I would like to quantitatively describe the relationship between the two variables shown in this plot, but I am not sure what would be the correct way to do so. More specifically, my aim would be to assess whether there is a relationship between both variables and to quantify the strength of such a relationship (e.g. quantifying the amount of V2 variance explained by V1 with the coefficient of determination R2 or any other suitable statistic)
I have tentatively used the GAM model available in the geom_smooth function of ggplot (method="gam",formula = y ~ s(x, bs = "cs") and I have reproduced the same model with the gam package (see code and output below). However, I am not sure if this would be enough to claim that there is a relevant relationship between V1 and V2 and, if so, what would be the crucial statistics that I should report to substantiate such a possible conclusion. On the other hand, I have calculated the Spearman correlation trying to assess the relationship at the ordinal level (rho=0.480, p= 7.221e-08).
Thanks in advance for any help/ advice
library(gam) v=gam(V1~s(V2),bs="cs", data=ff) Call: gam(formula = V1 ~ s(V2), data = ff, bs = "cs") Deviance Residuals: Min 1Q Median 3Q Max -3.36459 -0.31432 -0.14645 0.09815 7.18685 (Dispersion Parameter for gaussian family taken to be 1.7516) Null Deviance: 377.6946 on 115 degrees of freedom Residual Deviance: 194.4251 on 111.0001 degrees of freedom AIC: 401.1024 Number of Local Scoring Iterations: 2 Anova for Parametric Effects Df Sum Sq Mean Sq F value Pr(>F) s(V2) 1 71.352 71.352 40.736 4.159e-09 *** Residuals 111 194.425 1.752 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Anova for Nonparametric Effects Npar Df Npar F Pr(F) (Intercept) s(V2) 3 21.299 5.709e-11 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
############# ############ EDIT:
Following the advices received I have transformed V1 using log (both proposed transformations produced virtually identical results). Then, I have estimated the relationship through linear regression. However, the obtained fit does not seem to be much better...