Assignment 8

library(readr)
library(plotly)

## Loading required package: ggplot2

## 
## Attaching package: 'plotly'

## The following object is masked from 'package:ggplot2':
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##     last_plot

## The following object is masked from 'package:stats':
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##     filter

## The following object is masked from 'package:graphics':
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##     layout

library(dplyr)

## 
## Attaching package: 'dplyr'

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

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

hr <- read_csv('https://raw.githubusercontent.com/aiplanethub/Datasets/refs/heads/master/HR_comma_sep.csv')

## Rows: 14999 Columns: 10

## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr (2): Department, salary
## dbl (8): satisfaction_level, last_evaluation, number_project, average_montly...
## 
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.

Perform four (4) t-tests using any appropriate variables (continuous) by the variable left. Note that the variable left describes whether the employee left the company (left = 1), or not (left = 0).

For each of the four t-tests:

-Perform the t-test (.5 point) Choose any two appropriate variables from the data and perform the t-test, displaying the results.

-Interpret the results in technical terms (.5 point) For each t-test, explain what the test’s p-value means (significance).

-Interpret the results in non-technical terms (1 point) For each t-test, what do the results mean in non-techical terms.

-Create a plot that helps visualize the t-test (.5 point) For each t-test, create a graph to help visualize the difference between means, if any. The title must be the non-technical interpretation.

1

t_test_satisfaction <- t.test(hr$satisfaction_level ~ hr$left)
t_test_satisfaction

## 
##  Welch Two Sample t-test
## 
## data:  hr$satisfaction_level by hr$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

• Since the p-value is extremely low, the difference in satisfaction levels between employees who leftand those who stayed is statistically significant.
• The results show employees with lower satisfaction levels were more likely to leave the company.

plot_data <- hr %>%
  mutate(Attrition = as.factor(ifelse(left == 1, 'Left', 'Stayed')))

plot_ly(plot_data, x = ~Attrition, y = ~satisfaction_level, type = 'box') %>%
  layout(title = "Employees with lower satisfaction levels are more likely to leave")

2

 t_test_hours <- t.test(hr$average_montly_hours ~ hr$left)
t_test_hours

## 
##  Welch Two Sample t-test
## 
## data:  hr$average_montly_hours by hr$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

• Since the p-value is much lower than 0.05,there is a statistically significant difference in average monthly hours between employees who left and those who stayed.

• The results show employees with higher average monthly hours were more likely to leave the company.

plot_data <- hr %>%
  mutate(Attrition = as.factor(ifelse(left == 1, 'Left', 'Stayed')))

plot_ly(plot_data, x = ~Attrition, y = ~average_montly_hours, type = 'box') %>%
  layout(title = "Employees with higher average monthly hours are more likely to leave")

3

t_test_evaluation <- t.test(hr$last_evaluation ~ hr$left)
t_test_evaluation

## 
##  Welch Two Sample t-test
## 
## data:  hr$last_evaluation by hr$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

• The p-value is greater than 0.05, meaning there is no statistically significant difference in last evaluation scores between employees who left and those who stayed.

• The results show evaluation scores do not appear to be a reason in whether an employee stays or leaves.

plot_data <- hr %>%
  mutate(Attrition = as.factor(ifelse(left == 1, 'Left', 'Stayed')))

plot_ly(plot_data, x = ~Attrition, y = ~last_evaluation, type = 'box') %>%
  layout(title = "Evaluation scores do not seem to impact employee attrition")

4

t_test_time_spent <- t.test(hr$time_spend_company ~ hr$left)
t_test_time_spent

## 
##  Welch Two Sample t-test
## 
## data:  hr$time_spend_company by hr$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

• Since the p-value is extremely low, we conclude that the difference in time spent at the company between employees who left and those who stayed is statistically significant.

• The results show employees who have been at the company for more years are more likely to leave.

plot_data <- hr %>%
  mutate(Attrition = as.factor(ifelse(left == 1, 'Left', 'Stayed')))

plot_ly(plot_data, x = ~Attrition, y = ~time_spend_company, type = 'box') %>%
  layout(title = "Employees with longer tenure are more likely to leave")