You can look into the concept of Robust mean and Robust standard deviation in this regard.

I think, the topic of Robust statistics may be useful in finding answers to your question.

Hello Ami,

It would be helpful if you could tell us about your study design and your research question. You said you have 141 test cases. Are these cases being used to make claims about a larger group that you have not tested? Or are these cases the only ones you are interested in? If it is the latter, you do not need the machinery of statistical hypothesis testing (p-value, significance level, and such). On the other hand, if it is the former, you would need hypothesis testing. But what hypothesis are you testing?

Not only is your distribution skewed, but it has gaps. On a range of 0-200, only five cases are above 110. You, the expert, will have to explain this. Are these five cases errors or merely unusual cases?

Q1. The median (middle score) is the appropriate "average" or center when the data are skewed. The interquartile range is the measure of spread or variability for skewed data. You can find out more about the median here: https://sphweb.bumc.bu.edu/otlt/mph-modules/bs/bs704_summarizingdata/BS704_SummarizingData5.html#headingtaglink_4, and the interquartile range here: https://sphweb.bumc.bu.edu/otlt/mph-modules/bs/bs704_summarizingdata/BS704_SummarizingData7.html#headingtaglink_1.

Q2. If you use the median as the average, you would not need to compute the mean. Do you want to compute the mean with and and without outliers just to show how outliers affect the mean? Or do you want to test some hypothesis?

Hi Paul,

I will clarify better what the histogram says and what I am looking for.

The histogram shows a distribution of performance efficiencies of 141 test cases. These are the only test cases I'm interested in. In the histogram the efficiencies are expressed as a percentage. That is, an efficiency of 0.8 is shown as 80%. I can well justify in practical terms why values ​​greater than 100% and less than 50% are outliers. But my critics do not accept this and argue that the whole spectrum of values ​​from 0 to 200 should be taken into account.

Now, I want to find the average of all the values. But the distribution is not normal. Therefore, the average will also be "not normal". Therefore, I am looking for a statistical approach so that

1. I can calculate the average in a more informed way.

2. I can calculate the averages with and without outliers to know the magnitude of the error. In your words, my hypothesis is that the gap is negligible and insignificant and I want to convince that I am right or to convince myself that I am not right.

The mean is still the mean. You mention p value but don't say what hypothesis you want to test. Please provide information so people can address your question. In particular, provide a detailed research question.

I'd like to second Daniel Wright :

The average is the average. There is no "more informed way" to calculate the average.

You write: "These are the only test cases I'm interested in." This indicates that this is the complete data you like to describe. If this is correct, then all the information is on the table and there is no statistical error you can make.

If you want, you can calculate the average for the entire data set or just for a subset, like for the values within 50%-100% and report both averages.

If, in contrast, these 141 values are a sample of some statistical population and you want to use the sample information to infer the population mean, then (1) the sample mean is still the best estimate of the population mean, and (2) you can get an impression about the distribution of such sample means be bootstrapping new samples and calculating their sample means.

It needs to be based on the assumed distribution of your non normal variable.

Dear Daniel,

Perhaps, you could try the Sign test in case of the non-normal or skewed data set.

Hi Ami,

IF you like, you can send me your data and I will provide you a way to analyse them

• I asked a similar question about the (NON-normal) exponential distribution in the question Minitab T charts and Confidence Intervals

Nobody dared to give a “single” sensible answer!

Hi Massimo,

I attach here an excel file.

A graphical analysis could be performed using boxplot (based on median values) and Tukey's fences (based on interquartile ranges).

For non-normal data Non-parametric methods are useful. You should use non parametric tests.

Hi Ami,

This is my analysis:

Your data are a mixture of two Weibull pdf:

1. Beta=2.5 and eta=110, until 70

2. Beta=65 and eta=90, from 80 onward

The CHI2 test fitting shows that the two curves “agree”

📷

Hi Massimo,

Your analysis is interesting and enlightening.

Thank you.

The researcher may use non-parametric statistics for non-normal data that fit in to his/her hypothesis or research question.

Use non-parametric test.

