比较抽样分布与自助分布
R 中的抽样
Richie Cotton
Data Evangelist at DataCamp
聚焦咖啡子集
set.seed(19790801)
coffee_sample <- coffee_ratings %>%
select(variety, country_of_origin, flavor) %>%
rowid_to_column() %>%
slice_sample(n = 500)
glimpse(coffee_sample)
Rows: 500
Columns: 4
$ rowid <int> 10, 278, 458, 622, 131, 385, 1292, 47, 904, 1020, 5...
$ variety <chr> "Other", "Bourbon", NA, "Caturra", "Caturra", "Yell...
$ country_of_origin <chr> "Ethiopia", "Guatemala", "Colombia", "Thailand", "C...
$ flavor <dbl> 8.58, 7.75, 7.75, 7.50, 8.00, 7.83, 7.17, 8.08, 7.3...
咖啡风味均值的自助法
mean_flavors_1000 <- replicate(
n = 1000,
expr = coffee_sample %>%
slice_sample(prop = 1, replace = TRUE) %>%
summarize(mean_flavor = mean(flavor, na.rm = TRUE)) %>%
pull(mean_flavor)
)
bootstrap_distn <- tibble(
resample_mean = mean_flavors_1000
)
风味均值的自助分布
ggplot(bootstrap_distn, aes(resample_mean)) +
geom_histogram(binwidth = 0.0025)
样本、自助分布、总体的均值
样本均值
coffee_sample %>%
summarize(mean_flavor = mean(flavor)) %>%
pull(mean_flavor)
7.5163
总体均值的估计
bootstrap_distn %>%
summarize(mean_mean_flavor = mean(resample_mean)) %>%
pull(mean_mean_flavor)
7.5167
真实总体均值
coffee_ratings %>%
summarize(mean_flavor = mean(flavor)) %>%
pull(mean_flavor)
7.5260
解读均值
- 自助分布的均值通常与样本均值几乎相同。
- 它未必是总体均值的良好估计。
- 自助法无法纠正样本与总体差异导致的偏差。
样本sd 与自助分布sd
样本标准差
coffee_focus %>%
summarize(sd_flavor = sd(flavor)) %>%
pull(sd_flavor)
0.3525
总体标准差的估计？
bootstrap_distn %>%
summarize(sd_mean_flavor = sd(resample_mean)) %>%
pull(sd_mean_flavor)
0.01572
样本、自助分布、总体的标准差
样本标准差
coffee_focus %>%
summarize(sd_flavor = sd(flavor)) %>%
pull(sd_flavor)
0.3525
总体标准差的估计
standard_error <- bootstrap_distn %>%
summarize(sd_mean_flavor = sd(resample_mean)) %>%
pull(sd_mean_flavor)
standard_error * sqrt(500)
0.3515
真实标准差
coffee_ratings %>%
summarize(sd_flavor = sd(flavor)) %>%
pull(sd_flavor)
0.3414
"标准误"是目标统计量的标准差。
标准误乘以样本量的平方根可估计总体标准差。
解读标准误
- 估计的标准误是样本统计量自助分布的标准差。
- 自助分布的标准误乘以样本量的平方根可估计总体标准差。
Passons à la pratique !
R 中的抽样
