blog/correlation-coefficient-and-correlation-test-in-r/

[utterances-bot]

Correlation coefficient and correlation test in R - Stats and R

Learn how to compute a correlation coefficient (Pearson and Spearman) and perform a correlation test in R

https://statsandr.com/blog/correlation-coefficient-and-correlation-test-in-r/

[AntoineSoetewey]

Comment written by YZurbuchen on October 21, 2020 08:03:49:

Hi Antoine,

Thanks a lot for your article. I would just have a short question regarding dichotomous variables. If I have a matrix consisting of continuous and dichotomous variables would it still be correct to use the cor.test function in order to subsequently visualise it as a correlation plot? I was unsure about how to handle it, according to this https://stats.stackexchange.com/questions/226157/p-value-for-point-biserial-correlation-in-r article it should automatically compute the point-biserial correlation and would therefore be correct?

Thanks a lot for your help!

[AntoineSoetewey]

Comment written by YZurbuchen on October 21, 2020 08:03:49:

Hi Antoine,

Thanks a lot for your article. I would just have a short question regarding dichotomous variables. If I have a matrix consisting of continuous and dichotomous variables would it still be correct to use the cor.test function in order to subsequently visualise it as a correlation plot? I was unsure about how to handle it, according to this https://stats.stackexchange.com/questions/226157/p-value-for-point-biserial-correlation-in-r article it should automatically compute the point-biserial correlation and would therefore be correct?

Thanks a lot for your help!

Comment written by Antoine Soetewey on October 21, 2020 08:48:24:

Dear YZurbuchen,

Thanks for your question.

I confirm that you can use cor.test() with one continuous and one dichotomous variable, given that the dichotomous variable is encoded as numeric in R (if it is encoded as factor, as it is usually the case for dichotomous variables, it will throw an error).

It will produce the exact same results than computing the point-biserial correlation, as shown by the following example:

set.seed(42)        
x <- rnorm(100)          
y <- sample(c(0, 1), 100, replace = TRUE)          
cor.test(x, y)          
# Pearson's product-moment correlation          
#          
# data:  x and y     
# t = -0.32697, df = 98, p-value = 0.7444          
# alternative hypothesis: true correlation is not equal to 0          
# 95 percent confidence interval:          
#  -0.2279507  0.1644740          
# sample estimates:          
#         cor          
# -0.03301059          

# install.packages("ltm")          
library(ltm)          
biserial.cor(x, y, level = 2)          
# [1] -0.03301059       

In the StackExchange thread you mentioned, the user Sal Mangiafico suggested (on April 9, 2018) to use the function biserial.cor.new() which uses the population standard deviation. However, as from April 17, 2018, the developer of the {ltm} package updated his biserial.cor() function to use the population standard deviation instead of the sample standard deviation (see this issue).

So now the two functions biserial.cor() and cor.test() functions produce the same results.

Hope this helps.

Regards,
Antoine

[AntoineSoetewey]

Comment written by Antoine Soetewey on October 21, 2020 08:48:24:

Dear YZurbuchen,

Thanks for your question.

I confirm that you can use cor.test() with one continuous and one dichotomous variable, it will produce the exact same results than computing the point-biserial correlation, as shown by the following example:

set.seed(42)        
x <- rnorm(100)          
y <- sample(c(0, 1), 100, replace = TRUE)          
cor.test(x, y)          
# Pearson's product-moment correlation          
#          
# data:  x and y     
# t = -0.32697, df = 98, p-value = 0.7444          
# alternative hypothesis: true correlation is not equal to 0          
# 95 percent confidence interval:          
#  -0.2279507  0.1644740          
# sample estimates:          
#         cor          
# -0.03301059          
# install.packages("ltm")          
library(ltm)          
biserial.cor(x, y, level = 2)          
# [1] -0.03301059       

In the StackExchange thread you mentioned, the user Sal Mangiafico suggested (on April 9, 2018) to use the function biserial.cor.new() which uses the population standard deviation. However, as from April 17, 2018, the developer of the {ltm} package updated his biserial.cor() function to use the population standard deviation instead of the sample standard deviation (see this issue).

So now the two functions biserial.cor() and cor.test() functions produce the same results.

Hope this helps.

Regards,
Antoine

Comment written by YZurbuchen on October 21, 2020 14:43:17:

Thank you very much for your super quick answer, that's great!

