# Checking normality of a Likert scaling question

I have likert scale responses (1-5 rating) of 20 respondents for a set of 25 questions. So that means that each respondent has answered 25 questions (with 1-5 rating). If I need to check the normality distribution of all the 25 questions together for a particular respondent, how do i do that?

If by "together" you mean the sum of all 25 questions, then the sum still can't be exactly normal, because it is still discrete. But it could be close enough to normal that everything works out OK.

Rather than a test of normality (which are all highly dependent on sample size) I would look at the quantile normal plot, available in R with the qqnorm function.

• Hello...THanks for your help. Can we generate this plot on SPSS? Feb 24 '12 at 12:21
• For SPSS through the GUI interface if you go to Analyze -> Descriptive Statistics -> QQ-plots one can get a QQ plot of the observed against various theoretical distributions. Check out the EXAMINE command in help for tests of normality like Shapiro-Wilks and KS. Feb 24 '12 at 13:54
• The QQ plot says that the data indeed has normal distribution but when we go for Shapiro Wilks or Konglomerov test..the P value is >.05. What could be the reason of contradiction?\ Feb 25 '12 at 6:51
• That's agreement, not contradiction. If KS < .05 it means the hypothesis of normality is rejected. Over .05 means the hypothesis is not rejected Feb 25 '12 at 13:20
Analyze // Nonparametric // Legacy dialogs // 1- Sample KS


Well, I felt this one-line answer would be enough. Ok, so this option conducts a KS (Kolmogorov-Smirnov) Test of Normality. Essentially doing the same as the Shapiro-Wilk test previously mentioned. If you need the pros and cons of both tests, wikipedia might give a good answer.

In many cases Likert scales are skewed, since people tend to do yea-saying if the data is from the US. So you will probably standardize each Likert variable by subtracting mean and dividing by standard deviation. You can then conduct the KS test on the transformed variables.

Skewness you can already detect by looking at the descriptives.

• What software are you using?
– whuber
Feb 24 '12 at 22:43
• SPSS, since this is what Shantanu asked for ^^ Feb 25 '12 at 2:39
• Thanks. The reference to SPSS was buried in a comment to another reply. I'll add the spss tag to this thread.
– whuber
Feb 25 '12 at 17:53

If the outcome takes values in {1,2,3,4,5}, then this is clearly not normal because it is a discrete variable. If despite this you still want to use a normal approximation, in R you could use the Shapiro-Wilk normality test

x = rnorm(100,10,3)
shapiro.test(x)


and perhaps reject at a 0.05 level.

I hope this helps.

Best wishes.

• hey Thanks for your suggestion. But there could be a way to check this in SPSS as well rite? Feb 24 '12 at 12:21
• The problem with testing normality is we know the answer to the question (it's your first sentence); the more useful question of whether it's close enough to normal is simply not addressed by a hypothesis test. Jun 8 '15 at 7:26