**Cohen’s kappa** is a measure of the agreement between two raters who determine which category a finite number of subjects belong to whereby agreement due to chance is factored out. The two raters either a agree in their rating (i.e. the category that a subject is assigned to) or they disagree; there are no degrees of disagreement (i.e. no weightings).

We illustrate the technique via the following example.

**Example 1**: Two psychologists (judges) evaluate 50 patients as to whether they are psychotic, borderline or neither. The results are summarized in Figure 1.

**Figure 1 – Data for Example 1**

We use Cohen’s kappa to measure the reliability of the diagnosis by measuring the agreement between the two judges, subtracting out agreement due to chance, as shown in Figure 2.

**Figure 2 – Calculation of Cohen’s kappa**

The diagnoses in agreement are located on the main diagonal of the table in Figure 1. Thus the percentage of agreement is 34/50 = 68%. But this figure includes agreement that is due to chance. E.g. Psychoses represents 16/50 = 32% of Judge 1’s diagnoses and 15/50 = 30% of Judge 2’s diagnoses. Thus 32% ∙ 30% = 9.6% of the agreement about this diagnosis is due to chance, i.e. 9.6% ∙ 50 = 4.8 of the cases. In a similar way, we see that 11.04 of the Borderline agreements and 2.42 of the Neither agreements are due to chance, which means that a total of 18.26 of the diagnoses are due to chance. Subtracting out the agreement due to chance, we get that there is agreement 49.6% of the time, where

Some key formulas in Figure 2 are shown in Figure 3.

**Figure 3 – ****Key formulas for worksheet in Figure 2**

**Definition 1**: If *p _{a}* = the proportion of observations in agreement and

*p*= the proportion in agreement due to chance, then

_{ε}**Cohen’s kappa**is

Alternativelywhere *n* = number of subjects, *n _{a}* = number of agreements and

*n*= number of agreements due to chance.

_{ε}**Observation**: Another way to calculate Cohen’s kappa is illustrated in Figure 4, which recalculates kappa for Example 1.

**Figure 4 – ****Calculation of Cohen’s kappa**

**Property 1**: 1 ≥ *p _{a}* ≥

*κ*

Proof:

since 1 ≥ *p _{a}* ≥ 0 and

*1 ≥*0. Thus 1 ≥

*p*≥_{ε}*p*≥

_{a}*κ.*

**Observation**: Note that

Thus, *κ* can take any negative value, although we are generally interested only in values of kappa between 0 and 1. Cohen’s kappa of 1 indicates perfect agreement between the raters and 0 indicates that any agreement is totally due to chance.

There isn’t clear-cut agreement on what constitutes good or poor levels of agreement based on Cohen’s kappa, although a common, although not always so useful, set of criteria is: less than 0% no agreement, 0-20% poor, 20-40% fair, 40-60% moderate, 60-80% good, 80% or higher very good.

A key assumption is that the judges act independently, an assumption which isn’t easy to satisfy completely in the real world.

**Observation**: Provided and *np _{a}* and

*n*(1

*–p*) are large enough (usually > 5),

_{a}*κ*is normally distributed with an estimated standard error calculated as follows.

Let *n _{ij}* = the number of subjects for which rater A chooses category

*i*and rater B chooses category

*j*and

*p*=

_{ij}*n*/

_{ij}*n*. Let

*n*= the number of subjects for which rater A chooses category

_{i}*i*and

*m*= the number of subjects for which rater B chooses category

_{j}*j.*

The standard error is given by the formula

**Example 2**: Calculate the standard error for Cohen’s kappa of Example 1, and use this value to create a 95% confidence interval for kappa.

The calculation of the standard error is shown in Figure 5.

**Figure 5 – ****Calculation of standard error and confidence interval**

We see that the standard error of kappa is .10625 (cell M9), and so the 95% confidence interval for kappa is (.28767, .70414), as shown in cells O15 and O16.

**Observation**: In Example 1, ratings were made by people. The raters could also be two different measurement instruments, as in the next example.

**Example 3**: A group of 50 college students are given a self-administered questionnaire and asked how often they have used recreational drugs in the past year: Often (more than 5 times), Seldom (1 to 4 times) and Never (0 times). On another occasion the same group of students was asked the same question in an interview. The following table shows their responses. Determine how closely their answers agree.

**Figure 6 – Data for Example 3**

Since the figures are the same as in Example 1, once again kappa is .496.

