These are all ordinal variables -> Spearman

Are you wanting a single value for the association among the three, or just the three pairwise ones?

You have rank-ordered variables so the Spearman will be correct. For Duration and Body Posture you may count in strata determined by pain levels (adjusted). On the other hand you may aggregate across the strata of pain (one value).

Jochen Wilhelm Hanna Mielniczuk Thank you very much. May I ask when should I use chi-square?

Daniel Wright I want to find correlation in the three pairwise ones, but I just want to know 2 correlation:

- between duration and pain

- between posture and pain

I would use a Chi-square test with such data. Why would you use correlation?

Raid Amin I'm sorry but I don't know much about data analysis. May I ask why did you recommend chi-square test? Thank you

Chi² can be used with categorical variables with no ordinal information. It measures how much the frequency distributions of the categories differ. Spearman's rank correlation measures the monotone association of two at least ordinal variables (actually, the variables should be continuous, so that there are no ties in the ranks - but it's still not too wrong in the presence of ties). The monotone association gives you an information about the direction of the correlation, which is available for ordinal or continuous variables and which is not analyzed by comparing frequency distributions (Chi²).

Jochen Wilhelm Thank you very much.

I noticed that Nata Vitha included SPPS as one of the relevant topics. If that was a typo for SPSS, then Nata will see in the output from the CROSSTABS command a Chi-square test for "linear-by-linear" association. It does assume/require that both variables are ordinal. What is less well known is that the actual values used for the categories matter. Why? Because this particular Chi-square is equal to (N-1)r2. See these notes by the late David Howell for more info:

• https://www.uvm.edu/~statdhtx/StatPages/OrdinalChisq/OrdinalChiSq.html

See also these pages from Sal Mangiafico's R Companion site for more info and additional options.

• https://rcompanion.org/handbook/H_09.html
• https://rcompanion.org/handbook/H_11.html

HTH.

There exist Chi-square tests for ordered categories. Correlation analysis works best with continuous data.

Because you have ordinal variables that will have many ties, I think I would recommend using Kendall tau-b or tau-c correlation.

tau-c may also be called Stuart's tau-c.

Because the number of categories in e.g. Duration and Pain are not the same (3 vs. 4), tau-c is usually recommended.

Spearman correlation probably won't be much different, so if this is easier, it's probably fine to use that also. Likewise, tau-b and tau-c shouldn't be terribly different in this case.

I don't know how easy any of Spearman's rho, Kendalls tau-b, or Stuart's tau-c is to obtain in SPSS.

The following is example code in R. It can be run at https://rdrr.io/snippets/ without installing software. It produces the Kendalls tau-b, Stuart's tau-c, and Spearman's rho. And the 95% confidence intervals for each. And the p-value for hypothesis tests for Kendalls tau-b and Spearman's rho.

Data =read.table(header=TRUE, stringsAsFactors=TRUE, text="

Posture Pain

Safe Low

Safe Low

Safe Low

Safe Low

Safe Mid

Safe High

SmallRisk Low

SmallRisk Low

SmallRisk Mid

SmallRisk Mid

SmallRisk Mid

SmallRisk High

MiddleRisk Low

MiddleRisk Mid

MiddleRisk Mid

MiddleRisk Mid

MiddleRisk High

MiddleRisk VeryHigh

HighRisk Low

HighRisk Mid

HighRisk High

HighRisk High

HighRisk VeryHigh

HighRisk VeryHigh

")

Data$Posture = factor(Data$Posture, levels=c('Safe', 'SmallRisk', 'MiddleRisk', 'HighRisk'))

Data$Pain = factor(Data$Pain, levels=c('Low', 'Mid', 'High', 'VeryHigh'))

Table = xtabs(~ Posture + Pain, data=Data)

Table

library(DescTools)

KendallTauB(Table, conf.level=0.95)

StuartTauC(Table, conf.level=0.95)

cor.test(as.numeric(Data$Posture), as.numeric(Data$Pain), method="kendall")

SpearmanRho(Table, conf.level=0.95)

cor.test(as.numeric(Data$Posture), as.numeric(Data$Pain), method="spearman")

Raid Amin , which chi-square test are you thinking of ?