Best wishes,
Yves

[AntoineSoetewey]

Comment written by YZurbuchen on October 21, 2020 14:43:17:

Thank you very much for your super quick answer, that's great!

Best wishes,
Yves

Comment written by Antoine Soetewey on October 21, 2020 14:44:40:

You're welcome!

[AntoineSoetewey]

Comment written by madihah manggabarani on October 22, 2020 23:24:51:

Dear Antoine, I really appreciate your article, it is easily to be understood and to follow the practice using R !

[AntoineSoetewey]

Comment written by madihah manggabarani on October 22, 2020 23:24:51:

Dear Antoine, I really appreciate your article, it is easily to be understood and to follow the practice using R !

Comment written by Antoine Soetewey on October 23, 2020 05:11:57:

Glad you find it useful Madihah!

[AntoineSoetewey]

Comment written by vijayarajamanickam on November 30, 2020 22:55:44:

Dear Antoine, even after devtools::install_github("laresbernardo/lares") package installation, I am not able to access corr_cross function.

Could you please help me from that?

[AntoineSoetewey]

Comment written by vijayarajamanickam on November 30, 2020 22:55:44:

Dear Antoine, even after devtools::install_github("laresbernardo/lares") package installation, I am not able to access corr_cross function.

Could you please help me from that?

Comment written by Antoine Soetewey on December 01, 2020 12:50:39:

Hello,

I just tried on my side with the following code:

devtools::install_github("laresbernardo/lares")          
library(lares)          
corr_cross(mtcars)      

and it works smoothly.

So few things to make it work on your side:

• are you sure you executed devtools::install_github("laresbernardo/lares") and not devtools::installgithub("laresbernardo/lares")?
• when R asks you if you want to update packages when installing the {lares} package, choose the option 3: None (so don't update packages)
• make sure to load the package by running library(lares) after having installed it
• try again with a basic dataset

Let me know if this does not fix your issue.

Regards,
Antoine

[AntoineSoetewey]

Comment written by Antoine Soetewey on December 01, 2020 12:50:39:

Hello,

I just tried on my side with the following code:

devtools::install_github("laresbernardo/lares")          
library(lares)          
corr_cross(mtcars)      

and it works smoothly.

So few things to make it work on your side:

• are you sure you executed devtools::install_github("laresbernardo/lares") and not devtools::installgithub("laresbernardo/lares")?
• when R asks you if you want to update packages when installing the {lares} package, choose the option 3: None (so don't update packages)
• make sure to load the package by running library(lares) after having installed it
• try again with a basic dataset

Let me know if this does not fix your issue.

Regards,
Antoine

Comment written by vijayarajamanickam on December 02, 2020 15:26:04:

Dear Antoine,

Now it works fine.
Thanks a lot for your response.

kind regards
vijay

[AntoineSoetewey]

Comment written by vijayarajamanickam on December 02, 2020 15:26:04:

Dear Antoine,

Now it works fine.
Thanks a lot for your response.

kind regards
vijay

Comment written by Antoine Soetewey on December 02, 2020 15:59:28:

You're welcome, glad it worked!

[parksonpurity]

Hi Antoine;

Just a simple question that if I have a continuous and three dichotomous variables, and in that case, do you know how to calculate/test their correlation?

Cheers - Tao

[AntoineSoetewey]

Hi Antoine;

Just a simple question that if I have a continuous and three dichotomous variables, and in that case, do you know how to calculate/test their correlation?

Cheers - Tao

Dear Tao,

Assuming your three dichotomous variables are qualitative nominal variables, you cannot compute a correlation as defined in this post (even if your second variable is a quantitative continuous variable).

Remember that a Pearson, Spearman or Kendall correlation can only be computed between quantitative or qualitative ordinal variables.

What you can do:

• consider your continuous variable as dependent variable and any of your dichotomous variable as independent variable (but only one at a time). You therefore have two groups, and you can compare them via a Student's t-test or a Wilcoxon test (depending on the number of observations or on the normality of your data)
• if you need to use two dichotomous variables, you can test their independence thanks to a Chi-square test of independence (see how to do this test in R or by hand)
• consider your continuous variable as dependent variable and your dichotomous variables as independent variables and perform a linear regression. If you consider only one dichotomous variable it will be a simple linear regression, whereas if you consider two or more independent variables it will be a multiple linear regression

Last but not least, it is true that there is a lesser known correlation (not covered in this post), referred as point-biserial correlation which can be used when one variable is dichotomous and the other is continuous. Similar to the Pearson, Spearman and Kendall correlations, this coefficient ranges between –1 and +1 and values close to ±1 indicate a strong positive/negative relationship, and values close to 0 indicate no relationship between the two variables.

