The Mann-Kendall rank test for monotone trend with the Theil-Sen median-of-slopes estimator: no distributional assumption, resistant to a single aberrant period, and meaningful at the series lengths an annual administrative extract actually has.
Usage
trend_test(
y,
x = NULL,
value = NULL,
period = NULL,
exact = NULL,
alternative = c("two.sided", "increasing", "decreasing"),
conf_level = 0.95
)Arguments
- y
The series, in period order, or a data frame.
- x
The periods. Defaults to the position, which is right for an evenly spaced series.
- value, period
Column names, when
yis a data frame.- exact
Whether to compute the exact null distribution of Mann-Kendall's S by enumeration. Feasible and used by default up to
n = 8(40,320 orderings); above that the normal approximation with the tie and continuity corrections is used.- alternative
"two.sided","increasing"or"decreasing".- conf_level
Confidence level for the slope interval.
Value
A list with S, tau, p_value, slope
(Theil-Sen), intercept, slope_lower / slope_upper
(the distribution-free interval), n and method.
Details
Kendall's tau here is S over the number of comparable pairs, so it is the rank correlation between the value and the period.
The slope interval is the standard rank-based one: the pairwise slopes are sorted and the interval runs between the order statistics that Mann-Kendall's variance places at the chosen level, so it is consistent with the test rather than derived from a different model.
References
Sen, P. K. (1968). Estimates of the regression coefficient based on Kendall's tau. Journal of the American Statistical Association 63(324), 1379-1389, for the median-of-slopes estimator and the distribution-free interval.
Wilcox, R. R. Modern Statistics for the Social and Behavioral Sciences: A Practical Introduction treats Theil-Sen among the regression methods that carry no normality assumption, which is the reason for preferring it on a series this short.
Examples
# Five years of placements.
y <- c(402, 377, 190, 268, 331)
trend_test(y)
#> $S
#> [1] -4
#>
#> $tau
#> [1] -0.4
#>
#> $p_value
#> [1] 0.4833333
#>
#> $slope
#> [1] -21.375
#>
#> $intercept
#> [1] 419.75
#>
#> $slope_lower
#> [1] -187
#>
#> $slope_upper
#> [1] 78
#>
#> $n
#> [1] 5
#>
#> $var_S
#> [1] 16.66667
#>
#> $alternative
#> [1] "two.sided"
#>
#> $conf_level
#> [1] 0.95
#>
#> $method
#> [1] "Mann-Kendall, exact over all 120 orderings"
#>
# A monotone series is detected even at n = 5, where a regression's
# standard error would be nearly uninformative.
trend_test(c(1, 2, 3, 4, 5))
#> $S
#> [1] 10
#>
#> $tau
#> [1] 1
#>
#> $p_value
#> [1] 0.01666667
#>
#> $slope
#> [1] 1
#>
#> $intercept
#> [1] 0
#>
#> $slope_lower
#> [1] 1
#>
#> $slope_upper
#> [1] 1
#>
#> $n
#> [1] 5
#>
#> $var_S
#> [1] 16.66667
#>
#> $alternative
#> [1] "two.sided"
#>
#> $conf_level
#> [1] 0.95
#>
#> $method
#> [1] "Mann-Kendall, exact over all 120 orderings"
#>
# One aberrant period does not create a trend.
trend_test(c(100, 100, 100, 100, 900))$p_value
#> [1] 0.4833333
# From a data frame.
d <- data.frame(year = 2019:2023, n = y)
trend_test(d, value = "n", period = "year")$slope
#> [1] -21.375