using computer simulation. Based on examples from the infer package. Code for Quiz 13.
Load the R packages we will use.
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)
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")
skim(hr)
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
Response: hours (numeric)
# A tibble: 500 × 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
hr %>%
specify(response = hours) %>%
hypothesize(null = "point", mu = 48)
Response: hours (numeric)
Null Hypothesis: point
# A tibble: 500 × 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
hr %>%
specify(response = hours) %>%
hypothesize(null = "point", mu = 48) %>%
generate(reps = 1000, type = "bootstrap")
Response: hours (numeric)
Null Hypothesis: point
# A tibble: 500,000 × 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 _________
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 × 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
visualize(null_t_distribution)
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 × 1
stat
<dbl>
1 2.37
null_t_distribution %>%
get_p_value(obs_stat = observed_t_statistic, direction = "two-sided")
# A tibble: 1 × 1
p_value
<dbl>
1 0.014
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
hr_2_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")
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
hr_2 %>%
ggplot(aes(x = gender, y = hours)) +
geom_boxplot()
Response: hours (numeric)
Explanatory: gender (factor)
# A tibble: 500 × 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
hr_2 %>%
specify(response = hours, explanatory = gender) %>%
hypothesize(null = "independence")
Response: hours (numeric)
Explanatory: gender (factor)
Null Hypothesis: independence
# A tibble: 500 × 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
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 × 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
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 × 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
visualize(null_distribution_2_sample_permute)
_ Assign the output 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 × 1
stat
<dbl>
1 0.160
null_t_distribution %>%
get_p_value(obs_stat = observed_t_2_sample_stat, direction = "two-sided")
# A tibble: 1 × 1
p_value
<dbl>
1 0.924
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
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")
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
hr_anova %>%
ggplot(aes(x = status, y = hours)) +
geom_boxplot()
Response: hours (numeric)
Explanatory: status (factor)
# A tibble: 500 × 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
hr_anova %>%
specify(response = hours, explanatory = status) %>%
hypothesize(null = "independence")
Response: hours (numeric)
Explanatory: status (factor)
Null Hypothesis: independence
# A tibble: 500 × 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
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 × 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
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 × 2
replicate stat
<int> <dbl>
1 1 0.667
2 2 2.78
3 3 1.24
4 4 0.330
5 5 2.08
6 6 1.95
7 7 0.243
8 8 0.312
9 9 0.440
10 10 0.0281
# … with 990 more rows
visualize the simulated null distribution
visualize(null_distribution_anova)
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 × 1
stat
<dbl>
1 115.
null_t_distribution %>%
visualize() +
shade_p_value(obs_stat = observed_f_sample_stat, direction = "greater")
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
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