PSTAT 131 Homework 2 (Solution)

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Linear Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Linear Regression
The age of an abalone is typically determined by cutting the shell open and counting the number of rings with a microscope. The purpose of this data set is to determine whether abalone age (number of rings + 1.5) can be accurately predicted using other, easier-to-obtain information about the abalone.
The full abalone data set is located in the data subdirectory. Read it into R using read_csv(). Take a moment to read through the codebook (abalone_codebook.txt) and familiarize yourself with the variable definitions.
Make sure you load the tidyverse and tidymodels!
library(ggplot2) library(tidyverse) library(tidymodels) library(corrplot) library(ggthemes) tidymodels_prefer() tidymodels_packages()
## [1] “broom” “cli” “conflicted” “dials” “dplyr”
## [6] “ggplot2” “hardhat” “infer” “modeldata” “parsnip”
## [11] “purrr” “recipes” “rlang” “rsample” “rstudioapi”
## [16] “tibble” “tidyr” “tune” “workflows” “workflowsets”
## [21] “yardstick” “tidymodels”
abalone <- read.csv(“abalone.csv”) head(abalone)
## type longest_shell diameter height whole_weight shucked_weight viscera_weight
## 1 M 0.455 0.365 0.095 0.5140 0.2245 0.1010 ## 2 M 0.350 0.265 0.090 0.2255 0.0995 0.0485 ## 3 F 0.530 0.420 0.135 0.6770 0.2565 0.1415 ## 4 M 0.440 0.365 0.125 0.5160 0.2155 0.1140 ## 5 I 0.330 0.255 0.080 0.2050 0.0895 0.0395
## 6 I 0.425 0.300 0.095 0.3515 0.1410 0.0775
## shell_weight rings
## 1 0.150 15 ## 2 0.070 7 ## 3 0.210 9 ## 4 0.155 10 ## 5 0.055 7
## 6 0.120 8
Question 1
Your goal is to predict abalone age, which is calculated as the number of rings plus 1.5. Notice there currently is no age variable in the data set. Add age to the data set. Assess and describe the distribution of age.
abalone_new <- abalone %>%
mutate(age = rings + 1.5)
head(abalone_new)
## type longest_shell diameter height whole_weight shucked_weight viscera_weight
## 1 M 0.455 0.365 0.095 0.5140 0.2245 0.1010 ## 2 M 0.350 0.265 0.090 0.2255 0.0995 0.0485 ## 3 F 0.530 0.420 0.135 0.6770 0.2565 0.1415 ## 4 M 0.440 0.365 0.125 0.5160 0.2155 0.1140 ## 5 I 0.330 0.255 0.080 0.2050 0.0895 0.0395
## 6 I 0.425 0.300 0.095 0.3515 0.1410 0.0775
## shell_weight rings age
## 1 0.150 15 16.5 ## 2 0.070 7 8.5 ## 3 0.210 9 10.5 ## 4 0.155 10 11.5 ## 5 0.055 7 8.5
## 6 0.120 8 9.5
hist(abalone_new$age, xlab = “Age”,breaks =30, main = “Histogram of Abalone’s Age”, col = ‘orange’)
Histogram of Abalone’s Age

