(1) Create a .CSV file that includes all of the information above.

The CSV file is saved in the Github.

(2) Read the information from your .CSV file into R, and use tidyr and dplyr as needed to tidy and transform your data.

(3) Analysis ( I did (2) and (3) together)

library(tidyr)

## Warning: package 'tidyr' was built under R version 3.2.5

library(dplyr)

## Warning: package 'dplyr' was built under R version 3.2.5

## 
## Attaching package: 'dplyr'

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

require(knitr)

## Loading required package: knitr

## Warning: package 'knitr' was built under R version 3.2.5

salary1<- read.csv("https://raw.githubusercontent.com/xkong100/IS607/master/Project2/Salaries.csv", stringsAsFactors = FALSE, check.names = FALSE, na.strings = c("", "NA"))
kable(head(salary1))

	rank	discipline	yrs.since.phd	yrs.service	sex	salary
1	Prof	B	19	18	Male	139750
2	Prof	B	20	16	Male	173200
3	AsstProf	B	4	3	Male	79750
4	Prof	B	45	39	Male	115000
5	Prof	B	40	41	Male	141500
6	AssocProf	B	6	6	Male	97000

We want to analyze professors’ salary when they service under 20 years.

salary1 <- data.frame(salary1[,-4])
salary2 <- salary1 %>% filter(yrs.service <=20)
salary3 <- salary2 %>% arrange(salary)
kable(head(salary3))

Var.1	rank	discipline	yrs.service	sex	salary
238	AsstProf	A	6	Female	63100
227	AsstProf	A	1	Male	63900
65	AsstProf	B	3	Male	68404
241	AsstProf	A	3	Male	69200
235	AsstProf	A	3	Male	69700
228	AssocProf	A	7	Male	70000

salary3 %>% summarise (Mean_salary_under20yrs = mean(salary), min_salary_under20yrs = min(salary), max_salary_under20yrs= max(salary))

##   Mean_salary_under20yrs min_salary_under20yrs max_salary_under20yrs
## 1               108768.8                 63100                204000

The mean salary is $108768.8, the minimum salary is $63100, and the maximum salary is $204000 when the professors service under 20 years.

We also want to find out professor’s salary when they service more than 20 years.

salary4 <- salary1 %>% filter(yrs.service > 20)
salary5 <- salary4 %>% arrange(salary)
kable(head(salary5))

Var.1	rank	discipline	yrs.service	sex	salary
283	Prof	A	51	Male	57800
124	AssocProf	A	22	Female	62884
318	Prof	B	45	Male	67559
300	AssocProf	A	39	Male	70700
299	Prof	A	43	Male	72300
232	AssocProf	A	24	Female	73300

salary5 %>% summarise (Mean_salary_morethatn20yrs = mean(salary), min_salary_morethan20yrs = min(salary), max_salary_morethan20yrs= max(salary))

##   Mean_salary_morethatn20yrs min_salary_morethan20yrs
## 1                   122103.9                    57800
##   max_salary_morethan20yrs
## 1                   231545

The mean salary is $122103.9, the minimum salary is $57800, and the maximum salary is $231545 when the professors service more than 20 years.

By comparing the mean salary, we can conclude that, the service time and the salary has positive relationship.

We also want to find out the relationship between salary and genders.

salary6 <- salary1 %>% spread(sex, salary)
salary6$Female[is.na(salary6$Female)] <-0
salary6$Male[is.na(salary6$Male)] <- 0
salary7 <- within(salary6, {Female <- as.numeric(as.character(Female))
Male <-as.numeric(as.character(Male))})
kable(head(salary7))

Var.1	rank	discipline	yrs.service	Male
1	Prof	B	18	139750
2	Prof	B	16	173200
3	AsstProf	B	3	79750
4	Prof	B	39	115000
5	Prof	B	41	141500
6	AssocProf	B	6	97000

salary7 %>% summarise (Mean_salary_Female = mean(Female),  max_salary_Female= max(Female),Mean_salary_male=mean(Male), max_salary_male=max(Male))

##   Mean_salary_Female max_salary_Female Mean_salary_male max_salary_male
## 1           9922.151            161101         103784.3          231545

From our analysis, we can conclude that Female’s salary is lower than the Male’s salary by comparing their mean salaries and max salaries.

Project2 Professors’ salaries