I am not sure why you say that "lapply(ml, mean)" shows (incorrectly)
that the second year has a larger average; it is correct for the data:
> lapply(ml, my.func)
$y1
Count Mean SD Min Median 90% 95%
Max Sum
18.00000 16.83333 12.42980 4.00000 12.50000 37.20000 41.05000
47.00000 303.00000
$y2
Count Mean SD Min Median 90% 95%
Max Sum
15.00000 20.06667 25.27694 4.00000 11.00000 45.80000 70.40000
97.00000 301.00000
You have a larger "outlier" in the second year that causes the mean to
be higher. The median is lower, but I usually look at the 90th
percentile if I am looking at response time from a system and again
the second year has a higher value.
So exactly why do you not "trust" your data?
On Thu, Aug 18, 2011 at 7:49 AM, Petr PIKAL <petr.pi...@precheza.cz> wrote:
> Hallo all
>
> I try to find a way how to compare set of waiting times during different
> periods. I tried learn something from queueing theory and used also R
> search. There is plenty of ways but I need to find the easiest and quite
> simple.
> Here is a list with actual waiting times.
>
> ml <- structure(list(y1 = c(10, 9, 9, 10, 8, 20, 16, 47, 4, 7, 15,
> 18, 36, 5, 24, 15, 40, 10), y2 = c(97, 10, 26, 11, 11, 10, 5,
> 13, 19, 5, 5, 59, 4, 16, 10)), .Names = c("y1", "y2"))
>
> par(mfrow=c(1,2))
> lapply(ml, hist)
>
> shows that in the first year is more longer waiting times
>
> lapply(ml, mean)
>
> shows (incorrectly) that in the second year there is longer average
> waiting time.
>
> lapply(ml, mean)
>
> gives me completely reversed values.
>
> Can you please give me some hints what to use for "correct" and "simple"
> comparison of waiting times in two or more periods.
>
> Thank you
> Petr
>
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>
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
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