group = paste(Well, Period, sep = "."),
combine.groups = FALSE, stats.in.rows = TRUE, digits = 1)
# GW-1.Background GW-1.Compliance GW-2.Background GW-2.Compliance
#N 4 12 4 12
#Mean 8.9 8.1 7.4 9.5
#SD 2.4 1.8 3.9 2.5
#Median 8.8 8.6 7.4 9.8
#Min 6.3 4.9 3 5.6
#Max 11.9 11.2 12 13.8
# Group Shapiro-Wilk Test Shows p-values for Individual Wells
#------------------------------------------------------------
gofGroupTest(VC.ppb ~ Well, data = EPA.09.Ex.22.1.VC.df)
#
#Results of Group Goodness-of-Fit Test
#-------------------------------------
#
#Test Method: Wilk-Shapiro GOF (Normal Scores)
#
#Hypothesized Distribution: Normal
#
#Data: VC.ppb
#
#Grouping Variable: Well
#
#Data Source: EPA.09.Ex.22.1.VC.df
#
#Number of Groups: 2
#
#Sample Sizes: GW-1 = 16
# GW-2 = 16
#
#Test Statistic: z (G) = 2.901581
#
#P-values for
#Individual Tests: GW-1 = 0.9686679
# GW-2 = 0.9875157
#
#P-value for
#Group Test: 0.9981436
#
#Alternative Hypothesis: At least one group
# does not come from a
# Normal Distribution.
# Time Series Plot
#-----------------
Year <- EPA.09.Ex.22.1.VC.df$Year
Qtr <- EPA.09.Ex.22.1.VC.df$Quarter
n1 <- sum(Well == "GW-1")
n2 <- sum(Well == "GW-2")
windows()
plot(1:n1, VC[Well == "GW-1"], xaxt = "n",
type = "b", pch = 15, lwd = 2,
ylim = c(0, 20),
xlab = "Time (Years)",
ylab = "Vinyl Chloride Concentration (ppb)",
main = "Figure 22-1. Vinyl Chloride Time Series Plot")
axis(1, at = (1:n1)[Qtr == 1 & Well == "GW-1"],
labels = Year[Qtr == "1" & Well == "GW-1"])
points(1:n2, VC[Well == "GW-2"], type = "b", pch = 3, lty = 3, lwd = 2)
abline(h = 5, lty = 2, lwd = 2)
legend(1, 20, c("GW-1", "GW-2", "MCL = 5 ppb"),
pch = c(15, 3, 1), lty = c(1, 3, 2),
pt.cex = c(1, 1, 0), lwd = 2)
# Step 2: Determine Type I Error level necessary in order to achieve
# 80% power to detect an increase in mean VC of 2 times
# the 5 ppb GWPS based on n=4 observations.
# Then compute one-sided lower confidence limits
# for each well using the Year 2 data based on
# this Type I error level.
#--------------------------------------------------------------------
R <- 2
mu0 <- 5
# Approach 1:
# Based on Guidance Document Method That *Does Not* Require
# an Estimate of SD or CV. This method assumes:
# 1) sigma = mean under null hypothesis
# 2) delta/sigma = R - 1, where R is the ratio of the alternative mean
# to the null mean
#------------------------------------------------------------------
required.alpha <- tTestAlpha(n.or.n1 = 4,
delta.over.sigma = R-1, power = 0.8,
sample.type = "one.sample",
alternative = "greater")
required.alpha
#[1] 0.1633382
# Lower one-sided confidence limits for Well GW-1:
enorm(VC[Well == "GW-1" & Year == "2"], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#6.693358
# Lower one-sided confidence limits for Well GW-2:
enorm(VC[Well == "GW-2" & Year == "2"], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#8.469282
# Approach 2:
# Based on standard method that requires an estimate of SD or CV.
#----------------------------------------------------------------
# Estimate standard deviation of VC based on Background data.
# Pool estimate of SDs based on both wells (GW-1 and GW-2)
VC.fit <- lm(VC.ppb ~ Well, data = EPA.09.Ex.22.1.VC.df,
subset = Period == "Background")
SD.Background <- summary(VC.fit)$sigma
SD.Background
#[1] 3.200976
#NOTE: Note that the estimated SD of 3.2 ppb is much smaller than the
# assumed value of 5 ppb based on Approach 1 above.
tTestAlpha(n.or.n1 = 4,
delta.over.sigma = mu0 * (R - 1) / SD.Background,
power = 0.8,
sample.type = "one.sample",
alternative = "greater", approx = F)
#[1] 0.05767842
#NOTE: Note the required Type I Error level based on Approach 2
# is much smaller than that based on Approach 1.
# Step 3: Compute one-sided lower confidence limits for each well
# using the Year 2 data based on Type I Error Level = 0.01.
#-------------------------------------------------------------------
# Use alpha = 0.01
# Lower one-sided confidence limits for Well GW-1:
enorm(VC[Well == "GW-1" & Year == "2"], ci = T, ci.type = "lower",
conf.level = 0.99)$interval$limits["LCL"]
# LCL
#2.926725
# Lower one-sided confidence limits for Well GW-2:
enorm(VC[Well == "GW-2" & Year == "2"], ci = T, ci.type = "lower",
conf.level = 0.99)$interval$limits["LCL"]
