Confidence Level for Nonparametric Prediction Interval for Continuous Distribution
Compute the confidence level associated with a nonparametric prediction interval that should contain at least k out of the next m future observations for a continuous distribution.
predIntNparConfLevel(n, k = m, m = 1, lpl.rank = ifelse(pi.type == "upper", 0, 1), n.plus.one.minus.upl.rank = ifelse(pi.type == "lower", 0, 1), pi.type = "two.sided")
n |
vector of positive integers specifying the sample sizes.
Missing ( |
k |
vector of positive integers specifying the minimum number of future
observations out of |
m |
vector of positive integers specifying the number of future observations.
The default value is |
lpl.rank |
vector of positive integers indicating the rank of the order statistic to use for
the lower bound of the prediction interval. If |
n.plus.one.minus.upl.rank |
vector of positive integers related to the rank of the order statistic to use for
the upper bound of the prediction interval. A value of
|
pi.type |
character string indicating what kind of prediction interval to compute.
The possible values are |
If the arguments n
, k
, m
, lpl.rank
, and
n.plus.one.minus.upl.rank
are not all the same length, they are replicated
to be the same length as the length of the longest argument.
The help file for predIntNpar
explains how nonparametric prediction
intervals are constructed and how the confidence level
associated with the prediction interval is computed based on specified values
for the sample size and the ranks of the order statistics used for
the bounds of the prediction interval.
vector of values between 0 and 1 indicating the confidence level associated with the specified nonparametric prediction interval.
See the help file for predIntNpar
.
Steven P. Millard (EnvStats@ProbStatInfo.com)
See the help file for predIntNpar
.
# Look at how the confidence level of a nonparametric prediction interval # increases with increasing sample size: seq(5, 25, by = 5) #[1] 5 10 15 20 25 round(predIntNparConfLevel(n = seq(5, 25, by = 5)), 2) #[1] 0.67 0.82 0.87 0.90 0.92 #--------- # Look at how the confidence level of a nonparametric prediction interval # decreases as the number of future observations increases: round(predIntNparConfLevel(n = 10, m = 1:5), 2) #[1] 0.82 0.68 0.58 0.49 0.43 #---------- # Look at how the confidence level of a nonparametric prediction interval # decreases with minimum number of observations that must be contained within # the interval (k): round(predIntNparConfLevel(n = 10, k = 1:5, m = 5), 2) #[1] 1.00 0.98 0.92 0.76 0.43 #---------- # Look at how the confidence level of a nonparametric prediction interval # decreases with the rank of the lower prediction limit: round(predIntNparConfLevel(n = 10, lpl.rank = 1:5), 2) #[1] 0.82 0.73 0.64 0.55 0.45 #========== # Example 18-3 of USEPA (2009, p.18-19) shows how to construct # a one-sided upper nonparametric prediction interval for the next # 4 future observations of trichloroethylene (TCE) at a downgradient well. # The data for this example are stored in EPA.09.Ex.18.3.TCE.df. # There are 6 monthly observations of TCE (ppb) at 3 background wells, # and 4 monthly observations of TCE at a compliance well. # Look at the data #----------------- EPA.09.Ex.18.3.TCE.df # Month Well Well.type TCE.ppb.orig TCE.ppb Censored #1 1 BW-1 Background <5 5.0 TRUE #2 2 BW-1 Background <5 5.0 TRUE #3 3 BW-1 Background 8 8.0 FALSE #... #22 4 CW-4 Compliance <5 5.0 TRUE #23 5 CW-4 Compliance 8 8.0 FALSE #24 6 CW-4 Compliance 14 14.0 FALSE longToWide(EPA.09.Ex.18.3.TCE.df, "TCE.ppb.orig", "Month", "Well", paste.row.name = TRUE) # BW-1 BW-2 BW-3 CW-4 #Month.1 <5 7 <5 #Month.2 <5 6.5 <5 #Month.3 8 <5 10.5 7.5 #Month.4 <5 6 <5 <5 #Month.5 9 12 <5 8 #Month.6 10 <5 9 14 # If we construct the prediction limit based on the background well # data using the maximum value as the upper prediction limit, # the associated confidence level is only 82%. #----------------------------------------------------------------- predIntNparConfLevel(n = 18, m = 4, pi.type = "upper") #[1] 0.8181818 # We would have to collect an additional 18 observations to achieve a # confidence level of at least 90%: predIntNparN(m = 4, pi.type = "upper", conf.level = 0.9) #[1] 36 predIntNparConfLevel(n = 36, m = 4, pi.type = "upper") #[1] 0.9
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