knitr::opts_chunk$set(warning = FALSE, message = FALSE)
library(tidyverse)
library(ISLR)(1) For each of parts (a) through (d), indicate whether we would generally expect the performance of a flexible statistical learning method to be better or worse than an inflexible method. Justify your answer.
a. The sample size n is extremely large, and the number of predictors p is small.
A: A more flexible method should have better performance, because of the large sample size.
b. The number of predictors p is extremely large, and the number of observations n is small.
A: A more flexible method should have worse performance, because the small sample size n would lead to a high variance and overfitting.
c. The relationship between the predictors and response is highly non-linear.
A: A more flexible method should have better performance, because of the reduced bias.
d. The variance of the error terms, i.e. (\sigma^2 = Var(\epsilon)), is extremely high.
A: Worse performance, because a more flexible method would be more likely to overfit to the errors in the training data.
(2) Explain whether each scenario is a classification or regression problem, and indicate whether we are most interested in inference or prediction. Finally, provide n and p.
a. We collect a set of data on the top 500 firms in the US. For each firm we record profit, number of employees, industry and the CEO salary. We are interested in understanding which factors affect CEO salary.
A: Regression. Inference. n = 500 (firms in the US). p = 3 (profit, number of employees, and industry salary).
b. We are considering launching a new product and wish to know whether it will be a success or a failure. We collect data on 20 similar products that were previously launched. For each product we have recorded whether it was a success or failure, price charged for the product, marketing budget, competition price, and ten other variables.
A: Classification. Prediction. n = 20 (other products). p = 13 (price charged, marketing budget, competition price, and the 10 other variables).
c. We are interesting in predicting the % change in the US dollar in relation to the weekly changes in the world stock markets. Hence we collect weekly data for all of 2012. For each week we record the % change in the dollar, the % change in the US market, the % change in the British market, and the % change in the German market.
A: Regression. Prediction. n = 52 (weeks of 2012). p = 3 (% changes in US, UK and German markets).
(3) We now revisit the bias-variance decomposition.
a. Provide a sketch of typical (squared) bias, variance, training error, test error, and Bayes (or irreducible) error curves, on a single plot, as we go from less flexible statistical learning methods towards more flexible approaches. The x-axis should represent the amount of flexibility in the method, and the y-axis should represent the values for each curve. There should be five curves. Make sure to label each one.
b. Explain why each of the five curves has the shape displayed in part (a).
A: Bias goes down when flexibility increases, at the same time that variance goes up. The Bayes error is a constant, the minimum Test MSE we can theoretically achieve. Test MSE is the sum of the Bayes error, the bias and the variance, so this curve is always above the others, and goes downwards when the reductions in the bias (due to higher flexibility) are big enough to compensate the increases in variance, and viceversa. Train MSE can go down below the Bayes error and can even reach zero because, one way or another, all statistical learning methods try to minimize the Train MSE, and as we give them more degrees of freedom, they inevitably move towards that goal.
(4) You will now think of some real-life applications for statistical learning.
a. Describe three real-life applications in which classification might be useful. Describe the response, as well as the predictors. Is the goal of each application inference or prediction? Explain your answer.
A: 1. A bank may want to predict which customers are more likely to churn (close their accounts). The response variable would be if the customer has churned or not, and some posible predictors would be the amount of money spent by the customer, the number of transactions, their recency, among others. This could be either a prediction or a inference problem, since the bank may be interested not only in knowing which customers are likely to churn, but also why (so they can prevent churn).
-
There is a soccer match between Team A and Team B, and we want to know which of the 2 teams will win or if both will tie. The response variable is the match result (Team A wins, Team B wins, or it’s a tie), and some predictors we would like to use are: possession of the ball in previous matches, number of championships won, which team is local or visitor, and so on. The goal of this application is prediction.