# LCL
#4.34498
# Step 4: Determine Type I Error level necessary in order to achieve
# 80% power to detect an increase in mean VC of 2 times
# the 5 ppb GWPS based on n=8 observations.
# Then compute one-sided lower confidence limits
# for each well using the Year 1 and Year 2 data based on
# this Type I error level.
#--------------------------------------------------------------------
required.alpha <- tTestAlpha(n.or.n1 = 8,
delta.over.sigma = R-1, power = 0.8,
sample.type = "one.sample",
alternative = "greater")
required.alpha
#[1] 0.0458989
# Lower one-sided confidence limits for Well GW-1:
enorm(VC[Well == "GW-1" & Year %in% c("1", "2")], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#6.963887
# Lower one-sided confidence limits for Well GW-2:
enorm(VC[Well == "GW-2" & Year %in% c("1", "2")], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#6.403578
# Step 5: Compute one-sided lower confidence limits
# for each well using the Year 2 and Year 3 data based on
# Type I error level from Step 4.
#--------------------------------------------------------------------
# Lower one-sided confidence limits for Well GW-1:
enorm(VC[Well == "GW-1" & Year %in% c("2", "3")], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#6.227279
# Lower one-sided confidence limits for Well GW-2:
enorm(VC[Well == "GW-2" & Year %in% c("2", "3")], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#8.078256
# Step 6: Determine Type I Error level necessary in order to achieve
# 80% power to detect an increase in mean VC of 2 times
# the 5 ppb GWPS based on n=12 observations.
# Then compute one-sided lower confidence limits
# for each well using the Years 2-4 data based on
# this Type I error level.
#--------------------------------------------------------------------
required.alpha <- tTestAlpha(n.or.n1 = 12,
delta.over.sigma = R-1, power = 0.8,
sample.type = "one.sample",
alternative = "greater")
required.alpha
#[1] 0.01311214
# Lower one-sided confidence limits for Well GW-1:
enorm(VC[Well == "GW-1" & Year != "1"], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#6.798091
# Lower one-sided confidence limits for Well GW-2:
enorm(VC[Well == "GW-2" & Year != "1"], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#7.609904
rm(Year, Qtr, Period, Well, VC, n1, n2,
R, mu0, required.alpha, VC.fit, SD.Background)
#########################################################################################
# Example 22-2, pp. 22-11 to 22-17
#---------------------------------
EPA.09.Ex.22.2.Specific.Conductance.df
# Well Date Specific.Conductance.umho
#1 GW-12 1987-10-16 2100
#2 GW-12 1988-01-28 2550
# ...
#42 GW-13 1992-12-01 530
#43 GW-13 1993-03-24 625
Well <- EPA.09.Ex.22.2.Specific.Conductance.df$Well
Date <- EPA.09.Ex.22.2.Specific.Conductance.df$Date
SC <- EPA.09.Ex.22.2.Specific.Conductance.df$Specific.Conductance.umho
#Step 1: Time series plot for Well GW-12
#---------------------------------------
index <- Well == "GW-12"
windows()
plot(Date[index], SC[index], ylim = c(0, 3000),
xlab = "Sampling Date",
ylab = "Specific Conductance (umho)",
main = "Figure 22-3. Time Series Plot of \nSpecific Conductance Measurements at GW-12")
#Step 2: Fit regression line
#----------------------------
SC.fit <- lm(Specific.Conductance.umho ~ Date,
data = EPA.09.Ex.22.2.Specific.Conductance.df,
subset = Well == "GW-12")
summary(SC.fit)$coef
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) 9526.116147 1042.1592274 9.14075 3.493624e-08
#Date -1.087567 0.1384346 -7.85618 3.170120e-07
# NOTE: These estimates of slope and intercept differ from those in the
# Guidance Document because our time variable is in terms of
# days since Jan 1, 1970, where as the units of time in the
# Guidance Document is Years.
windows()
plot(Date[index], SC[index], ylim = c(0, 3000),
xlab = "Sampling Date",
ylab = "Specific Conductance (umho)",
main = "Figure 22-4. Regression of Specific Conductance vs.\nSampling Date at GW-12")
abline(SC.fit)
# Step 3: Examine residuals
#---------------------------
windows()
qqPlot(SC.fit$residuals, xlab = "Z-score", add.line = T,
ylab = "Specific Conductance Residuals",
main = "Figure 22-5. Probability Plot of \nSpecific Conductance Residuals at GW-12")
gofTest(SC.fit$residuals)
#
#Results of Goodness-of-Fit Test
#-------------------------------
#
#Test Method: Shapiro-Wilk GOF
#
#Hypothesized Distribution: Normal
#
#Estimated Parameter(s): mean = 3.538836e-16
# sd = 3.790014e+02
#
#Estimation Method: mvue
#
#Data: SC.fit$residuals
#
#Sample Size: 20
#
#Test Statistic: W = 0.960157
#
#Test Statistic Parameter: n = 20
#
#P-value: 0.5469918
#
#Alternative Hypothesis: True cdf does not equal the
# Normal Distribution.
windows()
plot(Date[index], SC.fit$residuals, xlab = "Sampling Date",
ylab = "Specific Conductance Residuals",
main = "Figure 22-6. Plot of Specific Conductance Residuals vs. \nSampling Date at GW-12")
abline(h = 0, lty = 2)
# Alternatively, you can use used the built-in plotting function for linear models:
#windows()
#plot(SC.fit, ask = FALSE)