**Observation**: Cohen’s kappa takes into account disagreement between the two raters, but not the degree of disagreement. This is especially relevant when the ratings are ordered (as they are in Example 2. A weighted version of Cohen’s kappa can be used to take the degree of disagreement into account. See Weighted Cohen’s Kappa for more details.

Another modified version of Cohen’s kappa, called Fleiss’ kappa, can be used where there are more than two raters. See Fleiss’ Kappa for more details.

**Real Statistics Function**: The Real Statistics Resource Pack contains the following function:

**WKAPPA**(R1) = where R1 contains the observed data (formatted as in range B5:D7 of Figure 2).

Thus for Example 1, WKAPPA(B5:D7) = .496.

Actually WKAPPA is an array function which also returns the standard error and confidence interval. This version of the function is described in Weighted Kappa. The full output from WKAPPA(B5:D7) is shown range AB8:AB11 of Figure 7.

**Real Statistics Data Analysis Tool**: The **Reliability** data analysis tool supplied by the Real Statistics Resource Pack can also be used to calculate Cohen’s kappa. To calculate Cohen’s kappa for Example 1 press **Ctrl-m** and choose the **Reliability** option from the menu that appears. Fill in the dialog box that appears (see Figure 7 of Cronbach’s Alpha) by inserting B4:D7 in the **Input Range** and choosing the **Cohen’s kappa **option.

**Figure 7 – ****Cohen’s Kappa data analysis**

If you change the value of alpha in cell AB6, the values for the confidence interval (AB10:AB11) will change automatically.

Hello Charles,

Thank you for this great guide.

I am having problems with calculating kappa scores between 4 reviewers (let’s call them A, B, C, D). First separated into pairs and did 25% of the screening work, then we switched pairs again, to do another 25% of the work.

So:

A+B = 25% of work

A+C = 25% of work

B +D = 25% of work

C+D = 25% of work

I have no idea how to calculate this score – could you please advise me what is the best way to calculate the combined kappa score?

Stella,

I don’t know a way to calculate kappa in this case.

Charles

Dear Charles

I am really impressed by your website, and would really appreciate your help please.

I have data for classification of medical cases, and i want to calculate the Inter-class & Intra-class agreement for it.

Would you able to help me with that please?

Many thanks

See Intraclass Correlation

Charles

Hi Charles,

I have the following results for my data set:

N=132, Comparing 2 screening tests with 3 risk category groups no risk, low-moderate risk and high risk.

Observed agreement 47/132 35.61%

Expected agreement by chance 46 / 132 35.06%

Kappa SE 95% CI

Kappa value (adjusted for chance) 0.0085 0.051 -0.092 to 0.109

Kappa value with linear weighting 0.082 0.0492 0.033 to 0.179

Maximum agreement of weighted kappa 60.6%

As I proposed in my hypothesis these 2 screening test do not agree!

Is there any use calculating a P value for the K and wK? if not, why not?

Jo,

I can’t see any real advantage of calculating the p-value. It will only tell you whether kappa is significantly different from zero. This is true if zero doesn’t lie in the confidence interval and the confidence interval gives you more information.

Charles

Hello,

Thanks very much for making this package freely available – it is very useful!

I’d like to include the the results in a report template and I was wondering if there is any way of making the result of the kappa calculation done using Ctrl-M can be made to appear on the same worksheet instead of it appearing automatically on a new sheet? Or for it to appear on a pre-designated blank sheet?

Thanks in advance for your advice.

Please don’t bother replying! Saw the output range field!! *smacking forehead*

Good day!

Can I use this formula for interval data? thanks

Shiela,

Yes, you can use Cohen’s kappa with interval data, but it will ignore the interval aspect and treat the data as nominal. If you want to factor the interval aspect into the measurement, then you can use the weighted version of Cohen’s kappa, assigning weights to capture the interval aspect. This is described at Weighted Kappa.

Charles

I would like to estimate sample size for my study. Here is the necessary details for your review/action.

One enumerator will conduct two different interviews (one at hospital and one interview at client’s home) with a same client. Which means, one client will be interviewed twice at different locations by same enumerator. Therefore, I would like to assess inter-rated reliability testing between the responses. Whether there is a change in response if clients give an interview at hospital and again interviewed at home?

