tolIntNormHalfWidth: Half-Width of a Tolerance Interval for a Normal Distribution

Description

Compute the half-width of a tolerance interval for a normal distribution.

Usage

tolIntNormHalfWidth(n, sigma.hat = 1, coverage = 0.95, cov.type = "content", 
    conf.level = 0.95, method = "wald.wolfowitz")

Arguments

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.

sigma.hat

numeric vector specifying the value(s) of the estimated standard deviation(s). The default value is sigma.hat=1.

coverage

numeric vector of values between 0 and 1 indicating the desired coverage of the tolerance interval. The default value is coverage=0.95.

cov.type

character string specifying the coverage type for the tolerance interval. The possible values are "content" ($\beta$-content; the default), and "expectation" ($\beta$-expectation).

conf.level

numeric vector of values between 0 and 1 indicating the confidence level of the prediction interval. The default value is conf.level=0.95.

method

character string specifying the method for constructing the tolerance interval. The possible values are "exact" (the default) and "wald.wolfowitz" (the Wald-Wolfowitz approximation).

Value

numeric vector of half-widths.

Details

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: $$HW = Ks \;\;\;\;\;\; (4)$$

References

See the help file for tolIntNorm.

Examples

Run this code

# NOT RUN {
  # 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 

  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

  #----------

  # Look at how the half-width of a tolerance interval decreases with 
  # increasing sample size:

  2:5 
  #[1] 2 3 4 5 

  round(tolIntNormHalfWidth(n = 2:5), 2) 
  #[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 

  round(tolIntNormHalfWidth(n = 10, sigma.hat = seq(0.5, 2, by = 0.5)), 2) 
  #[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 

  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

  #==========

  # 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

  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

  tolIntNormHalfWidth(n = c(4, 8, 16), sigma.hat = sigma.hat)
  #[1] 3.999681 2.343160 1.822759

  #==========

  # Clean up
  #---------
  rm(summary.stats, sigma.hat)
# }

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