# Step 4: Plot Specific Conductance at GW-12 vs. Date, along with
# regression line and two-sided lower and upper 90% confidence limits
# Note that the upper confidence limit considered by itself is
# equivalent to a one-sided upper 95% confidence limit.
#-----------------------------------------------------------------------------
Dates.for.prediction <- seq(min(Date[index]), max(Date[index]), length = 100)
CLs <- predict(SC.fit, newdata = data.frame(Date = Dates.for.prediction),
interval = "confidence", level = 0.9)
windows()
plot(Date[index], SC[index], ylim = c(0, 3000),
xlab = "Sampling Date",
ylab = "Specific Conductance (umho)",
main = "Figure 22-7. Two-sided 90% Confidence Bounds \nAround Trend Line at GW-12")
abline(SC.fit, lwd = 2)
lines(Dates.for.prediction, CLs[, "lwr"], lty = 3, lwd = 2, col = 4)
lines(Dates.for.prediction, CLs[, "upr"], lty = 3, lwd = 2, col = 4)
abline(h = 1000, lty = 2, lwd = 2, col = 2)
legend(as.numeric(as.Date("1988-01-01")), 500,
c("Regression Line", "90% Conf Limits", "Clean-Up Std"),
lty = c(1, 3, 2), lwd = 2, col = c(1, 4, 2))
# Step 5: Determine when the One-sided upper 95% confidence limit
# drops below the clean-up standard of 1000 umho.
min(Dates.for.prediction[CLs[, "upr"] < 1000])
#[1] "1991-12-18".
# Step 6: Time series plot for Well GW-13
# Also, remove first 4 observations
#------------------------------------------
index <- Well == "GW-13"
windows()
plot(Date[index], SC[index], type = "b",
ylim = c(0, 2500),
xlab = "Sampling Date",
ylab = "Specific Conductance (umho)",
main = "Figure 22-8. Time Series Plot of \nSpecific Conductance Measurements at GW-13")
new.index <- index & (Date > Date[index][4])
# Step 7: Test normality of GW-13 data
#-------------------------------------
# Using all the data
gof.obj <- gofTest(SC[index])
gof.obj[c("statistic", "p.value")]
#$statistic
# W
#0.5809712
#
#$p.value
#[1] 5.617319e-07
windows()
plot(gof.obj, plot.type = "Q-Q Plot",
ylab = "Specific Conductance (umho)",
main = "Figure 22-9. Probability Plot at GW-13\nUsing Entire Sampling Record")
# Omitting the first 4 observations
#----------------------------------
gof.obj <- gofTest(SC[new.index])
gof.obj[c("statistic", "p.value")]
#$statistic
# W
#0.9542246
#
#$p.value
#[1] 0.4647555
windows()
plot(gof.obj, plot.type = "Q-Q Plot",
ylab = "Specific Conductance (umho)",
main = "Figure 22-10. Probability Plot at GW-13\nExcluding First Four Measures")
# Step 8: Compute One-sided upper 95% confidence bounds for
# mean Specific Conductance at Well GW-13 using the first 8
# observations, and then omitting the first 4 observations
#--------------------------------------------------------------------
# Using the first 8 observations
enorm(SC[index][1:8], ci = T, ci.type = "upper")$interval$limits["UCL"]
# UCL
#1327.572
# Using observations 5 to 8
enorm(SC[index][5:8], ci = T, ci.type = "upper")$interval$limits["UCL"]
# UCL
#515.3008
# Step 9: Compute One-sided upper 95% confidence bounds for
# mean Specific Conductance at Well GW-13 using
# observations 5 to 12, and then using all observations
# except the first 4.
#--------------------------------------------------------------------
# Using observations 5 to 12
enorm(SC[index][5:12], ci = T, ci.type = "upper")$interval$limits["UCL"]
# UCL
#524.53
# Using all observations except the first 4
enorm(SC[new.index], ci = T, ci.type = "upper")$interval$limits["UCL"]
# UCL
#532.7065
# Step 10: Compute sample size necessary to detect a decrease in
# mean conductance from 1000 umho to 500 umho
# based on estimated SD using all the data at Well GW-13
# except the first 4 observations.
#-----------------------------------------------------------------
SD <- sd(SC[new.index])
SD
#[1] 74.27596
tTestN(delta.over.sigma = (500 - 1000) / SD,
alpha = 0.05, power = 0.95,
sample.type = "one.sample", alternative = "less", round.up = T)
#[1] 3
# NOTE: The Guidance Document states n=6 is necessary using
# it's second approach.
rm(Well, Date, SC, index, SC.fit, Dates.for.prediction,
CLs, new.index, gof.obj, SD)
###############################################################################################
# Example 22-3, p. 22-20
#-----------------------
# Based on normal approximation to the binomial
#----------------------------------------------
propTestN(p.or.p1 = 0.3, p0.or.p2 = 0.1, alpha = 0.05, power = 0.9,
sample.type = "one.sample", alternative = "greater",
approx = T, round.up = T)
#[1] 30
# Based on exact binomial distribution
#-------------------------------------
propTestN(p.or.p1 = 0.3, p0.or.p2 = 0.1, alpha = 0.05, power = 0.9,
sample.type = "one.sample", alternative = "greater",
approx = F, round.up = T)
#$n
#[1] 34
#
#$power
#[1] 0.9214717
#
#$alpha
#[1] 0.04814433
#
#$q.critical.upper
#[1] 6
propTestN(p.or.p1 = 0.3, p0.or.p2 = 0.1, alpha = 0.07, power = 0.9,
sample.type = "one.sample", alternative = "greater",
approx = F, round.up = T)
#$n
#[1] 29
#
#$power
#[1] 0.9068196
#
#$alpha
#[1] 0.06371744
#
#$q.critical.upper
#[1] 5
propTestN(p.or.p1 = 0.3, p0.or.p2 = 0.1, alpha = 0.1, power = 0.9,
sample.type = "one.sample", alternative = "greater",
approx = F, round.up = T)
