Question

1. This question uses the variables dis (the weighted mean of distances to five Boston employment centers) and nox(nitrogen oxides concentration in parts per 10 million) from the Boston data (library MASS). We will treat dis as the predictor and nox as the response.  

Use the ns() function to fit a natural cubic spline to predict nox using dis. Perform a 5-fold cross-validation in order to select the best degrees of freedom upto 10 degrees.

• Plot degrees of freedom and its cross validation error.
• Plot the resulting fits of the best degrees of freedom.

Please use set.seed(1) on the first line of your R code.

Answer 1

For cross-validation or another approach in order to select the best degrees of freedom for a regression spline on this data.

folds <- sample(1:10, size = 506, replace = TRUE) errors <- matrix(NA, 10, 9) models <- list() for (k in 1:10) { for (i in 1:9) { models[[i]] <- lm(nox ~ bs(nox, df = i), data = Boston[folds != k,]) pred <- predict(models[[i]], Boston[folds == k,]) errors[k, i] <- sqrt(mean((Boston$nox[folds == k] - pred)^2)) } } errors <- apply(errors, 2, mean) data_frame(RMSE = errors) %>% mutate(df = row_number()) %>% ggplot(aes(df, RMSE, fill = df == which.min(errors))) + geom_col() + theme_tufte() + guides(fill = FALSE) + scale_x_continuous(breaks = 1:9) + coord_cartesian(ylim = range(errors))

To plot degrees of freedom :

errors <- list() models <- list() pred_df <- data_frame(V1 = 1:506) for (i in 1:9) { models[[i]] <- lm(nox ~ bs(dis, df = i), data = Boston) preds <- predict(models[[i]]) pred_df[[i]] <- preds errors[[i]] <- sqrt(mean((Boston$nox - preds)^2)) } names(pred_df) <- paste(1:9, 'Degrees of Freedom') data_frame(RMSE = unlist(errors)) %>% mutate(df = row_number()) %>% ggplot(aes(df, RMSE, fill = df == which.min(errors))) + geom_col() + guides(fill = FALSE) + theme_tufte() + scale_x_continuous(breaks = 1:9) + coord_cartesian(ylim = range(errors))