This is my hypothesis and I wanted to collect xx number of sample in order to test the responses. The study is on “Assessment of Quality of Care related to Family Planning provided by doctor”.

Can you please let me know the desire sample size or how to estimate sample size for IRR?

One approach is to provide an estimate of what you think the measurement of Cohen’s Kappa will be (e.g. based on previous studies or the value you need to achieve). You then calculate the the confidence interval (as discussed on the referenced webpage). This calculation will depend on a sample size value n. The narrower the confidence interval, the higher the value of n needs to be. So you first need to determine how narrow you want this confidence interval to be (i.e. how accurate you need Cohen’s Kappa to be) and once you have done this, progressively guess at values of n that will achieve this sized confidence interval. You can actually use Excel’s Goal Seek capability to avoid the guessing part. I explain all of this in more detail for Cronbach’s alpha; the approach is similar.

Charles

I ran Cohen’s Kappa on a dataset to examine IRR and I am now questioning my choice of statistic. What I have are 100 different raters of two different types (Rater1 and Rater2). Each pair of raters assessed 6-8 people on 25 behaviors using a 4 point rating scale. So, I do not have multiple raters examining the same people (R1 and R2 assessing all people) which seems to be an assumption for Kappa. I have 50 R1s and 50 R2s (paired). The pair independently rated the same 6-8 people on those 25 behaviors using the same 4-point rating scale. I easily calculated percent agreement, but I need a more robust analysis. I would appreciate your insight and suggestions.

Teddy,

If each person is only assessed by one rater, then clearly Cohen’s Kappa can’t be used. In fact, all the measurements that I am familiar with require that each subject be rater by more than one rater.

Cohen’s kappa can be used when all the subjects are rated by both raters. If you have more than two raters then Fleiss’s kappa or the Intraclass correlation coefficient could be used. You might even find Kendall’s W to be useful. I suggest that you investigate each of these on the Real Statistics website.

Charles

Teddy,

If each person is only assessed by one rater, then clearly Cohen’s Kappa can’t be used. In fact, all the measurements that I am familiar with require that each subject be rater by more than one rater.

Cohen’s kappa can be used when all the subjects are rated by both raters. If you have more than one rater then Fleiss’s kappa or the Intraclass correlation coefficient could be used. You might even find Kendall’s W to be useful. I suggest that you investigate each of these on the Real Statistics website.

Charles

Hi Charles,

This is a really nice website and super helpful. I have a question. How can I run IRR when it is a mixture of ordinal and binary data? Some of my variables are coded as yes/no while others are pretty much like likert scale.

Thanks！

You need to provide additional details before I am able to respond. Also, why do you want to perform IRR when the rating scales are so different?

Charles

Thank you for this rich and fruitful demonstration, but I’d like to ask whether kappa statistic is useful in evaluating a diagnostic tool by agreement measurement with a golden standard tool?

I used to do so for my clients in addition to calculating sensitivity, specificity and predictive values. Thank you a lot!

You can use Cohen’s Kappa to measure agreement with a gold standard. Depending on the specific situation, Bland Altman can also be used. See the following webpage

Bland-Altman

The following is an interesting paper that describes many different approaches to reliability testing

https://orbi.ulg.ac.be/bitstream/2268/39575/1/vanbelle-thesis-5-5-2009.pdf

Charles

Hi,

how to compare two Cohen’s Kappa statistics based on the same sample of subjects but different raters? I wonder if there is a way to compare these two kappa statistics — that is, whether they are statistically different.

Thanks

Karina,

I found this paper online. Perhaps it will be useful.

http://hera.ugr.es/doi/16521778.pdf

Charles

Hi

Thank you for this page

I am unsure of how to calculate the sample size or if it required (i have read a lot of papers recently where no sample size is mentioned).

I am interested in testing a new scoring method for disease severity where severity is scored from say 1 to4 or 1 to3 from normal to most severe.

I will have two testers who are less experienced and two experts in the field doing this.

Unfortunately I don’t understand how to determine what sample size I might need to this. I intend to calculate kappa between the two ‘novices’ and the then the two ‘experts’ and then also test intra reader reliability for each.

All I have come across so far is very complicated mathematics !

Any help would be great

Hi,

The calculations do require some complicated mathematics. I will try to add software to calculate the sample size required for Cohen’s Kappa to the next release of the Real Statistics software.

Charles

Hi

Could you please provide it ? because its very important.

