R Markdown

## spc_tbl_ [14,999 × 10] (S3: spec_tbl_df/tbl_df/tbl/data.frame)
##  $ satisfaction_level   : num [1:14999] 0.38 0.8 0.11 0.72 0.37 0.41 0.1 0.92 0.89 0.42 ...
##  $ last_evaluation      : num [1:14999] 0.53 0.86 0.88 0.87 0.52 0.5 0.77 0.85 1 0.53 ...
##  $ number_project       : num [1:14999] 2 5 7 5 2 2 6 5 5 2 ...
##  $ average_montly_hours : num [1:14999] 157 262 272 223 159 153 247 259 224 142 ...
##  $ time_spend_company   : num [1:14999] 3 6 4 5 3 3 4 5 5 3 ...
##  $ Work_accident        : num [1:14999] 0 0 0 0 0 0 0 0 0 0 ...
##  $ left                 : num [1:14999] 1 1 1 1 1 1 1 1 1 1 ...
##  $ promotion_last_5years: num [1:14999] 0 0 0 0 0 0 0 0 0 0 ...
##  $ Department           : chr [1:14999] "sales" "sales" "sales" "sales" ...
##  $ salary               : chr [1:14999] "low" "medium" "medium" "low" ...
##  - attr(*, "spec")=
##   .. cols(
##   ..   satisfaction_level = col_double(),
##   ..   last_evaluation = col_double(),
##   ..   number_project = col_double(),
##   ..   average_montly_hours = col_double(),
##   ..   time_spend_company = col_double(),
##   ..   Work_accident = col_double(),
##   ..   left = col_double(),
##   ..   promotion_last_5years = col_double(),
##   ..   Department = col_character(),
##   ..   salary = col_character()
##   .. )
##  - attr(*, "problems")=<externalptr>

1. T-Test 1: Satisfaction Level by Employee Attrition

## 
##  Welch Two Sample t-test
## 
## data:  satisfaction_level by left
## t = 46.636, df = 5167, p-value < 2.2e-16
## alternative hypothesis: true difference in means between group 0 and group 1 is not equal to 0
## 95 percent confidence interval:
##  0.2171815 0.2362417
## sample estimates:
## mean in group 0 mean in group 1 
##       0.6668096       0.4400980

P-value interpretation: The p-value is less than 0.1, therefore the difference between means of satisfaction level by attrition status is significant.

T-test Interpretation: The difference in mean satisfaction level between employees who left and those who stayed is significant, with a gap of at least 0.12 in satisfaction level.

Non-Technical Interpretation: Employees who left were less satisfied

2. T-Test 2. Average Monthly Hours by Employee Attrition

## 
##  Welch Two Sample t-test
## 
## data:  average_montly_hours by left
## t = -7.5323, df = 4875.1, p-value = 5.907e-14
## alternative hypothesis: true difference in means between group 0 and group 1 is not equal to 0
## 95 percent confidence interval:
##  -10.534631  -6.183384
## sample estimates:
## mean in group 0 mean in group 1 
##        199.0602        207.4192

p-value interpretation: The p-value is less than 0.1, therefore the difference between means of average monthly hours by attrition status is significant.

T-test Interpretation: The difference in average monthly hours between employees who left and those who stayed is significant, with a difference of at least 18 hours.

Non-Technical Interpretation: Employees who left worked more hours monthly

3. T-Test 3: Last Evaluation Score by Employee Attrition

## 
##  Welch Two Sample t-test
## 
## data:  last_evaluation by left
## t = -0.72534, df = 5154.9, p-value = 0.4683
## alternative hypothesis: true difference in means between group 0 and group 1 is not equal to 0
## 95 percent confidence interval:
##  -0.009772224  0.004493874
## sample estimates:
## mean in group 0 mean in group 1 
##       0.7154734       0.7181126

p-value interpretation: The p-value is 0.4683, which is much larger than 0.01, so the difference between means of last evaluation score by attrition status is not statistically significant.

t-test interpretation: There is no significant difference in mean evaluation scores between employees who left the company and those who stayed. The observed difference is small (about 0.0026) and could be due to random chance.

non-technical interpretation: Employees who left had higher performance evaluations.

4. T-Test 4: Time Spent at Company by Employee Attrition

## 
##  Welch Two Sample t-test
## 
## data:  time_spend_company by left
## t = -22.631, df = 9625.6, p-value < 2.2e-16
## alternative hypothesis: true difference in means between group 0 and group 1 is not equal to 0
## 95 percent confidence interval:
##  -0.5394767 -0.4534706
## sample estimates:
## mean in group 0 mean in group 1 
##        3.380032        3.876505

p-value interpretation: The p-value is less than 0.1, therefore the difference between means of time spent at the company by attrition status is significant.

t-test interpretation: The difference in mean years at the company between employees who left and those who stayed is significant, with those who left spending at least 0.8 more years at the company.

non-technical interpretation: Employees who left had spent more time at the company.