Firm 2 Workforce Data Analysis
10/2/2019
firms<-read.csv("F2019_Section_5.csv")
my_firm<-firms %>%
filter(firmid==2)Background
How does the firm prepare and hire new employees based on our aging workforce for the foreseeable future? To match the targeted increase/ decrease of the workforce in each occupation over the next five years.
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Select/Create Variables
Created the age variable to better see the data without having to calculate the ages from the given birthyear within the date, changed age variable to show a nominal variable in the form of old or not, 1 being old, 0 being not old. Then made the nominal variable a factor variable as opposed to a numeric variable for ease of veiwing the data. Created age group in order to better see the distributiton of the ages within the form to better compare to the german national average. Then using the ggplot command to create a histogram of the age distribution.
my_firm<-my_firm %>%
mutate(age= 2019- birthyear) %>%
mutate(old = ifelse(age>=55, 1, 0))
my_firm$old <- factor(my_firm$old, labels = c("Young", "Old"))
view(my_firm)
my_firm$age[1:10]## [1] 39 39 57 63 54 39 53 63 33 49my_firm$old[1:10]## [1] Young Young Old Old Young Young Young Old Young Young
## Levels: Young Oldmy_firm<-my_firm %>%
mutate(age_group= cut(age, breaks = c(19, 24, 29, 34, 39, 44, 49, 54, 59, 64, 69, 74), labels=c("20-24", "25-29", "30-34", "35-39", "40-44", "45-49", "50-54", "55-59", "60-64", "65-69", "70-74")))
ggplot(my_firm, aes(x=age))+
geom_histogram()## `stat_bin()` using `bins = 30`. Pick better value with `binwidth`.
Filter
Using the group by and filter commands in order to determine how many workers are going to become older workers, as defined by being 55 years of age or older, within the next five years. How many are currently older workers, as well as how many workers will be retiring, using the normal retiring age of 65 years old, within the next five years.
my_firm%>%
group_by(age) %>%
filter(age>=50, age<55)## # A tibble: 561 x 22
## # Groups: age [5]
## firmid personid birthyear dailywage date hiredate engagement unit
## <int> <int> <int> <int> <fct> <fct> <int> <int>
## 1 2 584106 1965 219 25-F… 14-Dec-… 5 24
## 2 2 488991 1966 224 25-F… 23-Jul-… 2 14
## 3 2 164314 1969 291 25-F… 3-Oct-06 2 21
## 4 2 119400 1968 231 25-F… 16-Mar-… 3 54
## 5 2 584103 1965 299 25-F… 14-Dec-… 5 24
## 6 2 28903 1967 267 25-F… 8-Nov-17 5 24
## 7 2 118503 1969 90 25-F… 25-Feb-… 4 54
## 8 2 119593 1966 472 25-F… 16-Mar-… 1 54
## 9 2 119475 1967 314 25-F… 11-Mar-… 2 54
## 10 2 119183 1967 248 25-F… 30-Sep-… 2 54
## # … with 551 more rows, and 14 more variables: region <fct>, gender <fct>,
## # married <fct>, children <fct>, education <fct>, taskcomplexity <fct>,
## # workschedule <fct>, contract <fct>, industry <fct>, nationality <fct>,
## # occupation <fct>, age <dbl>, old <fct>, age_group <fct>##561 people will become 'older workers' within the next five yearsmy_firm%>%
group_by(age) %>%
filter(age>=55)## # A tibble: 1,207 x 22