Hi Ahmed,what do you want me to provide?

Charles

Charles,

Thanks for the wonderful software that you given free to us all.

I chanced upon your site while reading about Cohen’s kappa.

I tried an example given in the free chapter of the book Wonders of Biostatistics,

where cell a is 40, b is 15, c is 10 and d is 35. While the kappa calculated by your software and the result given in the book agree, the standard error doesn’t match.

The book gives a value of 0. 08617, while the value calculated by your software is 0.086746758. The formula given at your site and in the book are the same. I tried the same example through excel myself (not using your software) and got the result the book gave. Is there an error in calculation or is there any other explanation?

Thanks,ajith

Ajith,

I would like to duplicate your results, but I don’t have access to the original data. In particular, I can’t find the free chapter in the book you are referring to and don’t know what cell d you are referring to (nor have you provided me with the value of kappa).

If you can email me an Excel file with the data and your calculations, I will try to figure out what is going on. See Contact Us for my email address.

Charles

Pingback: Cohen’s Kappa value | UX.Data.Psychology.AR

There was result of my Cohen’s Kappa, κ = 0.593 in 95% CI, but in some books, results of Cohen’s Kappa, k= 0.593 (95% CI, 0.300 to 0.886), p < .0005. What does it mean? and the numbers 0.300 to 0.886 can be from where?

It means that Cohen-s kappa is .593, which is a value which indicates agreement, but not as high as some authors suggest shown satisfactory agreement, namely .7. But in any case, the actual value may not even be .593, but you are 95% confident that the value lies between .300 and .886. At the lower range agreement is poor, while at the upper range agreement is very good. p < .0005 indicates that you are very confident that Cohen's kappa is not zero. Charles

Sir,

How can we calculate 95% confidence interval from cohen’s kappa. PLease mail me

Example 2 of the referenced webpage shows you how to calculate a 95% confidence interval for Cohen’s kappa.

Charles

Hi Charles,

Thank you for your article on Cohen’s kappa. Is there a way to calculate Cohen’s kappa for each of the categories (i.e. disorders in your first example). In another post, you showed how Fleiss’ kappa is calculated for each disorders and I wonder if this can be done for Cohen’s kappa as well.

Thank you!

Maria,

I don’t know any way to do this.

Charles

Bonjour,

Je dois calculer, dans un premier temps, l’écart type de l’échelle d’évaluation appelé TRST (échelle comprenant 5 critères, de 1 à 5)

Sachant que mes données sont les suivantes :

TRST Nombre Pourcentage TRST Moyen

1 23 (patients) 6,12% 3,05

2 77 20,48%

3 105 27,92%

4 163 43,35%

5 8 2,13%

Dans un second temps, je voudrait trouver le Coefficient KAPPA pour chaque seuil du score TRST.

Après multiples calcul, aucune solutions n’a été trouvé…

See the following webpage to find the mean of data in the form of a frequency table

Frequency Tables

Sorry, but I don’t understand your question about the Kappa coefficient.

Charles

Hi,

For a sample of 150 decision units, percentage agreement was 82% (123 out of 150), however Cohen’s kappa using SPSS was 83%. Is it possible to have kappa greater than percentage agreement? This has baffled me so much as I would generally not expect this. However, there were close to 10 categories on which decision has to be made. I realise that expected agreement by chance decreases the higher the categories but to what extent?

Thanks

If I am understanding the situation properly, this should not happen. Please let me know the category each of the two judges placed the 150 sample members (e.g. as in range B5:D7 of Figure 2 of the referenced webpage), so that I can see what is going wrong. It would be best if you could put it in an Excel spreadsheet and send it to my email address (see Contact Us).

Charles

Hi Charles,

I have a data set of two Physicians classifying (from category 1 to 8) 29 people. From the example (Figure 1 for Example 1) you have given it seems both the judges saying 10 psychotic, 16 Borderline and 8 Neither (Diagonals). But, what if Judge 1 says 10 Psychotic and Judge 2 says 15 Psychotic (or any number other than 10). How to calculate Cohen’s Kappa here?

In my data, according to Physician 1 there are 5 people in category 1, but according to 2nd Physician there are 8 people in the same category.

Kindly explain me how to calculate Cohen’s kappa in this case?

Bharatesh

I think you are interpreting the figure in a way that wasn’t intended.

