Hypothesis Testing

Using computer simulation. Based on examples from the infer package. Code for Quiz 13.

Load the R package we will use.

library(tidyverse)
library(infer)
library(skimr)

Replace all the instances of ???. These are answers on your moodle quiz.

Run all the individual code chunks to make sure the answers in this file correspond with your quiz answers

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Question: t-test

• The data this quiz is a subset of HR

  • Look at the variable definitions

  • Note that the variables evaluation and salary have been recoded to be represented as words instead of numbers

• Set random seed generator to 123

set.seed(123)

hr_2_tidy.csv is the name of your data subset

• Read it into and assign to hr

• Note: col_types = “fddfff” defines the column types factor-double-double-factor-factor-factor

hr <- read_csv("https://estanny.com/static/week13/data/hr_2_tidy.csv", col_types = "fddfff") 

use the skim to summarize the data in hr

skim(hr)

Table 1: Data summary

Name	hr
Number of rows	500
Number of columns	6
_______________________
Column type frequency:
factor	4
numeric	2
________________________
Group variables	None

Variable type: factor

skim_variable	n_missing	complete_rate	ordered	n_unique	top_counts
gender	0	1	FALSE	2	mal: 256, fem: 244
evaluation	0	1	FALSE	4	bad: 154, fai: 142, goo: 108, ver: 96
salary	0	1	FALSE	6	lev: 95, lev: 94, lev: 87, lev: 85
status	0	1	FALSE	3	fir: 194, pro: 179, ok: 127

Variable type: numeric

skim_variable	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
age	0	1	39.86	11.55	20.3	29.60	40.2	50.1	59.9	▇▇▇▇▇
hours	0	1	49.39	13.15	35.0	37.48	45.6	58.9	79.9	▇▃▂▂▂

The mean hours worked per week is: 49.4

#Q: Is the mean number of hours worked per week 48? specify that hours is the variable of interest

hr %>% 
specify(response = hours)  %>% 
hypothesize(null = "point", mu = 48)

Response: hours (numeric)
Null Hypothesis: point
# A tibble: 500 x 1
   hours
   <dbl>
 1  78.1
 2  35.1
 3  36.9
 4  38.5
 5  36.1
 6  78.1
 7  76  
 8  35.6
 9  35.6
10  56.8
# ... with 490 more rows

#generate 1000 replicates representing the null hypothesis

hr %>% 
  specify(response = hours)  %>% 
  hypothesize(null = "point", mu = 48)  %>% 
  generate(reps = 1000, type = "bootstrap")

Response: hours (numeric)
Null Hypothesis: point
# A tibble: 500,000 x 2
# Groups:   replicate [1,000]
   replicate hours
       <int> <dbl>
 1         1  39.7
 2         1  44.3
 3         1  46.8
 4         1  33.7
 5         1  39.6
 6         1  39.5
 7         1  40.5
 8         1  55.8
 9         1  72.6
10         1  35.7
# ... with 499,990 more rows

The output has 1000 rows

calculate the distribution of statistics from the generated data

Assign the output null_t_distribution

Display null_t_distribution

null_t_distribution  <- hr  %>% 
  specify(response = age)  %>% 
  hypothesize(null = "point", mu = 48)  %>% 
  generate(reps = 1000, type = "bootstrap")  %>% 
  calculate(stat = "t")

null_t_distribution

# A tibble: 1,000 x 2
   replicate     stat
 *     <int>    <dbl>
 1         1  0.144  
 2         2 -1.72   
 3         3  0.404  
 4         4 -1.11   
 5         5  0.00894
 6         6  1.46   
 7         7 -0.905  
 8         8 -0.663  
 9         9  0.291  
10        10  3.09   
# ... with 990 more rows

null_t_distribution has 1000 t-stats

visualize the simulated null distribution

visualise(null_t_distribution)

calculate the statistic from your observed data

Assign the output observed_t_statistic

Display observed_t_statistic

observed_t_statistic <- hr  %>%
specify(response = hours)  %>% 
hypothesize(null = "point", mu = 48)  %>%
calculate(stat = "t")

observed_t_statistic

# A tibble: 1 x 1
   stat
  <dbl>
1  2.37

get_p_value from the simulated null distribution and the observed statistic

p_value <- null_t_distribution %>%
  get_p_value(obs_stat = observed_t_statistic, direction = "two-sided")

p_value

# A tibble: 1 x 1
  p_value
    <dbl>
1   0.014

shade_p_value on the simulated null distribution

null_t_distribution  %>%
  visualize() +
  shade_p_value(obs_stat =  observed_t_statistic, direction = "two-sided")