## # Groups: age [15]
## firmid personid birthyear dailywage date hiredate engagement unit
## <int> <int> <int> <int> <fct> <fct> <int> <int>
## 1 2 119872 1962 121 25-F… 10-Sep-… 5 54
## 2 2 164437 1956 803 25-F… 12-Nov-… 1 21
## 3 2 584032 1956 280 25-F… 18-Jun-… 4 24
## 4 2 355693 1960 247 25-F… 16-Apr-… 3 15
## 5 2 584530 1955 258 25-F… 6-Nov-89 4 24
## 6 2 119868 1962 106 25-F… 10-Sep-… 5 54
## 7 2 584185 1954 242 25-F… 12-Jul-… 3 24
## 8 2 584608 1952 264 25-F… 11-Feb-… 4 24
## 9 2 584252 1962 161 25-F… 12-Aug-… 2 24
## 10 2 119867 1962 41 25-F… 10-Sep-… 3 54
## # … with 1,197 more rows, and 14 more variables: region <fct>,
## # gender <fct>, married <fct>, children <fct>, education <fct>,
## # taskcomplexity <fct>, workschedule <fct>, contract <fct>,
## # industry <fct>, nationality <fct>, occupation <fct>, age <dbl>,
## # old <fct>, age_group <fct>##1207 workers are currently considered 'older workers'my_firm%>%
group_by(age) %>%
filter(age>=60)## # A tibble: 705 x 22
## # Groups: age [10]
## firmid personid birthyear dailywage date hiredate engagement unit
## <int> <int> <int> <int> <fct> <fct> <int> <int>
## 1 2 164437 1956 803 25-F… 12-Nov-… 1 21
## 2 2 584032 1956 280 25-F… 18-Jun-… 4 24
## 3 2 584530 1955 258 25-F… 6-Nov-89 4 24
## 4 2 584185 1954 242 25-F… 12-Jul-… 3 24
## 5 2 584608 1952 264 25-F… 11-Feb-… 4 24
## 6 2 164504 1949 760 25-F… 19-Oct-… 1 21
## 7 2 584022 1956 237 25-F… 18-Jun-… 4 24
## 8 2 583786 1956 228 25-F… 10-Nov-… 5 24
## 9 2 118325 1958 202 25-F… 11-Mar-… 3 54
## 10 2 119265 1952 153 25-F… 11-Dec-… 3 54
## # … with 695 more rows, and 14 more variables: region <fct>, gender <fct>,
## # married <fct>, children <fct>, education <fct>, taskcomplexity <fct>,
## # workschedule <fct>, contract <fct>, industry <fct>, nationality <fct>,
## # occupation <fct>, age <dbl>, old <fct>, age_group <fct>##705 workers will be retiring within the next 5 years```
Mutate
We use the mutate command to create a variable showing those who will be or wont’ be retiring within the next five years, 1 being those that are retiring, 0 being those that are not retring. Then giving those values a variable name of “Retired” and “Not Retired” respectively. Using the new variable and a view command to show a compressed table of the new variables. Then using the ggplot command to compare the the amount of retirees vs. non retirees based on their occcupations.
my_firm<- my_firm%>%
mutate(retired = ifelse(age>= 60, 1, 0))
my_firm$retired <- factor(my_firm$retired, labels = c("Not Retired", "Retired"))
view(my_firm)
my_firm$retired[1:10]## [1] Not Retired Not Retired Not Retired Retired Not Retired
## [6] Not Retired Not Retired Retired Not Retired Not Retired
## Levels: Not Retired Retiredmy_firm%>%
group_by(retired)## # A tibble: 2,852 x 23
## # Groups: retired [2]
## firmid personid birthyear dailywage date hiredate engagement unit
## <int> <int> <int> <int> <fct> <fct> <int> <int>
## 1 2 119844 1980 88 25-F… 25-Feb-… 4 54
## 2 2 119842 1980 75 25-F… 25-Feb-… 4 54
## 3 2 119872 1962 121 25-F… 10-Sep-… 5 54
## 4 2 164437 1956 803 25-F… 12-Nov-… 1 21
## 5 2 584106 1965 219 25-F… 14-Dec-… 5 24
## 6 2 355723 1980 37 25-F… 25-Feb-… 3 15
## 7 2 488991 1966 224 25-F… 23-Jul-… 2 14
## 8 2 584032 1956 280 25-F… 18-Jun-… 4 24
## 9 2 584992 1986 37 25-F… 7-May-16 3 24
## 10 2 584403 1970 169 25-F… 25-Feb-… 4 24
## # … with 2,842 more rows, and 15 more variables: region <fct>,
## # gender <fct>, married <fct>, children <fct>, education <fct>,
## # taskcomplexity <fct>, workschedule <fct>, contract <fct>,
## # industry <fct>, nationality <fct>, occupation <fct>, age <dbl>,
## # old <fct>, age_group <fct>, retired <fct>My_firmSpecialty <- my_firm%>%
filter(occupation == "Sales"|occupation== "Management, Business and Financial" | occupation == "Legal" | occupation == "Computer, Engineering and Science")