Judge 1 finds 16 of the patients to be psychotic. Of these 16 patients, judge 2 agrees with judge 1 on 10 of them, namely he/she too finds them psychotic. But judge 2 disagrees with judge 1 on 6 of these patients, finding them to be borderline (and not psychotic as judge does).

Judge 2 finds that 15 of the patients are psychotic. As stated above, he/she agrees with judge 1 on 10 of these. 4 of the patients that judge 2 finds psychotic are rated by judge 1 to be borderline, while judge 2 rates 1 of them to be neither.

I hope this helps.

Charles

Yes! Got it now.

Thank you

Bharatesh

Hi Charles,

Can you please show me how to transfer the dataset below into the table format (2×2 since there are 2 responses (Reject vs. Accept) and 2 appraisers with 50 samples). Thanks.

Part Appraiser Response Standard

1 Anne A A

1 Brian A A

2 Anne A A

2 Brian A A

3 Anne R R

3 Brian R R

4 Anne R R

4 Brian R R

5 Anne R R

5 Brian R R

6 Anne A A

6 Brian A A

7 Anne R R

7 Brian R R

8 Anne R R

8 Brian R R

9 Anne A A

9 Brian R A

10 Anne A A

10 Brian A A

11 Anne R R

11 Brian R R

12 Anne R R

12 Brian R R

13 Anne A A

13 Brian A A

14 Anne R R

14 Brian R R

15 Anne A A

15 Brian A A

16 Anne A A

16 Brian A A

17 Anne A A

17 Brian A A

18 Anne A A

18 Brian A A

19 Anne R R

19 Brian R R

20 Anne A R

20 Brian R R

21 Anne A A

21 Brian A A

22 Anne R R

22 Brian R R

23 Anne A A

23 Brian A A

24 Anne A A

24 Brian A A

25 Anne R R

25 Brian R R

26 Anne A R

26 Brian A R

27 Anne R R

27 Brian R R

28 Anne A A

28 Brian A A

29 Anne R R

29 Brian R R

30 Anne R R

30 Brian R R

31 Anne A A

31 Brian A A

32 Anne A A

32 Brian A A

33 Anne A A

33 Brian A A

34 Anne R R

34 Brian R R

35 Anne A R

35 Brian R R

36 Anne A A

36 Brian A A

37 Anne A A

37 Brian A A

38 Anne A R

38 Brian R R

39 Anne A A

39 Brian A A

40 Anne A A

40 Brian A A

41 Anne R R

41 Brian R R

42 Anne R R

42 Brian R R

43 Anne A A

43 Brian A A

44 Anne A A

44 Brian A A

45 Anne A R

45 Brian R R

46 Anne R R

46 Brian R R

47 Anne A A

47 Brian A A

48 Anne A R

48 Brian R R

49 Anne A A

49 Brian R A

50 Anne R R

50 Brian R R

Minnie,

I am not sure what the fourth column represents, but I will try to answer your question referencing the first three columns. The key is to pair the rating from Anne with that from Brian for each of the 50 subjects. I’ll assume that the input data occupies the range A1:C100. I then do the following

1. Insert the formula =IF(ISODD(ROW(C1)),C1,””) in cell E1

2. Insert the formula =IF(ISEVEN(ROW(C2)),C2,””) in cell F1

3. Highlight range range E1:F100 and press Ctrl-D

This accomplishes the pairing. Now you can build the 2 x 2 table as follows:

4. Insert “A” in cell H2, “R” in cell H3, “A” in cell I1 and “R” in cell J1 (the headings)

5. Insert =COUNTIFS($E$1:$E$100,$H2,$F$1:$F$100,I$1) in cell I2

6. Highlight range I2:J3 and press Ctrl-R and then Ctrl-D

Charles

Hi Charles,

For this dataset, do you think if it is right if transferred as followed:

AA AR RR

AA 24 0 0

AR 0 26 0

RR 0 5 19

Thank you!

Sorry, but I don’t understand your question.

Charles

For Minnie’s dataset(above), I transferred the data into frequency table as following,

AA AR RR

AA 24 0 0

AR 0 26 0

RR 0 5 19

Do you think it is right?

I don’t really know. It depends on how to interpret the original data. Since there were 50 subjects in the original data judged presumably by two raters, I would expect that the sum of the frequencies in the frequency table would be 50 or perhaps 100. I don’t see this in the table you have provided, but perhaps I am not interpreting things in the way they were intended.

