why don't you use a t-test for independent groups (or the anova if you compare more than to groups)? if you are saying that you cannot use the t-test (because those are ordinal variables, not normally distributed or whatever) then use the non-parametric equivalent of the t-test: the mann-whitney u-test to compare 2 groups, or the kruskal wallis test to compare more than 2 groups.

remember: 1 item cannot be a likert scale. i'm sure you meant rating scale. a likert scale consists of several items which measure the same construct/dimension/factor.

In addition to Manfred's response, it also depends on if you only have one independent variable (i.e. group), or if you have additional covariates you'd like to adjust for. If so, you should consider an ordinal logit model. The second part of your query appears to be about concordance between responses between groups. If so, then are you expecting the responses to be pairwise (among respondents in the two groups) concordant (correlated) or overall?  This needs clarification on your part.

Questions about creating scales from items with Likert scoring are so common on ResearchGAae that I have created a thread to provide resources on this issue:

https://www.researchgate.net/post/How_can_one_help_people_create_simple_scales_from_Likert-scored_items

Thanks for all your input. Let me try to refine what I'm trying to do. 

@Manfred, in this case I'm not considering the scale (the items related to the factor). I really would like to compare differences in individual items (e.g.: 'I enjoy working here', but not the 'organizational climate' factor). Do you see it as valid approach? What are the limitations? 

@Ariel Since I'm just comparing the differences, I was not considering a regression.  My second question deals with a contigency table like this (using the original example), where I should answer 'does these 2 groups have different agreement regarding this statement?'.

______________

                   Team1 | Team 2

Agree                46 |  34

Disagree           11 |  30

@DavidMorgan, this would be a invaluable resource. I think that a lot of research beginners (as I) have this same issue (analysis of likert data), since the majority of data analysis resources examples are for continuous data.

Thanks again. As a beginner researcher, all your feedback are precious to me.

Straightforward Chi-squared test for the question directed at me

Bruno, it just occured to me that you could also use the Chi-squared for the comparison of Likert level scores between the two groups as well. You'd simply put the counts of responses into a 2 X 6 table. You can then use several different tests to assess if there is a difference in distributions between the two groups. Below is a mock-up of a 2 X 6 table where there are two columns (treatment / conrtol) and 6 rows (6 ordinal scores on the Likert scale). Below that is a list of statistical tests and their values....

| col

row | 1 2 | Total

-----------+----------------------+----------

1 | 30 13 | 43

2 | 18 7 | 25

3 | 38 22 | 60

4 | 20 18 | 38

5 | 15 16 | 31

6 | 10 9 | 19

-----------+----------------------+----------

Total | 131 85 | 216

Pearson chi2(5) = 6.5170 Pr = 0.259

likelihood-ratio chi2(5) = 6.5549 Pr = 0.256

Cramér's V = 0.1737

gamma = 0.2315 ASE = 0.093

Kendall's tau-b = 0.1451 ASE = 0.059

Fisher's exact = 0.262

The main problem with reducing the full Likert scale into a simple "Agree-Disagree" dichotomy is that you "lose data" in the sense that you no longer use the ordinal information in the original measure. In addition, the results could very well differ depending on the point that you use to dichotomize the original variable.

Relying on a 2x6 Chi-square doesn't solve this problem because it still treats all six categories as nominal rather than ordinal, which is why it takes 5 degrees of freedom.

A different more practical problem is that you have to repeat each such analysis for every separate item in the overall scale. Add in the inability to use control variables and you are really losing a lot.

By comparison, creating an interval-level scale from the Likert-score items will use the full amount of information across the whole set of items, and it will allow you do a much broader range of analyses.

Hi @David, I agree with you about information loss, but in some cases (e.g. managerial decisions), all they want to know is a reduction like that - so they can prioritize the most critical items. Of course, with some limitations.

I see that a Likert scale is really the best option, but the individual items provide a finer-grained explanation for the problem. Do you think I should perform complementary analysis regarding those things you said (control variables, per-item repetition)?

Thanks @Ariel, at first, I used chi-square for both (2x2 and 2x6 for each item), but I found that this is not the best option. So, I tried to compare the results of both approaches for some items, (considering the 2x6 matrix for chi-square and the whole for Mann-Whitney).

But, for one item, the following occurs: p-value for ChiSquare is 0.030 and p-value for Mann-Whitney is 0.098. Thus, following ChiSquare, I would reject the null hypothesis, but not with Mann-Whitney.

How can I interpret this result?

By the way, in some cases, the difference is very big: p-value of 5,2% in ChiSquare and 50,6% in MW.

But, the opposite also occurs - p-value5% for ChiSquare, which makes more sense to me, since it is more difficult to differ in the order than in general.

You answered your own question. :-) The general Chi-squared makes no assumptions whereas the MW (which also goes by the name of rank-sum) does. If you use Stata I would recommend for you an additional test using a program I wrote called -alignedranks-. (you can download it from within Stata by typing: ssc install alignedranks

Thanks @Ariel! I'm using R (chisq.test() and wilcox.test functions()).

But can you help me figure out why this discrepancy happens (MW p-value so different and above ChiSq p-value)l leading to different hypothesis results?

Can you attach a graph of the distribution of your responses? I can only guess at this point, and I assume that there is a difference at one or more points on the continuum that are affecting the results differently using the rank-sum vs a straight-up Chi-squared. 

Are items independent?

The best option it to use polytomous IRT models and convert your frequencies to latentt variables at an interval scale. Afterwards, you can compare the latent variables.