ggplot(My_firmSpecialty, aes(x= occupation, fill= retired))+
geom_bar(position="dodge")+
coord_flip()
my_firmNonSpecialty <- my_firm%>%
filter(occupation == "Service" | occupation == "Production" | occupation == "Office and Administrative")
ggplot(my_firmNonSpecialty, aes(x= occupation, fill= retired))+
geom_bar(position="dodge")+
coord_flip()
Estimate
Using the same mutate and label commands in the last chunk, we now use the view and table commands in order to see the total values of retirees based on their occupation in order to find the percent of retirees within each occupation.
my_firm<- my_firm%>%
mutate(retired = ifelse(age>= 60, 1, 0))
my_firm$retired <- factor(my_firm$retired, labels = c("Not Retired", "Retired"))
view(my_firm)
my_firm$retired[1:10]## [1] Not Retired Not Retired Not Retired Retired Not Retired
## [6] Not Retired Not Retired Retired Not Retired Not Retired
## Levels: Not Retired Retiredtable(my_firm$occupation)##
## Computer, Engineering and Science Legal
## 16 14
## Management, Business and Financial Office and Administrative
## 244 322
## Production Sales
## 1043 708
## Service
## 505ret_firm <- my_firm %>%
filter(retired == "Retired")
table(ret_firm$occupation)##
## Computer, Engineering and Science Legal
## 13 14
## Management, Business and Financial Office and Administrative
## 124 28
## Production Sales
## 197 323
## Service
## 6Using ther ggplot command along with the coord_polar command we can create pie charts of our critical and non critical occupations in order to better visualize the difference between those who will be retired within the next five years, and those who won’t be.
NonCritDF <- data.frame(group = c("Retired", "Non-Retired"), value = c(231, 1639))
head(NonCritDF)## group value
## 1 Retired 231
## 2 Non-Retired 1639NonCriticalOccupations <- ggplot(NonCritDF, aes(x="", y=value, fill=group))+
geom_bar(width=1, stat="identity")
NonCriticalOccupations
NonCriticalPie <- NonCriticalOccupations + coord_polar("y", start = 0)
NonCriticalPie
Same as in commands in the last chunk but filtering for critical workers as opposed to non critical occupations.
CritDF <- data.frame(group = c("Retired", "Non-retired"), value = c(474, 508))
head(CritDF)## group value
## 1 Retired 474
## 2 Non-retired 508CriticalOccupations <- ggplot(CritDF, aes(x="", y=value, fill=group))+
geom_bar(width=1, stat="identity")
CriticalOccupations
CriticalOccupationPie <- CriticalOccupations + coord_polar("y", start = 0)
CriticalOccupationPie
Test
Now using a chi squared test of frequencies we can see if our firms data is similar or not to the national average of german workers age distribution, so we know that our firm is not an outlier within the german workforce. Using values gotten from the CIA World factbook workforce census data for Germany.
chisq.test(table(my_firm$age_group), p=c(.0778, .0825, .0935, .0928, .0865, .0887, .1142, .1164, .0995, .0825, .0656))##
## Chi-squared test for given probabilities
##
## data: table(my_firm$age_group)
## X-squared = 997.71, df = 10, p-value < 2.2e-16