Charles

Hi.

This a great site.

I submitted my research to a journal. I provided the intra and inter examiner’s kappa, but one of the reviewers asked for the “average kappa” for my codes.

I had 5 examiners, 100 sample and 3different coding categories for each sample. How can I do this in SPSS? and what is the value of the average kappa for the codes?

Thank you very much for your help

Max

Max,

I am not an expert on SPSS, but from what I have seen SPSS does not calculate an average value for kappa. I have seen a number of different ways of calculating the average kappa. Also Fleiss’ kappa is often used instead (the Real Statistics package provides Fleiss’ kappa).

Charles

Hello Charles,

Is it possible to run an inter-rater reliability or agreement test with 2 coders, when the categories that are being rated (1=yes, 0=no) are NOT mutually exclusive?

Thanks

Alessi,

With two raters and categorical ratings, Cohen’s kappa is a good fit.

What do you mean by the “categories that are being rated (1=yes, 0=no) are NOT mutually exclusive”. Yes and no seem pretty mutually exclusive to me.

Charles

Can this Kappa statistic be used without testing hypothesis? Like the diagnostic test? Thank you.

Christia,

Yes, Kappa is a measure of agreement and does not need to be used as part of hypothesis testing.

Charles

Hi Charles

This is a nice website, it has been great help, thank you. I have a question: i´m performing a study: I have two raters, ten samples and five categories (1 to 5). Each rater measured each sample. All samples were scored five, i mean:

Rater B

Rater A 1 2 3 4 5

1 0 0 0 0 0

2 0 0 0 0 0

3 0 0 0 0 0

4 0 0 0 0 0

5 0 0 0 0 10

In this case, p0 = 1 and Pe = 1, and cohen´s kappa is K = 1-1/1-1 = ? , meaning that the proportion of observations in agreement is equal to the proportion in agreement due to chance. What does it mean? In this case I´m sure each rater answered correctly cause really all samples have a 5 category. The agreement was perfect but kappa does not indicate this level of agreement.

I appreciate your help.

Hi Alexander,

This is one of the oddities about Cohen’s Kappa. If both raters choose the same rating for all subjects then the value of kappa is undefined. If you shift only one of samples to agreement in another category then the value of kappa is 1 as expected. Fortunately this problem doesn’t occur often, but unfortunately there are other counterintuitive thuns about this measurement.

Charles

Hi, thank you so much for creating this post!

If it’s not too much to ask, could you kindly help me determine how I can apply Cohen’s Kappa for my research?

I have a very big set of statements (more than 2500) and I’ve asked 2 raters to identify the emotions being evoked in these sentences. There are 10 predefined clusters of emotions, so the coders just need to choose among these 10 emotions. But each statement can be coded with more than just 1 emotion, to a maximum of 3 emotions.

What would you advise as the best way to compute reliability in this case?

Thanks in advance!

Auee,

I have been trying to figure out a clean way of using Cohen’s Kappa. One idea, although not so elegant, is to assume that your categories are all subsets of the ten emotions with 1, 2 or 3 emotions and use Weighted Cohen’s Kappa. You then assign the (disagreement) weights so that if the two raters don’t agree on any emotions the weight is high (e.g. rater A picks emotions 1, 2 and 3 while rater B picks emotions 5 and 6), if they agree on 1 emotion the weight is lower (e.g. rater A picks emotions 1 and 2, while rater B picks emotions 1 and 3), if they agree on 2 emotions the weight is even lower (e.g. rater A picks emotions 1, 2 and 3 and rater B picks 1 and 2 only) and finally if they agree completely then the weight is zero (e.g. both raters pick emotions 1 and 2 or both raters pick only emotion 4). There are more possibilities, and you would need to determine the appropriate weight based on some ordering of disagreement. This is the best I can think of at the moment, although maybe someone else has a better idea.

Charles

I have two raters, and two categories (1 and 0). I need a Kappa calculation for each of my codes from a data set. My 2×2 table calculator worked perfectly until I came across a code that had too many zeros in the table.

The values are: 0 2

0 97

Obviously, the agreement is high ( only 2 / 99 are disagreements). However, the kappa calculator breaks down when zeros are multiplied, and I’m curious if there’s a way around this. Thanks for your help.

Cameron,

If you send me a spreadsheet with your data I will see whether I can find a way around the problem.

