Hypothesis Testing

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

Load the R packages 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

• After you check all your code chunks run then you can knit it. It won’t knit untill the ??? are replaced

• Safe a plot to be your preview plot

Question: t-test

• The data this quiz uses 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 123

set.seed(123)

hr_1_tidy.cvs

• 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_1_tidy.csv", 
                col_types = "fddfff")

Use 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	fem: 260, mal: 240
evaluation	0	1	FALSE	4	bad: 153, fai: 142, goo: 106, ver: 99
salary	0	1	FALSE	6	lev: 93, lev: 92, lev: 91, lev: 84
status	0	1	FALSE	3	fir: 185, pro: 162, ok: 153

Variable type: numeric

skim_variable	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
age	0	1	40.60	11.58	20.2	30.37	41.00	50.82	59.9	▇▇▇▇▇
hours	0	1	49.32	13.13	35.0	37.55	45.25	58.45	79.7	▇▂▃▂▂

Q: Is the mean number of hours worked per week 48?

specify that hours is the variable interest

hr  %>% 
  specify(response = hours)

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

hypothesize that the average hours worked is 48

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

Response: hours (numeric)
Null Hypothesis: point
# A tibble: 500 x 1
   hours
   <dbl>
 1  36.5
 2  55.8
 3  35  
 4  52  
 5  35.1
 6  36.3
 7  40.1
 8  42.7
 9  66.6
10  35.5
# ... 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  33.7
 2         1  34.9
 3         1  46.6
 4         1  33.8
 5         1  61.2
 6         1  34.7
 7         1  37.9
 8         1  39.0
 9         1  62.8
10         1  50.9
# ... with 499,990 more rows

The output has 500,000 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

Response: age (numeric)
Null Hypothesis: point
# A tibble: 1,000 x 2
   replicate   stat
       <int>  <dbl>
 1         1  0.802
 2         2 -0.706
 3         3  1.33 
 4         4 -0.245
 5         5 -1.11 
 6         6  0.382
 7         7 -0.904
 8         8  0.816
 9         9  0.968
10        10  0.979
# ... with 990 more rows

• null_t_distribution has 1000 t-stats

visualize the simulated null distribution

visualize(null_t_distribution)

calculate the static from your observed data

• Assign the output observed_t_static

• Display observed_t_static

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

observed_t_statistic

Response: hours (numeric)
Null Hypothesis: point
# A tibble: 1 x 1
   stat
  <dbl>
1  2.25

get_p_value from the simulated null distribution and the observed statistic

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

# A tibble: 1 x 1
  p_value
    <dbl>
1   0.022

shade_p_value on the simulated null distribution

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

Is 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

hr_3_tidy.csv 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?

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)  %>% 
  hypothesize(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

Response: hours (numeric)
Explanatory: gender (factor)
Null Hypothesis: independence
# 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 1,000 t-stats

visualize the simulated null distribution

visualize(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

Response: hours (numeric)
Explanatory: gender (factor)
# 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   0.892

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")

is 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

hr_2_tidy.csv 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_2_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	promoted	0	1	FALSE	2	mal: 90, fem: 89
gender	fired	0	1	FALSE	2	fem: 101, mal: 93
gender	ok	0	1	FALSE	2	mal: 73, fem: 54
evaluation	promoted	0	1	FALSE	4	goo: 70, ver: 62, fai: 24, bad: 23
evaluation	fired	0	1	FALSE	4	bad: 78, fai: 72, goo: 25, ver: 19
evaluation	ok	0	1	FALSE	4	bad: 53, fai: 46, ver: 15, goo: 13
salary	promoted	0	1	FALSE	6	lev: 42, lev: 42, lev: 39, lev: 34
salary	fired	0	1	FALSE	6	lev: 54, lev: 44, lev: 34, lev: 24
salary	ok	0	1	FALSE	6	lev: 32, lev: 31, lev: 26, lev: 19

Variable type: numeric

skim_variable	status	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
age	promoted	0	1	40.63	11.25	20.4	30.75	41.10	50.25	59.9	▆▇▇▇▇
age	fired	0	1	40.03	11.53	20.3	29.45	40.40	50.08	59.9	▇▅▇▆▆
age	ok	0	1	38.50	11.98	20.3	28.15	38.70	49.45	59.9	▇▆▅▅▆
hours	promoted	0	1	59.21	12.66	35.0	49.75	58.90	70.65	79.9	▅▆▇▇▇
hours	fired	0	1	41.67	8.37	35.0	36.10	38.45	43.40	77.7	▇▂▁▁▁
hours	ok	0	1	47.35	10.86	35.0	37.10	45.70	54.50	78.9	▇▅▃▂▁

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

• Employees that were ok worked an average of 48.0 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  78.1 promoted
 2  35.1 fired   
 3  36.9 fired   
 4  38.5 fired   
 5  36.1 fired   
 6  78.1 promoted
 7  76   promoted
 8  35.6 fired   
 9  35.6 ok      
10  56.8 promoted
# ... 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")

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

generate 1000 replicates representing the null hypothesis

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  41.9 promoted         1
 2  36.7 fired            1
 3  35   fired            1
 4  58.9 fired            1
 5  36.1 fired            1
 6  39.4 promoted         1
 7  54.3 promoted         1
 8  59.2 fired            1
 9  40.2 ok               1
10  35.3 promoted         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 = status)  %>% 
  hypothesize(null = "independence")  %>% 
  generate(reps = 1000, type = "permute")  %>% 
  calculate(stat = "F")

null_distribution_anova

Response: hours (numeric)
Explanatory: status (factor)
Null Hypothesis: independence
# A tibble: 1,000 x 2
   replicate  stat
       <int> <dbl>
 1         1 0.312
 2         2 2.85 
 3         3 0.369
 4         4 0.142
 5         5 0.511
 6         6 2.73 
 7         7 1.06 
 8         8 0.171
 9         9 0.310
10        10 1.11 
# ... with 990 more rows

• 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

Response: hours (numeric)
Explanatory: status (factor)
# A tibble: 1 x 1
   stat
  <dbl>
1  128.

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

shape_p_value on the simulated nll distribution

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

Save the previous plot to preview.png and add to the yaml chunk at the top

ggsave(filename = "preview.png", 
       path = here::here("_posts", "2022-04-30-hypotheses-testing"))

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