Hope that helps.

Regards,
Antoine

[rafabelokurows]

Thanks for sharing all that, I learned a lot from this post. I'll use some methods from package "correlation" in my master's thesis :)

[AntoineSoetewey]

Thanks for sharing all that, I learned a lot from this post. I'll use some methods from package "correlation" in my master's thesis :)

Thanks for your feedback Rafael, and good luck with your thesis!

[sfpravelo]

Hi. When I did a combination of correlation, I get this error "Error in tetra.mat(x, y = y, correct = correct, smooth = smooth, global = global, : Tetrachoric correlations require dictomous data". Im pretty much new to this so I wonder what I can do about it. Thank you!

[AntoineSoetewey]

Hello Stephanie,

I'm not sure what function you are referring to, so can you show the code you use?

Regards,
Antoine

[sfpravelo]

Here's the code of the entire run:

> setwd("/Users/faithravelo/Desktop/Trial")
> cor.data = read.csv("4.Biliran.csv", header = TRUE, row.names = 1)
> round(cor(cor.data), digits = 2)
> res <- rcorr(as.matrix(cor.data))
> round(res$P, 3)
> correlation::correlation(cor.data, include_factors = TRUE, method = "auto")
Error in tetra.mat(x, y = y, correct = correct, smooth = smooth, global = global,  : 
  Tetrachoric correlations require dictomous data

[AntoineSoetewey]

Thanks.

There are several lines and functions here.
On which line exactly you encounter an error?

[MontseGG]

Hi Antonie,
Thank you for this article, I learned and implement some functions, but I realised that maybe is an error in note 2.

says:
3 asterisks means that the coefficient is significant at the 5% level, 2 is at the 1% significance level, and 3 is at the 0.1% significance level.

could be:
3 asterisks means that the coefficient is significant at the 5% level, 2 is at the 1% significance level, and 1 is at the 0.1% significance level.
dice

I think the last 3 , actually would be one. right?

Thanks.

[AntoineSoetewey]

Hello Maria,

Thanks for pointing it out, there was indeed a typo!

I have corrected the footnote with the following:

One asterisk means that the coefficient is significant at the 5% level, 2 is at the 1% significance level, and 3 is at the 0.1% significance level.

The more asterisks, the more it is significant (at least in R).

In doubt, run any linear regression and R will remind you. See Signif. codes in the following output:

Hope it is clearer now.

Regards,
Antoine

[MontseGG]

Thanks !! Everything clear.

[Reineany]

Dear Antoine, thank you so much for this I really learned a lot! May I ask: I inferred from your explanation that to read these scatterplots one reads from left to right or top-down which two variables are linked (that is my confusion and it is such a stupid question that I did not dare openly ask it :/)?
Because you say that mpg and hp are negatively correlated as are mpg and wt, one would know that because the panels right in first row from mpg are representing a) link between mpg and hp b) link between mpg and wt --- similarly, the two panels of the left column under mpg represent the link between mpg and hp and mpg and wt?
It would be good to have a proper simple rule of thumb on how to best understand which two variables are linked? Thank you in advance!

[AntoineSoetewey]

Dear @Reineany,

This might help:

Remember also that scatterplot matrices are symmetric.

Hope this helps.

Regards,
Antoine

[Reineany]

Antoine:genius! Thank you so much! Maybe you could consider doing something on network analyses in R and betweenness, closeness and distances and so on! :) I bet that would turn out just as fabulous. Thank you so much for statsandr. !

[lu2608]

Hi Antoine,

this article was very helpful as well as giving plenty of insights in understanding the main principles of R.

However, I have a further question as part of Case Studies we are given we have to check for the correlation of two variables: yi and xi, with xi being the independent variable of which we assume that it is normal distributed. Now we are given an equation of the form yi=a+b*xi+ei, so the relation is clearly non-linear. In order to perform a correlation test can I rely on the one built-in by R? Or do I have to check for the distance correlation? I do know there is an expansion for computing such but would there also be a method to do it in basic R?

Thank you so much in advance,

Have a great evening,

Luisa

[AntoineSoetewey]

Hello @lu2608,

Now we are given an equation of the form yi=a+b*xi+ei, so the relation is clearly non-linear.

