Hi Nasir Al-Allawi, our following paper could help you to better understand the concept of normally distributed.

Majeed, H., Kayani, U. N., & Haider, S. A. (2021). The Project Communication and Trust Nexus as an Antecedents of Project Success: Moderating Role of Authentic Leadership. International Journal of Business Communication, 23294884211019098.

Hello Nasir Al-Allawi. OLS regression does not require a normally distributed outcome variable. The normality assumption applies to the errors, not to the dependent variable. Furthermore, normality of the errors is a sufficient condition, but not a necessary condition. The necessary condition is approximate1 normality of the sampling distributions of the parameter estimates (i.e., the coefficients). If the errors are normally distributed, those sampling distributions will also be normal (hence normality of the errors being sufficient). But even if the errors are not normally distributed, those sampling distributions will become approximately normal as n increases. See the attached slides, paying attention to slides 9 & 10 especially.

Also, statistical tests of normality as precursors to OLS models are generally ill-advised. They tend to be under-powered when n is small and become over-powered as n increases. This has been discussed many times on RG. Examining your model residuals graphically is generally a much better approach.

HTH.

1 The reason I say approximate normality is that I believe George Box was correct when he noted that nothing in the real world is normally distributed--see the attached image.

I would try the Box-Cox transformation which is probably in your current software . For those that talk about normal errors and not normal DVs I would love to see a proof that one doesn't imply the other. So far such a proof is beyond me and those others that I have asked. Best wishes, David Booth

BTW the attached search gives many ways to do Box-Cox including SPSS. BEST WISHES, David Booth

Conversely, I would love to see a proof that normally distributed errors (logically) imply a normally distributed DV.

Hello,

Nasir Al-Allawi I strongly recommend visually inspecting the residuals of the model instead of the outcome variable (see checking assumptions using SPSS here: https://www.open.ac.uk/socialsciences/spsstutorial/files/tutorials/assumptions.pdf). Note also that normality (of residuals) only needs to be approximate and is arguably the least important of the assumptions for the linear model (it's quite robust). If deviation is severe, other transformations might work depending on the type of outcome variable. A Generalized Linear Model might be even better and also more flexible to satisfy the more important variance assumptions (https://www.ibm.com/docs/en/spss-statistics/28.0.0?topic=SSLVMB_28.0.0/statistics_mainhelp_ddita/spss/advanced/idh_idd_genlin_typeofmodel.html)

The following is by no means a general statement, but an example to fuel the discussion (David Eugene Booth , Bruce Weaver). A skewed independent variable can result in a skewed response variable despite normally distributed residuals (I don't have time to do the reverse at the moment).

A skewed independent variable would not be typical of controlled experiments but could be common in "field" research (e.g. concentration of chemicals in the environment, income of survey respondents, age of patients, etc.)

Any thoughts?

### R code starts

set.seed(321)

# regression slope

beta1

Well Bruce you cited George Box and I assume he developed Box-cox for some reason. In any case a proof of either assertion will suffice to answer the question and I look forward to seeing one. Best wishes, David Booth

Hello Gi-Mick Wu. Thanks for providing that example. Here is another example showing residuals that approximate a normal distribution quite closely despite Y being quite far from normal. (To my eye, Y more closely approximates a uniform distribution.) Find attached the data (in the CSV file), SPSS syntax (.sps file), and SPSS output (in the Excel file). I've also attached a PNG file showing key distribution plots for Y and the residuals (in case anyone is leery of opening Excel files). (The syntax file is just plain text, so can be viewed with any text editor.)

Please note that I do not recommend relying on statistical tests of normality to justify use of OLS models, because they generally have too little power when n is low and too much power as n increases. Having said that, however, notice that in the attached example, both tests of normality SPSS EXAMINE implements (KS & WS) are far from statistical significance for the residuals, even though n = 1000.

David, would you refrain from using OLS regression on these data because Y is so far from being normally distributed? Thank you for clarifying.