If the p-value < 0.05? yes

Does your analysis support the null hypothesis that the true mean number of hours worked was 48? No

Question: 2 sample t-test

SEE QUIZ is the name of your data subset

• Read it into and assign to hr_2

• Note: col_types = “fddfff” defines the column types factor-double-double-factor-factor-factor

hr_2 <- read_csv("https://estanny.com/static/week13/data/hr_3_tidy.csv", col_types = "fddfff") 

Q: Is the average number of hours worked the same for both genders in hr_2?

use skim to summarize the data in hr_2 by gender

hr_2 %>% 
  group_by(gender)  %>% 
  skim()

Table 2: Data summary

Name	Piped data
Number of rows	500
Number of columns	6
_______________________
Column type frequency:
factor	3
numeric	2
________________________
Group variables	gender

Variable type: factor

skim_variable	gender	n_missing	complete_rate	ordered	n_unique	top_counts
evaluation	male	0	1	FALSE	4	bad: 72, fai: 67, goo: 61, ver: 47
evaluation	female	0	1	FALSE	4	bad: 76, fai: 71, goo: 61, ver: 45
salary	male	0	1	FALSE	6	lev: 47, lev: 43, lev: 43, lev: 42
salary	female	0	1	FALSE	6	lev: 51, lev: 46, lev: 45, lev: 43
status	male	0	1	FALSE	3	fir: 98, pro: 81, ok: 68
status	female	0	1	FALSE	3	fir: 98, pro: 91, ok: 64

Variable type: numeric

skim_variable	gender	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
age	male	0	1	38.23	10.86	20	28.9	37.9	47.05	59.9	▇▇▇▇▅
age	female	0	1	40.56	11.67	20	31.0	40.3	50.50	59.8	▆▆▇▆▇
hours	male	0	1	49.55	13.11	35	38.4	45.4	57.65	79.9	▇▃▂▂▂
hours	female	0	1	49.80	13.38	35	38.2	45.6	59.40	79.8	▇▂▃▂▂

Females worked an average of 49.8 hours per week

Males worked an average of 49.6 hours per week

Use geom_boxplot to plot distributions of hours worked by gender

hr_2 %>% 
  ggplot(aes(x = gender, y = hours)) + 
  geom_boxplot()

specify the variables of interest are hours and gender

hr_2 %>% 
  specify(response = hours, explanatory = gender)

Response: hours (numeric)
Explanatory: gender (factor)
# A tibble: 500 x 2
   hours gender
   <dbl> <fct> 
 1  49.6 male  
 2  39.2 female
 3  63.2 female
 4  42.2 male  
 5  54.7 male  
 6  54.3 female
 7  37.3 female
 8  45.6 female
 9  35.1 female
10  53   male  
# ... with 490 more rows

hypothesize that the number of hours worked and gender are independent

hr_2 %>% 
  specify(response = hours, explanatory = gender)  %>% 
  hypothesise(null = "independence")

Response: hours (numeric)
Explanatory: gender (factor)
Null Hypothesis: independence
# A tibble: 500 x 2
   hours gender
   <dbl> <fct> 
 1  49.6 male  
 2  39.2 female
 3  63.2 female
 4  42.2 male  
 5  54.7 male  
 6  54.3 female
 7  37.3 female
 8  45.6 female
 9  35.1 female
10  53   male  
# ... with 490 more rows

generate 1000 replicates representing the null hypothesis

hr_2 %>% 
  specify(response = hours, explanatory = gender)  %>% 
  hypothesize(null = "independence")  %>% 
  generate(reps = 1000, type = "permute") 