Age
As the result we got above, we can see that the age of abalone is not distributed evenly. The shape of the distribution is skewed to the left with a clear mode around age of 11, and most of abalone’s age are around 8 to 13 and there is an outlier at around 30.
Question 2
Split the abalone data into a training set and a testing set. Use stratified sampling. You should decide on appropriate percentages for splitting the data.
Remember that you’ll need to set a seed at the beginning of the document to reproduce your results.
set.seed(2000)
abalone_split <- initial_split(abalone_new, prop = 0.80, strata = age) abalone_train <- training(abalone_split) abalone_test <- testing(abalone_split) head(abalone_train)
## type longest_shell diameter height whole_weight shucked_weight
## 2 M 0.350 0.265 0.090 0.2255 0.0995 ## 5 I 0.330 0.255 0.080 0.2050 0.0895 ## 36 M 0.465 0.355 0.105 0.4795 0.2270 ## 38 F 0.450 0.355 0.105 0.5225 0.2370 ## 43 I 0.240 0.175 0.045 0.0700 0.0315 ## 45 I 0.210 0.150 0.050 0.0420 0.0175 ## viscera_weight shell_weight rings age
## 2 0.0485 0.070 7 8.5 ## 5 0.0395 0.055 7 8.5 ## 36 0.1240 0.125 8 9.5 ## 38 0.1165 0.145 8 9.5 ## 43 0.0235 0.020 5 6.5
## 45 0.0125 0.015 4 5.5
head(abalone_test)
## type longest_shell diameter height whole_weight shucked_weight
## 6 I 0.425 0.300 0.095 0.3515 0.1410
## 11 F 0.525 0.380 0.140 0.6065 0.1940
## 17 I 0.355 0.280 0.085 0.2905 0.0950
## 19 M 0.365 0.295 0.080 0.2555 0.0970
## 21 M 0.355 0.280 0.095 0.2455 0.0955
## 24 F 0.550 0.415 0.135 0.7635 0.3180
## viscera_weight shell_weight rings age
## 6 0.0775 0.120 8 9.5
## 11 0.1475 0.210 14 15.5
## 17 0.0395 0.115 7 8.5
## 19 0.0430 0.100 7 8.5
## 21 0.0620 0.075 11 12.5
## 24 0.2100 0.200 9 10.5
Question 3
Using the training data, create a recipe predicting the outcome variable, age, with all other predictor variables. Note that you should not include rings to predict age. Explain why you shouldn’t use rings to predict age.
Steps for your recipe:
1. dummy code any categorical predictors
2. create interactions between
• type and shucked_weight,
• longest_shell and diameter,
• shucked_weight and shell_weight 3. center all predictors, and
4. scale all predictors.
You’ll need to investigate the tidymodels documentation to find the appropriate step functions to use.
abalone_recipe <- abalone_train %>%
recipe(age ~ type +longest_shell + diameter + height + whole_weight + shucked_weight + viscera_weight + shell_weight) %>%
step_dummy(all_nominal_predictors()) %>%
step_interact(terms = ~ starts_with(“type”):shucked_weight+
longest_shell:diameter+ shucked_weight:shell_weight) %>%
step_center(all_predictors()) %>% step_scale(all_predictors())
Question 4
Create and store a linear regression object using the “lm” engine.
lm_model <- linear_reg() %>%
set_engine(“lm”)
Question 5 Now:
1. set up an empty workflow,
2. add the model you created in Question 4, and
3. add the recipe that you created in Question 3.
lm_wflow <- workflow() %>% add_model(lm_model) %>% add_recipe(abalone_recipe)
Question 6
Use your fit() object to predict the age of a hypothetical female abalone with longest_shell = 0.50, diameter = 0.10, height = 0.30, whole_weight = 4, shucked_weight = 1, viscera_weight = 2, shell_weight = 1.
#fitting the model lm_fit <- fit(lm_wflow, abalone_train)
#predicting
abalone_hypo <- data.frame(type=”F”,longest_shell = 0.50,
diameter = 0.10, height = 0.30, whole_weight = 4, shucked_weight = 1, viscera_weight = 2, shell_weight = 1)
predict(lm_fit,new_data = abalone_hypo)
## # A tibble: 1 x 1 ## .pred ## <dbl> ## 1 24.8
Question 7
Now you want to assess your model’s performance. To do this, use the yardstick package:
1. Create a metric set that includes R2, RMSE (root mean squared error), and MAE (mean absolute error).
abalone_metrics <- metric_set(rsq, rmse, mae)
2. Use predict() and bind_cols() to create a tibble of your model’s predicted values from the training data along with the actual observed ages (these are needed to assess your model’s performance).
abalone_train_res <- predict(lm_fit, new_data = abalone_train %>% select(-age))
abalone_train_res <- bind_cols(abalone_train_res, abalone_train %>% select(age))
abalone_train_res %>% head()
## # A tibble: 6 x 2
## .pred age ## <dbl> <dbl> ## 1 9.48 8.5 ## 2 8.10 8.5 ## 3 10.1 9.5 ## 4 11.0 9.5 ## 5 6.30 6.5 ## 6 5.94 5.5
3. Finally, apply your metric set to the tibble, report the results, and interpret the R2 value.
abalone_metrics(abalone_train_res, truth = age, estimate = .pred)
## # A tibble: 3 x 3
## .metric .estimator .estimate
## <chr> <chr> <dbl> ## 1 rsq standard 0.559
## 2 rmse standard 2.13
## 3 mae standard 1.53
The Rˆ2 represents how well the regression model fits the observed data. As the we getting rsq about 0.56, the model is not fitting the observed data well.
The root mean squared error is the standard deviation of the residuals (prediction errors), As we are predicting the age of abalone, about 2.13 of rmse is a bit to big as a prediction error.
mae: The mean absolute error indicates that the magnitude of difference between the prediction of an observation and the true value of that observation is about 1.53.

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