PS- The data used for this example were inspired by a post by Alexis in this StackExchange discussion:

https://stats.stackexchange.com/questions/105239/regression-on-a-non-normal-dependent-variable

Bruce thanks for the update but did you try a Box- cox transformation to see if that works. Further did your eye tell you that wasn't a normal with very large standard deviation and

.do you notice the behavior at the tales of the scatter plot. It suggests that something is going on there that needs further investigation and by the way the axes are not labeled so I have no clue what I'm looking at here. I would suggest that your example needs work before publication

Even further this seems by your admission to be contrived data and it is well known that fake data is not like real data because it's generated by a non probability process unlike that produced by nature . Finally without some of the missing information I have no idea what I would do. Perhaps you can enlighten us on some of this. BTW if you are offering this as some kind of counterexample, and if you are please note that you have much more verification to do before this can be considered as an answer to the question as posed. A good source to look at is Bickel and Doksum Mathematical statistics available I believe from the z-library

Best wishes David Booth

PS apologies for the typos but it is late here

Oops Bruce I forgot. Something. Since you don't quote a p-value for the K-S test I must remind you that large p-values for this test Favor the normal hypothesis. See any reference, the easiest is Wikipedia. This is a common error for many doctoral students. So please do give us a p-value so we can be sure. That your example is as you claim. Best wishes David Booth

David, I am in a meeting right now, but I did take a moment to add the results for the two tests of normality to the plot. (Those results are in the Excel file I attached last time, by the way.) Plots on the left show the raw Y-variable, and plots on the right the residuals from a simple linear regression model.

EDIT

PS- I deliberately uploaded the raw data as a CSV file so anyone who wished to could estimate the model, make plots, etc. using their own software of choice.

EDIT 2

The axes are labeled. To see the full image, you must click on it.

Thanks Bruce that is lovely but all I asked was that you supply the p-value for YOUR K-S test. A p-value greater than 0.05 would imply that the distribution was normal. See any good K-S references. Best wishes for a great day Bruce. PS looking forward to the p-value. and BTW HAVE A GREAT MEETING

Apologies, David. You may not have remembered that SPSS reports p-values in the column headed Sig. If you look again at the image attached to my last post, you will find the p-values you seek. But let me repeat them here:

For the Y-variable:

KS test statistic = 0.059, p < .001

SW test statistic = 0.968, p < .001

For the residuals:

KS test statistic = 0.016, p = .200

SW test statistic = 0.999, p = .770

Try square root or inverse transformation.

Dear Anuraj.. thank you for your suggestion ... we tried this earlier but unfortunately it did not work.. my best

Hello Nasir Al-Allawi. If you have not already done so, I suggest that you estimate the model using the original outcome variable, save the residuals, and plot the residuals--e.g., a histogram, and more importantly, a QQ-plot. See an example of how to do that in the syntax file I attached to an earlier post. Or see the instructions on this page:

• https://libguides.bates.edu/spss/data#histresid

And then upload the figures here please.

HTH.

EDIT

Nasier, as an SPSS user, you may also find this page helpful:

https://stats.oarc.ucla.edu/spss/seminars/introduction-to-regression-with-spss/introreg-lesson2/

Notice especially the first sentence in section 2.2:

"In linear regression, a common misconception is that the outcome has to be normally distributed, but the assumption is actually that the residuals are normally distributed."

Late to the party, but one point and one question.

point. Linear regression does NOT make any distributional assumptions. How you estimate may, but that is a different issue.

question. Are you interesting in the linear association between each of the six and the response variable, or how (in some way) values for these six can be combined so that some aggregation of them may have a linear(ish) relationship with the response variable?

Seems this discussion skipped these points, which I think are worth addressing.

ps. And for stats procedures, start with scatterplots.

Nasir Al-Allawi , Can you post us a histogram and maybe also tabulate the data?

