# NOT RUN {
# Look at how the required sample size for a tolerance interval increases
# with increasing coverage:
seq(0.5, 0.9, by = 0.1)
#[1] 0.5 0.6 0.7 0.8 0.9
tolIntNormN(half.width = 3, coverage = seq(0.5, 0.9, by = 0.1))
#[1] 4 4 5 6 9
#----------
# Look at how the required sample size for a tolerance interval decreases
# with increasing half-width:
3:6
#[1] 3 4 5 6
tolIntNormN(half.width = 3:6)
#[1] 15 8 6 5
tolIntNormN(3:6, round = FALSE)
#[1] 14.199735 7.022572 5.092374 4.214371
#----------
# Look at how the required sample size for a tolerance interval increases
# with increasing estimated standard deviation for a fixed half-width:
seq(0.5, 2, by = 0.5)
#[1] 0.5 1.0 1.5 2.0
tolIntNormN(half.width = 4, sigma.hat = seq(0.5, 2, by = 0.5))
#[1] 4 8 24 3437
#----------
# Look at how the required sample size for a tolerance interval increases
# with increasing confidence level for a fixed half-width:
seq(0.5, 0.9, by = 0.1)
#[1] 0.5 0.6 0.7 0.8 0.9
tolIntNormN(half.width = 3, conf.level = seq(0.5, 0.9, by = 0.1))
#[1] 3 4 5 7 11
#==========
# Example 17-3 of USEPA (2009, p. 17-17) shows how to construct a
# beta-content upper tolerance limit with 95% coverage and 95%
# confidence using chrysene data and assuming a lognormal distribution.
# The data for this example are stored in EPA.09.Ex.17.3.chrysene.df,
# which contains chrysene concentration data (ppb) found in water
# samples obtained from two background wells (Wells 1 and 2) and
# three compliance wells (Wells 3, 4, and 5). The tolerance limit
# is based on the data from the background wells.
# Here we will first take the log of the data and then estimate the
# standard deviation based on the two background wells. We will use this
# estimate of standard deviation to compute required sample sizes for
# various half-widths on the log-scale.
head(EPA.09.Ex.17.3.chrysene.df)
# Month Well Well.type Chrysene.ppb
#1 1 Well.1 Background 19.7
#2 2 Well.1 Background 39.2
#3 3 Well.1 Background 7.8
#4 4 Well.1 Background 12.8
#5 1 Well.2 Background 10.2
#6 2 Well.2 Background 7.2
longToWide(EPA.09.Ex.17.3.chrysene.df, "Chrysene.ppb", "Month", "Well")
# Well.1 Well.2 Well.3 Well.4 Well.5
#1 19.7 10.2 68.0 26.8 47.0
#2 39.2 7.2 48.9 17.7 30.5
#3 7.8 16.1 30.1 31.9 15.0
#4 12.8 5.7 38.1 22.2 23.4
summary.stats <- summaryStats(log(Chrysene.ppb) ~ Well.type,
data = EPA.09.Ex.17.3.chrysene.df)
summary.stats
# N Mean SD Median Min Max
#Background 8 2.5086 0.6279 2.4359 1.7405 3.6687
#Compliance 12 3.4173 0.4361 3.4111 2.7081 4.2195
sigma.hat <- summary.stats["Background", "SD"]
sigma.hat
#[1] 0.6279
tolIntNormN(half.width = c(4, 2, 1), sigma.hat = sigma.hat)
#[1] 4 12 NA
#Warning message:
#In tolIntNormN(half.width = c(4, 2, 1), sigma.hat = sigma.hat) :
# Value of 'half.width' is too smallfor element3.
# Try increasing the value of 'n.max'.
# NOTE: We cannot achieve a half-width of 1 for the given value of
# sigma.hat for a tolerance interval with 95% coverage and
# 95% confidence. The default value of n.max is 5000, but in fact,
# even with a million observations the half width is greater than 1.
tolIntNormHalfWidth(n = 1e6, sigma.hat = sigma.hat)
#[1] 1.232095
#==========
# Clean up
#---------
rm(summary.stats, sigma.hat)
# }
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