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I have 20 samples of network latencies. Each sample consists of 5000 numbers (in milliseconds). I would like to find a 95% confidence interval for average (Mean), 90th percentile, 95th percentile and 99th percentile for network latencies.

I have measured the above as follows:

For a 95% confidence interval for average, I need to find average for each sample. So, I will get 20 values, each represents an average for one sample. Then, I calculate Mean (M) and Standard Deviation (SD) for those 20 values and apply the following formula to obtain a 95% confidence interval for average network latencies:

a 95% confidence interval = M ± 1.96 (SD/√20)

For a 95% confidence interval for 90th percentile. First I need to find ascending order for each sample. Then, the 90th value for each sample is picked. So, we will have 20 values, each represents 90th percentile for one sample. Then, I calculate Mean (M) and Standard Deviation (SD) for those 20 values and apply the following formula to obtain a 95% confidence interval for 90th percentile network latencies:

a 95% confidence interval = M ± 1.96 (SD/√20)

Is it true?

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CI for mean: It seems you are not taking full advantage of the available information.

CI for 90th percentile: Also, not taking full advantage of the available information, and assuming 90th percentiles are normally distributed, which is not necessarily the case. (Percentiles do converge to normal in sufficiently large samples, extreme ones such as the 95th sometimes quite slowly, so I wouldn't want to guess without checking.)


Example: I generated 10,000 samples each of size 5000 from an exponential distribution (severely right-skewed), here is a histogram of the resulting 5000 95th percentiles, along with the best fitting normal density curve. These percentiles are sufficiently right-skewed to decisively fail a Shapiro-Wilk normality test, but the plot shows they are not very far from normal.

enter image description here


Before speculating further, I have some questions. (a) Do you think network latencies are roughly normally distributed, or are they come from a right-skewed distribution? A histogram of a sample of 5000 should give a good clue, and there are formal tests you could use. (b) Do you think the 20 samples are from the same distribution? Put another way, would you feel comfortable merging them into one huge sample of size $20 \times 5000 ?$ (c) Do you use R software; if not what?

It is not reasonable to transmit all 100,000 observations as part of your Question. But it would be helpful to see just the means and SDs of each of the 20 samples. Also the 90th percentiles of the 20 samples.


Addendum. Looking at data from Comment of 3/24.

There are some difficulties. Among the 20 averages the four values above 10 show a severe departure from normality, based on a normal probability plot (also called a normal quantile plot) and a strong rejection of normality by the Shapiro-Wilk test. Similarly, among the 20 95th percentiles, the four values above 17 show a sever departure from normality (based on the same criteria). Departures are so far from normal, that I would not use t confidence intervals to describe the center of the populations from which these two samples of 20 were chosen.

Normal probability plots are shown below. For normal data one expects to see points (at least roughly) in a straight line.

enter image description here

First, you might want to verify that these values are correct. If so, you might want to consider whether the samples involved are somehow markedly different in other respects from others among the 20.

The Wilcoxon one-sample ('signed rank') procedure can provide 95% confidence intervals for the population 'medians' of of the averages and 95th percentiles. It does not rely on the assumption that data are normal, but does assume a roughly symmetrical sample.

For the averages, the Wilcoxon CI is $(8.16, 9.45).$ If the top four values are omitted for a more-nearly symmetrical sample, then the CI is $(8.00, 8.63).$

For the 95th percentiles, the Wilcoxon CI is $(11.56, 15.17).$ Without the top four values, $(11.52,11.67).$

But if you are trying to give assurances that latencies will rarely be above a certain point, it may be counterproductive to discard high outliers.

If (after pondering the several datasets out of 20 that are giving unusual values) you still feel that the 20 datasets are still equivalent, relevant samples from some population of interest, then you might consider merging them into one dataset of size $20 \times 5000 = 100,000,$ finding the 95th percentile of the 100,000 and reporting that.

Without seeing your data and understanding them better, that is about all I can responsibly say. But I can say that the values you provided in your Comment are so far from normal that I cannot agree that t confidence intervals would be a reasonable choice for analysis.

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  • $\begingroup$ (a) I drow a histogram of a sample 5000 but it does not give me a clue, I am not able to attach the histogram here, is there any way to attach here? (b) Yes, samples are from the same distribution. (c) No, I do not use R software. I just use Java code. @BruceET $\endgroup$ –  Ibrahim EL-Sanosi Mar 20 '17 at 17:02
  • $\begingroup$ Can you at least show me the twenty 95th percentiles and your resulting CI? $\endgroup$ – BruceET Mar 20 '17 at 17:16
  • $\begingroup$ (1) twenty Averages =[7.643, 7.676, 8.455, 9.457, 8.288, 7.515, 8.657, 9.259, 10.441, 10.981, 8.627, 8.119 ,8.690, 8.472, 7.995, 8.156, 8.242, 7.879, 10.599, 12.771], The result (average of the averages) = 8.896 and CI = (8.284,9.508). (2) twenty 95th percentiles = [11.376, 11.533, 11.404, 11.518, 11.603, 11.555, 11.709, 11.641, 17.491, 19.921, 11.515, 11.597, 11.808, 11.823, 11.506, 11.667, 11.770, 11.578, 18.802, 21.101]. The result (average of the 95th percentiles) = 13.146 and CI = (11.676,14.616). @BruceET $\endgroup$ –  Ibrahim EL-Sanosi Mar 24 '17 at 12:43
  • $\begingroup$ Good to have these. Please see Addendum to my Answer. $\endgroup$ – BruceET Mar 24 '17 at 19:40

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