Thank you for your interest in my question Bruce Weaver and Anuraj Nayarisseri

I appreciate your feedback ...Please find attached the SPSS plots by Multivariate linear regression.

Appreciate your thoughts.

Thank you Nasir Al-Allawi. Even though your graphs say "Dependent Variable" at the top, I believe they are showing the standardized residuals from your model. And they show that the residuals are not close to normally distributed. What is the dependent variable? If you told us previously, I missed it. I ask because depending on what it is, some other kind of model might be more suitable--e.g., Poisson or negative binomial regression if it is a count. Thanks for clarifying.

@Bruce I apologize I have noticed that you did give the p-value for the K-S test on the residuals . I noticed that it's greater than 0.2. That p-value favors the alternative hypothesis which is the residuals are normal see the appropriate Wikipedia entry for the test for an explanation of the test hypotheses. Would you care to revisit your statements above. ?Best wishes David Booth

@Bruce .. thank you for ur response .. Dependent variable is age at first presentation ...

David, please quote the statements you think I should revisit. Thank you.

This looks a bit like a mixture of two (approximately normal) distributions with considerably different variances. I would look into the theory/background of these data if there is some variable or aspect missing in the model that could be able to explain this mixture.

Nasir Al-Allawi Yeah I had to work with a variable like that, it had a long tail. In biological data this seems to usually indicate people in a disease state. The best I was able to do with my variable was log transform it.

@Bruce i would start with the ones that say that the residuals aren't normal because of your misinterpreted K-S test. As I said to you before In other contexts READ THE REFERENCES. DAVID BOOTH

I like Jochen's suggestion that they may be at least two different types of observations as that treats the non-normality as a substantive issue not a technical problem. Clearly there a quite a few responses with high values that are not being modelled adequately by your current specification. There also looks to be a 'floor' value below which the response cannot/does not go.

In addition to thinking and looking quite hard at those cases; I would also try an avowedly exploratory procedure to see if I can gain some insight. In particular, and especially if I was trying to do predictions, I would consider using Tree regression in addition to standard regression as the latter is likely to miss both non-linearities and interactions and you really need multiple hold-out samples so as not to over-fit to the data you have. Here is a worked example -

http://www2.stat.duke.edu/~rcs46/lectures_2017/08-trees/08-tree-regression.pdf

The cross validation is important as it gives some idea of how complex a model can be supported by the data you have. As you increase the number of terms in the model usually you get a better and better fit to the observed data but a worse and worse fit to the holdout sample - you are then modelling noise and what you have found will not reproduce, and can be very misleading.

Modified slightly from my previous answer to https://www.researchgate.net/post/R-squared_adjusted_values_from_12-15

David Eugene Booth, please find and "quote" any statements of mine that show I misunderstood the null and alternative hypotheses for the KS test. Thank you.

Bruce I did that 2 days ago and I gave you references to document my position.i now attached the SPSS tutorial that says if p is less than 0.05 the distribution is not normal..you show in your output p equals or greater than 0.2. and hence the conclusion must be that you cannot reject the residuals are normal. THIS IS THE CRUX OF THE QUESTION THE WRITER PROPOSED AND YOU MADE THE MISTAKE THAT THE WRITER DID. THE ANSWER TO THE QUESTION IS YOU AND THE WRITER MADE A ERROR. THUS THE RESIDUALS ARE ESSENTIALLY NORMAL. AND THIS DISCUSSION IS ONLY OF ACADEMIC INTEREST. YOUR'S IN COMRADSHIP DAVID BOOTH. BTW BRUCE THE CANADIAN STATISTICAL ASSOCIATION OFFERS A VERY GOOD PSTAT QUALIFICATION. I SUGGEST YOU CONSIDER GETTING IT. D BOOTH PSTAT

@Bruce you may find the attachment to be of further assistance. David Booth

David Eugene Booth may it possible that you misread Bruce's example? He showed that the raw Y variable was not normally distributed according to KS and SW tests (ps < .05), wheras, by contrast, the residuals of the model could be considered as normally distributed (ps > .20). So where is he wrong in his statement? Additionally, he asked, if you would not use the OLS regression because the raw Y variable was not normally distributed.

