You shouldn't be looking at p values for this. (does whatever package you are using have quantile quantile plots, so in R try qqnorm(x) ). If you are worried about the effect of outliers, see Rand Wilcox's textbooks on robust methods (or other robust statistics books). You can still explore differences in the means, just change don't use least squares if that is what you are worried about.

Hi Daniel Wright thanks for your response.

Why is it that I shouldn't be looking at p-values (apologies if this is a silly question)? I was under the impression that when examining normality, there are two ways to do so; one being statistical tests and the other being visualisation. Then, if the statistical tests return as non-normal, you can complete a visual inspection to see if the plots are also showing results of non-normality.

Also, yes, I am using SPSS and it has Q-Q plots. Thanks as well for the recommended resources.

Thank you!

Hello James,

Given the sample size, I don't think that violation of the normality assumption is likely to be that profound an issue (b/c of the central limit theorem). As well, be sure that you are checking normality separately by group, as it is the model errors/residuals that are to be checked for conformance, not the collective single pool of observed scores.

An alternative method would be to use bootstrapped (standard errors) estimates for the anova (ordinary, one-way design). This should work quite well. A number of statistical software packages offer this option (c.f., ANOVA.boot in R; SPSS).

If you're concerned about whether scores are genuinely interval strength, then the rank analog would be the Kruskal-Wallis one-way anova by ranks.

Good luck with your work.

David Morse I believe you are correct in your application of the CLT in this case but rate of convergence does matter. Some people apply it at a questionable n. A good reference I think is this link: https://www.bing.com/search?q=central+limit+theorem+speed+of+convergence&qs=n&form=QBRE&sp=-1&pq=central+limit+theorem+speed+of+convergence&sc=0-42&sk=&cvid=A63B32A87866466387CCD4F36EF56DB9

James Phillip White Please note that a p-value less than 0.05 in a test for normality may not be what to are looking for because the H0 is not always the same. In fact in many p

James Phillip White, the data presented does not show what procedure / statistical test you used to test normality.

Please provide some additional details about the procedure used. Thank you!

@James Phillip White I am assuming that you have checked normality with Kolmogorov Smirnov. The question does not give enough detail on the exact nature of data collected. Though it is apparently unlikely that the data violates normality assumptions, given the huge sample size, you could probably use a Kruskal Wallis ANOVA test to determine differences between the four groups.

Viswa Chandu If he did use k-s and got a p-value < 0.05 then his popn is very unlikely to be normal. See my earlier link repeated here: https://www.bing.com/search?q=What+is+the+p+value+for+normality+test%3F&qs=n&form=QBRE&sp=-1&pq=what+is+the+p+value+for+normality+test%3F&sc=0-39&sk=&cvid=42EAA666A84B41519C84C8EE040A25BE He had better get what he's done right before doing anything else. Best David Booth

Dear James,

Indeed - a good question - I also have faced similar issues -

As per Field (2013) - there is no need for normality tests for a big sample size. Read the reasons for this.

Just cite this and report the outcomes of the normality test if you wish.

Here is the solution for you. Please read Page -7.

http://www.discoveringstatistics.com/docs/exploringdata.pdf

Look at page 22 of (Field, 2013) and other relevant sections as per your design.

References

Field, A. (2013). Discovering statistics using IBM SPSS statistics. sage.

I think that you could use Kruskal-Wallis test to be more robust if you want to se more conservative, but I agree that, with your sample size, you do not need to do non-parametrics, and you can use some citations above for that.

BTW for those that think a large sample size protects a normality condition under any circumstances please read the central limit theorem. I have attached these links for easy reading: https://www.google.com/search?sxsrf=ALeKk03guxXdlE5pYjcsvUhM27FJzQhXYg%3A1588385436939&ei=nNasXtHtOOHptQbHibOgDA&q=Central+limit+theorem&oq=Central+limit+theorem&gs_lcp=CgZwc3ktYWIQAzICCAAyAggAMgIIADICCAAyAggAMgIIADICCAAyAggAMgIIADICCAA6BAgAEEc6BwgjEK4CECc6BwgjEOoCECc6BQgAEJECOgQIABBDOgUIABCDAVDvhkFY8N9BYPTlQWgBcAJ4BIABkQGIAaIakgEEMjcuOZgBAKABAaoBB2d3cy13aXqwAQo&sclient=psy-ab&ved=0ahUKEwjRq_DtjJTpAhXhdM0KHcfEDMQQ4dUDCAw&uact=5

and

https://www.bing.com/search?q=central+limit+theorem+speed+of+convergence&qs=n&form=QBRE&sp=-1&pq=central+limit+theorem+speed+of+convergence&sc=0-42&sk=&cvid=A63B32A87866466387CCD4F36EF56DB9

SPSS Books notwithstanding.