Charles

We have a data set where we a running kappa. We are trying to differentiate the appropriate use of “overall” & “average” kappa.

For example:

We had two raters that rated x number of variables. We ran this data in SPSS for overall kappa and received .445. When we did kappa for each variable and summed the results to get an “average” kappa, we received .377.

What are the rules about “overall” kappa and “average” kappa for x number of variables with two raters? Which one is more appropriate to use, the overall that the output provides in SPSS or is an average adequate?

Comments?

Michael,

I have only used overall kappa and have not tried to average kappa values, and so I don’t know how average values perform.

Charles

Hi! I need help to find an inter-rater reliability. i have 2 raters examining writing essays. how do i put the scores (continuous data) by the different raters into the formula?

for example, rater 1 marked essay A: 60/100

rater 2 marked essay A: 55/100

Thanks for your help.

Mimi,

The usual approach is estimate the continuous ratings by discrete numbers. This seems to work well in your situation since you can use 11 ratings 0, 10, 20, 30, 40, 50, 60, 70, 80, 90 and 100. If you like, instead you can use 0, 5, 10, 15, 20, etc. or even 0, 1, 2, 3 , etc. The important thing is that you probably want to weight differences between the raters. E.g. in Cohen’s kappa if the two judges rate an essay 60 and 70 this has the same impact as if they rate them 0 and 100. Clearly the difference in ratings in the first case is not as bad as in the second case, but Cohen’s Kappa doesn’t take this into account.

Instead you should use Cohen’s Weighted Kappa as explained on the webpage http://www.real-statistics.com/reliability/weighted-cohens-kappa/.

Another approach is to use the intraclass coefficient as explained on the webpage http://www.real-statistics.com/reliability/intraclass-correlation/.

Charles

Hi Charles, to clarify can I use Fliess with over 100 raters? If I have two model approach with two options each (yes or no) for agreement,is there another procedure you suggest?

Hi Jorge,

Fleiss’s Kappa should work with over 100 raters.

Charles

Thanks for the simple explanation.

My question is: what do you do if the categories are not distinct? For example, I have two raters coding interview statements by a series of categories (e.g. learning style, educational content, multi-sensory learning). We want to establish a sense of inter-rater agreement but are somewhat confused as the categories are not distinct – each statement can be coded into multiple categories. What should we do?

Thanks!

Daniel,

One approach is to create one value of Cohen’s kappa for each category (using 1 if the category is chosen and 0 if the cateory is not chosen). A better approach might be to combine all the categories and use a weighted Cohen’s kappa. E.g. with the three cateories you described (which I will label L, E and M) you would have 8 different ratings none, L, E, M, LE, LM, EM, LEM, which would be coded as 0 through 7. You can now establish the desired pairwise ratings and use weighted kappa as described on the webpage http://www.real-statistics.com/reliability/weighted-cohens-kappa/

Charles

Could you explain how to calculate one value of Cohen’s kappa for each category? I have 5 categories and 2 raters.

Thank you.

Kelly,

If I understand your situation correctly (namely that the two raters are assigning a category to each subject), you create one value for Cohen’s kappa which cuts across all 5 categories. You can’t get a separate kappa for each category.

Charles

This is very useful site thanks for explanation about looking kappa statistic in different scenarios.

I have a question to do in my assignment there are 125 questions in rows, three vendors was rated by 10 raters in three scales 0-not good,3-good,6- very good, can you pl. explain how can I see agreement by raters have any bias or reliable.

Is Kappa statistic is valid for this hypothesis testing or ICC ?

Also, pl. send me the link to calculation worksheet or methodology.

I look forward to hearing from you soon.

Rita,

I don’t understand the relationship between (a) the 125 questions and (b) the 10 raters who rate 3 vendors (using ratings 0, 3 6). Please explain this more completely.

Depending on this explanation, I see three possible measures: ICC, Fleiss’s Kappa or Kendall’s W. All three are explained on the website.

You can download the software which calcuates all three at the webpage http://www.real-statistics.com/free-download/.

You can also see spreadsheets with the examples shown on the website at http://www.real-statistics.com/free-download/real-statistics-examples-workbook/

Charles

(second attempt)

There seems to be a mistake in

‘Observation: Since 0 ≤ pe ≤ pa and 0 ≤ pa ≤ 1, it follows that…’

The expected probability of agreement pe is not necessarily less than the observed probability pa (otherwise Kappa could never be negative).