#$n
#[1] 25
#
#$power
#[1] 0.909528
#
#$alpha
#[1] 0.09799362
#
#$q.critical.upper
#[1] 4
# So using the exact Binomial test, you would need a minimum sample size of 34, NOT 30,
# to detect an increase from 10% to 30% with 90% power and a Type I error of
# no greater than 5% (4.8%) in this case.
# Allowing a Type I Error of 6.3% would let us use a sample size of n=29.
############################################################################################
# Example 22-4, pp. 22-21 to 22-22
#---------------------------------
# Step 1: Compute sample size necessary to detect a change in proprtion of
# exceedances from 0.05 to (0.25 * 0.05) = 0.0125 using
# a Type I error of 20% or 10% and a power of 80%
#--------------------------------------------------------------------------
k <- 0.25
p0 <- 0.05
# Based on normal approximation to the binomial
#----------------------------------------------
propTestN(p.or.p1 = k * p0, p0.or.p2 = p0,
alpha = 0.2, power = 0.8,
sample.type = "one.sample", alternative = "less",
approx = T, round.up = T)
#[1] 55
#Warning message:
#In propTestN(p.or.p1 = k * p0, p0.or.p2 = p0, alpha = 0.2, :
# The computed sample size 'n' is too small, relative to the given value of 'p', for the normal approximation to work well for the following element indices:
# 1
propTestN(p.or.p1 = k * p0, p0.or.p2 = p0,
alpha = 0.1, power = 0.8,
sample.type = "one.sample", alternative = "less",
approx = T, round.up = T)
#[1] 99
#Warning message:
#In propTestN(p.or.p1 = k * p0, p0.or.p2 = p0, alpha = 0.1, :
# The computed sample size 'n' is too small, relative to the given value of 'p', for the normal approximation to work well for the following element indices:
# 1
# Step 2a: Compute sample size necessary to detect a change in proprtion of
# exceedances from 0.1 to (0.25 * 0.1) = 0.025 using
# a Type I error of 10% and a power of 80%
#----------------------------------------------------------------------------
p0 <- 0.1
propTestN(p.or.p1 = k * p0, p0.or.p2 = p0,
alpha = 0.1, power = 0.8,
sample.type = "one.sample", alternative = "less",
approx = T, round.up = T)
#[1] 48
#Warning message:
#In propTestN(p.or.p1 = k * p0, p0.or.p2 = p0, alpha = 0.1, power = 0.8, :
# The computed sample size 'n' is too small, relative to the given value of 'p', for the normal approximation to work well for the following element indices:
# 1
# Step 2b: Compute sample size necessary to detect a change in proprtion of
# exceedances from 0.05 to (0.1 * 0.05) = 0.005 using
# a Type I error of 10% and a power of 80%
#----------------------------------------------------------------------------
k <- 0.1
p0 <- 0.05
propTestN(p.or.p1 = k * p0, p0.or.p2 = p0,
alpha = 0.1, power = 0.8,
sample.type = "one.sample", alternative = "less",
approx = T, round.up = T)
#[1] 57
#Warning message:
#In propTestN(p.or.p1 = k * p0, p0.or.p2 = p0, alpha = 0.1, power = 0.8, :
# The computed sample size 'n' is too small, relative to the given value of 'p', for the normal approximation to work well for the following element indices:
# 1
# Step 2c: Compute sample size necessary to detect a change in proprtion of
# exceedances from 0.05 to (0.1 * 0.05) = 0.005 using
# a Type I error of 10% and a power of 60%
#----------------------------------------------------------------------------
k <- 0.25
p0 <- 0.05
propTestN(p.or.p1 = k * p0, p0.or.p2 = p0,
alpha = 0.1, power = 0.6,
sample.type = "one.sample", alternative = "less",
approx = T, round.up = T)
#[1] 68
#Warning message:
#In propTestN(p.or.p1 = k * p0, p0.or.p2 = p0, alpha = 0.1, power = 0.6, :
# The computed sample size 'n' is too small, relative to the given value of 'p', for the normal approximation to work well for the following element indices:
# 1
rm(k, p0)