Thanks everyone for the assistance thus far... I've provided a few responses below in relation to your questions, and also for some further clarification I would like...

Viswa Chandu and Marius Babici I used both Shapiro-Wilk and Kolmogorov tests with both p-values returning as very similar (and both significant).

David Eugene Booth thanks for your detailed explanations and the resources you've provided. Correct me if I am wrong, but are you saying the hypotheses I am question may be incorrectly addressing what I am investigating? For example, I should be looking for differences in groups and therefore the null hypothesis returning as significant in contrast to non-significant is a positive outcome? I'm not sure what exactly you are saying here sorry. I think though now after reading your earlier comment again is in relation to what test I've applied e.g., kolmogorov, Shapiro, or even Anderson-darling as another example?

Yubaraj Adhikari have you seen many publications that have cited the work of Andy field as provided which have simply noted that due to the sample size, the results of the normality tests can be presented, and there is no need to do any further tests?

Pau Garcia Grau in relation to your comment which also conforms with Yubaraj, are you concluding that non-parametric tests are not required due to sample size? What about parametric tests? Can they be conducted, and if so, are they even required?

James Phillip White, for large samples like yours, only the value of p on the Kolmogorov test is relevant. Given that the value of p is significant, you correctly rejected the hypothesis of normality of the distribution.

In my opinion, before you decide to use non-parametric tests, I consider that you have to check if the sample of the values used comes from an asymmetrical or normally vaulted population. In other words, if skewness and kurtosis are so different from 0 (zero) that you reject the hypothesis that the sample values come from a normal distribution.

To test this hypothesis I recommend using the standard error to calculate the confidence interval of Skewness and Kurtosis. The standard error for the two indicators can be calculated in the Descriptive Procedure.

If the limits of the 95% confidence interval include the value 0 then you will have to accept the hypothesis that the population distribution has normal symmetry or vaulting and use parametric tests.

The confidence limits for Skewness are:

95% CI = Skewness +/- 1.96 * standard error Skewness

The confidence limits for Kurtosis are:

95% CI = Kurtosis +/- 1.96 * standard error Kurtosis.

Parametric tests are more sensitive to asymmetry than to the existence of the vault, which is why I recommend using non-parametric tests if 0 at Skewness is beyond the 95% confidence interval.

Best Regards,

I am agreeing with Marius Babici that you interpreted the p-value on the normality test correctly. Your population is non-normal. Good job! D. B.

Thanks again David Eugene Booth and Marius Babici for your comments :)

Marius Babici to examine the standard error of the skewness and kurtosis values in SPSS, is it correct to use the following process (see below)?

Analyse - Descriptive Statistics - Descriptive Statistics (again) - then add the variable/s in - select the skewness and kurtosis boxes - press ok to run the output?

The std error alone for the variable I did a practice run on for skewness returned as .032, meanwhile the std error for the kurtosis std error returned as .064.

So does this mean that I can conclude that the sample has come from a normal distribution, and therefore parametric tests can be conducted?

Another note - I have previously been under the impression that the skewness and kurtosis scores are recorded by dividing the skewness statistic by the skewness std error score - is that incorrect, and do I just simply use the std error score on its own for both skewness and kutosis?

James Phillip White, your procedure is correct.

Additionally, you need to calculate the 95% confidence interval limits for Skewness and Kurtosis, respectively the lower limit and the upper limit. To calculate the limits please use the formulas:

95% CI = Skewness value +/- 1.96 * standard error Skewness

95% CI = Kurtosis value +/- 1.96 * standard error Kurtosis.

If the standard errors for Skewness and Kurtosis are within the confidence interval then you can assume that the sample comes from a normal distribution.

Best Regards,

I'd like add a little to Marius Babici 's answer. There are several ways to calculate the confidence intervals for these (e.g., see Article Comparing Measures of Sample Skewness and Kurtosis

Article Problematic standard errors and confidence intervals for ske...

I figured it would be nice to give more detail on those measures. Here are my lecture notes for skewness and kurtosis. Later in the notes I talk about confidence intervals, and suggest bootstrapping for quantile based ones.

Dear James Phillip White,

Field (2013) is widely cited, I have read some of the other articles but could not recall the citation (need to revisit). The logic is simple when we have more than adequate data from a larger sample (hopefully, you have already committed outliers), then the normality is not an issue. Some of the other researchers already outlined and suggested to read - Central limit theorem.

Field suggests reporting - Histograms, box-plots, scatter-plots (distribution check), and freely presenting the value of Skewness and Kurtosis. Please explain the reason for skewed data and a big number of data as a reason for this.

I personally believe that there is no need to go for a non-parametric test if you have a large sample.

Kindly discuss with your supervisor, this is important (or suggest your views to your supervisor) - if you would like to avoid any pitfall at the end.

with best regards, Yubaraj