Hypothesis Testing

A short description of the post.

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

SEE QUIZ 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 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 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 worked per week is: 49.7

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)  %>% 
  hypothesise(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  40.8
 2         1  33.4
 3         1  54.6
 4         1  51.5
 5         1  33.3
 6         1  44.6
 7         1  34.1
 8         1  47.0
 9         1  34.2
10         1  55.1
# ... with 499,990 more rows

The output has 500000 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 = "mean")
null_t_distribution

# A tibble: 1,000 x 2
   replicate  stat
 *     <int> <dbl>
 1         1  47.3
 2         2  47.3
 3         3  48.7
 4         4  47.9
 5         5  47.9
 6         6  47.5
 7         7  47.7
 8         8  48.0
 9         9  48.2
10        10  48.2
# ... with 990 more rows

null_t_distribution has 10 t-stats

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  2.83

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

shade_p_value on the simulated null distribution

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

Question 2

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? 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  55.2 male           1
 2  72.6 female         1
 3  59.1 female         1
 4  58.7 male           1
 5  66.8 male           1
 6  35.1 female         1
 7  50.6 female         1
 8  35.1 female         1
 9  49.2 female         1
10  35   male           1
# ... with 499,990 more 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  0.379 
 2         2 -0.0940
 3         3 -0.943 
 4         4  0.0654
 5         5 -0.0785
 6         6 -0.203 
 7         7 -1.28  
 8         8 -0.750 
 9         9  0.467 
10        10 -0.181 
# ... with 990 more rows
visualize(null_distribution_2_sample_permute)

calculate the stst 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_statistic, direction = "two-sided")

# 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_t_distribution, direction = "two-sided")

no, p-value is not less than .05

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_anova <- 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 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 ▇▃▃▂▂
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  41.2 promoted         1
 2  36.5 fired            1
 3  35.1 promoted         1
 4  36.2 promoted         1
 5  35.1 promoted         1
 6  58.7 fired            1
 7  35.9 fired            1
 8  51.3 promoted         1
 9  42.3 fired            1
10  66.2 promoted         1
# ... with 499,990 more 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 6.58  
 2         2 0.150 
 3         3 0.203 
 4         4 0.0991
 5         5 0.246 
 6         6 0.178 
 7         7 1.09  
 8         8 1.96  
 9         9 1.20  
10        10 0.587 
# ... with 990 more rows

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 = null_distribution_anova , direction = "greater")

# A tibble: 1 x 1
  p_value
    <dbl>
1   0.291

shade p-value

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