6 Data manipulation

Exam PA will have questions about the data. That is a fact. Every version of Exam PA such as December 2018, June 2019, and December 2019 have all had questions about data manipulation. Putting in extra practice in this area is garanteed to give you a better score because it will free up time that you can use elsewhere.

This lesson is based on the first task of the December 2019 Exam PA.

TASK 1 (12 points)

Examine each variable and make appropriate adjustments.

Examine each predictor variable other than cap_gain both on its own and with respect to value_flag. Make appropriate adjustments. Do not make any adjustments to the cap_gain variable at this time.

There should be no further variable adjustments unless specifically requested.

All data for this book can be accessed from the package ExamPAData. In the real exam, you will read the file from the Prometric computer.

Because the data is already loaded, simply use the below code to access the data.

library(ExamPAData)

To save keystrokes, give the data a short name such as df for “data frame”.

df <- customer_value %>% mutate_if(is.character, as.factor)

6.1 Garbage in; garbage out 🗑

This is a common saying when working with predictive models. No matter how complex the model, if the data is bad then the entire result will be bad. For this exam, master the art of data manipulation and everything becomes easier!

Begin by looking at a summary.

summary(df)

##       age        education_num                 marital_status     occupation   
##  Min.   :17.00   Min.   : 1.00   Divorced             : 6633   Group 1 : 7237  
##  1st Qu.:28.00   1st Qu.: 9.00   Married-AF-spouse    :   37   Group 2 :10123  
##  Median :37.00   Median :10.00   Married-civ-spouse   :22379   Group 3 :13971  
##  Mean   :38.64   Mean   :10.08   Married-spouse-absent:  628   Group 4 : 2444  
##  3rd Qu.:48.00   3rd Qu.:12.00   Never-married        :16117   Group 5 :12258  
##  Max.   :90.00   Max.   :16.00   Separated            : 1530   Group NA: 2809  
##                                  Widowed              : 1518                   
##     cap_gain     hours_per_week      score       value_flag  
##  Min.   :    0   Min.   : 1.00   Min.   :43.94   High:11687  
##  1st Qu.:    0   1st Qu.:40.00   1st Qu.:57.50   Low :37155  
##  Median :    0   Median :40.00   Median :60.24               
##  Mean   : 1079   Mean   :40.42   Mean   :60.23               
##  3rd Qu.:    0   3rd Qu.:45.00   3rd Qu.:62.95               
##  Max.   :99999   Max.   :99.00   Max.   :76.53               
##

str(df)

## Classes 'spec_tbl_df', 'tbl_df', 'tbl' and 'data.frame': 48842 obs. of  8 variables:
##  $ age           : num  39 50 38 53 28 37 49 52 31 42 ...
##  $ education_num : num  13 13 9 7 13 14 5 9 14 13 ...
##  $ marital_status: Factor w/ 7 levels "Divorced","Married-AF-spouse",..: 5 3 1 3 3 3 4 3 5 3 ...
##  $ occupation    : Factor w/ 6 levels "Group 1","Group 2",..: 2 5 1 1 5 5 1 5 5 5 ...
##  $ cap_gain      : num  2174 0 0 0 0 ...
##  $ hours_per_week: num  40 13 40 40 40 40 16 45 50 40 ...
##  $ score         : num  59 55.8 62.8 60.1 53.3 ...
##  $ value_flag    : Factor w/ 2 levels "High","Low": 2 2 2 2 2 2 2 1 1 1 ...

6.2 Be a detective 🔍

Sherlock Holmes is famous for saying “You see, Watson, but you do not observe!”

Just like detectives, actuaries need to collect data and make observations. Each exam has a few “anomalies” in the data which they expect candidates to mention. These could be

Any value that doesn’t match the Data Dictionary in the Project Statement
Variables that have NA values
Variables that have a lot of factor levels
Incorrect data types
- Factors that are read as characters or vice versa
- Numeric variables that are factor/character
Extreme values (Numeric values are too high or low)

You’ll get very good at spotting these with practice. Just from looking at the above summary, we can observe the following:

Observations

The data consists of 48,842 obs. of 8 variables.
The lowest age is 17 but the Project Statement says to only include records with age >= 25.
The age distribution is right skewed because the median (37) is less than the mean (38).
education_num takes integer values between 1 and 16. There are a lot of values that are low.
There are missing values when occupation is group NA, which means that the person’s occupation is unknown.
The amount that people work per work, hours_per_week, varies by a lot. The lowest is 1 hour and the highest is 99. Most people work 40 hours per week.

