tolIntNormK(n, df = n - 1, coverage = 0.95, cov.type = "content",
ti.type = "two-sided", conf.level = 0.95, method = "exact",
rel.tol = 1e-07, abs.tol = rel.tol)df=n-1.coverage=0.95."content" ($\beta$-content; the default), and
"expectation" ($\beta$-expectation). See the help file for "two-sided" (the default), "lower", and
"upper".conf.level=0.95.ti.type="lower" or
ti.type="upper". The possible values are
ti.type="two-sided" and method="exact", the argument
rel.tol is passed to the function integrate. The default value is
rel.tol=1e-07ti.type="two-sided" and method="exact", the argument
abs.tol is passed to the function integrate. The default value is the
value of rti.type="upper" or ti.type="lower", the constant $K$ for a
$100 \beta %$ $\beta$-content tolerance interval with associated
confidence level $100(1 - \alpha)%$ is given by:
$$K = t(n-1, 1 - \alpha, z_\beta \sqrt{n}) / \sqrt{n}$$
where $t(\nu, p, \delta)$ denotes the $p$'th quantile of a non-central
t-distribution with $\nu$ degrees of freedom and noncentrality parameter
$\delta$ (see the help file for TDist), and $z_p$ denotes the
$p$'th quantile of a standard normal distribution.
Two-Sided Case
When ti.type="two-sided" and method="exact", the exact formula for
the constant $K$ for a $100 \beta %$ $\beta$-content tolerance interval
with associated confidence level $100(1-\alpha)%$ requires numerical integration
and has been derived by several different authors, including Odeh (1978),
Eberhardt et al. (1989), Jilek (1988), Fujino (1989), and Janiga and Miklos (2001).
Specifically, for given values of the sample size $n$, degrees of freedom $\nu$,
confidence level $(1-\alpha)$, and coverage $\beta$, the constant $K$ is the
solution to the equation:
$$\sqrt{\frac{n}{2 \pi}} \, \int^\infty_{-\infty} {F(x, K, \nu, R) \, e^{(-nx^2)/2}} \, dx = 1 - \alpha$$
where $F(x, K, \nu, R)$ denotes the upper-tail area from $(\nu \, R^2) / K^2$ to
$\infty$ of the chi-squared distribution with $\nu$ degrees of freedom, and
$R$ is the solution to the equation:
$$\Phi (x + R) - \Phi (x - R) = \beta$$ where
$\Phi()$ denotes the standard normal cumulative distribuiton function.
When ti.type="two-sided" and method="wald.wolfowitz", the approximate formula
due to Wald and Wolfowitz (1946) for the constant $K$ for a $100 \beta %$
$\beta$-content tolerance interval with associated confidence level
$100(1-\alpha)%$ is given by:
$$K \approx r \, u$$
where $r$ is the solution to the equation:
$$\Phi (\frac{1}{\sqrt{n}} + r) - \Phi (\frac{1}{\sqrt{n}} - r) = \beta$$
$\Phi ()$ denotes the standard normal cumulative distribuiton function, and $u$ is
given by:
$$u = \sqrt{\frac{n-1}{\chi^{2} (n-1, \alpha)}}$$
where $\chi^{2} (\nu, p)$ denotes the $p$'th quantile of the chi-squared
distribution with $\nu$ degrees of freedom.
The Derivation of $K$ for a $\beta$-Expectation Tolerance Interval
As stated above, a $\beta$-expectation tolerance interval with coverage $100 \beta %$ is
equivalent to a prediction interval for one future observation with associated confidence level
$100 \beta %$. This is because the probability that any single future observation will fall
into this interval is $100 \beta %$, so the distribution of the number of $N$ future
observations that will fall into this interval is binomial with parameters size = $N$ and
prob = $\beta$ (see the help file for Binomial). Hence the expected proportion
of future observations that will fall into this interval is $100 \beta %$ and is independent of
the value of $N$. See the help file for predIntNormK for information on
how to derive $K$ for these intervals.tolIntNorm, predIntNorm, Normal,
estimate.object, enorm, eqnorm,
Tolerance Intervals, Prediction Intervals,
Estimating Distribution Parameters,
Estimating Distribution Quantiles.# Compute the value of K for a two-sided 95% beta-content
# tolerance interval with associated confidence level 95%
# given a sample size of n=20.
#----------
# Exact method
tolIntNormK(n = 20)
#[1] 2.760346
#----------
# Approximate method due to Wald and Wolfowitz (1946)
tolIntNormK(n = 20, method = "wald")
# [1] 2.751789
#--------------------------------------------------------------------
# Compute the value of K for a one-sided upper tolerance limit
# with 99% coverage and associated confidence level 90%
# given a samle size of n=20.
tolIntNormK(n = 20, ti.type = "upper", coverage = 0.99,
conf.level = 0.9)
#[1] 3.051543
#--------------------------------------------------------------------
# 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 sample size is n = 8 observations from
# the two compliance wells. Here we will compute the
# multiplier for the log-transformed data.
tolIntNormK(n = 8, ti.type = "upper")
#[1] 3.187294Run the code above in your browser using DataLab