Regression analysis or Anova? - r

Question

I hope to be the clearest I can.
Let's say I have a dataset with 10 variables, where 4 of them represent for me a certain phenomenon that I call Y.
The other 6 represent for me another phenomenon that I call X.
Each one of those variables (10) contains 37 units. Those units are just the respondents of my analysis (a survey).
Since all the questions are based on a Likert scale, they are qualitative variables. The scale is from 0 to 7 for all of them, but there are "-1" and "-2" values where the answer is missing. Hence the scale goes actually from -2 to 7.
What I want to do is to calculate the regression between my Y (which contains 4 variables in this case and 37 answers for each variable) and my X (which contains 6 variables instead and the same number of respondents). I know that for qualitative analyses I should use Anova instead of the regression, although I have read somewhere that it is even possible
to make the regression.
Until now I have tried to act this way:
> apply(Y, 1, function(Y) mean(Y[Y>0])) #calculate the average per rows (respondents) without considering the negative values
> Y.reg<- c(apply(Y, 1, function(Y) mean(Y[Y>0]))) #create the vector Y, thus it results like 1 variable with 37 numbers
> apply(X, 1, function(X) mean(X[X>0]))
> X.reg<- c(apply(X, 1, function(X) mean(X[X>0]))) #create the vector X, thus it results like 1 variable with 37 numbers
> reg1<- lm(Y.reg~ X.reg) #make the first regression
> summary(reg1) #see the results
Call:
lm(formula = Y.reg ~ X.reg)
Residuals:
Min 1Q Median 3Q Max
-2.26183 -0.49434 -0.02658 0.37260 2.08899
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.2577 0.4986 8.539 4.46e-10 ***
X.reg 0.1008 0.1282 0.786 0.437
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.7827 on 35 degrees of freedom
Multiple R-squared: 0.01736, Adjusted R-squared: -0.01072
F-statistic: 0.6182 on 1 and 35 DF, p-value: 0.437
But as you can see, although I do not use Y as composed by 4 variables and X by 6, and I do not consider the negative values too, I get a very low score as my R^2.
If I act with anova instead I have this problem:
> Ymatrix<- as.matrix(Y)
> Xmatrix<- as.matrix(X) #where both this Y and X are in their first form, thus composed by more variables (4 and 6) and with negative values as well.
> Errore in UseMethod("anova") :
no applicable method for 'anova' applied to an object of class "c('matrix', 'integer', 'numeric')"
To be honest, a few days ago I succeeded in using anova, but unfortunately I do not remember how and I did not save the commands anywhere.
What I would like to know is:
First of all, am I wrong in how I approach to my problem?
What do you think about the regression output?
Finally, how can I do to make the anova? If I have to do it.

Answer 1

If your response (Y) and predictor (x) are numeric scale, you can use regression.
If your response (Y) is numeric scale with predictor (x) is categoric scale, you can use ANOVA.
Suggested:
you have to use validity and reliability test to know if the answers (indicators) are valid and reliable for response and predictor, before you use regression method.

Answer 2

I disagree with Denny's answer. You can use either approach regardless of the type of data that you have. If you have categorical data you can express it as numeric using dummy encoding. For example given a feature x with 3 options, say 1, 2, and 3, you can encode this as numeric by creating 3 new additional variables x1, x2, and x3. If x is 1 x1 will be 1, x2 will be 0, and x3 will be zero. If x is missing, the three new x values will all be zero.
In your case I would recommend that you try regression first because of the amount of features that you have and because it tends to be straight forward. ANOVA can become complicated as the number of the features increases. Both should work, assuming your data meets the assumptions required by both techniques.