The SOA’s solution recommends leaving comments in your Rmd file. This helps to give you partial credit on questions that you may answer incorrectly.

Good comments

#I observe that there are no missing values other than those indicated by Group NA for occupation.

#I removed the code provided by my assistant. It is embedded in later chunks as needed.

#I excluded people under the age of 25
df <- df[df$age >= 25,]

#I convert the target to 0-1.
df$value_flag <- ifelse(df$value_flag == "High",1,0)

Useless comments

#this is a comment

#this loads a library called ggplot2
library(ggplot2)

6.3 A picture is worth a thousand words 📷

What’s your favorite type of graph? Mine is a radar chart This is a graphical method of displaying multivariate data in the form of a two-dimensional chart of three or more quantitative variables represented on axes staring from the same point.

Isn’t this pretty?

Don’t waste time trying to make the graphs perfect!

Only change code that you need to change. The SOA is trying to help you save time by giving you templates. That being said, if you can do it faster on your own, then do it your own way.

This is the code template that they give you. You only need to change the “variable” names.

# This code makes a histogram for a continuous variable.
ggplot(df, aes(x = variable)) + 
  geom_histogram(bins = 30) +
  labs(x = "variable") +
  theme(axis.text.x = element_text(angle = 90, hjust = 1))

# This code makes a bar chart for a factor variable.
ggplot(df, aes(x = variable)) +
  geom_bar() +
  labs(x = "variable") +
  theme(axis.text.x = element_text(angle = 90, hjust = 1))

Okay, that wasn’t 100% true. It’s common for the code to require small changes. For instance, the histogram that they give you has bins that are too narrow forage, which causes these spikes to show up in the graph.

The best candidates altered the code to create more appropriate plots and summary tables.

# This code makes a histogram for a continuous variable.
ggplot(df, aes(x = age)) + 
  geom_histogram(bins = 30) + #not the right number of bins
  labs(x = "age") +
  theme(axis.text.x = element_text(angle = 90, hjust = 1)) + 
  ggtitle("Bad: Default histogram with spikes")

#Histograms of continuous variables
ggplot(df, aes(x = age)) +
  geom_histogram(breaks = seq(24.5,99.5, by = 5)) +  #make bins wider and set range 
  labs(x = "Age") +
  theme(axis.text.x = element_text(angle = 90, hjust = 1)) + 
  ggtitle("Good: After adjusting the breaks")

How do you know when a variable should a factor and when it should be numeric?

6.4 Factor or numeric ❓

Which variables should be converted to factors and which should be numeric?

Questions of this sort have come up twice. On Hospital Readmissions, there was a Length of Stay variable that was numeric, but had only a few values and so some candidates treated it as a factor. The education_num variable here is also numeric but only has 16 unique values. So should this be a numeric or a factor?

table(df$education_num)

## 
##     1     2     3     4     5     6     7     8     9    10    11    12    13 
##    68   231   446   877   618   995  1042   382 13320  7812  1862  1411  7298 
##    14    15    16 
##  2621   834   593

Ask yourself this question: is there a way of comparing two values of the variable together?

If yes, then use numeric
If no, then use a factor

For exmaple, we can say that education_num = 2 is less than education_num = 4, which means that there’s a natural order. This is also known as an ordinal.

If the factor is say, color, which can be red, blue, or green, then there is no way of comparing values together. Is red greater than blue? This question has no meaning.

ggplot(df, aes(x = education_num)) +
  geom_histogram(bins = 30) +
  labs(x = "Education") +
  theme(axis.text.x = element_text(angle = 90, hjust = 1)) + 
  ggtitle("Default number of bins (30)")

ggplot(df, aes(x = education_num)) +
  geom_histogram(bins = 16) +
  labs(x = "Education") +
  theme(axis.text.x = element_text(angle = 90, hjust = 1)) + 
  ggtitle("Set number of bins to number of factor levels (16) ")

# Set bins equal to number of levels, could have made bar chart.

We could also use a bar plot.

ggplot(df, aes(x = education_num)) + 
  geom_bar(stat = "count") + 
  ggtitle("Bar plot")

Lastly, read the Project Statement carefully and only do what it tells you to do.

Candidates were not required to make a plot for cap_gain. This solution has the plot made here rather than in Task 6.