@Rainer

Bruce also said the distribution of importance is the that of the RESIDUALS. Further he said originally that the distribution for y doesn't matter.Thus it would appear that somewhere Bruce is at least confused. Further he was asked by me what p-value he was quoting and twice refused to say. How is that helpful to anyone.? Best wishes David Booth

Dear David Eugene Booth ,

I am not sure if we talked cross-purposes, so here in detail what has been written so far.

On Dec 12th you wrote;“For those that talk about normal errors and not normal DVs I would love to see a proof that one doesn't imply the other.“ I think Bruce demonstrated that the distribution of the DV can be absolutely non-normal, but at the same time, the residuals can. In a simpler example, you can demonstrate that the distribution of the DV can be bimodal, e.g. if the data is generated by two groups with different means (take any survey on body height for men and women for real data). Yet, the distributions around the conditional mean values could be normal and hence also the residuals. Therefore, there is no need for the DV to be normally distributed. But I am pretty sure that is what you meant and just misread what Bruce wrote.

On the other hand on Dec 13th Bruce wrote:“I would love to see a proof that normally distributed errors (logically) imply a normally distributed DV.“ I think this is still open to be shown by anyone.

On dec 15th Bruce provided an example including all data and mentioned in text that with N=1000 the residuals did not differ significantly from a normal distribution according to KS and SW tests (hence p > .05). Although the sample data was provided (which would enable everyone to verify the p-values), Bruce additionally provided them on Dec 16th and it lined up with his previous comments and the conclusions, i.e. the DV was not normally distributed, but the residuals were. On Dec. 17th you asked again for the p-values (which have been delivered the day before in the table) and reiterated „A p-value greater than 0.05 would imply that the distribution was normal.“ Bruce did not claim anything else, but provided again the p-values on the 17th.

On Dec 22nd you wrote:“I noticed that it's greater than 0.2. That p-value favors the alternative hypothesis which is the residuals are normal see the appropriate Wikipedia entry for the test for an explanation of the test hypotheses.“ Maybe you didn’t notice, Bruce provided two distributions, one for the DV (with KS p < .05) and one for the residuals (with KS p > .2). Maybe you missed that detail.

I also cannot follow your objection from Dec 16th. You write „Even further this seems by your admission to be contrived data and it is well known that fake data is not like real data because it's generated by a non probability process unlike that produced by nature.“ So? If you can show in a perfectly controlled environment (i.e. simulation study) that the DV may be not normally distributed, but the residuals are, I think it is a quite good argument. By contrast, with natural data, I would rather expect that neither the DV, nor the residuals are normally distributed, therefore, I really do not get the argument. Please elaborate on that.

In your last post it sounds as if you would claim that for the OLS regression the distribution of the raw DV is of importance. Can you show evidence for this claim? Maybe I am wrong, but I thought it was about the conditional distributions which should be approx. normal. You referred to the Bickel & Doksum book. On page 10 of volume 1 they define the linear regression as Yi = μ(zi) + ei where μ(z) denote the expected value of a response with given covariate vector z. Isn’t that exactly the conditional distribution, so that Y ∼ Nn(Zβ, σ2J). As far as I understand this, the DV has not to be normally distributed, but the errors around the vector of expected values Zβ.

Many thanks for clarification!! Have a nice holiday.

@Rainer.Thank you for your note First, I will only reply to the portions I believe I understand. Given that I will cut to the chase I see. 1. Bruce did not tell me which test was the one he thought was important by identifying the appropriate p-value. Certainly the residual one is very important so I assumed it's that he was talking about. This would seem to be important since 2 distributions are discussed here I had to make a decision. If I chose the wrong one he could have told me that value that I asked for.