Here is SPSS code for Sal's example. The NONPAR CORR command computes both tau-B and Spearman's rho. For CROSSTABS, the CHISQ option is needed to see the so-called test of linear-by-linear association. The other key words on the STATISTICS subcommand are (likely) self-explanatory.

NEW FILE.

DATASET CLOSE ALL.

DATA LIST LIST / Posture Pain (2F1).

BEGIN DATA

0 0

0 0

0 0

0 0

0 1

0 2

1 0

1 0

1 1

1 1

1 1

1 2

2 0

2 1

2 1

2 1

2 2

2 3

3 0

3 1

3 2

3 2

3 3

3 3

END DATA.

VALUE LABELS

Posture 0 "Safe" 1 "Low risk" 2 "Medium risk" 3 "High risk" /

Pain 0 "Low" 1 "Mid" 2 "High" 3 "Very high".

CROSSTABS

/TABLES=Posture BY Pain

/FORMAT=AVALUE TABLES

/STATISTICS=CHISQ GAMMA D BTAU CTAU

/CELLS=COUNT

/COUNT ROUND CELL.

NONPAR CORR

/VARIABLES=Posture Pain

/PRINT=BOTH TWOTAIL SIG LOWER

/MISSING=PAIRWISE.

Thanks, Bruce Weaver .

It turns out I have access to SPSS through the university (they keep changing this), so I was able to compare results.

At the time of writing, PSPP, will give you the values for tau-c and tau-b without the p-values, but it does give you the T values and the standard error, so you could easily calculate the p-values and confidence intervals. (Not sure why the last step isn't implemented).

For R users, there's an answer here that will give you the p-value for tau-c. It's pretty simple. (But, as written, you have to take the absolute value of tau and its confidence interval limits if tau is negative).

https://stackoverflow.com/questions/51918648/p-value-for-kendalls-tau-c-in-r

P.S.

For PSPP, I had to change the first couple of lines to the following. I really don't understand SPSS language, but this worked.

DATA LIST / Posture 1 (A) Pain 3 (A).

BEGIN DATA.

And then you can change the labels command.

VALUE LABELS

Posture "0" "Safe" "1" "Low risk" "2" "Medium risk" "3" "High risk" /

Pain "0" "Low" "1" "Mid" "2" "High" "3" "Very high".

There are many ways to go about it. First, you could find the average or sum of all items per variable to obtain continuous level data. With the data, Pearson correlation could be performed on the aggregated data.

Salvatore: I recall from many years ago having seen a report on a Chi-square test for ordered categories. I do not remember which exact name the test has.

Hi, Raid Amin . Probably this is what Agresti calls the "linear-by-linear" test. This called "Linear-by-Linear Association" by SPSS and in some packages in R. Confusingly, I think SAS and some R packages use terms like "Cochran-Mantel-Haenszel" or "Mantel-Haenszel" for this test.

If the data are paired, the McNemar test can be used. The CMH-test is a generalization where the pairing is done via stratification (within each stratum an odds ratio is calculated from a 2x2 table, and these are weighted and combined). I think McNemar's test is the one that applies to Nata's problem.

Raid Amin, in my Nov 13 post, I gave this link to the late David Howell's notes regarding the ordinal Chi-square you are talking about:

• https://www.uvm.edu/~statdhtx/StatPages/OrdinalChisq/OrdinalChiSq.html

Notice that this statistic = (N-1)r2. Therefore, the numeric values used for the different categories matter. E.g., suppose you have a 5-point item with labels Strongly disagree, Disagree, Neutral, Agree, Strongly agree. People often use numeric values 1-5 to code these categories. But suppose you believe that these numbers better reflect the "distances" between the categories: SD=1, D=2, N=3.5, A=5, SA=6. Pearson's Chi-square would be the same for both sets of numeric codes, but the ordinal Chi-square would not. Make up an example and try it! HTH.

