You could do a χ2 test for independence. Basically you would be testing whether or not the location has any affect on the distribution of soil type.

To perform the test you would first estimate the distribution of soil type under the null hypothesis where the distributions are identical. To do this you'd want to find the marginal probability of each soil type, ignoring location (use the total of each soil type divided by the total count). Then, within each cell find the squared difference between the observed and expected counts, divide by the expected counts, and sum over all the cells. Under the null this will approximately follow a χ2 distribution with (r−1)(c−1) degrees of freedom where r and c are the number of rows and columns respectively.

I would not worry too much about the statistics, but rather how to justify that these three corpora of three different languages can be used for direct comparisons. Are the types of data the same (assuming it is spoken data)? Are the genres the same (say, structured interview on the same topic with the same questions)? Are they taken from similar sources (say, transcribed data from participants of certain age group / education background / language ability)? Are the number of words, or at least the percentage of contribution of each genre, balanced? If any one of these is not controlled carefully or justified, then whatever the statistical methods will only lead to results that has little meaning. (Imagine obtaining statistically significant correlations between eating cabbage and having an inner bellybutton.)

Hi,

the numbers look very similar...

I would go for the chi-square-test but with the absolute numbers of occurrences of reformulation markers and the absence of reformulation markers for the three corpora. I am not sure how you would count the absence, though, so this is a bit tricky (maybe count possible places of reformulation markers, such as phrases?). If you were able to count the number of occurrences and absences, you could simply add them into an online chi-square test, e.g. http://www.quantpsy.org/chisq/chisq.htm.

As for the interpretation of the outcome, I agree with Lucky Law.

Best,

Bettina

I would use 2-sample test for equality of proportions with continuity correction. In R, the function is prop.test(), which essentially uses chi-square statistics in the background. You would need to use absolute numbers, as Bettina Braun also suggested. However, I am not sure if the total number of words in each corpus would be the respective sample size. This is because reformulation markers can be phrases ("I mean", "that is") or individual words ("or", "namely"). In some languages, perhaps word-like reformulation markers may be more common (e.g., "yani" in Turkish, which means "that is to say"). If you treat all of your reformulation markers as special word types (i.e., as "chunks" or "multi-word units"), then perhaps you can justify that you will use the total number of words in each corpus as your sample size.

Hi Elyas,

There are two options for you:

1) chi-square test based on raw frequency counts of reformulators in each corpus since chi-square test can be performed regardless of different corpus sizes;

2) one-way ANOVA based on normalized frequency counts of reformulators in each file/script of three corpora. But to do this statistical analysis, there are some assumptions that your data should meet (fairly easy to examine in SPSS, JASP, etc.), for example, dependent variable should be approximately normally distributed, homogeneity of variances among groups. If these assumptions are violated, alternative statistical tests should be performed instead. Please refer to here for more information https://statistics.laerd.com/spss-tutorials/one-way-anova-using-spss-statistics.php. If finally the one-way ANOVA is statistically significant, since ANOVA is an omnibus test statistic, a follow-up post-hoc test should be performed to see in which group or groups (i.e., English vs. Persian, English vs. Arabic, Persian vs. Arabic) the statistical difference lies.

Hope my reply is of help to you.

All the best,

Gang