-
A one-stop-shop retailer is sending marketing e-mails to their customers, and they would like to send each e-mail only to the customers who will be interested in the respective products promoted. The response variable could be if the customer reads the e-mail or not, and some posible predictors are if the customer has read previous e-mails about the product categories included in the next e-mail, if the customer has bought products of those categories, age, amount of money spent in the retailer, etc.
b. Describe three real-life applications in which regression might be useful. Describe the response, as well as the predictors. Is the goal of each application inference or prediction? Explain your answer.
ANSWER:
-
A retailer wants to know how the demand for products will change as they change the price. The goal of this application is inference, and the response variable are the sales of each product (measured in money or units). Some predictors could be the proportion of high income customers for each product, the weight of each product in the customers’ monthly expenditure, the price of the product in competitor retailers, and, of course, the change of the price of the product in the same retailer.
-
An e-commerce website wants to know how much traffic will come to their page this christmas, so they can provisionate the appropriate web server capacity, avoiding slow downs for their customers. The goal of this application is purely prediction. The response variable would be the number of page views in the chistmas season, and some predictors could be the amount of page views in the previous 3 months, the amount of page views in last year christmas, the economic growth year over year, and the confidence of customers in the future of the economy.
-
A researcher would like to know which factors affect the score obtained by students in a standarized math test. This is a inference application, since the researcher would like to give recommendations to improve the scores in the future, not to accurately predict such scores. The response variable is the score of each student in the test, and some explanatory variables could be: study hours per week for each students, years of experience of the teacher, percent of asistance to classes, etc.
c. Describe three real-life applications in which cluster analysis might be useful.
ANSWER:
-
Segmenting customers based on purchase behavior.
-
Segmenting app users based on how the use the app.
-
Segmenting twitter users based on the topics that they talk about.
(5) What are the advantages and disadvantages of a very flexible (versus a less flexible) approach for regression or classification? Under what circumstances might a more flexible approach be preferred to a less flexible approach? When might a less flexible approach be preferred?
a. Disadvantages
ANSWER: A very flexible model could be more difficult or costly to compute.
Also it requieres a lot of data (big samples) to be reliable, since it has higher variance (higher risk of overfitting).
It’s less interpretable, we may get an accurate prediction, but probably we will not know why the model is doing such prediction for each observation.
b. Advantages
ANSWER: If the “true model” which generated the data is complex, a more flexible model will make better predictions (will have lower Test MSE).
Generally, a more flexible approach is prefered if we have a large n,
and we suspect that the relationship between the explanatory variables
is complex (non-linear). If these conditions doesn’t hold, we may want a
less flexible approach.
(6) Describe the differences between a parametric and a non-parametric statistical learning approach. What are the advantages of a parametric approach to regression or classification (as opposed to a non-parametric approach)? What are its disadvantages?
ANSWER: A parametric approach assumes some functional form to estimate the relationship between the predictors and the response variable. Because of that, the problem is simplified to just estimate the parameters of the assumed functional form. A non-parametric approach allows more flexibility, since it doesn’t imposses a pre-defined functional form, but it alsow requieres more data to obtain a reliable estimation. The advantages and disadvantages of non-parametric approaches are pretty much the same as those of the more flexible models.
(7) The table below provides a training data set containing six observations, three predictors, and one qualitative response variable.
| Obs | X1 | X2 | X3 | Y |
|---|---|---|---|---|
| 1 | 0 | 3 | 0 | Red |
| 2 | 2 | 0 | 0 | Red |
| 3 | 0 | 1 | 3 | Red |
| 4 | 0 | 1 | 2 | Green |
| 5 | -1 | 0 | 1 | Green |
| 6 | 1 | 1 | 1 | Red |
Suppose we wish to use this data set to make a prediction for (Y) when (X1 = X2 = X3 = 0) using K-nearest neighbors.
a. Compute the Euclidean distance between each observation and the test point, (X1 = X2 = X3 = 0).
distance_to_p1 <- function(x1, x2, x3) {
sqrt(x1^2 + x2^2 + x3^2)
}
df1 %>%
mutate(distance = distance_to_p1(X1, X2, X3))## # A tibble: 6 x 6
## Obs X1 X2 X3 Y distance
## <int> <dbl> <dbl> <dbl> <chr> <dbl>
## 1 1 0 3 0 Red 3
## 2 2 2 0 0 Red 2
## 3 3 0 1 3 Red 3.16
## 4 4 0 1 2 Green 2.24
## 5 5 -1 0 1 Green 1.41
## 6 6 1 1 1 Red 1.73
b. What is our prediction with (K = 1)? Why?