Thank you very much for the information. Really appreciate the feedback and it really helps Bruce Weaver Raid Amin Sal Mangiafico Valentine Joseph Owan Jochen Wilhelm

Bruce Weaver I apologize, I didn't realize until I read your answer. Yes, what I mean is SPSS. Once again, thank you very much for your helps!

Hi Sal Mangiafico. I found an SPSS syntax file from 2016 in which I did a demonstration of the effect different coding schemes on the linear-by-linear Chi2. I cleaned up the code a bit to make it more presentable--see attached. For those who do not have SPSS, the output can be viewed in the attached PDF.

And before anyone busts my chops, yes, I know it is a bad idea to use age groups when one has the actual ages. Age group was just a convenient variable to use to demonstrate the effect of coding on the ordinal (or linear-by-linear) Chi2.

In my opinion correlation is not a suitable word when used for categorical variables. It suits and was historically introduced for continuous variables.

When talking about categorical variables, it would be better use the word association. I believe also that what the poster has in mind is question of association rather than correlation.

Hi Mehmet Sinan Iyisoy. I do agree with you that it is generally better practice to only say correlation when one is talking about Pearson r or one of the other common correlation coefficients. This can help to prevent confusion. However, there is a common expression most of us know that uses the word correlation in a much more general sense (IMO):

• Correlation does not imply causation.

IMO, this statement translates to:

• Statistical covariation on its own does not logically imply causation.

In other words, it is not referring specifically and only to specific correlation coefficients that measure linear association. YMMV. ;-)

Probably for categorical data, "association" is a more typical term than "correlation". Though Kendall's tau- b or c is often used for categorical data and is often called "correlation". In contrast, the linear-by-linear test is often called "association". For a 2 x 2 table, phi is sometimes called a "correlation" coefficient, but is usually called a measure of "association". Upshot: "association" is probably a safer term for categorical data, and won't confuse the reader into thinking that Pearson correlation was used.

This is why I asked the question why do we need correlation here. We need some type of the Chi-square test. Right?

Raid Amin Yes. Your very post was a reason for my post above.

Hi,

There is a PhD thesis with the name of “Essays on Market Reaction and Cryptocurrency” of Toan Luu Duc Huynh (January 25, 2022). This thesis may be usefull for your needs.

S. Ç.

Internet link: https://opus4.kobv.de/opus4-whu/frontdoor/index/index/docId/900

Hi, Salih Çelebioğlu . Can you let us know how that thesis applies to this question ? I looked at it briefly. I didn't find anything related to ordinal variables. There is the use of wavelet correlation, but I don't see how this would apply to this situation.

Dear Mangiafico,

I just saw the title of the question on ResearchGate today. Sorry, I don't know the details of your discussion. I have shown this source (thesis) in case it may be of use to you. I can't be of any further help as I'm busy in these days. But sometimes I also face such problems. As an example, I can give the question of how I can calculate a correlation between the rankings of the champion teams in the super football leagues of the countries over time. For now, I will not be able to continue this discussion because of my excuses. I hope you don't mind me.

Hi, Salih Çelebioğlu . No worries. ~ Sal

To find the correlation between three variables, you can use multiple correlation analysis. Multiple correlation analysis is used to determine the relationship between a dependent variable and two or more independent variables1.

Chuck A Arize , in a correlation analysis there is no "dependent" and "independent" variable. The question remains: what is the correlation between three (or more) variables? And shouldn't your "multiple correlation analysis" (whatever this is meant to be!) take care of the kind of values (nominal, ordered, metric) and functional relationships between the variables (monotone, linear, periodic, etc.)?