predIntNormN.RdCompute the sample size necessary to achieve a specified half-width of a prediction interval for the next \(k\) observations from a normal distribution.
predIntNormN(half.width, n.mean = 1, k = 1, sigma.hat = 1,
method = "Bonferroni", conf.level = 0.95, round.up = TRUE,
n.max = 5000, tol = 1e-07, maxiter = 1000)numeric vector of (positive) half-widths.
Missing (NA), undefined (NaN), and infinite (Inf, -Inf) values are not allowed.
numeric vector of positive integers specifying the sample size associated with
the \(k\) future averages. The default value is
n.mean=1 (i.e., individual observations). Note that all future averages
must be based on the same sample size.
numeric vector of positive integers specifying the number of future observations
or averages the prediction interval should contain with confidence level
conf.level. The default value is k=1.
numeric vector specifying the value(s) of the estimated standard deviation(s).
The default value is sigma.hat=1.
character string specifying the method to use if the number of future observations
(k) is greater than 1. The possible values are method="Bonferroni"
(approximate method based on Bonferonni inequality; the default), and method="exact" (exact method due to Dunnett, 1955).
This argument is ignored if k=1.
numeric vector of values between 0 and 1 indicating the confidence level of the
prediction interval. The default value is conf.level=0.95.
logical scalar indicating whether to round up the values of the computed sample
size(s) to the next smallest integer. The default value is round.up=TRUE.
positive integer greater than 1 indicating the maximum possible sample size. The
default value is n.max=5000.
numeric scalar indicating the tolerance to use in the uniroot
search algorithm. The default value is tol=1e-7.
positive integer indicating the maximum number of iterations to use in the
uniroot search algorithm. The default value is
maxiter=1000.
If the arguments half.width, k, n.mean, sigma.hat, 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 predIntNorm and predIntNormK
give formulas for a two-sided prediction interval based on the sample size, the
observed sample mean and sample standard deviation, and specified confidence level.
Specifically, the two-sided prediction 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, the number of future averages \(k\), and the
sample size associated with the future averages, \(m\) (see the help file for
predIntNormK). Thus, the half-width of the prediction interval is
given by:
$$HW = Ks \;\;\;\;\;\; (4)$$
The function predIntNormN uses the uniroot search algorithm to
determine the sample size for specified values of the half-width, number of
observations used to create a single future average, number of future observations or
averages, the sample standard deviation, and the confidence level. Note that
unlike a confidence interval, the half-width of a prediction interval does not
approach 0 as the sample size increases.
numeric vector of sample sizes.
See the help file for predIntNorm.
See the help file for predIntNorm.
# Look at how the required sample size for a prediction interval increases
# with increasing number of future observations:
1:5
#> [1] 1 2 3 4 5
#[1] 1 2 3 4 5
predIntNormN(half.width = 3, k = 1:5)
#> [1] 6 9 11 14 18
#[1] 6 9 11 14 18
#----------
# Look at how the required sample size for a prediction interval decreases
# with increasing half-width:
2:5
#> [1] 2 3 4 5
#[1] 2 3 4 5
predIntNormN(half.width = 2:5)
#> [1] 86 6 4 3
#[1] 86 6 4 3
predIntNormN(2:5, round = FALSE)
#> [1] 85.567387 5.122911 3.542393 2.987861
#[1] 85.567387 5.122911 3.542393 2.987861
#----------
# Look at how the required sample size for a prediction 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
#[1] 0.5 1.0 1.5 2.0
predIntNormN(half.width = 4, sigma.hat = seq(0.5, 2, by = 0.5))
#> [1] 3 4 7 86
#[1] 3 4 7 86
#----------
# Look at how the required sample size for a prediction 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
#[1] 0.5 0.6 0.7 0.8 0.9
predIntNormN(half.width = 2, conf.level = seq(0.5, 0.9, by = 0.1))
#> [1] 2 2 3 4 9
#[1] 2 2 3 4 9
#==========
# The data frame EPA.92c.arsenic3.df contains arsenic concentrations (ppb)
# collected quarterly for 3 years at a background well and quarterly for
# 2 years at a compliance well. Using the data from the background well,
# compute the required sample size in order to achieve a half-width of
# 2.25, 2.5, or 3 times the estimated standard deviation for a two-sided
# 90% prediction interval for k=4 future observations.
#
# For a half-width of 2.25 standard deviations, the required sample size is 526,
# or about 131 years of quarterly observations! For a half-width of 2.5
# standard deviations, the required sample size is 20, or about 5 years of
# quarterly observations. For a half-width of 3 standard deviations, the required
# sample size is 9, or about 2 years of quarterly observations.
EPA.92c.arsenic3.df
#> Arsenic Year Well.type
#> 1 12.6 1 Background
#> 2 30.8 1 Background
#> 3 52.0 1 Background
#> 4 28.1 1 Background
#> 5 33.3 2 Background
#> 6 44.0 2 Background
#> 7 3.0 2 Background
#> 8 12.8 2 Background
#> 9 58.1 3 Background
#> 10 12.6 3 Background
#> 11 17.6 3 Background
#> 12 25.3 3 Background
#> 13 48.0 4 Compliance
#> 14 30.3 4 Compliance
#> 15 42.5 4 Compliance
#> 16 15.0 4 Compliance
#> 17 47.6 5 Compliance
#> 18 3.8 5 Compliance
#> 19 2.6 5 Compliance
#> 20 51.9 5 Compliance
# Arsenic Year Well.type
#1 12.6 1 Background
#2 30.8 1 Background
#3 52.0 1 Background
#...
#18 3.8 5 Compliance
#19 2.6 5 Compliance
#20 51.9 5 Compliance
mu.hat <- with(EPA.92c.arsenic3.df,
mean(Arsenic[Well.type=="Background"]))
mu.hat
#> [1] 27.51667
#[1] 27.51667
sigma.hat <- with(EPA.92c.arsenic3.df,
sd(Arsenic[Well.type=="Background"]))
sigma.hat
#> [1] 17.10119
#[1] 17.10119
predIntNormN(half.width=c(2.25, 2.5, 3) * sigma.hat, k = 4,
sigma.hat = sigma.hat, conf.level = 0.9)
#> [1] 526 20 9
#[1] 526 20 9
#==========
# Clean up
#---------
rm(mu.hat, sigma.hat)