Hypothesis Testing

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

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(???) 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

• use the skim to summarize the data in 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

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

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

Response: hours (numeric)
Null Hypothesis: point
# A tibble: 500,000 x 2
# Groups:   replicate [1,000]
   replicate hours
       <int> <dbl>
 1         1  49.9
 2         1  37.0
 3         1  33.4
 4         1  45.9
 5         1  50.4
 6         1  36.6
 7         1  52.7
 8         1  49.6
 9         1  52.7
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

# A tibble: 1,000 x 2
   replicate     stat
 *     <int>    <dbl>
 1         1  1.96   
 2         2  0.532  
 3         3 -1.04   
 4         4 -0.00975
 5         5  1.32   
 6         6  0.177  
 7         7  0.550  
 8         8  0.517  
 9         9  0.492  
10        10 -0.821  
# … with 990 more rows

null_t_distribution has 1000 t-stats

• visualize the simulated null distribution

• calculate the statistic from your observed data

• Assign the output observed_t_statistic

• Display 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

# A tibble: 1 x 1
  p_value
    <dbl>
1   0.008

shade_p_value on the simulated null distribution

If 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_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

• Q: Is the average number of hours worked the same for both genders?

use skim to summarize the data in hr_2 by gender

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

• specify the variables of interest are hours and 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

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

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  60.8 male           1
 2  36.4 female         1
 3  62.6 female         1
 4  61.9 male           1
 5  48.2 male           1
 6  35.2 female         1
 7  64.1 female         1
 8  42.9 female         1
 9  47.2 male           1
10  65.8 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

# A tibble: 1,000 x 2
   replicate   stat
 *     <int>  <dbl>
 1         1  1.30 
 2         2  1.56 
 3         3  1.46 
 4         4  0.884
 5         5  0.145
 6         6 -0.684
 7         7 -1.20 
 8         8 -0.378
 9         9  0.642
10        10  0.288
# … with 990 more rows

• null_t_distribution has 1000 t-stats

• visualize the simulated null distribution

• calculate the statistic from your observed data

  • Assign the output observed_t_2_sample_stat

  • Display observed_t_2_sample_stat

# A tibble: 1 x 1
   stat
  <dbl>
1 0.160

• get_p_value from the simulated null distribution and the observed statistic

# A tibble: 1 x 1
  p_value
    <dbl>
1   0.862

• shade_p_value on the simulated null distribution

If the p-value < 0.05? ??? (yes)

Does your analysis support the null hypothesis that the true mean number of hours worked by female and male employees was the same? (no)

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

• 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

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 42.7 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

• specify the variables of interest are hours and 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

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

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  46.2 fired            1
 2  65.1 ok               1
 3  40   fired            1
 4  48   promoted         1
 5  56.4 ok               1
 6  40.5 ok               1
 7  39.6 promoted         1
 8  59.7 fired            1
 9  56.7 promoted         1
10  35.2 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

# A tibble: 1,000 x 2
   replicate   stat
 *     <int>  <dbl>
 1         1 0.365 
 2         2 2.30  
 3         3 0.166 
 4         4 2.00  
 5         5 0.496 
 6         6 0.0308
 7         7 1.18  
 8         8 0.394 
 9         9 0.0437
10        10 1.23  
# … with 990 more rows

null_distribution_anova has 1000 F-stats

• visualize the simulated null distribution

• calculate the statistic from your observed data

• Assign the output observed_f_sample_stat

• Display observed_f_sample_stat

# A tibble: 1 x 1
   stat
  <dbl>
1  115.

• get_p_value from the simulated null distribution and the observed statistic

# A tibble: 1 x 1
  p_value
    <dbl>
1       0

• shade_p_value on the simulated null distribution

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? ??? (yes)