# How to use Spearman’s correlation with two Likert scales?

Let’s say that my survey has 2 sections that I want to find the correlation between, they each have 5 questions with each question being a 5-point Likert scale. What do I do after? What are the steps to doing this in Excel? I can't seem to find any good guides on how to do correlation with Likert scale variables. I was also told that I should test for normality and check for linearity with a scatter plot before I begin with correlation, but I can't find a guide about it for Likert scale variables either.

• Just calculating the statistic doesn't require any assumption of normality. But if you are trying to perform a null hypothesis test or estimate a posterior distribution for the statistic then you will want to consider what joint distribution you want to assume over the original variables. Commented Mar 16, 2023 at 1:29
• Linearity and normality have nothing to do with Spearman correlation. Could you have been getting guidance about Pearson correlation?
– Dave
Commented Mar 16, 2023 at 1:41
• I see you have added nonparametric. Note that rank-based statistics are usually nonparametric when the original variables are continuous because the resulting distribution of ranks will be almost-surely uniform. A Likert variable is not continuous, so in what sense would you like to consider a nonparametric analysis involving this rank-based statistic? Commented Mar 16, 2023 at 1:43
• Spearman's Rho(X, Y) = Pearson's Correlation(rank(X), rank(Y)). The relevant functions exist in Excel. Commented Mar 16, 2023 at 7:40
• Two levels of small correction: Likert was Rensis Likert, and I have followed the admirable convention that proper names should be capitalised. Flagged but not corrected here: many distinguish, IIUC, between individual Likert items (such as grades 1 to 5) and Likert scales, which are composite variables. Commented Aug 22, 2023 at 9:23

Spearman's rho ranks values in your data based on how they compare between each-other. In order to rank them uniquely, we need unique values. A Likert scale only has 5 or 7 possible values. Therefore, you are not supposed to correlate two Likert scales with Spearman's rho.

• This is incorrect, because it's possible and even useful to rank datasets that have tied values. This was discussed in comments to a previous answer that unfortunately had been deleted. I have undeleted that answer to give you an opportunity to read that discussion.
– whuber
Commented Aug 26, 2023 at 14:58
• Aren't you effectively editing the data when you rank the tied values? And also, if my anwear is incorrect, then why has no one given a proper answear? I don't understand. None of the examples I've seen online of spearman's rho have used ordinal data. Commented Aug 26, 2023 at 15:49
• @brostats Every function on data in some sense 'edits'. On the modelling side this is just a function of random variables which induces a change in variables for the distribution. Many people assume that the Spearman correlation coefficient is synonymous with a particular null hypothesis significance testing procedure taught in many frequentist courses on statistics. What the math allows is much broader than what most people use as examples. Commented Aug 26, 2023 at 17:06
• The entire point of Spearman's correlation coefficient is that it is applicable to ordinal data. When one represents data by their ranks, all information about relative sizes and differences between the original data are lost, and the only information this correlation provides is about the ordinal characteristics of the data. There can be many reasons for the absence of a good answer. One is clarified in comments to the post that ask for context and clarification that were never provided, making unclear what a good answer might even be.
– whuber
Commented Aug 26, 2023 at 18:07
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– whuber
Commented Aug 27, 2023 at 17:11

The reason why you have not found a good guide about Spearman's rho for Likert scales is because you do not actually calculate Spearman's rho with ordinal data.

Spearman's rho takes in quantitative (metric) data, as its input, and ranks it based on N (number of responses). For example, if you have 20 respondents, that means you have 20 ranks. If you have 1200 responses you have 1200 ranks.

A Likert scale by its definition is ordinal (qualitative) data, meaning A < B < C, but A + A does not equal B. Or another example: A + B does not equal C.

Furthermore, if we ignore the rule for ordinal data for Likert scale (which many people do, apparently), and treat it as a quantitative variable you still end up with 5 unique values for your variable.

Spearman expects each value to be unique so we can rank them uniquely: 23 is rank 1; 25 is rank 2; 34 is rank 3; 55 is rank 4; 64 is rank 5; 68 is rank 6; 75 is rank 7; 99 is rank 8; 110 is rank 9; 255 is rank 10;

How do you assign unique rank to data that looks like this:? 1; 1; 1; 2; 2; 3; 3; 4; 5; 5;

• I've corrected several typos here but left the main thread as you wrote it. But the main idea that "Spearman expects each value to be unique" is quite incorrect in respect of any explanation or implementation I've ever encountered. It's standard that tied values are allocated the mean of the ranks that would have otherwise been given, So, 1 1 1 2 2 3 3 4 5 5 would be ranked 2 2 2 4.5 4.5 6.5 6.5 8 9.5 9.5. What I sense to be widely agreed is that the more ties you have, the more problematic the Spearman correlation estimate would be, even descriptively. Commented Aug 22, 2023 at 9:16
• My own stance is that in principle Spearman correlation could be calculated for Likert items, but the researcher should be circumspect about doing that because of the probability of many, many ties. There is a literature on ordinal measures of association, FWIW. Commented Aug 22, 2023 at 9:18
• Hence I have to regard this answer as more wrong than right. Commented Aug 22, 2023 at 9:20
• You're implying that lack of uniqueness is fatal for applications of Spearman correlation. I am stating that lack of uniqueness is awkward for such applications, and the awkwardness is a matter of degree. There is a difference between fatal and awkward, not a contradiction. Suppose there are many ties but all grades on two Likert items agree. Then the Spearman correlation is identically 1. I am happy that makes sense. Many applications of Spearman to Likert variables will not be very helpful, but they will not meaningless or invalid. Commented Aug 22, 2023 at 14:06
• StatsAnon, you might appreciate reading Lord's (in)famous "football numbers" paper. (Google it: you can find copies online.) It ought to help you change your mind. BTW, there are various ways to assign ranks to tied data. First, the ranks needn't be unique: the only requirement on the data for computing any correlation coefficient is that they must not be constant. Second, you can resolve ties randomly if you really must have unique ranks. There is an issue concerning what that coefficient might mean for ordinal data: that's where Lord's paper provides insight.
– whuber
Commented Aug 22, 2023 at 15:47