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I used the Sens slope from the library: trend

https://www.rdocumentation.org/packages/trend/versions/1.1.4/topics/sens.slope

The sens slope is the median of all values.

If I am estimating the trend slope for 50 years of data (so we have 50 values).

  v=runif(50)
   sens.slope(v)

And the sens slope estimate was

Sen's slope 0.005153406

Can we say that there is a change of O.OO5153406 per year?

What shall I do to get the change per 5 years?

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That is not really a reproducible example (a seed would have helped) and it points to a median slope essentially $0$. Suppose instead you had data with an annual of trend of $0.3$

library(trend)
set.seed(37518)
v = runif(50) + (1:50)*0.3
sens.slope(v)

giving

        Sen's slope

data:  v
z = 9.8873, n = 50, p-value < 2.2e-16
alternative hypothesis: true z is not equal to 0
95 percent confidence interval:
 0.2992930 0.3106585
sample estimates:
Sen's slope 
  0.3051533 

so the trend slope has been spotted reasonably correctly.

The trend over $5$ years would be $5$ times that so $1.525767$ in this case, close to the designed trend over $5$ years of $5 \times 0.3=1.5$

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  • $\begingroup$ Thanks, So we simply multiply the sens slope by 5? if we need the change per 10 years, we multiply by 10 ,right? $\endgroup$
    – Tpellirn
    Mar 17 at 14:03
  • $\begingroup$ @Tpellirn Slope is change vertically per unit change horizontally, which I was taught as "rise over run". So yes: a slope $\frac{\Delta y}{\Delta t}$ corresponds to a change of $\Delta y= 10\frac{\Delta y}{\Delta t}$ when $\Delta t =10$, i.e. ten times the slope $\endgroup$
    – Henry
    Mar 17 at 14:09

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