Response: hours (numeric)
Explanatory: gender (factor)
Null Hypothesis: independence
# A tibble: 500,000 x 3
# Groups:   replicate [1,000]
   hours gender replicate
   <dbl> <fct>      <int>
 1  55.7 male           1
 2  35.5 female         1
 3  35.1 female         1
 4  44.2 male           1
 5  52.8 male           1
 6  46   female         1
 7  41.2 female         1
 8  52.9 female         1
 9  35.6 female         1
10  35   male           1
# ... with 499,990 more rows

The output has 500,000 rows.

calculate the distribution of statistics from the generated data

Assign the output null_distribution_2_sample_permute

Display null_distribution_2_sample_permute

null_distribution_2_sample_permute  <- hr_2 %>% 
  specify(response = hours, explanatory = gender)  %>% 
  hypothesize(null = "independence")  %>% 
  generate(reps = 1000, type = "permute")  %>% 
  calculate(stat = "t", order = c("female", "male"))

null_distribution_2_sample_permute

# A tibble: 1,000 x 2
   replicate    stat
 *     <int>   <dbl>
 1         1 -1.81  
 2         2 -1.29  
 3         3  0.0525
 4         4 -0.793 
 5         5  0.826 
 6         6  0.429 
 7         7  0.0843
 8         8 -0.264 
 9         9  2.42  
10        10  0.603 
# ... with 990 more rows

null_t_distribution has 1000 t-stats

visualize the simulated null distribution

visualise(null_distribution_2_sample_permute)

calculate the statistic from your observed data

Assign the output observed_t_2_sample_stat

Display observed_t_2_sample_stat

observed_t_2_sample_stat <- hr_2 %>%
  specify(response = hours, explanatory = gender)  %>% 
  calculate(stat = "t", order = c("female", "male"))

observed_t_2_sample_stat

# A tibble: 1 x 1
   stat
  <dbl>
1 0.208

get_p_value from the simulated null distribution and the observed statistic

null_t_distribution  %>% 
  get_p_value(obs_stat = observed_t_2_sample_stat , direction = "two-sided")

# A tibble: 1 x 1
  p_value
    <dbl>
1   0.878

shade_p_value on the simulated null distribution

null_t_distribution  %>% 
  visualize() +
  shade_p_value(obs_stat = observed_t_2_sample_stat , direction = "two-sided")

If the p-value < 0.05? No

Does your analysis support the null hypothesis that the true mean number of hours worked by female and male employees was the same? Yes

Question: ANOVA

SEE QUIZ is the name of your data subset

Read it into and assign to hr_anova

Note: col_types = “fddfff” defines the column types factor-double-double-factor-factor-factor

hr_anova <- read_csv("https://estanny.com/static/week13/data/hr_1_tidy.csv", 
                col_types = "fddfff") 

Q: Is the average number of hours worked the same for all three status (fired, ok and promoted) ?

use skim to summarize the data in hr_anova by status

hr_anova %>% 
  group_by(status)  %>% 
  skim()

Table 3: Data summary

Name	Piped data
Number of rows	500
Number of columns	6
_______________________
Column type frequency:
factor	3
numeric	2
________________________
Group variables	status

Variable type: factor

skim_variable	status	n_missing	complete_rate	ordered	n_unique	top_counts
gender	fired	0	1	FALSE	2	fem: 96, mal: 89
gender	ok	0	1	FALSE	2	fem: 77, mal: 76
gender	promoted	0	1	FALSE	2	fem: 87, mal: 75
evaluation	fired	0	1	FALSE	4	bad: 65, fai: 63, goo: 31, ver: 26
evaluation	ok	0	1	FALSE	4	bad: 69, fai: 59, goo: 15, ver: 10
evaluation	promoted	0	1	FALSE	4	ver: 63, goo: 60, fai: 20, bad: 19
salary	fired	0	1	FALSE	6	lev: 41, lev: 37, lev: 32, lev: 32
salary	ok	0	1	FALSE	6	lev: 40, lev: 37, lev: 29, lev: 23
salary	promoted	0	1	FALSE	6	lev: 37, lev: 35, lev: 29, lev: 23