6.5 73.6% of statistics are false 😲

Really? No, but statistics can help you see patterns that data visualization by itself can miss. Along with the ggplot codes, there will be code to look at summary statistics. Here’s a refresher on what these statistics mean (no pun intended).

Mean: The average. This gets skewed by outliers easily. If the mean is greater than the median, then the distribution is right skewed.
Median: The “middle” value. This is an average that reduces the impact of outliers.
Variance: The amount by which each observation differs from the mean.
Standard Deviation: The square root of the variance.
n(): The number of observations. Always take note of groups that don’t have many observations.

# This code provides, for each level of a factor variable, the number for which value_flag is zero, the number for which it is one, the total number of observations, and the proportion of ones.
# Note that the variable name should not be enclosed in quotation marks.
df %>%
  group_by(variable) %>%
  dplyr::summarise(
    zeros = sum(value_flag == 0),
    ones = sum(value_flag == 1),
    n = n(),
    proportion = mean(value_flag)
  )

Factors levels should be simplified. If a group has only a few observations then there will be problems with the model. In our data, take a look at the marital_status column. Do you observe anything unusual?

table(df$marital_status)

## 
##              Divorced     Married-AF-spouse    Married-civ-spouse 
##                  6498                    31                 21661 
## Married-spouse-absent         Never-married             Separated 
##                   573                  8697                  1438 
##               Widowed 
##                  1512

Only 31 records have Married-AF-spouse. This is because the sample size n = 31 is too small. In modeling jargon, this is “statistical insignificant” and will cause the p-value on marital_status to be large. You can fix this in a few different ways

Delete these records (Not recommended)
Group these records together with Married-spouse (Simplest method)

Let’s use the second method.

First, look at the levels of the factor variable.

levels(df$marital_status)

## [1] "Divorced"              "Married-AF-spouse"     "Married-civ-spouse"   
## [4] "Married-spouse-absent" "Never-married"         "Separated"            
## [7] "Widowed"

Now look at the profitability across marital status. For Married-AF-spouse and Married-civ-spouse the proportion of high profit customers is high, but for Married-spouse-absent it is low. Even though these are all “married”, it would be a bad idea to combine them because the profitability is so different.

#Proportion of ones by category of factor variable
df %>%
  group_by(marital_status) %>%
  dplyr::summarise(
    zeros = sum(value_flag == 0),
    ones = sum(value_flag == 1),
    n = n(),
    proportion = mean(value_flag)
  )

## # A tibble: 7 x 5
##   marital_status        zeros  ones     n proportion
##   <fct>                 <int> <int> <int>      <dbl>
## 1 Divorced               5829   669  6498     0.103 
## 2 Married-AF-spouse        19    12    31     0.387 
## 3 Married-civ-spouse    11727  9934 21661     0.459 
## 4 Married-spouse-absent   515    58   573     0.101 
## 5 Never-married          8000   697  8697     0.0801
## 6 Separated              1342    96  1438     0.0668
## 7 Widowed                1384   128  1512     0.0847

Then create a vector that has the simpler levels that you want. The order needs to be the same.

simple_levels <- c("Divorced", "Married-spouse", "Married-spouse", "Married-spuouse-absent", "Neber-married", "Separated", "Widowed")

The SOA likes to use the function mapvalues. This takes in three arguments. You can read about this by typing ?mapvalues into the console.

x : the factor or vector to modify

from : a vector of the items to replace

to : a vector of replacement values

Then map the old values to the simpler values.

#Combine the two marital status levels
var.levels <- levels(df$marital_status)
df$marital_status <- mapvalues(x = df$marital_status,
                               from = var.levels, 
                               to = simple_levels)

Now, when you look at the marital_status levels, you will see the simpler levels.

levels(df$marital_status)

## [1] "Divorced"               "Married-spouse"         "Married-spuouse-absent"
## [4] "Neber-married"          "Separated"              "Widowed"

You can also check that the number of records is what you expect.

table(df$marital_status)

## 
##               Divorced         Married-spouse Married-spuouse-absent 
##                   6498                  21692                    573 
##          Neber-married              Separated                Widowed 
##                   8697                   1438                   1512

There’s an automatic way of doing this as well. And everyone likes to automate things because it gives us more free time 😪.

Create dummy variables for each category and then use a variable selection method to decide which levels to keep. This could be from a LASSO, a GLM with Step AIC/BIC, or any tree-based method.