I don't see why you assume it is clearly non-linear based on the equation yi=a+b*xi+ei (which is the equation for a simple linear regression).

You can check the linearity by drawing a scatterplot of the two variables, for instance:

plot(data$x, data$y)

In order to perform a correlation test can I rely on the one built-in by R?

You mentioned that variables are normally distributed, so I assume both variables are quantitative. If the link between the two variables is linear, you can use the built-in cor.test(data$y, data$x) function (the default method being the Pearson correlation).

If at least one variable is qualitative ordinal and the link between the two variables is at least partially linear, you can still use cor.test(data$y, data$x, method = "spearman").

If the link is not linear, yes you could use the distance correlation, as it measures both linear and nonlinear association between two random variables. I am not aware of any built-in function in R for this type of correlation, but there must be some available via a package.

Hope this helps.

Regards,
Antoine

[lu2608]

Dear Antoine,

thank you for the quick reply!
Sorry I made a small mistake by not typing in the equation correctly, it should say: yi=a+b*xi^2+ei.

Therefore, it is definitely not linear.

As for the distance correlation, would it also be possible to do it by hand or compute it as a chunk of code in R without a package?

Thank you!

[AntoineSoetewey]

Sorry I made a small mistake by not typing in the equation correctly, it should say: yi=a+b*xi^2+ei.

Makes sense!

As for the distance correlation, would it also be possible to do it by hand or compute it as a chunk of code in R without a package?

I am not an expert in distance correlation, but based on these two sources (Wikipedia and a paper), it seems possible to compute it by hand. And if it is possible by hand, it is definitely possible in R without using any package.

[riscas14]

Hello Antoine,

Thanks for your blog. I am performing a multiple regression analysis and now I am assessing the correlation between the explanatory/independent variables. I have some continuous variables and a few binary variables as well, however my data does not follow a normal distribution. Which tests should I perform in order to evaluate the correlation to make sure they are independent from each other?

[AntoineSoetewey]

Thanks for this detailed explanation.

since one of the assumptions for regression analysis is to make sure the explanatory variables are independent from each other (to avoid multicollinearity), I have to do correlation tests between them.

You’re right, one assumption of multiple linear regression is no high multicollinearity (which occurs when independent variables are strongly correlated). However, you don’t have to do correlation tests between them. Computing the VIF (variance inflation factor) to assess their correlation is enough: a VIF below 10 (some authors say 5) is usually considered low enough to say that multicollinearity is not an issue. (And I would say it’s even better to check it via the VIF instead of the correlation tests because correlation tests only show the bivariate relationship between the independent variables, whereas VIF can show the correlation of a variable with a group of other variables.)

However, none of my explanatory variables follow a normal distribution. I tested the normality using ggqqplots() and performing Shapiro Tests to each one of them.

That’s another assumption of linear regression, and I recommend:

• to check the normality on the residuals instead of on each variable separately
• check it only with a histogram and/or a QQ-plot because normality tests (such as the Shapiro-Wilk test) are quite powerful and will easily reject the null hypothesis of normality with large sample sizes (+ the normality assumption may be relaxed with large sample sizes if there is only a small deviation from normality, see more in this section)

My problem is that I don't know which tests I should perform for assessing the correlation between a pair of a binary and a continuous variable, and a pair of two continuous variables.

As explained above, if it’s in the context of multiple linear regression, I would only compute VIF. If you’d like to do correlation tests for something else than assessing the relationship between your independent variables of your linear model, that’s another story. In that case, let me know.

[riscas14]

Hello Antoine,

Thanks for your explanation.

In fact I was trying to run correlation tests because I have a database with a lot of independent variables and I have to select the ones that fit better, so basically to select the "best variables". To do so I was trying to run all the possible combinations (all subsets) to get the "best" set of variables, however since some variables are highly correlated, I get an error from RStudio because of that. So I wanted to know which are the "problematic variables" and then decide which ones I should exclude from my model or maintain.

Of course I already have a good guess of which variables are highly correlated, but my supervisor advised me to do the correlation tests basically to corroborate my decisions when it comes to variable selection.

[AntoineSoetewey]

Ok I understand your point.

For the pairs with two binary variables I used Fisher Test because the some frequencies were lower than 5.

I agree.

