Assume that we have a 5-likert scale questionnaire, we want to measure cheating behavior (plagiarism, use of crib notes, etc.) among students. We have 0-disagree and 1-agree. So I’m assuming

0.25-partially disagree, 0.5-sometimes, 0.75-partially agree.

I’ve read in the book that, formally, this type of variable is a discrete ordinal variable;

however, under certain assumptions, it is possible to treat this type of variable ‘as if ’ it was continuously measured at interval level. This is possible when the number of categories is large enough (at least five), the variable is normally distributed, and the sample size is adequate. (1)

I don’t understand how this variable is said to be discrete ordinal since the numbers have order (0,0.25,0.5,0.75,1) and there are equal intervals between adjacent categories (1-0.75=0.25, 0.75-0.5=0.25,…). Therefore we have interval type.

What am I missing here?

Actually, me thinks it wouldn’t matter what would we be measuring as long as we have order and equal intervals between adjacent categories, it’s always a interval type.

2nd question.

The variable will be always normally distributed because we have ’tests scores' on a test to a large class (assuming we have more than 30 data). But then this doesn’t makes sense with (1). Why?

Please help me to understand.

Thanks in advance


2 Answers 2


A Likert scale attempts to measure opinions. If someone has thought about an issue at all, then it's likely they will be able to say whether they're in strong agreement or slight disagreement. Maybe less likely to say whether their agreement is strong or weak. So it is commonly agreed that Likert scales can be considered as ordinal.

However, they are not numerical--not even at the 'interval' level. Is it clear that the difference between strong and weak agreement is quantitatively the same as the difference between neutrality and weak agreement, or as the difference between weak and strong disagreement? Consequently, the treatment of Likert scores as numerical is controversial. Some reasonable arguments for treating them as numerical have been advanced.

But I have to say I find your quote to be especially unpersuasive. Personally, I find I am not as able to do an honest job of distinguishing between categories on a seven point scale as on a five point scale. And I'd often prefer just to say Disagree, Agree, or Undecided. So I think it is delusionary to suppose more detailed scales fetch more useful data. What normality has to do with the issue is truly a puzzle. Similarly, it is unclear how having more ordinal responses tends to make them more numerical.

Statistical methods for analyzing ordinal data can be more challenging to create, understand, and compute than methods for numerical data. So it may be tempting to suppose that pretending assumptions for t tests are reasonable is a valid excuse to use t tests---instead of more appropriate methods of testing developed for ordinal data.

Finally, the "rule" that $n > 30$ assures normality is only very approximate--at best. Specifically, it is unquestionably true of data are uniform $(n = 15$ is enough), probably OK if data are pretty much restricted to an interval such as $\bar X\pm 3S,$ and completely unreasonable if data are exponential (typically, with many 'outliers' on the high side and none on the low side).

Suppose you want to know if there is a difference of opinion about the the use of crib notes on an exam between students in the College of Engineering and the College of Business.

If 53 out of 200 engineering students think crib notes are OK, and 85 out of 250 business students think so, then you can test whether $\hat p_e = 53/200 = 0.265$ is significantly different from $\hat p_b = 85/250 = 0.360,$ using prop.test in R, as follows:

prop.test(c(53,85), c(200,250), cor=F)

        2-sample test for equality of proportions 
        without continuity correction

data:  c(53, 85) out of c(200, 250)
X-squared = 2.9395, df = 1, p-value = 0.08644
alternative hypothesis: two.sided
95 percent confidence interval:
 -0.159789116  0.009789116
sample estimates:
prop 1 prop 2 
 0.265  0.340 

The data do not show a significant difference at the 5% level; P-value $= 0.086 > 0.05.$ (If there is a true difference in proportions of this size, you would need greater numbers of subjects in order to establish that.)

Because it uses ranks, which make sense for ordinal data, a two-sample Wilcoxon rank sum test might be used to see if Likert scores from engineering students differ significantly from those of business students.

For small sample sizes, the Wilcoxon test can have trouble with ties, and there will be many ties for Likert data. But recent implementations of the Wilcoxon test, such as wilcox.test in R, handle ties in large datasets agreeably.

Here are some fake Likert scores along with results for them from a Wilcoxon test. These data show no significant difference.

E = sample(1:5, 200, rep=T, p=c(7,8,10,3,2))
B = sample(1:5, 250, rep=T, p=c(6,7,10,4,3))
[1] 46 62 60 16 16  # Likert scores 1,2,3,4,5, resp
[1] 62 51 85 35 17

        Wilcoxon rank sum test 
      with continuity correction

data:  E and B
W = 23493, p-value = 0.2557
alternative hypothesis: 
   true location shift is not equal to 0
  • $\begingroup$ Thank you for your answer. I've never heard of wilcox.test. I am only familiar with ANOVA. I still don't get why the book stated that about discrete ordinal to pass to interval. From what I can see, as an outcome from the data I'd have some of these [1] 46 62 60 16 16 [1] 62 51 85 35 17 which would be the input on ANOVA. Am I correct? tbh I feel like something is missing $\endgroup$ May 11, 2020 at 23:45

Not sure what the questionnaire is... is it many Likert scale questions? Do the agree/disagree scores signify a calculation or are they the actual responses? How does this relate to the ANOVA tag?

1) You're not missing anything – there is no single right answer. It turns out you can encode the variable as ordinal or numeric depending on your use case. I'm not exactly sure what you want to do with "measuring cheating behavior." Make a visualization? Manually investigate the most similar instances? It shouldn't matter for those use cases.

If you want to build supervised machine learning model predicting if a student is cheating, I'd first try the feature as numeric, but then also try the feature as categorical.

2) Sorry if my answers are off topic, but I think the normality/makeup of the variable depends on the use case. If you were building a linear supervised machine learning model like logistic regression, you would want to make sure numeric "interval" features are standardized first. If it was categorical, this should be less important.

  • $\begingroup$ Thanks for your answer. "Do the agree/disagree scores signify a calculation or are they the actual responses?" wdym? It's related to the ANOVA tag, because I wanted to understand how from a likert scale you can pass to analyze the data (comparison among groups) on ANOVA. I've read that one needs normal data as a requirement to use ANOVA. $\endgroup$ May 11, 2020 at 23:34

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