HELP! T-test, normality assumption. Great difference between Shapiro Wilk and Q-Qplot/ Z-score calculation

#1
Hey there!

I have a question about the difference in results between the Shapiro Wilk test and the calculation of a Z score with skweness and kurtosis (divided by SE).
I have a sample size of N = 143. It is divided into two groups (Western candidates N = 71 and non-Western candidates N = 72). To assess the normality, I did the following:
Calculate a z-score by dividing the values for skewness and kurtosis by their respective standard errors (SE).
If I use this calculation, I only have 1 non-normal distribution. However, when I look at the Shapiro wilk test of normality within the output, all analyzes are NOT normally distributed (except one). So what confuses me is that If I take a look at the normal Q-Q Plots, I only see an obvious non-normality in the applicant trait importance rating for the non-Western candidate (just as I found in the , z- value calculation of the skewness and kurtosis).

Basically, the only thing I need to know is in which cases I have violated the assumption of normality (according to Z-values and Q-Q plots only in 1 case, while according to Shapiro Wilk in 5 cases).

In my thesis I wrote the following about this: (this is the part I have doubts about whether I have used the correct method):

Various methods for normality testing were used to assess whether the applicant rating scores for each level of the candidate (Western and non-Western) were normally distributed. To assess normality within the total sample (N = > 50), Normal Q-Q Plots were visually inspected, and z-scores were calculated by dividing the skewness and kurtosis values by their respective standard errors. Mayers (2013) suggested that a cutoff of ±2.58 should be used for samples from 51 to 100 participants. Applicant rating scores for each level of the candidate (Western and non-Western) were normally distributed, except for the applicant trait importance rating of the non-Western applicants with a skewness of -.92 (SE = .28) and kurtosis of 1.68 (SE = .56).

So I only want to know whether I should write down the results like I did in the paragraph above. OR that I should say something like:

Within the total sample, all applicant rating scores for each level of the candidate (Western and non-Western) violated the assumption of normality, as assessed by Shapiro-Wilk’s test (p < .05), except for the credential rating of the non-Western candidate.


That's all I want to know. Thanks!

I will attach the screenshots of the Q-Q plots and also the Shapiro Wilk outcome and the Skewness and Kurtosis numbers.
 

Attachments

Karabiner

TS Contributor
#3
I am not quite sure what you want to achieve. With a total sample size as large as yours,
is is not required that each sample is from a nornally distributed population, or that the
residuals are normally distributed in the population, for the t-test or an F-test to be valid
(central limit theorem). So why all that fuss about normality of... whatever?

With kind regards

Karabiner
 
#4
I have to test for normality since it's an assumption of the T-test (according to my text book that I need to follow through)...
I only wonder though if I should look at the Shapiro Wilk output or at the Z-score (Z= Skweness/Se, and Z= Kurtosis/SE ), and Q-Q plots to determine normality.
 
#6
I am not quite sure what you want to achieve. With a total sample size as large as yours,
is is not required that each sample is from a nornally distributed population, or that the
residuals are normally distributed in the population, for the t-test or an F-test to be valid

(central limit theorem). So why all that fuss about normality of... whatever?

With kind regards

Karabiner
You're right about this. The thing is I need to write down the interpretation of the assumptions I've tested prior to my T-test. That's why I want to know whether I should interpret the normality output of the Shapiro Wilk or of the Q-Q plots/ Z-score calculations
 

Karabiner

TS Contributor
#7
I have to test for normality since it's an assumption of the T-test (according to my text book that I need to follow through)...
I do not know what exactely is written there. Often readers misunderstand a
detail. Or a book might spread pure BS. The t-test assumes that each
group is sampled from a normally distributed population, but if n is large
enough (n > 30 or so), then the random sample distribution of the mean
difference is normal, regardless of how the populations are distributed.
The F-test (in the 2-group case it is equivalent to the t-test, t²=F) assumes
that model residuals are from a normally distributed populkation, but
if n > 30 or so, the test is valid regardless of the residuals' distribution.

With kind regards

Karabiner

References:
https://www.annualreviews.org/doi/pdf/10.1146/annurev.publhealth.23.100901.140546
https://journals.physiology.org/doi/full/10.1152/advan.00064.2017 see "Practical considerations"
Stevens, J. (1999). Intermediate Statistics. A Modern Approach. London: Erlbaum, 75-76.
Everitt, B.S. (1996). Making Sense of Statistics in Psychology. Oxford: Oxford University Press, 55.