ANSWER: Green, since observation n°5 is Green and is their closest neighbor.
c. What is our prediction with (K = 3)? Why?
ANSWER: Red, because 2 of their 3 closest neighbors are red.
d. If the Bayes decision boundary in this problem is highly nonlinear, then would we expect the best value for (K) to be large or small? Why?
ANSWER: Small. If the Bayes decision boundary is highly nonlinear, then more flexibility in our estimation is desired, and low values for K provide more flexibility.
(8) This exercise relates to the College data set, which can be found
in the file College.csv. It contains a number of variables for 777
different universities and colleges in the US. Descriptions of the
variables can be found on page 54 of the book.
a. Read the data into R.
(college <- read_csv("College.csv"))## # A tibble: 777 x 19
## X1 Private Apps Accept Enroll Top10perc Top25perc F.Undergrad
## <chr> <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 Abil~ Yes 1660 1232 721 23 52 2885
## 2 Adel~ Yes 2186 1924 512 16 29 2683
## 3 Adri~ Yes 1428 1097 336 22 50 1036
## 4 Agne~ Yes 417 349 137 60 89 510
## 5 Alas~ Yes 193 146 55 16 44 249
## 6 Albe~ Yes 587 479 158 38 62 678
## 7 Albe~ Yes 353 340 103 17 45 416
## 8 Albi~ Yes 1899 1720 489 37 68 1594
## 9 Albr~ Yes 1038 839 227 30 63 973
## 10 Alde~ Yes 582 498 172 21 44 799
## # ... with 767 more rows, and 11 more variables: P.Undergrad <dbl>,
## # Outstate <dbl>, Room.Board <dbl>, Books <dbl>, Personal <dbl>,
## # PhD <dbl>, Terminal <dbl>, S.F.Ratio <dbl>, perc.alumni <dbl>,
## # Expend <dbl>, Grad.Rate <dbl>
Now we’re converting the first column (X1, which contain college
names) to row names, since we don’t really want to use it as data.
college <- remove_rownames(college) %>%
column_to_rownames(var = "X1")Use the summary() function to produce a numerical summary of the
variables in the data set.
summary(college)## Private Apps Accept Enroll
## Length:777 Min. : 81 Min. : 72 Min. : 35
## Class :character 1st Qu.: 776 1st Qu.: 604 1st Qu.: 242
## Mode :character Median : 1558 Median : 1110 Median : 434
## Mean : 3002 Mean : 2019 Mean : 780
## 3rd Qu.: 3624 3rd Qu.: 2424 3rd Qu.: 902
## Max. :48094 Max. :26330 Max. :6392
## Top10perc Top25perc F.Undergrad P.Undergrad
## Min. : 1.00 Min. : 9.0 Min. : 139 Min. : 1.0
## 1st Qu.:15.00 1st Qu.: 41.0 1st Qu.: 992 1st Qu.: 95.0
## Median :23.00 Median : 54.0 Median : 1707 Median : 353.0
## Mean :27.56 Mean : 55.8 Mean : 3700 Mean : 855.3
## 3rd Qu.:35.00 3rd Qu.: 69.0 3rd Qu.: 4005 3rd Qu.: 967.0
## Max. :96.00 Max. :100.0 Max. :31643 Max. :21836.0
## Outstate Room.Board Books Personal
## Min. : 2340 Min. :1780 Min. : 96.0 Min. : 250
## 1st Qu.: 7320 1st Qu.:3597 1st Qu.: 470.0 1st Qu.: 850
## Median : 9990 Median :4200 Median : 500.0 Median :1200
## Mean :10441 