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 until the ??? are replaced
• Save 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 re coded to be represented as words instead of numbers
• Set random seed generator to 123

set.seed(123)

hr_3_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_3_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: 253, mal: 247
evaluation	0	1	FALSE	4	bad: 148, fai: 138, goo: 122, ver: 92
salary	0	1	FALSE	6	lev: 98, lev: 87, lev: 87, lev: 86
status	0	1	FALSE	3	fir: 196, pro: 172, ok: 132

Variable type: numeric

skim_variable	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
age	0	1	39.41	11.33	20	29.9	39.35	49.1	59.9	▇▇▇▇▆
hours	0	1	49.68	13.24	35	38.2	45.50	58.8	79.9	▇▃▃▂▂

The mean hours 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)

Response: hours (numeric)
# A tibble: 500 x 1
   hours
   <dbl>
 1  49.6
 2  39.2
 3  63.2
 4  42.2
 5  54.7
 6  54.3
 7  37.3
 8  45.6
 9  35.1
10  53  
# ... 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  49.6
 2  39.2
 3  63.2
 4  42.2
 5  54.7
 6  54.3
 7  37.3
 8  45.6
 9  35.1
10  53  
# ... 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  34.5
 2         1  33.6
 3         1  35.6
 4         1  78.2
 5         1  52.7
 6         1  77.0
 7         1  37.1
 8         1  41.9
 9         1  62.7
10         1  38.8
# ... 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

# A tibble: 1,000 x 2
   replicate    stat
 *     <int>   <dbl>
 1         1  0.929 
 2         2  0.480 
 3         3 -0.0136
 4         4  0.435 
 5         5 -0.810 
 6         6 -1.06  
 7         7 -0.0470
 8         8  0.809 
 9         9  0.986 
10        10  0.199 
# ... with 990 more rows

null_t_distribution has 1,000 t-stats.

---

Visualize the simulated null distribution

visualize(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  2.83

---

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   0.012

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

# 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

# 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   0.796

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")

ggsave(filename = "preview.png", 
       path = here::here("_posts", "2021-05-16-hypothesis-testing"))

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_3_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_3_tidy.csv", 
                col_types = "fddfff")

---

Is 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	fem: 91, mal: 81
gender	fired	0	1	FALSE	2	mal: 98, fem: 98
gender	ok	0	1	FALSE	2	mal: 68, fem: 64
evaluation	promoted	0	1	FALSE	4	goo: 79, ver: 52, fai: 21, bad: 20
evaluation	fired	0	1	FALSE	4	bad: 77, fai: 64, ver: 30, goo: 25
evaluation	ok	0	1	FALSE	4	fai: 53, bad: 51, goo: 18, ver: 10
salary	promoted	0	1	FALSE	6	lev: 42, lev: 37, lev: 36, lev: 28
salary	fired	0	1	FALSE	6	lev: 59, lev: 40, lev: 39, lev: 25
salary	ok	0	1	FALSE	6	lev: 33, lev: 29, lev: 28, lev: 23

Variable type: numeric

skim_variable	status	n_missing	complete_rate	mean	sd	p0	p25	p50	p75	p100	hist
age	promoted	0	1	40.22	11.11	20.1	31.67	41.00	48.82	59.7	▆▇▇▇▇
age	fired	0	1	38.95	11.23	20.0	29.82	38.80	48.75	59.9	▇▆▇▇▅
age	ok	0	1	39.03	11.77	20.0	28.28	38.75	49.92	59.7	▇▇▆▇▆
hours	promoted	0	1	59.29	12.53	35.0	49.90	58.65	70.35	79.9	▅▆▇▆▇
hours	fired	0	1	42.37	9.15	35.0	36.20	39.20	43.80	79.6	▇▁▁▁▁
hours	ok	0	1	47.99	11.55	35.0	37.45	45.75	55.23	75.7	▇▃▃▂▂

• Employees that were fired worked an average of 42.4 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  49.6 promoted
 2  39.2 fired   
 3  63.2 promoted
 4  42.2 promoted
 5  54.7 promoted
 6  54.3 fired   
 7  37.3 fired   
 8  45.6 promoted
 9  35.1 fired   
10  53   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  49.6 promoted
 2  39.2 fired   
 3  63.2 promoted
 4  42.2 promoted
 5  54.7 promoted
 6  54.3 fired   
 7  37.3 fired   
 8  45.6 promoted
 9  35.1 fired   
10  53   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  38.4 promoted         1
 2  41   fired            1
 3  66.1 promoted         1
 4  46.1 promoted         1
 5  65.1 promoted         1
 6  43.7 fired            1
 7  35.1 fired            1
 8  40.9 promoted         1
 9  38.4 fired            1
10  67.7 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 = gender)  %>% 
  hypothesize(null = "independence")  %>% 
  generate(reps = 1000, type = "permute")  %>% 
  calculate(stat = "F")

null_distribution_anova

# A tibble: 1,000 x 2
   replicate     stat
 *     <int>    <dbl>
 1         1 0.00878 
 2         2 0.359   
 3         3 1.16    
 4         4 0.00524 
 5         5 2.25    
 6         6 0.259   
 7         7 2.61    
 8         8 0.0457  
 9         9 0.0830  
10        10 0.000339
# ... with 990 more rows

null_distribution_anova has 1,000 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  110.

---

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 = observed_f_sample_stat, direction = "greater")

Is 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