3. Perhaps of all the questions you asked the most important is what I Would Do in this situation? Given that I didn't collect the data,nor do I know the research questions the best I can do is describe my approach in a couple of similar situations. I am attaching two published examples. Perhaps you can ask Bruce for a similar example or perhaps two. I hope that this clarified my position.

Best wishes to you for a happy holiday season. David Booth BTW I had to do this on my phone which is somewhat difficult.

Rainer Duesing after seeing some of the comments that you quoted I still think that somehow this forum is used to iterate misconceptions and gross errors.

Students (& others) will most likely read some of what has been written here and take it at face value.

First of all, there are so many normality tests available that different people might come to different conclusions and use different tools depending on that preliminary test on the same exact data.

This could be an initial argument against using preliminary tests

On this note:

This being said, Nasir Al-Allawi do we really think that variables in real life CAN be normally distributed. If you do, then... heaven help us

Do we still think that a pvalue smaller than 5% show that a variable is normal...? With real data with enough cases you will almost always reject normality.

@David Eugene Booth: can we stop perpetuating the mechanical use of statistics and can we avoid to say that data is not normally distributed if p

I would like to thank all contributors who put time and effort to address the issue I raised. One important point that most of you stressed is that normality for the residuals and not outcome variable is required for linear regression. The latter assumption was clearly not met by the data at hand. Accordingly, I believe that alternative statistical methods are required to assess the data. Thank you all for your kind input and discussions.. wishing you all a happy prosperous 2022.

Thank you, Rainer Duesing, for your very helpful summary. Regarding my Dec 15 post, which seems to be at the centre of the controversy, let me add one small detail. The Excel file that is attached to that post shows all of the results for the demonstration, including the K-S and S-W normality tests on both the original (raw) Y-variables and on the residuals. In other words, the p-values that I "refused to share" on two or more occasions (according to some) were there from the very beginning. (I've attached that Excel to this post too so that anyone who wants to look at it to confirm that what I have said is true does not have to go hunting for the Dec 15 post. And for those who are leery of Excel files, I've attached a PDF too.)

Let me also emphasize one other thing about that Dec 15 post. I said I do not recommend relying on a test of normality to decide whether the residuals are (approximately) normally distributed or not, because when n is large, a small inconsequential departure from normality can easily be statistically significant, leading one to reject the null hypothesis for the test of normality (i.e., to conclude that the sample has not been drawn from a normally distributed population). Like Stefano Nembrini, I think that graphical methods for assessing normality of the residuals are much better than statistical tests of normality.

Nasir Al-Allawi, I'm sorry that your thread has become hi-jacked in the manner it has, and apologize if I have contributed to the problem. I wish you well, and hope you figure out a suitable means of analyzing your data. Quantile regression comes to mind as a possibility. I'm curious to know what Jochen Wilhelm, Kelvyn Jones, Daniel Wright, Stefano Nembrini and Rainer Duesing think about that.

Cheers, & Happy New Year everyone.

@bruce thanks for your comment. Since you did two K-S tests I made an assumption about which you were discussing. If I made an error in choosing which of your tests I apologize to you for the error.

@ Stefano. Thanks for your comments. The language I used in the post is the same as that used in the SPSS tutorial as attached.

At this point I would be happy to address questions on my positions as taken in.my two published examples as attached in my last reply to Rainer. Other than that I plan to exit this thread. Best wishes to all. David Booth

Bruce Weaver I would rather address the problem than change the analysis.

Quantile regression is a great tool and I would personally use at least both methods for an analysis where the data has already been generated (i.e. no power involved). Possibly more

My general concern is that regardless of using a model which is a conditional expectation, a conditional quantile or conditional mode the problem still remains.

The problem:

Say we have a dependent variable which is skewed or not unimodal, how would we interpret the results of a conditional mean or quantile?: maybe the mean is not even located where there is any mass at all.