Mean :4358 Mean : 549.4 Mean :1341
## 3rd Qu.:12925 3rd Qu.:5050 3rd Qu.: 600.0 3rd Qu.:1700
## Max. :21700 Max. :8124 Max. :2340.0 Max. :6800
## PhD Terminal S.F.Ratio perc.alumni
## Min. : 8.00 Min. : 24.0 Min. : 2.50 Min. : 0.00
## 1st Qu.: 62.00 1st Qu.: 71.0 1st Qu.:11.50 1st Qu.:13.00
## Median : 75.00 Median : 82.0 Median :13.60 Median :21.00
## Mean : 72.66 Mean : 79.7 Mean :14.09 Mean :22.74
## 3rd Qu.: 85.00 3rd Qu.: 92.0 3rd Qu.:16.50 3rd Qu.:31.00
## Max. :103.00 Max. :100.0 Max. :39.80 Max. :64.00
## Expend Grad.Rate
## Min. : 3186 Min. : 10.00
## 1st Qu.: 6751 1st Qu.: 53.00
## Median : 8377 Median : 65.00
## Mean : 9660 Mean : 65.46
## 3rd Qu.:10830 3rd Qu.: 78.00
## Max. :56233 Max. :118.00
Use the pairs() function to produce a scatterplot matrix of the first
ten columns or variables of the data.
college %>%
select(2:11) %>%
pairs()
Produce a side-by-side boxplot of Outstate versus Private.
ggplot(college, aes(Private, Outstate)) +
geom_boxplot()
Create a new qualitative variable, called Elite, by binning the
Top10perc variable. We are going to divide universities into two
groups based on whether or not the proportion of students coming from
the top 10% of their high school classes exceeds 50%.
college <- college %>%
mutate(Elite = if_else(Top10perc > 50,
true = "Yes",
false = "No"),
Elite = as.factor(Elite))Use the summary() function to see how many elite universities there
are.
summary(college$Elite)## No Yes
## 699 78
Now produce side-by-side boxplots of Outstate versus Elite.
ggplot(college, aes(Elite, Outstate)) +
geom_boxplot()
Produce some histograms with differing numbers of bins for a few of the quantitative variables.
par(mfrow = c(2, 2))
hist(college$Apps, breaks = 30)
hist(college$Top10perc, breaks = 20)
hist(college$Expend, breaks = 30)
hist(college$PhD, breaks = 40)
(9) This exercise involves the Auto data set studied in the lab. Make
sure that the missing values have been removed from the data.
Auto <- as_tibble(ISLR::Auto)
is.na(Auto) %>% mean()## [1] 0
a. Which of the predictors are quantitative, and which are qualitative?
str(Auto)## Classes 'tbl_df', 'tbl' and 'data.frame': 392 obs. of 9 variables:
## $ mpg : num 18 15 18 16 17 15 14 14 14 15 ...
## $ cylinders : num 8 8 8 8 8 8 8 8 8 8 ...
## $ displacement: num 307 350 318 304 302 429 454 440 455 390 ...
## $ horsepower : num 130 165 150 150 140 198 220 215 225 190 ...
## $ weight : num 3504 3693 3436 3433 3449 ...
## $ acceleration: num 12 11.5 11 12 10.5 10 9 8.5 10 8.5 ...
## $ year : num 70 70 70 70 70 70 70 70 70 70 ...
## $ origin : num 1 1 1 1 1 1 1 1 1 1 ...
## $ name : Factor w/ 304 levels "amc ambassador brougham",..: 49 36 231 14 161 141 54 223 241 2 ...
All of them are quantitative, except the name.
b. What is the range of each quantitative predictor?