6.6 How to save time with dplyr

You may have noticed that writing code for data manipulation can be slow. Fortunately, there is a faster, 100%-legal, way of doing data manipulation that has worked for hundreds of exam candidates (the author included) who have taken Exam PA.

Up to this point we have been using old R libraries. By making use of newer R libraries we can save ourselves time. These will all be provided for you at Prometric within the tidyverse library.

Suggested reading of R for Data Science (https://r4ds.had.co.nz/index.html):

Chapter	Topic
9	Introduction
10	Tibbles
12	Tidy data
15	Factors
17	Introduction
18	Pipes
19	Functions
20	Vectors

6.7 Look at the data

Let’s look at the health insurance data set. This contains information on patients along with their annual health care costs.

The descriptions of the columns are below.

age: Age of the individual
sex: Sex
bmi: Body Mass Index
children: Number of children
smoker: Is this person a smoker?
region: Region
charges: Annual health care costs.

head() shows the top n rows. head(20) shows the top 20 rows.

head(health_insurance)

## # A tibble: 6 x 7
##     age sex      bmi children smoker region    charges
##   <dbl> <chr>  <dbl>    <dbl> <chr>  <chr>       <dbl>
## 1    19 female  27.9        0 yes    southwest  16885.
## 2    18 male    33.8        1 no     southeast   1726.
## 3    28 male    33          3 no     southeast   4449.
## 4    33 male    22.7        0 no     northwest  21984.
## 5    32 male    28.9        0 no     northwest   3867.
## 6    31 female  25.7        0 no     southeast   3757.

Using a pipe is an alternative way of doing this.

health_insurance %>% head()

Shortcut: Use CTRL + SHFT + M to create pipes %>%

The glimpse function is similar to str().

health_insurance %>% glimpse()

## Observations: 1,338
## Variables: 7
## $ age      <dbl> 19, 18, 28, 33, 32, 31, 46, 37, 37, 60, 25, 62, 23, 56, 27...
## $ sex      <chr> "female", "male", "male", "male", "male", "female", "femal...
## $ bmi      <dbl> 27.900, 33.770, 33.000, 22.705, 28.880, 25.740, 33.440, 27...
## $ children <dbl> 0, 1, 3, 0, 0, 0, 1, 3, 2, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0...
## $ smoker   <chr> "yes", "no", "no", "no", "no", "no", "no", "no", "no", "no...
## $ region   <chr> "southwest", "southeast", "southeast", "northwest", "north...
## $ charges  <dbl> 16884.924, 1725.552, 4449.462, 21984.471, 3866.855, 3756.6...

One of the most useful data science tools is counting things. The function count() gives the number of records by a categorical feature.

health_insurance %>% count(children)

## # A tibble: 6 x 2
##   children     n
##      <dbl> <int>
## 1        0   574
## 2        1   324
## 3        2   240
## 4        3   157
## 5        4    25
## 6        5    18

Two categories can be counted at once. This creates a table with all combinations of region and sex and shows the number of records in each category.

health_insurance %>% count(region, sex)

## # A tibble: 8 x 3
##   region    sex        n
##   <chr>     <chr>  <int>
## 1 northeast female   161
## 2 northeast male     163
## 3 northwest female   164
## 4 northwest male     161
## 5 southeast female   175
## 6 southeast male     189
## 7 southwest female   162
## 8 southwest male     163

The summary() function is shows a statistical summary. One caveat is that each column needs to be in it’s appropriate type. For example, smoker, region, and sex are all listed as characters when if they were factors, summary would give you count info.

With incorrect data types

health_insurance %>% summary()

##       age            sex                 bmi           children    
##  Min.   :18.00   Length:1338        Min.   :15.96   Min.   :0.000  
##  1st Qu.:27.00   Class :character   1st Qu.:26.30   1st Qu.:0.000  
##  Median :39.00   Mode  :character   Median :30.40   Median :1.000  
##  Mean   :39.21                      Mean   :30.66   Mean   :1.095  
##  3rd Qu.:51.00                      3rd Qu.:34.69   3rd Qu.:2.000  
##  Max.   :64.00                      Max.   :53.13   Max.   :5.000  
##     smoker             region             charges     
##  Length:1338        Length:1338        Min.   : 1122  
##  Class :character   Class :character   1st Qu.: 4740  
##  Mode  :character   Mode  :character   Median : 9382  
##                                        Mean   :13270  
##                                        3rd Qu.:16640  
##                                        Max.   :63770