If the dependent variable is at least unimodal than we 'd have a chance of extracting some meaningful information from a model., otherwise it's just garbage and probably not a good prediction tool.

What do you think?

David Eugene Booth I totally understand what you are coming from, but we should know better than SPSS manuals that basically say "if p>0.05 then Y is normal", otherwise as statisticians we are just perpetrating this nonsense.

I think we need to use their brain first and test second, i.e. rather than applying a bunch of tests meaninglessly, we should evaluate what a model (an OLS for instance) does and see if it is appropriate (in this thread it would sound something like if we decide not to use OLS, what can we use?).

I wish that one day we could use this forum in a more constructive matter, rather than engaging with people that are so strongly opinionated about dependent variables when using OLS or using a Mann–Whitney when data is not normal right away without thinking of the cost-benefit of using a model that possibly has more assumptions and does not really do what we think it does.

Good morning Stefano Nembrini. One possibility I was pondering with quantile regression was simultaneous estimation of multiple quantiles, as in the -sqreg- example on this page that estimates q25, q50 and q75 simultaneously:

https://www.stata.com/features/overview/quantile-regression/

It seems to me that could provide a sharper picture of how the explanatory variables are associated with the outcome.

Nasir Al-Allawi, apologies if you already told us this: What are the explanatory variables in your model? Can you provide a list and indicate whether they are quantitative or categorical (with number of levels for the latter)? Plots (e.g., box plots) of residuals by categorical explanatory variables (if there are any) might help you figure out how to proceed.

@Bruce quantile regression is certainly a good way to get a more complete picture (more robust than ols for instance)

I have my reservations on conditional quantile regression (sqreg) because we are usually interested in the unconditional distribution (with ols you can go from conditional to unconditional using the law of iterated expectations)

My point was that instead of convincing ourselves that one single method is the only way to go we should educate ourselves to use more tools to get a better picture

Second, the distribution we are studying can be unexpectedly behaving in a weird way ( maybe it's a mixture ) and focusing on values where there's little mass can be a useless statistical exercise.

Nasir -

Back to the original question, yes it is only the residuals that we'd like to be close to normal, but no they do not have to be. You get better efficiency over all unbiased estimators, but you do not need normality. If you do prediction intervals, a larger sample size helps the model part of the variance become smaller, but the sigma for the residuals part, a standard deviation, is not made smaller by increased sample size. The shape of your prediction interval could be a problem. But estimating the variance of the prediction error is OK.

But looking at your graphical residual analysis, I wonder if you have an omitted variable problem. You don't want too many variables, or too few, but just the right ones. If you were to review your subject matter knowledge for this situation, what would you consider to be the best set of variables, working together, to form predicted-y? You could compare alternative models on the same scatterplot, for the same sample, using graphical residual analysis, but you might want to look at a cross-validation if your result is 'too good' for some model. You mentioned six variables. You noted multivariate, but I guess you meant multiple regression? Your scatterplot graphical residual analysis was for that? Do you have any other variables? It looks like maybe an omitted variable, I'm guessing, but maybe not. Please try some other competing models.

I looked through the other contributions quickly and may have missed something that negates the above, but I think trying other models might be helpful.

Cheers - Jim

Nasir -

I strongly recommend that you review the contributions of Jochen Wilhelm and Kelvyn Jones above.

Best wishes - Jim

For me, I found this academic interchange to be helpful. However, when Bruce Weaver presented this table

For the Y-variable:

KS test statistic = 0.059, p < .001

SW test statistic = 0.968, p < .001

For the residuals:

KS test statistic = 0.016, p = .200

SW test statistic = 0.999, p = .770

I believe the whole argument ended. The dependent variable (Y) does not need to be normal and can be measured with error so OLS is BLUE and transformation experimentation can be conducted. Normality is a theoretical assumption! Increasing N and relying on CLT and sampling distribution theory says we are close to a steady-state.