Auto %>%
select_if(is.numeric) %>%
map(range)## $mpg
## [1] 9.0 46.6
##
## $cylinders
## [1] 3 8
##
## $displacement
## [1] 68 455
##
## $horsepower
## [1] 46 230
##
## $weight
## [1] 1613 5140
##
## $acceleration
## [1] 8.0 24.8
##
## $year
## [1] 70 82
##
## $origin
## [1] 1 3
c. What is the mean and standard deviation of each quantitative predictor?
Auto %>%
select_if(is.numeric) %>%
map(mean)## $mpg
## [1] 23.44592
##
## $cylinders
## [1] 5.471939
##
## $displacement
## [1] 194.412
##
## $horsepower
## [1] 104.4694
##
## $weight
## [1] 2977.584
##
## $acceleration
## [1] 15.54133
##
## $year
## [1] 75.97959
##
## $origin
## [1] 1.576531
Auto %>%
select_if(is.numeric) %>%
map(sd)## $mpg
## [1] 7.805007
##
## $cylinders
## [1] 1.705783
##
## $displacement
## [1] 104.644
##
## $horsepower
## [1] 38.49116
##
## $weight
## [1] 849.4026
##
## $acceleration
## [1] 2.758864
##
## $year
## [1] 3.683737
##
## $origin
## [1] 0.8055182
Alternative using list-columns to find mean and sd:
(numeric_variables <- ISLR::Auto %>%
select_if(is.numeric) %>%
as.list() %>%
enframe())## # A tibble: 8 x 2
## name value
## <chr> <list>
## 1 mpg <dbl [392]>
## 2 cylinders <dbl [392]>
## 3 displacement <dbl [392]>
## 4 horsepower <dbl [392]>
## 5 weight <dbl [392]>
## 6 acceleration <dbl [392]>
## 7 year <dbl [392]>
## 8 origin <dbl [392]>
(numeric_variables <-
numeric_variables %>%
mutate(mean = map_dbl(value, mean),
sd = map_dbl(value, sd)))## # A tibble: 8 x 4
## name value mean sd
## <chr> <list> <dbl> <dbl>
## 1 mpg <dbl [392]> 23.4 7.81
## 2 cylinders <dbl [392]> 5.47 1.71
## 3 displacement <dbl [392]> 194. 105.
## 4 horsepower <dbl [392]> 104. 38.5
## 5 weight <dbl [392]> 2978. 849.
## 6 acceleration <dbl [392]> 15.5 2.76
## 7 year <dbl [392]> 76.0 3.68
## 8 origin <dbl [392]> 1.58 0.806
d. Now remove the 10th through 85th observations. What is the range, mean, and standard deviation of each predictor in the subset of the data that remains?
summary_mean_sd_range <- function(x) {
x %>%
select_if(is.numeric) %>%
as.list() %>%
enframe() %>%
mutate(mean = map_dbl(value, mean),
sd = map_dbl(value, sd),
min = map_dbl(value, min),
max = map_dbl(value, max))
}
Auto_filtered <- Auto[-c(10:85), ]
summary_mean_sd_range(Auto_filtered)## # A tibble: 8 x 6
## name value mean sd min max
## <chr> <list> <dbl> <dbl> <dbl> <dbl>
## 1 mpg <dbl [316]> 24.4 7.87 11 46.6
## 2 cylinders <dbl [316]> 5.37 1.65 3 8
## 3 displacement <dbl [316]> 187. 99.7 68 455
## 4 horsepower <dbl [316]> 101. 35.7 46 230
## 5 weight <dbl [316]> 2936. 811. 1649 4997
## 6 acceleration <dbl [316]> 15.7 2.69 8.5 24.8
## 7 year <dbl [316]> 77.1 3.11 70 82
## 8 origin <dbl [316]> 1.60 0.820 1 3
e. Using the full data set, investigate the predictors graphically, using scatterplots or other tools of your choice. Create some plots highlighting the relationships among the predictors. Comment on your findings.