With correct data types

This tells you that there are 324 patients in the northeast, 325 in the northwest, 364 in the southeast, and so fourth.

health_insurance <- health_insurance %>% 
  modify_if(is.character, as.factor)

health_insurance %>% 
  summary()

##       age            sex           bmi           children     smoker    
##  Min.   :18.00   female:662   Min.   :15.96   Min.   :0.000   no :1064  
##  1st Qu.:27.00   male  :676   1st Qu.:26.30   1st Qu.:0.000   yes: 274  
##  Median :39.00                Median :30.40   Median :1.000             
##  Mean   :39.21                Mean   :30.66   Mean   :1.095             
##  3rd Qu.:51.00                3rd Qu.:34.69   3rd Qu.:2.000             
##  Max.   :64.00                Max.   :53.13   Max.   :5.000             
##        region       charges     
##  northeast:324   Min.   : 1122  
##  northwest:325   1st Qu.: 4740  
##  southeast:364   Median : 9382  
##  southwest:325   Mean   :13270  
##                  3rd Qu.:16640  
##                  Max.   :63770

6.8 Transform the data

Transforming, manipulating, querying, and wrangling are synonyms in data terminology.

R syntax is designed to be similar to SQL. They begin with a SELECT, use GROUP BY to aggregate, and have a WHERE to remove records. Unlike SQL, the ordering of these does not matter. SELECT can come after a WHERE.

R to SQL translation

select() -> SELECT
mutate() -> user-defined columns
summarize() -> aggregated columns
left_join() -> LEFT JOIN
filter() -> WHERE
group_by() -> GROUP BY
filter() -> HAVING
arrange() -> ORDER BY

health_insurance %>% 
  select(age, region) %>% 
  head()

## # A tibble: 6 x 2
##     age region   
##   <dbl> <fct>    
## 1    19 southwest
## 2    18 southeast
## 3    28 southeast
## 4    33 northwest
## 5    32 northwest
## 6    31 southeast

Tip: use CTRL + SHIFT + M to create pipes %>%.

Let’s look at only those in the southeast region. Instead of WHERE, use filter.

health_insurance %>% 
  filter(region == "southeast") %>% 
  select(age, region) %>% 
  head()

## # A tibble: 6 x 2
##     age region   
##   <dbl> <fct>    
## 1    18 southeast
## 2    28 southeast
## 3    31 southeast
## 4    46 southeast
## 5    62 southeast
## 6    56 southeast

The SQL translation is

SELECT age, region
FROM health_insurance
WHERE region = 'southeast'

Instead of ORDER BY, use arrange. Unlike SQL, the order does not matter and ORDER BY doesn’t need to be last.

health_insurance %>% 
  arrange(age) %>% 
  select(age, region) %>% 
  head()

## # A tibble: 6 x 2
##     age region   
##   <dbl> <fct>    
## 1    18 southeast
## 2    18 southeast
## 3    18 northeast
## 4    18 northeast
## 5    18 northeast
## 6    18 southeast

The group_by comes before the aggregation, unlike in SQL where the GROUP BY comes last.

health_insurance %>% 
  group_by(region) %>% 
  summarise(avg_age = mean(age))

## # A tibble: 4 x 2
##   region    avg_age
##   <fct>       <dbl>
## 1 northeast    39.3
## 2 northwest    39.2
## 3 southeast    38.9
## 4 southwest    39.5

In SQL, this would be

SELECT region, 
       AVG(age) as avg_age
FROM health_insurance
GROUP BY region

Just like in SQL, many different aggregate functions can be used such as SUM, MEAN, MIN, MAX, and so forth.

health_insurance %>% 
  group_by(region) %>% 
  summarise(avg_age = mean(age),
            max_age = max(age),
            median_charges = median(charges),
            bmi_std_dev = sd(bmi))

## # A tibble: 4 x 5
##   region    avg_age max_age median_charges bmi_std_dev
##   <fct>       <dbl>   <dbl>          <dbl>       <dbl>
## 1 northeast    39.3      64         10058.        5.94
## 2 northwest    39.2      64          8966.        5.14
## 3 southeast    38.9      64          9294.        6.48
## 4 southwest    39.5      64          8799.        5.69

To create new columns, the mutate function is used. For example, if we wanted a column of the person’s annual charges divided by their age

health_insurance %>% 
  mutate(charges_over_age = charges/age) %>% 
  select(age, charges, charges_over_age) %>% 
  head(5)

## # A tibble: 5 x 3
##     age charges charges_over_age
##   <dbl>   <dbl>            <dbl>
## 1    19  16885.            889. 
## 2    18   1726.             95.9
## 3    28   4449.            159. 
## 4    33  21984.            666. 
## 5    32   3867.            121.