For non-parametric data you can use non-parametric methods like Chi-square, Kruskal Willis or Mann Whitney methods. It depends on what do you exactly want to test. For more information take a look at this page. I found it really helpful.

https://www.statisticshowto.com/probability-and-statistics/statistics-definitions/parametric-and-non-parametric-data/

Use non-parametric tests and be done.

Question to the people who gave the answer: "use non-parametric tests". Which non-parametric test compares the means of two samples, as asked by the OP and what is the "nonparametric" analog to the average? I am curious to know.

For non-normal distribution, you should use a non-parametric test.

Rainer Duesing : maybe the median?

I agree with Massimi Sivo. İnstead of a histogram, you should use a Weibull probability plot to mix the two distributions and see how well the fits are.You can find examples in one of my papers (Weibull analysis of....., part 2). Good luck.

non-Parametric methods

Following your answers, I did my homework. Below are the actions I performed and questions that remained open. Feel free to comment.

----

This histogram shows a distribution of experiment results and includes outliers.

The outliers cannot be investigated to ascertain whether they are real outliers or not. The observations that were marked as outliers determined based on expert knowledge.

The distribution with outliers (group A) show the following statistics:

Range (3-200), number of observations (141), mean (76), median (78), standard deviation (28), outliers (32, 22%)

The distribution without outliers (group B) show the following statistics:

Range (50-100), number of observations (109), mean (81), median (81), standard deviation (13)

The problem

Although the statistics indicate that the medians of groups A and B are identical and the means are close to each other, we are interested in performing statistical procedures that will strengthen the claim that the difference between the means is not statistically significant. And the question being asked is: what procedures are appropriate for such a case.

Most of the responses received argued that non-parametric hypotheses should be used without specifying which one. We will present the actions we have taken and the questions that remain open.

1. Shapiro-Wilk test showed that the distributions are not normal

Group A (P-value = 1.297e-7 < 0.05), group B (P-value = 0.00095 < 0.05).

2. 1-Sample Sign showed (75,83) 95% confidence interval.

3. 1-Sample t test showed (72,81) 95% confidence interval.

4. Bootstrapping test showed (76,83) 95% confidence interval.

5. Box-Cox transformation: The idea is to perform a transformation in order to get a normal distribution, then to calculate the mean and then perform a back-transformation of the result. The problem is that I know how to calculate the inverse function when it comes to the simplest transformation, but not that of the more accurate and modified inverse function.

See: https://www.spcforexcel.com/knowledge/basic-statistics/box-cox-transformation

6. Massimo Sivo and Murat Tiryakioḡlu suggested using a Weibull probability plot to mix the two distributions and see how well the fits are. Massimo Sivo presented a solution here. I have read several articles of Murat Tiryakioḡlu and others and I want to ask them:

How does mix distributions solve the problem? Will the mean that is obtained be more correct? Massimo Sivo, how did you find the scale and shape parameters? Did you use a particular tool? In excel? And how exactly did you decide on the two distributions ranges? Arbitrarily? And how did you calculate the mix of distributions you presented here?

Thanks in advance,

Ami

Can you please provide the Weibull probability plot for your data?

Hi Ami,

I beg your pardon BUT I did not find my previous analysis.

So I tried to do it again… and I did not get the same results I got 3 months ago.

I try to answer your question

• · how did you find the scale and shape parameters?

I used the Maximum Likelihood Method

• · how exactly did you decide on the two distributions ranges? Arbitrarily?

The previous time I remember that I divided the data in two parts: 1) the first 41 data, 2) the other data.

• · how did you calculate the mix of distributions you presented?

It was a linear combination of the two pdfs, taking into account the number of data.

Now I give you the results I found yesterday:

• Ø the 141 data provide the following estimates:

Beta=2.75 and mean=76.20, with 95% Confidence Interval 71.93----81.13

• Ø the first 132 data (exclusion of “outliers”) provide the following estimates:

Beta=3.78 and mean=72.37, with 95% Confidence Interval 69.31----75.87

SINCE 76.20 is outside the SECOND CI we could conclude that the “outliers” could be real…

Sorry for the inconvenience…

Massimo

@ Murat Tiryakioḡlu

Why did you not provide YOUR analysis of the data?

Go ahead with non-parametric method