Auto %>%
select_if(is.numeric) %>%
pairs()
Average miles per gallon increases over time, especially since 1980.
ggplot(Auto, aes(as.factor(year), mpg)) +
geom_boxplot() 
At the same time, the average weight decreases.
ggplot(Auto, aes(as.factor(year), weight)) +
geom_boxplot() 
¿Maybe both variables are correlated?
ggplot(Auto, aes(mpg, weight, color = as.factor(cut_width(year, width = 4)))) +
geom_point() +
geom_smooth()
The greater efficiency in modern cars is partly due to the fact that they weigh less, but even keeping the weight constant, modern cars are more efficient (in miles per gallon) than older cars.
f. Suppose that we wish to predict gas mileage (mpg) on the basis of
the other variables. Do your plots suggest that any of the other
variables might be useful in predicting mpg? Justify your answer.
ANSWER: Yes. As seen above, year and weight should be useful
variables to predict mpg. Also, as seen in the plots produced by
pairs(), displacement, horsepower and acceleration should also
have good predictive power over mpg.
(10) This exercise involves the Boston data set, from the MASS
package.
a. Load the Boston data set.
Boston <- as_tibble(MASS::Boston)b. Make some pairwise scatterplots of the predictors (columns) in this data set. Describe your findings.
pairs(Boston)
Some relationships/correlations we can find in the data:
- Median value of homes decreases as pupil-teacher ratio increases (people value better access to education?).
- Higher value homes are more far away of employment centers than cheap ones.
- Median value of homes increases as the number of rooms increases.
- Median value of home decreases when there is more contamination (measured as nitrogen oxides concentration).
c. Are any of the predictors associated with per capita crime rate? If so, explain the relationship.
ANSWER: Yes. Median value of houses is negatively correlated with crime rates (maybe people are less willing to pay for homes in high crime sectors). Also, there is a positive correlation between crime rates and the proportion of old buildings, and a negative correltation with the distance to employment centers (areas located further away from employment centers have less crime).
d. Do any of the suburbs of Boston appear to have particularly high crime rates? Tax rates? Pupil-teacher ratios? Comment on the range of each predictor.
First we obtain the ranges (min and max) for each of these variables:
Boston %>%
summary_mean_sd_range() %>%
filter(name %in% c("crim", "tax", "ptratio"))## # A tibble: 3 x 6
## name value mean sd min max
## <chr> <list> <dbl> <dbl> <dbl> <dbl>
## 1 crim <dbl [506]> 3.61 8.60 0.00632 89.0
## 2 tax <dbl [506]> 408. 169. 187 711
## 3 ptratio <dbl [506]> 18.5 2.16 12.6 22
- Crime rates: there are towns with near-zero crime rates, but on the other end, some have a crime rate of almost 90.
- Property tax rates goes from $187 (per $10,000) up to $711 in the heaviest taxed zone.
- Pupil-teacher ratio: fluctuates between 12.6 and 22.
e. How many of the suburbs in this data set bound the Charles river?
Boston %>%
filter(chas == 1) %>%
tally()## # A tibble: 1 x 1
## n
## <int>
## 1 35
f. What is the median pupil-teacher ratio among the towns in this data set?
Boston %>%
summarise(median_ptratio = median(ptratio))## # A tibble: 1 x 1
## median_ptratio
## <dbl>
## 1 19.0
g. Which suburb of Boston has lowest median value of owner occupied homes? What are the values of the other predictors for that suburb, and how do those values compare to the overall ranges for those predictors? Comment on your findings.
These are the suburbs with lowest median value of houses.
