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set split, you are selecting 16 observations for the training set and 16 for the testing set: train <- sample(1:nrow(data),nrow(data)*0.5) test <- setdiff(1:nrow(data),train) This means your linear regression … models are being fitted with 16 observations and 10 independent variables; this almost certainly leads to significant overfitting to the training data, which is why your out-of-sample MSE is poor for the regression …

No, just because a coefficient value is close to 1 in multiple linear regression does not imply that the variable is almost equal to the outcome. … 0.1 2 # 3 3.0997 0.1 3 # 4 4.0001 0.0 4 # 5 4.9999 0.0 5 In this dataset, the outcome variable y is basically equal to the sum of the two independent variables x1 and x2; as expected the linear regression …

By definition, $$ \bar x = \frac{\sum_{j=1}^n x_j}{n} $$ So multiplying through by $n$ yields the result that $\sum_{j=1}^n x_j = n\bar x$.

Mathematical optimization is a tool that is often used to optimize a decision given some objective function and some constraints. In your case, the decision variables would be the amount of each ingr …

You might consider instead using segmented regression, which models $y$ as a piecewise linear function of $x$. … Segmented regression is implemented in R in the segmented package. …

From the Stata documentation, the cholesky function returns a lower triangular matrix $G$ such that $A = GG'$ given input matrix $A$. Meanwhile, from the R documentation, chol returns an upper triangu …

An update formula in a similar vein to the one you provided for updating MAE would be: $$ RMSE_t = \sqrt{\frac{t-1}{t}RMSE_{t-1}^2 + \frac{(y_t^{true}-y_t^{pred})^2}{t}} $$ Here I assume you have ha …

The warning messages you received are informative here: 2: Some predictor variables are on very different scales: consider rescaling So it seems like a good approach is to rescale some variabl …

To see this, consider a linear regression model predicting the petal width of an iris (in centimeters) given its petal length (in centimeters): summary(lm(Petal.Width~Petal.Length, data=iris)) # Call … model predicting Z with X has R^2 value 0.2065, while the linear regression model predicting Z with Y has R^2 value 0.0511): summary(lm(Z~X, data=dat)) # Call: # lm(formula = Z ~ X, data = dat) # # …

Consider two regression models. …