The correct statement that Kappa can never fall below -1 arises from the fact that, in the case of minimum observed agreement (pa=0) the expected agreement pe cannot be more than one-half; therefore; the absolute magnitude of (pa – pe), i.e. the numerator in Kappa, is always less than or equal to (1 – pe), the denominator.

Jeremy,

I updated the referenced webpage a few weeks ago. I hope that I have now corrected the errors that you identified. Thanks for your help.

Charles

There seems to be a mistake in:

“Observation: Since 0 ≤ pε ≤ pa and 0 ≤ pa ≤ 1, it follows that…”.

pε can be larger than pa (otherwise Kappa could never be negative). The (correctly given) lower limit of -1 for the possible range of Kappa arises from the fact that (negative) deviations of the observed frequencies from expected are limited in size, since negative frequencies are impossible.

Jeremy,

I believe that you are correct, although I haven’t had the opportunity to check this yet. I plan to do so shortly and will update the webpage appropriately. Thanks for bringing this to my attention.

Charles

I want to compare the new test to the old one (but not the gold standard). Can I use Kappa test for this research? If yes, what is the formula to calculate the sample size for Kappa?

Connie,

You haven’t given me enough inofrmation to answe your question. The usual test for comparing two samples is the t test or one of the non-parametric equivalents, and not Cohen’s kappa. The t test and the sample size requirements for this test are described at http://www.real-statistics.com/students-t-distribution/. If you need help in using Cohen’s kappa, you need to provide some additional inoformation.

Charles

Two key formulas in Fig.3 are incorrect.

E10 should be SUM(B10:D10)

E11 should be SUM (B11:D11)

Jeremy,

Thanks for catching this mistake. I have just made the correction on the referenced webpage. Fixing errors such as this one makes it easier for people to understand the concepts and learn the material. I really appreciate your help in this regard.

Charles

Hi,

Can this be used if there are 2 raters rating only 2 items, using a 5 point ordinal scale? I keep getting errors/-1 values! Thank you

Hi Randeep,

I’m not sure how useful it is is, but I just tried using Cohen’s kappa with 2 raters and 2 items using a 5 point scale and it worked fine. The kappa value was negative, but that is possible. If you send me a copy of your worksheet I can investigate further to see if there is an error.

Charles

Nice website. Straight kappa is fine for a binomial situation (and a 2 X 2 cross comparison), but a weighted kappa often would be best for 3 x 3 or higher, wherein the degree of disagreement is taken into consideration (rather than just being a zero-one choice). Such as Cohen, J., 1968. Weighted kappa: nominal scale agreement with

provision for scaled disagreement or partial credit. Psychological

Bulletin 70, 213–220. Keep up the good work, Excel is a terrific tool and generally underappreciated.

Thanks Rick for your suggestion. I will look into the weighted kappa.

Charles

Update (4 Dec 2013): I plan to add weighted kappa to the next release of the Real Statistics Resource Pack due out in the next few days

Is that formula correct? Carletta (1996) and many others list it as (P(a)-P(e)) / (1-P(e)). The denominator in your formula has it as (1-P(a)).

Ryan,

Thanks for catching the typo. I have now corrected the webpage. The solution to Example 1 was correct, since it used the correct formula.

Charles

There might have a typo in example 1:

Judge 1’s diagnoses and 15/30 = 30% of Judge 2’s diagnoses, the denominator should be 50 instead of 30, please kindly take a note.

Thanks Stephen for catching the typo. I have now corrected this on the website.

Charles

Thanks, Charles, I have a case and not sure which model is more appropriate to apply, could you please enlighten me on this:

There is a group of inspectors and the team size is 20, they are required to inspect 10 samples, and each sample might contain ‘no defect’, ‘minor defect’ and ‘major defect, each inspector will repeat to inspect each sample randomly but repeatedly (at least 2 times per each sample) and record the result, the inspection result would be compared against the judge’s result (standard), and it is required to draw conclusion about the inspector’s performance (individual and whole team) to rate their performance. So, please advise which model would be appropriate to apply this scenario?

Thanks for the great examples, if there are more than 2 judges, is the same Kappa method applicable?

Stephen,

Cohen’s kappa is designed for two judges. If you have more than two judges you can use the intraclass correlation. See webpage

http://www.real-statistics.com/reliability/intraclass-correlation/

Charles