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#####
# File: Script - EPA09, Chapter 22 Examples.R
#
# Purpose: Reproduce Examples in Chapter 22 of the EPA Guidance Document
#
# USEPA. (2009). "Statistical Analysis of Ground Water
# Monitoring Data at RCRA Facilities, Unified Guidance."
# EPA 530-R-09-007.
# Office of Resource Conservation and Recovery,
# Program Information and Implementation Division.
# March 2009.
#
# Errata are listed in:
# USEPA. (2010). "Errata Sheet - March 2009 Unified Guidance."
# EPA 530/R-09-007a.
# Office of Resource Conservation and Recovery,
# Program Information and Implementation Division.
# August 9, 2010.
#
# Author: Steven P. Millard
#
# Last
# Updated: October 4, 2012
#####
#NOTE: Unless otherwise noted, differences between what is shown in the
# EPA Guidance Document and results shown here are due to the
# EPA Guidance Document rounding intermediate results prior to
# the final result.
#######################################################################################
library(EnvStats)
# Example 22-1, pp. 22-6 to 22-8
#-------------------------------
EPA.09.Ex.22.1.VC.df
# Year Quarter Period Well VC.ppb
#1 1 1 Background GW-1 6.3
#2 1 2 Background GW-1 9.5
#3 1 3 Background GW-1 8.1
#4 1 4 Background GW-1 11.9
#5 2 1 Compliance GW-1 7.3
#6 2 2 Compliance GW-1 11.2
#7 2 3 Compliance GW-1 6.0
#8 2 4 Compliance GW-1 7.5
#9 3 1 Compliance GW-1 8.4
#10 3 2 Compliance GW-1 6.4
#11 3 3 Compliance GW-1 8.9
#12 3 4 Compliance GW-1 4.9
#13 4 1 Compliance GW-1 9.6
#14 4 2 Compliance GW-1 9.7
#15 4 3 Compliance GW-1 8.7
#16 4 4 Compliance GW-1 8.7
#17 1 1 Background GW-2 5.9
#18 1 2 Background GW-2 3.0
#19 1 3 Background GW-2 8.8
#20 1 4 Background GW-2 12.0
#21 2 1 Compliance GW-2 11.2
#22 2 2 Compliance GW-2 8.6
#23 2 3 Compliance GW-2 12.6
#24 2 4 Compliance GW-2 7.2
#25 3 1 Compliance GW-2 13.8
#26 3 2 Compliance GW-2 5.6
#27 3 3 Compliance GW-2 11.0
#28 3 4 Compliance GW-2 9.8
#29 4 1 Compliance GW-2 6.3
#30 4 2 Compliance GW-2 10.4
#31 4 3 Compliance GW-2 7.5
#32 4 4 Compliance GW-2 9.7
# Step 1: Summary Statistics and Time Series Plots
#--------------------------------------------------
# Summary Statistics by Well
#---------------------------
summaryStats(VC.ppb ~ Well, data = EPA.09.Ex.22.1.VC.df,
combine.groups = TRUE, digits = 1, stats.in.rows = TRUE)
# GW-1 GW-2 Combined
#N 16 16 32
#Mean 8.3 9 8.6
#SD 1.9 2.9 2.4
#Median 8.6 9.2 8.7
#Min 4.9 3 3
#Max 11.9 13.8 13.8
# Summary Statistics by Period and Well
#--------------------------------------
VC <- EPA.09.Ex.22.1.VC.df$VC.ppb
Period <- EPA.09.Ex.22.1.VC.df$Period
Well <- EPA.09.Ex.22.1.VC.df$Well
summaryStats(VC, group = paste(Well, Period, sep = "."),
combine.groups = FALSE, stats.in.rows = TRUE, digits = 1)
# GW-1.Background GW-1.Compliance GW-2.Background GW-2.Compliance
#N 4 12 4 12
#Mean 8.9 8.1 7.4 9.5
#SD 2.4 1.8 3.9 2.5
#Median 8.8 8.6 7.4 9.8
#Min 6.3 4.9 3 5.6
#Max 11.9 11.2 12 13.8
# Group Shapiro-Wilk Test Shows p-values for Individual Wells
#------------------------------------------------------------
gofGroupTest(VC.ppb ~ Well, data = EPA.09.Ex.22.1.VC.df)
#
#Results of Group Goodness-of-Fit Test
#-------------------------------------
#
#Test Method: Wilk-Shapiro GOF (Normal Scores)
#
#Hypothesized Distribution: Normal
#
#Data: VC.ppb
#
#Grouping Variable: Well
#
#Data Source: EPA.09.Ex.22.1.VC.df
#
#Number of Groups: 2
#
#Sample Sizes: GW-1 = 16
# GW-2 = 16
#
#Test Statistic: z (G) = 2.901581
#
#P-values for
#Individual Tests: GW-1 = 0.9686679
# GW-2 = 0.9875157
#
#P-value for
#Group Test: 0.9981436
#
#Alternative Hypothesis: At least one group
# does not come from a
# Normal Distribution.
# Time Series Plot
#-----------------
Year <- EPA.09.Ex.22.1.VC.df$Year
Qtr <- EPA.09.Ex.22.1.VC.df$Quarter
n1 <- sum(Well == "GW-1")
n2 <- sum(Well == "GW-2")
windows()
plot(1:n1, VC[Well == "GW-1"], xaxt = "n",
type = "b", pch = 15, lwd = 2,
ylim = c(0, 20),
xlab = "Time (Years)",
ylab = "Vinyl Chloride Concentration (ppb)",
main = "Figure 22-1. Vinyl Chloride Time Series Plot")
axis(1, at = (1:n1)[Qtr == 1 & Well == "GW-1"],
labels = Year[Qtr == "1" & Well == "GW-1"])
points(1:n2, VC[Well == "GW-2"], type = "b", pch = 3, lty = 3, lwd = 2)
abline(h = 5, lty = 2, lwd = 2)
legend(1, 20, c("GW-1", "GW-2", "MCL = 5 ppb"),
pch = c(15, 3, 1), lty = c(1, 3, 2),
pt.cex = c(1, 1, 0), lwd = 2)
# Step 2: Determine Type I Error level necessary in order to achieve
# 80% power to detect an increase in mean VC of 2 times
# the 5 ppb GWPS based on n=4 observations.
# Then compute one-sided lower confidence limits
# for each well using the Year 2 data based on
# this Type I error level.
#--------------------------------------------------------------------
R <- 2
mu0 <- 5
# Approach 1:
# Based on Guidance Document Method That *Does Not* Require
# an Estimate of SD or CV. This method assumes:
# 1) sigma = mean under null hypothesis
# 2) delta/sigma = R - 1, where R is the ratio of the alternative mean
# to the null mean
#------------------------------------------------------------------
required.alpha <- tTestAlpha(n.or.n1 = 4,
delta.over.sigma = R-1, power = 0.8,
sample.type = "one.sample",
alternative = "greater")
required.alpha
#[1] 0.1633382
# Lower one-sided confidence limits for Well GW-1:
enorm(VC[Well == "GW-1" & Year == "2"], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#6.693358
# Lower one-sided confidence limits for Well GW-2:
enorm(VC[Well == "GW-2" & Year == "2"], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#8.469282