My problem is that I don't know which tests I should perform for assessing the correlation between a pair of a binary and a continuous variable, and a pair of two continuous variables.

For a binary variable (= nominal variable with two levels) and a continuous variable, I suggest using the Student’s t-test (or Mann-Whitney test for the non-parametric version). We don’t compute the correlation in this case (at least the 3 types of correlation mentioned in this post), but we compare two groups in terms of the continuous variable.

For a pair of two continuous variables, you can use the Pearson correlation (if there is a linear relationship) or the Spearman correlation (if the link is partially linear). All information can be found in this post.

Last but not least:

To do so I was trying to run all the possible combinations (all subsets) to get the "best" set of variables

I don’t know how you did it, so I’d like to highlight the step() function just in case you are unaware of it. See more information here. If this is what you did, you can ignore this paragraph.

[riscas14]

Hello Antoine,

Thanks again for helping me.

You have said that we don't compute the correlation between a binary and a continuous variable, but what about the Point-Biserial correlation? What is the reason why we cannot calculate the correlation in this case? Is it because the normality assumption is not met?

I read some articles about stepwise regressions and its disadvantages, so I was actually using regsubsets() function from the leaps package to apply the "Best Subsets" methodology

[AntoineSoetewey]

You have said that we don't compute the correlation between a binary and a continuous variable, but what about the Point-Biserial correlation?

Sorry I was perhaps not clear enough. I meant that if you choose to do a t-test for independent samples, it’s not about computing the correlation but about comparing two groups (just to use the appropriate wording). But, yes you can compute the point biserial correlation if you prefer that test over the t-test.

However, note that the two tests are very similar. In fact, the p-value obtained from a point biserial correlation will be the exact same as the p-value for an independent samples t-test.

I suggested the t-test over the point biseral correlation simply because it is more common when we have two groups and a continuous variable. But it’s not an obligation and in the end, the two tests will give you the same conclusion (you will gain the same understanding with both tests).

[Filippo-94]

A question:
why if I run

library(correlation)
correlation(mtcars)

are there a lot of > .999 p-value?

For example:
qsec and gear > .999
but if I run
correlation(mtcars, select=c("qsec", "gear"))
the result is 0.243

[AntoineSoetewey]

Dear @Filippo-94,

It is because with the correlation() function the p-values are adjusted by default with the Holm method (see the documentation here).

Try correlation(mtcars, p_adjust = "none") and you should obtain the same results.

Regards,
Antoine

[liux2250]

Dear Antoine,

You can't image how much that your blog helps me with my research. My gratefulness is totally beyond words!

Here there are a couple of questions that I don't understand well. Could you please help me with them?

One of my projects is to calculate the correlation between a two-level nominal variable and two-level ordinal variable. I searched online and found someone used the chisq_test function in "coin" package, but I am not sure if it is a good choice.

Besides, in your response to YZurbuchen that you confirmed that cor.test() can be used for one continuous and one dichotomous variable. However, in your response to Tao you specified that correlation can only be computed between quantitative or qualitative ordinal variables and you recommended a linear regression for the case of nominal and continuous variables. I am confused, as dichotomous variable is a two-level nominal variable.

If I have a dependent nominal variable and an independent continuous variable, should I use the cor.test() or use linear regression to calculate the correlation between them?

Looking forward to your response.

Best,

Yang

[AntoineSoetewey]

Dear @liux2250,

You can't image how much that your blog helps me with my research. My gratefulness is totally beyond words!

Thanks for your kind words!

One of my projects is to calculate the correlation between a two-level nominal variable and two-level ordinal variable. I searched online and found someone used the chisq_test function in "coin" package, but I am not sure if it is a good choice.

The chisq_test() function from the {coin} package is similar to the chisq.test() function without the Yate's continuity correction:

set.seed(42)
library(coin)
x <- as.factor(sample(c(0, 1), 100, replace = TRUE))  
y <- as.factor(sample(c(0, 1), 100, replace = TRUE))  

chisq_test(table(x, y))
# chi-squared = 0.53129, df = 1, p-value = 0.4661

chisq.test(table(x, y), correct = FALSE)
# X-squared = 0.53129, df = 1, p-value = 0.4661

Both functions give the results of the Chi-square test of independence, which is a test for the association (and not for the correlation) between two variables. I concede that in the end it is the same; we test the relationship between two variables, but formally we do not refer to correlation with a Chi-square test. So if you are asked explicitly to compute the correlation between a two-level nominal variable and two-level ordinal variable, I would not use the Chi-square test. If your goal is to test the relationship/association between these two variables, a Chi-square test is perfectly fine.

