Locates a single change-point in an annual series based on the Pettitt test.

pett(x, plot = FALSE, order = FALSE, pval = 0.05, pchs = c(19, 21), ...)

## Arguments

x a numeric vector, matrix or data frame with no missing interior values Should the trends be plotted when x is a matrix? Should the plotted trends be ordered by size? p-value for significance Plot symbols for significant and not significant trend estimates, respectively Other arguments to pass to plotting function

## Value

A list of the following if x is a vector:

pettitt.K

Pettitt's statistic

p.value

significance probability for statistic

change.point

last position preceding change to new level

change.time

if available, time of change.point position

change.size

median of all differences between points after and up to change.point

or a matrix with corresponding columns if x is a matrix or data frame.

## Details

Pettitt's (1979) method is a rank-based nonparametric test for abrupt changes in a time series. It uses the Mann-Whitney statistic for testing that two samples (before and after the change-point) come from the same distribution, choosing the change-point that maximizes the statistic. The p-value is approximate but accurate to 0.01 for $$p \le$$ 0.5. Serial correlation is ignored, so the interval between points should be long enough to avoid strong serial correlation. The size of the change is estimated as the median difference between all pairs of observations in which the first one is after the change-point and the second is up to the change-point.

Missing values are allowed at the beginning or end of each variable but interior missing values will produce an NA. Otherwise the change-point might not be meaningful.

If plot = TRUE, a dot plot of change.times is shown. If sort = TRUE, the dots are sorted by change.time. The plot symbols indicate, respectively, that the trend is significant or not significant. The plot can be customized by passing any arguments used by dotchart such as xlab, as well as graphical parameters described in par.

## Note

The change.point returned by these functions is the last position before the series actually changes, for consistency with the original Pettitt test. But for reporting purposes, the following position might be more appropriate to call the “change-point”.

The Pettitt test produces a supposed change-point, even when the trend is smooth, or when the abrupt change is smaller than the long-term smooth change. Remove any smooth, long-term trend before applying this test.

Pettitt, A. N. (1979) A non-parametric approach to the change-point problem. Journal of the Royal Statistical Society. Series C (Applied Statistics) 28(2), 126--135.

## Author

Alan Jassby, James Cloern

## Examples


# data from Pettitt (1979, Table 1):
y <- c(-1.05, 0.96, 1.22, 0.58, -0.98, -0.03, -1.54, -0.71, -0.35, 0.66,
0.44, 0.91, -0.02, -1.42, 1.26, -1.02, -0.81, 1.66, 1.05, 0.97, 2.14, 1.22,
-0.24, 1.60, 0.72, -0.12, 0.44, 0.03, 0.66, 0.56, 1.37, 1.66, 0.10, 0.80,
1.29, 0.49, -0.07, 1.18, 3.29, 1.84)
pett(y) # K=232, p=0.0146, change-point=17, the same results as Pettitt
#> $pettitt.K #>  232 #> #>$p.value
#>  0.0146
#>
#> $change.point #>  17 #> #>$change.time
#>  17
#>
#> $change.size #>  1.07 #> # identify the year of a change-point in an annual time series: pett(Nile) #>$pettitt.K
#>  1617
#>
#> $p.value #>  3.59e-07 #> #>$change.point
#>  28
#>
#> $change.time #>  1898 #> #>$change.size
#>  -260
#> # apply to a matrix time series:
y <- ts.intersect(Nile, LakeHuron)
pett(y)
#>           pettitt.K  p.value change.point change.time change.size
#> Nile           1362 7.83e-06           24        1898     -255.50
#> LakeHuron      1532 2.88e-07           46        1920       -1.54pett(y, plot = TRUE, xlab = "Change-point")
legend("topleft", legend = "p < 0.05", pch = 19, bty="n") # note how a smooth trend can disguise a change-point:
# smooth trend with change-point at 75
y <- 1:100 + c(rep(0, 75), rep(10, 25))
pett(y) # gives 50, erroneously
#> $pettitt.K #>  2500 #> #>$p.value
#>  1.5e-16
#>
#> $change.point #>  50 #> #>$change.time
#>  50
#>
#> $change.size #>  55 #> pett(residuals(lm(y~I(1:100)))) # removing trend gives 75, correctly #>$pettitt.K
#>  1743
#>
#> $p.value #>  2.9e-08 #> #>$change.point
#>  75
#>
#> $change.time #>  75 #> #>$change.size
#>  4.374437
#>