# Approach 2:
# Based on standard method that requires an estimate of SD or CV.
#----------------------------------------------------------------
# Estimate standard deviation of VC based on Background data.
# Pool estimate of SDs based on both wells (GW-1 and GW-2)
VC.fit <- lm(VC.ppb ~ Well, data = EPA.09.Ex.22.1.VC.df,
subset = Period == "Background")
SD.Background <- summary(VC.fit)$sigma
SD.Background
#[1] 3.200976
#NOTE: Note that the estimated SD of 3.2 ppb is much smaller than the
# assumed value of 5 ppb based on Approach 1 above.
tTestAlpha(n.or.n1 = 4,
delta.over.sigma = mu0 * (R - 1) / SD.Background,
power = 0.8,
sample.type = "one.sample",
alternative = "greater", approx = F)
#[1] 0.05767842
#NOTE: Note the required Type I Error level based on Approach 2
# is much smaller than that based on Approach 1.
# Step 3: Compute one-sided lower confidence limits for each well
# using the Year 2 data based on Type I Error Level = 0.01.
#-------------------------------------------------------------------
# Use alpha = 0.01
# Lower one-sided confidence limits for Well GW-1:
enorm(VC[Well == "GW-1" & Year == "2"], ci = T, ci.type = "lower",
conf.level = 0.99)$interval$limits["LCL"]
# LCL
#2.926725
# Lower one-sided confidence limits for Well GW-2:
enorm(VC[Well == "GW-2" & Year == "2"], ci = T, ci.type = "lower",
conf.level = 0.99)$interval$limits["LCL"]
# LCL
#4.34498
# Step 4: Determine Type I Error level necessary in order to achieve
# 80% power to detect an increase in mean VC of 2 times
# the 5 ppb GWPS based on n=8 observations.
# Then compute one-sided lower confidence limits
# for each well using the Year 1 and Year 2 data based on
# this Type I error level.
#--------------------------------------------------------------------
required.alpha <- tTestAlpha(n.or.n1 = 8,
delta.over.sigma = R-1, power = 0.8,
sample.type = "one.sample",
alternative = "greater")
required.alpha
#[1] 0.0458989
# Lower one-sided confidence limits for Well GW-1:
enorm(VC[Well == "GW-1" & Year %in% c("1", "2")], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#6.963887
# Lower one-sided confidence limits for Well GW-2:
enorm(VC[Well == "GW-2" & Year %in% c("1", "2")], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#6.403578
# Step 5: Compute one-sided lower confidence limits
# for each well using the Year 2 and Year 3 data based on
# Type I error level from Step 4.
#--------------------------------------------------------------------
# Lower one-sided confidence limits for Well GW-1:
enorm(VC[Well == "GW-1" & Year %in% c("2", "3")], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#6.227279
# Lower one-sided confidence limits for Well GW-2:
enorm(VC[Well == "GW-2" & Year %in% c("2", "3")], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#8.078256
# Step 6: Determine Type I Error level necessary in order to achieve
# 80% power to detect an increase in mean VC of 2 times
# the 5 ppb GWPS based on n=12 observations.
# Then compute one-sided lower confidence limits
# for each well using the Years 2-4 data based on
# this Type I error level.
#--------------------------------------------------------------------
required.alpha <- tTestAlpha(n.or.n1 = 12,
delta.over.sigma = R-1, power = 0.8,
sample.type = "one.sample",
alternative = "greater")
required.alpha
#[1] 0.01311214
# Lower one-sided confidence limits for Well GW-1:
enorm(VC[Well == "GW-1" & Year != "1"], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#6.798091
# Lower one-sided confidence limits for Well GW-2:
enorm(VC[Well == "GW-2" & Year != "1"], ci = T, ci.type = "lower",
conf.level = 1 - required.alpha)$interval$limits["LCL"]
# LCL
#7.609904
rm(Year, Qtr, Period, Well, VC, n1, n2,
R, mu0, required.alpha, VC.fit, SD.Background)
#########################################################################################
# Example 22-2, pp. 22-11 to 22-17
#---------------------------------
EPA.09.Ex.22.2.Specific.Conductance.df
# Well Date Specific.Conductance.umho
#1 GW-12 1987-10-16 2100
#2 GW-12 1988-01-28 2550
# ...
#42 GW-13 1992-12-01 530
#43 GW-13 1993-03-24 625
Well <- EPA.09.Ex.22.2.Specific.Conductance.df$Well
Date <- EPA.09.Ex.22.2.Specific.Conductance.df$Date
SC <- EPA.09.Ex.22.2.Specific.Conductance.df$Specific.Conductance.umho
#Step 1: Time series plot for Well GW-12
#---------------------------------------
index <- Well == "GW-12"
windows()
plot(Date[index], SC[index], ylim = c(0, 3000),
xlab = "Sampling Date",
ylab = "Specific Conductance (umho)",
main = "Figure 22-3. Time Series Plot of \nSpecific Conductance Measurements at GW-12")
#Step 2: Fit regression line
#----------------------------
SC.fit <- lm(Specific.Conductance.umho ~ Date,
data = EPA.09.Ex.22.2.Specific.Conductance.df,
subset = Well == "GW-12")