Besides, in your response to YZurbuchen that you confirmed that cor.test() can be used for one continuous and one dichotomous variable. However, in your response to Tao you specified that correlation can only be computed between quantitative or qualitative ordinal variables and you recommended a linear regression for the case of nominal and continuous variables. I am confused, as dichotomous variable is a two-level nominal variable.

You're right, I was not precise enough in these comments. I have slightly edited the comments to make it more clear.

The reasons I wrote that a correlation can only be computed between quantitative or qualitative variables is because I was thinking about the 3 types of correlation mentioned in my post (i.e., Pearson, Spearman and Kendall). These 3 correlations cannot be computed with nominal/dichotomous variable. Furthermore, I assume that a dichotomous variable is encoded as factor in R, and R will throw an error if you try to compute one of these 3 correlations with a factor variable. So, by default I would say that a correlation cannot be computed on dichotomous variable. Now, it is true that there is the point-biserial correlation which can be used with a nominal variable. Since YZurbuchen asked about it, I confirmed that it can be used with such variables. Nonetheless, this type of correlation is much less known and usually not covered in introductory statistics classes; with one continuous and one nominal variable, it is much more frequent to learn about the Student's t-test (for a nominal variable with 2 groups) or ANOVA (for a nominal variable with 3 or more groups).

If I have a dependent nominal variable and an independent continuous variable, should I use the cor.test() or use linear regression to calculate the correlation between them?

If you have a dependent nominal variable and an independent continuous variable, you can use the cor.test() function if you really want to, but in that case you must have the nominal variable encoded as numeric and not as factor in R. Otherwise, if your goal is to compare the 2 groups in terms of a continuous variable, I would recommend the Student's t-test.

Last but not least, if you want to use modeling, you should use a binary logistic regression rather than a linear regression. The difference between these two models is:

• binary logistic regression: the dependent variable is binary
• simple linear regression: the dependent variable is continuous

You'll also find more information about choosing the most appropriate measure of association depending on the type of variable in this text.

Hope this helps.

Regards,
Antoine

[liux2250]

Dear Antoine,

Thank you for your prompt and detailed explanation. Now, it is very clear and really helpful. Your time is greatly appreciated!

Best regards,

Yang

[kostas-moschonas]

Hello Antoine,

Thank you so much for this neat and informative article. May I please ask a question? I have two dataframes of which I want correlation matrix, but without calculating coefficient within the same dataframe.
E.g. I have a biomarker dataset another dataset with clinical information. I want to correlate each biomarker to each clinical variable, but not clinical variables or biomarkers within themselves.

The rcorr function in the Hmis package just does all possible combinations, and then I cannot extract easily.
I thought of manually running for loops with all possible combinations between the two datasets but I am sure there is a package/function out there for sth like this, so I am asking before doing it manually.

Do you have any advice please? Thank you so much!
Kostas

[AntoineSoetewey]

Hello @kostas-moschonas,

It can be achieved with the cor() function. Here is an example:

# biomarker dataset
biomarkers_df <- data.frame(
  biomarker1 = rnorm(100),
  biomarker2 = rnorm(100),
  biomarker3 = rnorm(100)
)

# clinical information dataset
clinical_df <- data.frame(
  clinical1 = rnorm(100),
  clinical2 = rnorm(100),
  clinical3 = rnorm(100)
)

# correlation matrix between the two dataframes
cor(biomarkers_df, clinical_df, use = "complete.obs")

Hope this helps.

Regards,
Antoine

[kostas-moschonas]

It does Antoine, thank you! It was simpler than I thought.
However, cor() does not output the p values for those correlations. Is there an alternative for this?

[Zerihunm12]

please help me how to calculate prediction in r if the variable is more than two?

[AntoineSoetewey]

Dear @Zerihunm12,

What is the type of the variable you would like to predict?

• If it is a quantitative variable, you can predict it based on other variables (called explanatory variables) thanks to a linear regression.
• If it is a categorical variable, you can predict it based on other variables thanks to a logistic regression (binary logistic regression if your dependent variable has two levels only, multinomial logistic regression if it has > 2 unordered levels, and ordinal logistic regression if it has > 2 ordered levels).

Feel free to add more details to your questions!

Regards,
Antoine