tolIntNormHalfWidth.RdCompute the half-width of a tolerance interval for a normal distribution.
tolIntNormHalfWidth(n, sigma.hat = 1, coverage = 0.95, cov.type = "content",
conf.level = 0.95, method = "wald.wolfowitz")numeric vector of positive integers greater than 1 indicating the sample size upon
which the prediction interval is based.
Missing (NA), undefined (NaN), and infinite (Inf, -Inf)
values are not allowed.
numeric vector specifying the value(s) of the estimated standard deviation(s).
The default value is sigma.hat=1.
numeric vector of values between 0 and 1 indicating the desired coverage of the
tolerance interval. The default value is coverage=0.95.
character string specifying the coverage type for the tolerance interval. The
possible values are "content" (\(\beta\)-content; the default), and
"expectation" (\(\beta\)-expectation).
numeric vector of values between 0 and 1 indicating the confidence level of the
prediction interval. The default value is conf.level=0.95.
character string specifying the method for constructing the tolerance interval.
The possible values are "exact" (the default) and "wald.wolfowitz"
(the Wald-Wolfowitz approximation).
If the arguments n, sigma.hat, coverage, and
conf.level are not all the same length, they are replicated to be the
same length as the length of the longest argument.
The help files for tolIntNorm and tolIntNormK
give formulas for a two-sided tolerance interval based on the sample size, the
observed sample mean and sample standard deviation, and specified confidence level
and coverage. Specifically, the two-sided tolerance interval is given by:
$$[\bar{x} - Ks, \bar{x} + Ks] \;\;\;\;\;\; (1)$$
where \(\bar{x}\) denotes the sample mean:
$$\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i \;\;\;\;\;\; (2)$$
\(s\) denotes the sample standard deviation:
$$s^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2 \;\;\;\;\;\; (3)$$
and \(K\) denotes a constant that depends on the sample size \(n\), the
confidence level, and the coverage (see the help file for
tolIntNormK). Thus, the half-width of the tolerance interval is
given by:
$$HW = Ks \;\;\;\;\;\; (4)$$
numeric vector of half-widths.
See the help file for tolIntNorm.
See the help file for tolIntNorm.
In the course of designing a sampling program, an environmental scientist may wish
to determine the relationship between sample size, confidence level, and half-width
if one of the objectives of the sampling program is to produce tolerance intervals.
The functions tolIntNormHalfWidth, tolIntNormN, and
plotTolIntNormDesign can be used to investigate these relationships
for the case of normally-distributed observations.
# Look at how the half-width of 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
#[1] 0.5 0.6 0.7 0.8 0.9
round(tolIntNormHalfWidth(n = 10, coverage = seq(0.5, 0.9, by = 0.1)), 2)
#> [1] 1.17 1.45 1.79 2.21 2.84
#[1] 1.17 1.45 1.79 2.21 2.84
#----------
# Look at how the half-width of a tolerance interval decreases with
# increasing sample size:
2:5
#> [1] 2 3 4 5
#[1] 2 3 4 5
round(tolIntNormHalfWidth(n = 2:5), 2)
#> [1] 37.67 9.92 6.37 5.08
#[1] 37.67 9.92 6.37 5.08
#----------
# Look at how the half-width of a tolerance interval increases with
# increasing estimated standard deviation for a fixed sample size:
seq(0.5, 2, by = 0.5)
#> [1] 0.5 1.0 1.5 2.0
#[1] 0.5 1.0 1.5 2.0
round(tolIntNormHalfWidth(n = 10, sigma.hat = seq(0.5, 2, by = 0.5)), 2)
#> [1] 1.69 3.38 5.07 6.76
#[1] 1.69 3.38 5.07 6.76
#----------
# Look at how the half-width of a tolerance interval increases with
# increasing confidence level for a fixed sample size:
seq(0.5, 0.9, by = 0.1)
#> [1] 0.5 0.6 0.7 0.8 0.9
#[1] 0.5 0.6 0.7 0.8 0.9
round(tolIntNormHalfWidth(n = 5, conf = seq(0.5, 0.9, by = 0.1)), 2)
#> [1] 2.34 2.58 2.89 3.33 4.15
#[1] 2.34 2.58 2.89 3.33 4.15
#==========
# 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 the half-widths of
# future tolerance intervals on the log-scale for various sample sizes.
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
# 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
# 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
# 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
#[1] 0.6279
tolIntNormHalfWidth(n = c(4, 8, 16), sigma.hat = sigma.hat)
#> [1] 3.999681 2.343160 1.822759
#[1] 3.999681 2.343160 1.822759
#==========
# Clean up
#---------
rm(summary.stats, sigma.hat)