summary(SC.fit)$coef
# Estimate Std. Error t value Pr(>|t|)
#(Intercept) 9526.116147 1042.1592274 9.14075 3.493624e-08
#Date -1.087567 0.1384346 -7.85618 3.170120e-07
# NOTE: These estimates of slope and intercept differ from those in the
# Guidance Document because our time variable is in terms of
# days since Jan 1, 1970, where as the units of time in the
# Guidance Document is Years.
windows()
plot(Date[index], SC[index], ylim = c(0, 3000),
xlab = "Sampling Date",
ylab = "Specific Conductance (umho)",
main = "Figure 22-4. Regression of Specific Conductance vs.\nSampling Date at GW-12")
abline(SC.fit)
# Step 3: Examine residuals
#---------------------------
windows()
qqPlot(SC.fit$residuals, xlab = "Z-score", add.line = T,
ylab = "Specific Conductance Residuals",
main = "Figure 22-5. Probability Plot of \nSpecific Conductance Residuals at GW-12")
gofTest(SC.fit$residuals)
#
#Results of Goodness-of-Fit Test
#-------------------------------
#
#Test Method: Shapiro-Wilk GOF
#
#Hypothesized Distribution: Normal
#
#Estimated Parameter(s): mean = 3.538836e-16
# sd = 3.790014e+02
#
#Estimation Method: mvue
#
#Data: SC.fit$residuals
#
#Sample Size: 20
#
#Test Statistic: W = 0.960157
#
#Test Statistic Parameter: n = 20
#
#P-value: 0.5469918
#
#Alternative Hypothesis: True cdf does not equal the
# Normal Distribution.
windows()
plot(Date[index], SC.fit$residuals, xlab = "Sampling Date",
ylab = "Specific Conductance Residuals",
main = "Figure 22-6. Plot of Specific Conductance Residuals vs. \nSampling Date at GW-12")
abline(h = 0, lty = 2)
# Alternatively, you can use used the built-in plotting function for linear models:
#windows()
#plot(SC.fit, ask = FALSE)
# Step 4: Plot Specific Conductance at GW-12 vs. Date, along with
# regression line and two-sided lower and upper 90% confidence limits
# Note that the upper confidence limit considered by itself is
# equivalent to a one-sided upper 95% confidence limit.
#-----------------------------------------------------------------------------
Dates.for.prediction <- seq(min(Date[index]), max(Date[index]), length = 100)
CLs <- predict(SC.fit, newdata = data.frame(Date = Dates.for.prediction),
interval = "confidence", level = 0.9)
windows()
plot(Date[index], SC[index], ylim = c(0, 3000),
xlab = "Sampling Date",
ylab = "Specific Conductance (umho)",
main = "Figure 22-7. Two-sided 90% Confidence Bounds \nAround Trend Line at GW-12")
abline(SC.fit, lwd = 2)
lines(Dates.for.prediction, CLs[, "lwr"], lty = 3, lwd = 2, col = 4)
lines(Dates.for.prediction, CLs[, "upr"], lty = 3, lwd = 2, col = 4)
abline(h = 1000, lty = 2, lwd = 2, col = 2)
legend(as.numeric(as.Date("1988-01-01")), 500,
c("Regression Line", "90% Conf Limits", "Clean-Up Std"),
lty = c(1, 3, 2), lwd = 2, col = c(1, 4, 2))
# Step 5: Determine when the One-sided upper 95% confidence limit
# drops below the clean-up standard of 1000 umho.
min(Dates.for.prediction[CLs[, "upr"] < 1000])
#[1] "1991-12-18".
# Step 6: Time series plot for Well GW-13
# Also, remove first 4 observations
#------------------------------------------
index <- Well == "GW-13"
windows()
plot(Date[index], SC[index], type = "b",
ylim = c(0, 2500),
xlab = "Sampling Date",
ylab = "Specific Conductance (umho)",
main = "Figure 22-8. Time Series Plot of \nSpecific Conductance Measurements at GW-13")
new.index <- index & (Date > Date[index][4])
# Step 7: Test normality of GW-13 data
#-------------------------------------
# Using all the data
gof.obj <- gofTest(SC[index])
gof.obj[c("statistic", "p.value")]
#$statistic
# W
#0.5809712
#
#$p.value
#[1] 5.617319e-07
windows()
plot(gof.obj, plot.type = "Q-Q Plot",
ylab = "Specific Conductance (umho)",
main = "Figure 22-9. Probability Plot at GW-13\nUsing Entire Sampling Record")
# Omitting the first 4 observations
#----------------------------------
gof.obj <- gofTest(SC[new.index])
gof.obj[c("statistic", "p.value")]
#$statistic
# W
#0.9542246
#
#$p.value
#[1] 0.4647555
windows()
plot(gof.obj, plot.type = "Q-Q Plot",
ylab = "Specific Conductance (umho)",
main = "Figure 22-10. Probability Plot at GW-13\nExcluding First Four Measures")
# Step 8: Compute One-sided upper 95% confidence bounds for
# mean Specific Conductance at Well GW-13 using the first 8
# observations, and then omitting the first 4 observations
#--------------------------------------------------------------------
# Using the first 8 observations
enorm(SC[index][1:8], ci = T, ci.type = "upper")$interval$limits["UCL"]
# UCL
#1327.572
# Using observations 5 to 8
enorm(SC[index][5:8], ci = T, ci.type = "upper")$interval$limits["UCL"]
# UCL
#515.3008
# Step 9: Compute One-sided upper 95% confidence bounds for
# mean Specific Conductance at Well GW-13 using
# observations 5 to 12, and then using all observations
# except the first 4.
#--------------------------------------------------------------------
# Using observations 5 to 12
enorm(SC[index][5:12], ci = T, ci.type = "upper")$interval$limits["UCL"]
# UCL
#524.53
# Using all observations except the first 4
enorm(SC[new.index], ci = T, ci.type = "upper")$interval$limits["UCL"]
