比較抽樣與自助法分佈
R 的抽樣
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...
咖啡風味平均值的自助法
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
)
風味平均值的自助法分佈
ggplot(bootstrap_distn, aes(resample_mean)) +
geom_histogram(binwidth = 0.0025)
樣本、自助法分佈、母體的平均
樣本平均
coffee_sample %>%
summarize(mean_flavor = mean(flavor)) %>%
pull(mean_flavor)
7.5163
母體平均的估計值
bootstrap_distn %>%
summarize(mean_mean_flavor = mean(resample_mean)) %>%
pull(mean_mean_flavor)
7.5167
真實母體平均
coffee_ratings %>%
summarize(mean_flavor = mean(flavor)) %>%
pull(mean_flavor)
7.5260
解讀平均值
- 自助法分佈的平均通常與樣本平均幾乎相同。
- 但不一定是母體平均的良好估計。
- 自助法無法修正樣本與母體差異所造成的偏差。
樣本 sd 與自助法分佈 sd
樣本標準差
coffee_focus %>%
summarize(sd_flavor = sd(flavor)) %>%
pull(sd_flavor)
0.3525
母體標準差的估計？
bootstrap_distn %>%
summarize(sd_mean_flavor = sd(resample_mean)) %>%
pull(sd_mean_flavor)
0.01572
樣本、自助法分佈、母體的標準差
樣本標準差
coffee_focus %>%
summarize(sd_flavor = sd(flavor)) %>%
pull(sd_flavor)
0.3525
母體標準差的估計
standard_error <- bootstrap_distn %>%
summarize(sd_mean_flavor = sd(resample_mean)) %>%
pull(sd_mean_flavor)
standard_error * sqrt(500)
0.3515
真實標準差
coffee_ratings %>%
summarize(sd_flavor = sd(flavor)) %>%
pull(sd_flavor)
0.3414
Standard error 是目標統計量的標準差。
Standard error 乘上樣本大小的平方根可用來估計母體標準差。
解讀標準誤
- 「標準誤」的估計值，是樣本統計量之自助法分佈的標準差。
- 將「自助法分佈的標準誤」乘上樣本大小的平方根，可估計母體的標準差。
一起來練習吧！
R 的抽樣