Variable type: numeric

skim_variable	status	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
age	fired	0	1	38.64	11.43	20.2	28.30	38.30	47.60	59.6	▇▇▇▅▆
age	ok	0	1	41.34	12.11	20.3	31.00	42.10	51.70	59.9	▆▆▆▆▇
age	promoted	0	1	42.13	10.98	21.0	33.40	42.95	50.98	59.9	▆▅▆▇▇
hours	fired	0	1	41.67	7.88	35.0	36.10	38.90	43.90	75.5	▇▂▁▁▁
hours	ok	0	1	48.05	11.65	35.0	37.70	45.60	56.10	78.2	▇▃▃▂▁
hours	promoted	0	1	59.27	12.90	35.0	51.12	60.10	70.15	79.7	▆▅▇▇▇

Employees that were fired worked an average of 41.7 hours per week

Employees that were ok worked an average of 48 hours per week

Employees that were promoted worked an average of 59.3 hours per week

Use geom_boxplot to plot distributions of hours worked by status

hr_anova %>% 
ggplot(aes(x =  status, y = hours)) + 
geom_boxplot()

specify the variables of interest are hours and status

hr_anova %>% 
specify(response = hours, explanatory = status) 

Response: hours (numeric)
Explanatory: status (factor)
# A tibble: 500 x 2
   hours status  
   <dbl> <fct>   
 1  36.5 fired   
 2  55.8 ok      
 3  35   fired   
 4  52   promoted
 5  35.1 ok      
 6  36.3 ok      
 7  40.1 promoted
 8  42.7 fired   
 9  66.6 promoted
10  35.5 ok      
# ... with 490 more rows

hypothesize that the number of hours worked and status are independent

hr_anova %>% 
  specify(response = hours, explanatory = status)  %>% hypothesize(null = "independence")  %>% 
generate(reps = 1000, type = "permute") 

Response: hours (numeric)
Explanatory: status (factor)
Null Hypothesis: independence
# A tibble: 500,000 x 3
# Groups:   replicate [1,000]
   hours status   replicate
   <dbl> <fct>        <int>
 1  40.3 fired            1
 2  40.3 ok               1
 3  37.3 fired            1
 4  50.5 promoted         1
 5  35.1 ok               1
 6  67.8 ok               1
 7  39.3 promoted         1
 8  35.7 fired            1
 9  40.2 promoted         1
10  38.4 ok               1
# ... with 499,990 more rows

The output has 500,000 rows.

calculate the distribution of statistics from the generated data

Assign the output null_distribution_anova

Display null_distribution_anova

null_distribution_anova <- hr_anova %>% 
  specify(response = hours, explanatory = gender)  %>% 
  hypothesize(null = "independence")  %>% 
  generate(reps = 1000, type = "permute")  %>% 
  calculate(stat = "F")

null_distribution_anova has 1000 F-stats

visualize the simulated null distribution

visualize(null_distribution_anova)

calculate the statistic from your observed data

Assign the output observed_f_sample_stat

Display observed_f_sample_stat

observed_f_sample_stat <- hr_anova %>%
specify(response = hours, explanatory = status)  %>% 
calculate(stat = "F")

observed_f_sample_stat

# A tibble: 1 x 1
   stat
  <dbl>
1  115.

get_p_value from the simulated null distribution and the observed statistic

null_distribution_anova  %>% 
  get_p_value(obs_stat = observed_f_sample_stat , direction = "greater")

# A tibble: 1 x 1
  p_value
    <dbl>
1       0

shade_p_value on the simulated null distribution

null_t_distribution  %>% 
visualize() +
shade_p_value(obs_stat = null_distribution_anova, direction = "greater")

If the p-value < 0.05? Yes.

Does your analysis support the null hypothesis that the true means of the number of hours worked for those that were “fired”, “ok” and “promoted” were the same? No.