Normality of the residuals is a crucial issue. In time-series analysis, we are taught to plot a graph of each regressor including Y as a first step (Much better to work with a double-log model for inference reasons). Nevertheless, as a second step, plot a graph of the residuals, perform a Jarque-Bera test of normality.

An insignificant JB test value implies doing nothing. A significant JB means that one could examine standardized residual (RESID/SEE) to find the outlying data point for the possible introduction of dummy variables. Carefully done should result in a low JB below 5.99. (Note that we can have an outlier that could influence kurtosis, however, this can be dummied out without affecting the inference).

At this point, one reestimates the OLS model, and to handle nonspherical errors one uses Newey-West estimator with Bartlett weights to produce efficient estimates OR White standard errors assuming unequal variances ( heteroskedastic errors) are implemented.

***We should all bear in mind that there is no perfect model even when it has passed all diagnostic tests. Such a model can be variance-dominated by another model as new data appear, but that doesn't mean that the dominated model is statistically unstable. Sometimes we argue as if there is only one way to New York city! Sometimes, with talk about p-values > 0.05 when we should say a large prob. value implies FTR (i.e., the null may be true). We cannot prove the null hypothesis.

The problem is specialised ! - I shall be happy the following link will solve if any - https://www.researchgate.net/post/Is_linear_regression_valid_when_the_outcome_dependant_variable_not_normally_distributed

Thank you Jaydip Datta for bringing this to our attention.

Did you try the Box-Cox (Atkinson et al., 2021) transformation (with one pr two parameters) ?

You may also have a look on this paper if you may read french : https://arxiv.org/abs/2112.03703

You can use a polytomous universal model (PLUM) ordinal regression procedure. If the data is not normally distributed, then you can only use non-parametric testing methods like a generalized linear model ordinal logistic or probit model or a multi-level modelling (MLM) equation so as to address the problem of clustering of the dataset.

Attached is Nasir's graphical residual analysis from above (Dec 22, 2021). Please see my comments above (Dec 27, 2021). As I noted, I think this may be an omitted variable problem. At any rate, other options for the set of predictors used should be explored. (The best set of predictors may produce 'essential heteroscedasticity.' See https://www.researchgate.net/publication/354854317_WHEN_WOULD_HETEROSCEDASTICITY_IN_REGRESSION_OCCUR.)

PS -

Nasir -

Your graphical residual analysis appears to show a great deal of heteroscedasticity (a feature which can be good, depending on if it is the natural kind to be expected), and a weird truncation at the bottom. It looks like something artificial in your data, or a model with some perhaps odd specification. I'd look at the model and see if there is something different from usual. I've seen odd things when, say a large intercept was reduced by the variables used (a - b1x1 - b2x2), and here there might be something unusual to consider as well. ???

Perhaps you could write out your model for us?

Anything restrictive about the data?

Thanks - Jim

PSS -

The tight lower near demarcation (almost a "line") for the residuals, which is not horizontal, and the points drifting above it shows heteroscedasticity, but it is not desirable to see them pile up that way. Something must be causing that.

Can you explain this from a subject matter perspective?

I still advise that you try some other predictor combinations.

Transformations are definitely not the solution, see:

Method Non-linear regression -Why you shouldn't take the logarithms...

Can you show us a graph of your data?

I doubt that it is the right thing to depend solely on Shapiro Wilk test to decide on whether a data set is normally distributed or not. I find plotting graphs (e.g. Histograms, normality plots, stem and leaf, scatter plots) more useful most times. Doing more readings is always a plus.

I would like to thank all contributors for their constructive comments .. I decided not to use linear regression and I am trying to look at other ways of analysing data ... and other ways to assess outcome with the help of the statician at our university.. I thank you all for your time and feedback.. which have really been quite informative and gave me better insight into the complexities of linear regression.

https://www.statsimprove.com/en/linear-regression-should-dependent-and-independent-variables-be-distributed-normally/

following up link