Boston %>%
mutate(rank_lowvalue = dense_rank(medv)) %>%
filter(rank_lowvalue == 1)## # A tibble: 2 x 15
## crim zn indus chas nox rm age dis rad tax ptratio black
## <dbl> <dbl> <dbl> <int> <dbl> <dbl> <dbl> <dbl> <int> <dbl> <dbl> <dbl>
## 1 38.4 0 18.1 0 0.693 5.45 100 1.49 24 666 20.2 397.
## 2 67.9 0 18.1 0 0.693 5.68 100 1.43 24 666 20.2 385.
## # ... with 3 more variables: lstat <dbl>, medv <dbl>, rank_lowvalue <int>
And these are the ranges (min-max) and median of the variables for the whole dataset.
Boston %>%
summary_mean_sd_range() %>%
select(-sd)## # A tibble: 14 x 5
## name value mean min max
## <chr> <list> <dbl> <dbl> <dbl>
## 1 crim <dbl [506]> 3.61 0.00632 89.0
## 2 zn <dbl [506]> 11.4 0 100
## 3 indus <dbl [506]> 11.1 0.46 27.7
## 4 chas <int [506]> 0.0692 0 1
## 5 nox <dbl [506]> 0.555 0.385 0.871
## 6 rm <dbl [506]> 6.28 3.56 8.78
## 7 age <dbl [506]> 68.6 2.9 100
## 8 dis <dbl [506]> 3.80 1.13 12.1
## 9 rad <int [506]> 9.55 1 24
## 10 tax <dbl [506]> 408. 187 711
## 11 ptratio <dbl [506]> 18.5 12.6 22
## 12 black <dbl [506]> 357. 0.32 397.
## 13 lstat <dbl [506]> 12.7 1.73 38.0
## 14 medv <dbl [506]> 22.5 5 50
Remarkable facts about the suburbus with the lowest median value:
- They have 100% of units built prior to 1940.
- Relatively high crime rate.
- High proportion of non-retail business (industries).
- High contamination.
- Rooms per dwelling are below average.
- Very close to employment centers.
- High property tax rate.
- High pupil-teacher ratio (few teachers for many pupils).
- One of them is the zone with highest proportion of black population.
h. In this data set, how many of the suburbs average more than seven rooms per dwelling? More than eight rooms per dwelling? Comment on the suburbs that average more than eight rooms per dwelling.
Boston %>%
mutate(rm_ranges = cut(rm, breaks = 3:9)) %>%
count(rm_ranges)## # A tibble: 6 x 2
## rm_ranges n
## <fct> <int>
## 1 (3,4] 2
## 2 (4,5] 14
## 3 (5,6] 157
## 4 (6,7] 269
## 5 (7,8] 51
## 6 (8,9] 13
64 suburbs have 7 or more average rooms per dwelling, and 13 suburbus have 8 or more.
Boston %>%
filter(rm >= 8) %>%
summary_mean_sd_range()## # A tibble: 14 x 6
## name value mean sd min max
## <chr> <list> <dbl> <dbl> <dbl> <dbl>
## 1 crim <dbl [13]> 0.719 0.902 0.0201 3.47
## 2 zn <dbl [13]> 13.6 26.3 0 95
## 3 indus <dbl [13]> 7.08 5.39 2.68 19.6
## 4 chas <int [13]> 0.154 0.376 0 1
## 5 nox <dbl [13]> 0.539 0.0924 0.416 0.718
## 6 rm <dbl [13]> 8.35 0.251 8.03 8.78
## 7 age <dbl [13]> 71.5 24.6 8.4 93.9
## 8 dis <dbl [13]> 3.43 1.88 1.80 8.91
## 9 rad <int [13]> 7.46 5.33 2 24
## 10 tax <dbl [13]> 325. 111. 224 666
## 11 ptratio <dbl [13]> 16.4 2.41 13 20.2
## 12 black <dbl [13]> 385. 10.5 355. 397.
## 13 lstat <dbl [13]> 4.31 1.37 2.47 7.44
## 14 medv <dbl [13]> 44.2 8.09 21.9 50
Suburbs with 8 or more average rooms per dwelling have:
- Less crime than average.
- Less proportion of acres with non-retail business.
- Slightly less property tax rate than average.
- A median value two times higher than average.