# UCL
#532.7065
# Step 10: Compute sample size necessary to detect a decrease in
# mean conductance from 1000 umho to 500 umho
# based on estimated SD using all the data at Well GW-13
# except the first 4 observations.
#-----------------------------------------------------------------
SD <- sd(SC[new.index])
SD
#[1] 74.27596
tTestN(delta.over.sigma = (500 - 1000) / SD,
alpha = 0.05, power = 0.95,
sample.type = "one.sample", alternative = "less", round.up = T)
#[1] 3
# NOTE: The Guidance Document states n=6 is necessary using
# it's second approach.
rm(Well, Date, SC, index, SC.fit, Dates.for.prediction,
CLs, new.index, gof.obj, SD)
###############################################################################################
# Example 22-3, p. 22-20
#-----------------------
# Based on normal approximation to the binomial
#----------------------------------------------
propTestN(p.or.p1 = 0.3, p0.or.p2 = 0.1, alpha = 0.05, power = 0.9,
sample.type = "one.sample", alternative = "greater",
approx = T, round.up = T)
#[1] 30
# Based on exact binomial distribution
#-------------------------------------
propTestN(p.or.p1 = 0.3, p0.or.p2 = 0.1, alpha = 0.05, power = 0.9,
sample.type = "one.sample", alternative = "greater",
approx = F, round.up = T)
#$n
#[1] 34
#
#$power
#[1] 0.9214717
#
#$alpha
#[1] 0.04814433
#
#$q.critical.upper
#[1] 6
propTestN(p.or.p1 = 0.3, p0.or.p2 = 0.1, alpha = 0.07, power = 0.9,
sample.type = "one.sample", alternative = "greater",
approx = F, round.up = T)
#$n
#[1] 29
#
#$power
#[1] 0.9068196
#
#$alpha
#[1] 0.06371744
#
#$q.critical.upper
#[1] 5
propTestN(p.or.p1 = 0.3, p0.or.p2 = 0.1, alpha = 0.1, power = 0.9,
sample.type = "one.sample", alternative = "greater",
approx = F, round.up = T)
#$n
#[1] 25
#
#$power
#[1] 0.909528
#
#$alpha
#[1] 0.09799362
#
#$q.critical.upper
#[1] 4
# So using the exact Binomial test, you would need a minimum sample size of 34, NOT 30,
# to detect an increase from 10% to 30% with 90% power and a Type I error of
# no greater than 5% (4.8%) in this case.
# Allowing a Type I Error of 6.3% would let us use a sample size of n=29.
############################################################################################
# Example 22-4, pp. 22-21 to 22-22
#---------------------------------
# Step 1: Compute sample size necessary to detect a change in proprtion of
# exceedances from 0.05 to (0.25 * 0.05) = 0.0125 using
# a Type I error of 20% or 10% and a power of 80%
#--------------------------------------------------------------------------
k <- 0.25
p0 <- 0.05
# Based on normal approximation to the binomial
#----------------------------------------------
propTestN(p.or.p1 = k * p0, p0.or.p2 = p0,
alpha = 0.2, power = 0.8,
sample.type = "one.sample", alternative = "less",
approx = T, round.up = T)
#[1] 55
#Warning message:
#In propTestN(p.or.p1 = k * p0, p0.or.p2 = p0, alpha = 0.2, :
# The computed sample size 'n' is too small, relative to the given value of 'p', for the normal approximation to work well for the following element indices:
# 1
propTestN(p.or.p1 = k * p0, p0.or.p2 = p0,
alpha = 0.1, power = 0.8,
sample.type = "one.sample", alternative = "less",
approx = T, round.up = T)
#[1] 99
#Warning message:
#In propTestN(p.or.p1 = k * p0, p0.or.p2 = p0, alpha = 0.1, :
# The computed sample size 'n' is too small, relative to the given value of 'p', for the normal approximation to work well for the following element indices:
# 1
# Step 2a: Compute sample size necessary to detect a change in proprtion of
# exceedances from 0.1 to (0.25 * 0.1) = 0.025 using
# a Type I error of 10% and a power of 80%
#----------------------------------------------------------------------------
p0 <- 0.1
propTestN(p.or.p1 = k * p0, p0.or.p2 = p0,
alpha = 0.1, power = 0.8,
sample.type = "one.sample", alternative = "less",
approx = T, round.up = T)
#[1] 48
#Warning message:
#In propTestN(p.or.p1 = k * p0, p0.or.p2 = p0, alpha = 0.1, power = 0.8, :
# The computed sample size 'n' is too small, relative to the given value of 'p', for the normal approximation to work well for the following element indices:
# 1
# Step 2b: Compute sample size necessary to detect a change in proprtion of
# exceedances from 0.05 to (0.1 * 0.05) = 0.005 using
# a Type I error of 10% and a power of 80%
#----------------------------------------------------------------------------
k <- 0.1
p0 <- 0.05
propTestN(p.or.p1 = k * p0, p0.or.p2 = p0,
alpha = 0.1, power = 0.8,
sample.type = "one.sample", alternative = "less",
approx = T, round.up = T)
#[1] 57
#Warning message:
#In propTestN(p.or.p1 = k * p0, p0.or.p2 = p0, alpha = 0.1, power = 0.8, :
# The computed sample size 'n' is too small, relative to the given value of 'p', for the normal approximation to work well for the following element indices:
# 1
# Step 2c: Compute sample size necessary to detect a change in proprtion of
# exceedances from 0.05 to (0.1 * 0.05) = 0.005 using
# a Type I error of 10% and a power of 60%
#----------------------------------------------------------------------------
k <- 0.25
p0 <- 0.05
propTestN(p.or.p1 = k * p0, p0.or.p2 = p0,
alpha = 0.1, power = 0.6,
sample.type = "one.sample", alternative = "less",
approx = T, round.up = T)
#[1] 68
#Warning message:
#In propTestN(p.or.p1 = k * p0, p0.or.p2 = p0, alpha = 0.1, power = 0.6, :
# The computed sample size 'n' is too small, relative to the given value of 'p', for the normal approximation to work well for the following element indices:
# 1
rm(k, p0)
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