We can create as many new columns as we want.

health_insurance %>% 
  mutate(age_squared  = age^2,
         age_cubed = age^3,
         age_fourth = age^4) %>% 
  head(5)

## # A tibble: 5 x 10
##     age sex     bmi children smoker region charges age_squared age_cubed
##   <dbl> <fct> <dbl>    <dbl> <fct>  <fct>    <dbl>       <dbl>     <dbl>
## 1    19 fema~  27.9        0 yes    south~  16885.         361      6859
## 2    18 male   33.8        1 no     south~   1726.         324      5832
## 3    28 male   33          3 no     south~   4449.         784     21952
## 4    33 male   22.7        0 no     north~  21984.        1089     35937
## 5    32 male   28.9        0 no     north~   3867.        1024     32768
## # ... with 1 more variable: age_fourth <dbl>

The CASE WHEN function is quite similar to SQL. For example, we can create a column which is 0 when age < 50, 1 when 50 <= age <= 70, and 2 when age > 70.

health_insurance %>% 
  mutate(age_bucket = case_when(age < 50 ~ 0,
                                age <= 70 ~ 1,
                                age > 70 ~ 2)) %>% 
  select(age, age_bucket)

## # A tibble: 1,338 x 2
##      age age_bucket
##    <dbl>      <dbl>
##  1    19          0
##  2    18          0
##  3    28          0
##  4    33          0
##  5    32          0
##  6    31          0
##  7    46          0
##  8    37          0
##  9    37          0
## 10    60          1
## # ... with 1,328 more rows

SQL translation:

SELECT CASE WHEN AGE < 50 THEN 0
       ELSE WHEN AGE <= 70 THEN 1
       ELSE 2
FROM health_insurance

6.9 Exercises

Run this code on your computer to answer these exercises.

The data actuary_salaries contains the salaries of actuaries collected from the DW Simpson survey. Use this data to answer the exercises below.

actuary_salaries %>% glimpse()

## Observations: 138
## Variables: 6
## $ industry    <chr> "Casualty", "Casualty", "Casualty", "Casualty", "Casual...
## $ exams       <chr> "1 Exam", "2 Exams", "3 Exams", "4 Exams", "1 Exam", "2...
## $ experience  <dbl> 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 5, 5, 5, 5, 5, 5, 5...
## $ salary      <chr> "48 - 65", "50 - 71", "54 - 77", "58 - 82", "54 - 72", ...
## $ salary_low  <dbl> 48, 50, 54, 58, 54, 57, 62, 63, 65, 70, 72, 85, 55, 58,...
## $ salary_high <chr> "65", "71", "77", "82", "72", "81", "87", "91", "95", "...

How many industries are represented?
The salary_high column is a character type when it should be numeric. Change this column to numeric.
What are the highest and lowest salaries for an actuary in Health with 5 exams passed?
Create a new column called salary_mid which has the middle of the salary_low and salary_high columns.
When grouping by industry, what is the highest salary_mid? What about salary_high? What is the lowest salary_low?
There is a mistake when salary_low == 11. Find and fix this mistake, and then rerun the code from the previous task.
Create a new column, called n_exams, which is an integer. Use 7 for ASA/ACAS and 10 for FSA/FCAS. Use the code below as a starting point and fill in the _ spaces
Create a column called social_life, which is equal to n_exams/experience. What is the average (mean) social_life by industry? Bonus question: what is wrong with using this as a statistical measure?

actuary_salaries <- actuary_salaries %>% 
  mutate(n_exams = case_when(exams == "FSA" ~ _,
                             exams == "ASA" ~ _,
                             exams == "FCAS" ~ _,
                             exams == "ACAS" ~ _,
                             TRUE ~ as.numeric(substr(exams,_,_))))

Create a column called social_life, which is equal to n_exams/experience. What is the average (mean) social_life by industry? Bonus question: what is wrong with using this as a statistical measure?

6.10 Answers to exercises

Answers to these exercises, along with a video tutorial, are available at ExamPA.net. Answers to these exercises, along with a video tutorial, are available at ExamPA.net.