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Regression using absoulte residuals

manphoolfageria

I have yearly precipitation extremes data (data in frequency, 0 to 15) from 1989 to 2019. My hypothesis is that climate extremes (precipitation) increased in last 16 years (2004 to 2019) than previous 15 years (1980 to 2003).

I have performed regression with this data and my results show that the regression line is downward. Then I performed regression using the absolute residuals and this time the results show that regression line of the absolute residuals is upward.

Am I doing right?

Why two kinds of results?

Which one is right?

Accepted answers

Mark_Bailey

First of all, why not enter your questions directly instead of putting them into an Excel workbook that we must open?

Second, there is not JMP data table. I could make one from the columns in the Excel workbook, but you do not seem to be using JMP. This Discussion forum is for JMP users.

Here is my answer to your questions.

Use of permutation test. I already answered this question previously.
Use of 'absolute' regression for non-normal data. I already answered this question previously.
Compare spread of two periods of time. I already answered this question previously.
Suggest another method. I already answered this question previously.

We seem to be 'going around in circles.'

manphoolfageria

Mark,

Thanks for your reply. I use JMP on a regular basis. Sorry, my mistake that I forgot to attach JMP file. I have attached it now that shows that I do use it

The dept. I work with have three licenses/users and they are frequent users.

For Mark Apil 25.jmp

All comments

Mark_Bailey

I am not sure that linear regression is the correct method to compare two time periods. Regression requires matching a X and Y value, like Precipitation versus Year for a trend. What is your response and what is your regressor?

How did you perform both regressions in JMP? Knowing the methods that you used will aid us in helping your with a correct answer.

If you are using counts as a response, then a Poisson loglinear regression might be more satisfactory. But I still don't see how you compared the two time periods. Just by comparing the parameter estimates? If so, there are more comprehensive methods for such a case. Assuming that the trend is linear over time, then you might have a linear predictor like Precipitation = constant + beta(1)*Year + beta(Period) + beta(2)*Period*Year. This way, you can test if there is an average difference between the two periods and if the trend is different between the two periods.

statman

Listen to Mark. I have some other comments:

1. You are regressing what? Extreme Precipitation vs. Time? Or are there other variables missing in your explanation?

2. Residuals (actual-predicted) is the delta between the prediction model and actual results. If this is increasing over time, then the model may have an unequal variance issue (it falls apart as time increases). Not sure why you would take the absolute value of residuals (OK, magnitude), but why wouldn't you care about sign? NID(0,variance) basic assumption of quantitative analysis.

3. Realize the regression model is dependent on what Y you are fitting, what terms are in the model and how you got your data.

manphoolfageria

Hi Mark,

Thanks so much. I have added a excel file which will show you how I have analyzed the data. Your answer will be highly appreciated.

Regression April 9 2020.xlsx

manphoolfageria

Thanks so much. I have attached a file containing data, and output files.

This will explain you how the data look like and I am sure you will be able to suggest me the right analysis to perform.

Mark_Bailey

You are comparing counts of occurrences of extreme precipitation during two periods of time. The permutation test is a good way to consider the observed difference.
A permutation test or ANOVA should answer your question. Other non-parametric tests are valid if you are concerned about any difference between the two periods.
The simple linear regression is testing if the mean response depends on the predictor Year.
1. The trend is not significant because the apparent increase in occurrences over time is not unusual while assuming that there is no trend.
2. The regression of the abs(residuals) versus the original counts is a completely different analysis. You cannot compare the two regressions.
3. The second regression indicates that your response might be homoscedastic. There is not a mean trend, there is a variation trend.
Yes, the permutation test is appropriate for a simple comparison of the two periods.
1. The statement based on the one-sided permutation test (counts are greater in the second period) is simply that it is estimated that 15% of samples had a difference at least as large as the observed difference.
2. Where did the operational definition for counting come from? That is occurrence of more than 40 mm rain in 4 consecutive days.

I ran my own permutation test with the data that you supplied. I consistently get a different result. Here is one of them:

So my one-sided p-value would be 1185 / 2501 = 0.4738. That is, it is estimated that 47% of the samples produce a difference at least as large as the observed difference assuming the null hypothesis is no difference (basis for permutations).

Mark_Bailey

In addition, the assumption of linear regression is that the response is unbounded and may vary between negative infinity and positive infinity. Counts are bounded by zero. The regression he performed is not the best model.

statman

Why don't you attach the JMP file?

manphoolfageria

Hi Mark,

Thank you so much for your kind support and advise. The info you provided is very useful. I have done a couple of analyses and looking for some additional advise. Pleased see the attached file. I would appreciate your advsie. Regards

For JMP community April 10.docx

manphoolfageria

Statman,

I have attached the JMP file. In my previous attachment, I have added a file with some analyses and questions in response of Mark's advise.

Thanks

JMP regression File.jmp

manphoolfageria

Hi Mark,

I performed the permutation test again a few times. Please see the attached file. My results are totall different than yours, means, I am doing a mistake. I run permutation while I think you did bootstarap. Please advise.

Before I contacted you a Mathematics Prof suggsted me using permutation than bootstrap. Therefore I used permutationt. Also, I do not have an access of Bootstarap.

Also, how you got value of 1185 / 2501 = 0.4738.

Regards

Caribou precipitation 40 mm 4 days permutation.jmp

Caribou precipitation.jmp

For Mark.xlsx