queue waiting times comparison

Hallo Jim

Thank you and see within text.

jim holtman <jholtman at gmail.com> napsal dne 18.08.2011 14:09:11:

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?

Well. I trust them, however mean is "correct" central value only when data
are normally distributed or at least symmetrical. As the values are
heavily ?distorted I feel that I shall not use mean for comparison of such
sets. Anyway t.test tells me that there is no difference between y2 and
y1.

t.test(ml[[1]], ml[[2]])

? ? ? ?Welch Two Sample t-test

data: ?ml[[1]] and ml[[2]]
t = -0.452, df = 19.557, p-value = 0.6563
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
?-18.17781 ?11.71115
sample estimates:
mean of x mean of y
?16.83333 ?20.06667

So based on this I probably will never get conclusive result as sd due to
"outliers" will be quite high.

When I do
plot(ecdf(ml[[2]]))
plot(ecdf(ml[[1]]), add=T, col=2)

it seems to me that both sets are almost the same and they differ
substantially only with those "outlier" values.

If I decreased small values of y2 (e.g.)

ml[[2]][ml[[2]]<20] <- ml[[2]][ml[[2]]<20]/2

I get same mean

lapply(ml, mean)
$y1
[1] 16.83333

$y2
[1] 16.1

and t.test tells me that there is no difference between those two sets,
although I know that most events take half of the time and only few last
longer so for me such set is better (we improved performance for most of
the time however there are still scarce events which take a long time).

plot(ecdf(ml[[2]]))
plot(ecdf(ml[[1]]), add=T, col=2)

So still the question stays - what procedure to use for comparison of two
or more sets with such long tailed distribution? - Trimmed mean?, Median?,
...

Thanks.

Regards
Petr

On Thu, Aug 18, 2011 at 7:49 AM, Petr PIKAL <petr.pikal at 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