Hypothesis Testing

Code for Quiz 13

Load the R packages we will use.

-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 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 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

specify that hours is the variable of interest

hr  %>% 
  specify(response = hours)
Response: hours (numeric)
# 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

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  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 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  0.144  
 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

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
Response: hours (numeric)
Null Hypothesis: point
# A tibble: 1 x 1
   stat
  <dbl>
1  2.37

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   0.014

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_1_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_2_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: 79, fai: 68, goo: 61, ver: 48
evaluation female 0 1 FALSE 4 bad: 75, fai: 74, ver: 48, goo: 47
salary male 0 1 FALSE 6 lev: 49, lev: 48, lev: 48, lev: 44
salary female 0 1 FALSE 6 lev: 47, lev: 46, lev: 41, lev: 39
status male 0 1 FALSE 3 fir: 93, pro: 90, ok: 73
status female 0 1 FALSE 3 fir: 101, pro: 89, ok: 54

Variable type: numeric

skim_variable gender n_missing complete_rate mean sd p0 p25 p50 p75 p100 hist
age male 0 1 38.63 11.57 20.3 28.50 37.85 49.52 59.6 ▇▇▆▆▆
age female 0 1 41.14 11.43 20.3 31.30 41.60 50.90 59.9 ▆▅▇▇▇
hours male 0 1 49.30 13.24 35.0 37.35 46.00 59.23 79.9 ▇▃▂▂▂
hours female 0 1 49.49 13.08 35.0 37.68 45.05 58.73 78.4 ▇▃▃▂▂

-Females worked an average of 49.5 hours per week -Males worked an average of 49.3 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  78.1 male  
 2  35.1 female
 3  36.9 female
 4  38.5 male  
 5  36.1 male  
 6  78.1 female
 7  76   female
 8  35.6 female
 9  35.6 male  
10  56.8 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  78.1 male  
 2  35.1 female
 3  36.9 female
 4  38.5 male  
 5  36.1 male  
 6  78.1 female
 7  76   female
 8  35.6 female
 9  35.6 male  
10  56.8 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  47.8 male           1
 2  60.3 female         1
 3  46.5 female         1
 4  37.2 male           1
 5  74.1 male           1
 6  35.9 female         1
 7  35.6 female         1
 8  54.5 female         1
 9  55.6 male           1
10  44.1 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  0.505
 2         2 -0.650
 3         3  0.279
 4         4  0.435
 5         5  1.73 
 6         6 -0.139
 7         7 -2.14 
 8         8  0.274
 9         9  0.766
10        10  1.52 
# ... with 990 more rows

null_t_distribution has 1000 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 0.160

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   0.924

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_1_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_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 48.0 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  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")
Response: hours (numeric)
Explanatory: status (factor)
Null Hypothesis: independence
# 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

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  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 = "t")

null_distribution_anova
Response: hours (numeric)
Explanatory: gender (factor)
Null Hypothesis: independence
# A tibble: 1,000 x 2
   replicate   stat
       <int>  <dbl>
 1         1  0.606
 2         2 -0.808
 3         3  0.430
 4         4 -0.136
 5         5  0.404
 6         6  0.140
 7         7 -2.21 
 8         8  1.46 
 9         9 -0.585
10        10 -0.927
# ... 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  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 = observed_f_sample_stat, direction = "greater")

Save preview

ggsave(filename = "preview.png", 
       path = here::here("_posts", "2022-04-02-hypothesis-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