社會科學統計方法
隨機實驗與因果關係
隨機實驗與因果關係
蔡佳泓
3/1/2024
1 課程目標
- 測量：我們統計一個國家民眾對於政治的興趣、一個地區的空氣品質、擲無數次銅板得到正面的比例。有一篇文章探討擲硬幣的實驗結果。
- 預測：我們收集資料，建立一個模型，然後預測下一個事件發生的機率。最近流行的人工智慧(AI)，屬於預測的一種，不過生成式AI是產生新的資料而不是預測，MIT有一篇文章Explained: Generative AI討論這些概念。
- 因果關係：我們可以透過隨機實驗，排除所有可能的干擾因素，估計出兩個變數的因果關係。例如對一群人隨機抽樣，有人抽到籤去服役，有人不需要去服役，然後在若干年後詢問這些人對於國家的認同感，這些人之間唯一的差別是有沒有去服役，就可以估計出服役與否對國家認同感的作用。
2 因果關係的實例
Figure 2.1: 學前教育
Figure 2.2: 餐廳員工數與最低時薪
3 實驗設計的原理
3.1 Potential Outcome Model
\(D_{i}\): Indicator of treatment for unit \(i\) \[ \mathrm{D}_{i} = \begin{cases} 1, & \text{if unit} \hspace{3pt} i \hspace{3pt} \text{receives treatment} \\ % & is your "\tab"-like command (it's a tab alignment character) 0, & \text{otherwise} \end{cases} \]
\(Y_{i}\): Observed outcome of interest for unit \(i\)
\[ \tag{1} \mathrm{Y}_{di}=\begin {cases} Y_{0i}, & \text{Potential outcome for unit} \hspace{3pt}i\hspace{3pt} \text{without treatment} \\ Y_{1i}, & \text{Potential outcome for unit } \hspace{3pt} i \hspace{3pt} \text{with treatment} \end{cases} \]
\(Y_{i}=D_{i}\cdot Y_{1i}+(1-D_{i})\cdot Y_{0i}\)
\[\begin{equation} \tag{2} \mathrm{Y}_{i}=\begin {cases} Y_{0i}, & \text{if} \hspace{3pt}D_{i}=0 \\ Y_{1i}, & \text{if} \hspace{3pt} D_{i}=1 \end{cases} \end{equation}\]
\(\alpha_{i} = Y_{1i}-Y_{0i}\)
\[ Y_{i}(D_{1}, D_{2},\ldots,D_{n})=Y_{i}(D'_{1}, D'_{2},\ldots,D'_{n})\quad \text{if}\quad D_{i}=D'_{i} \]
\[\begin{align} \tag{3} ATE & = E[Y_{1}-Y_{0}]\\ & =E[Y_{1}]-E[Y_{0}]\\ & =E[Y_{1}\mid D=1]-E[Y_{0}\mid D=0] \\ & =E[Y_{1}-Y_{0}]\\ & =E[Y_{1}]-E[Y_{0}] \end{align}\]
| \(i\) | \(Y_{1i}\) | \(Y_{0i}\) | \(D_{i}\) | \(\alpha_{i}\) | \(Y_{i}\) |
|---|---|---|---|---|---|
| 1 | 3 | 0 | 1 | 3 | 3 |
| 2 | 1 | 0 | 1 | 0 | 1 |
| 3 | 1 | 0 | 1 | 0 | |
| 4 | 1 | 1 | 0 | 0 | |
| \(E[Y_{1}]\) | 1.5 | ||||
| \(E[Y_{0}]\) | 0.5 | ||||
| \(E[Y_{1}-Y_{0}]\) |
\[ E[Y_{1}] = \frac{1}{N}\Sigma Y_{1i}=1.5 \] \[ E[Y_{0}] = \frac{1}{N}\Sigma Y_{0i}=0.5 \] \[ E[Y_{1}]-E[Y_{0}] = 1 \] \[ \alpha_{ATE}=E[Y_{1}-Y_{0}]=\frac{1}{4}\cdot(3+0+1+0)=1 \]
\[ \alpha_{ATT}=E[Y_{1i}]=\frac{1}{2}\cdot(3+0)=1.5 \]
4 實際操作：小班制與成績
star <- read.csv("~/Dropbox/EastAsia2024/data/DSS/STAR.csv")
head(star)## classtype reading math graduated
## 1 small 578 610 1
## 2 regular 612 612 1
## 3 regular 583 606 1
## 4 small 661 648 1
## 5 small 614 636 1
## 6 regular 610 603 0star <- star %>% mutate(D = as.factor(classtype)) %>%
mutate(graduated = recode_factor(graduated,
'1'='Yes', '0'='No'))
class(star$D)## [1] "factor"levels(star$D)## [1] "regular" "small"mean(star$reading[star$D=='small'])## [1] 632.7mean(star$reading[star$D=='regular'])## [1] 625.5library(gtsummary)
star %>% tbl_summary(
by = D,
statistic = list(
all_continuous() ~ "{mean} ({sd})",
graduated ~ "{n} / {N} ({p}%)"
),
digits = all_continuous() ~ 2) %>%
add_overall() %>%
add_n() %>%
modify_header(label ~ "**Variable**") | Variable | N | Overall, N = 1,2741 | regular, N = 6891 | small, N = 5851 |
|---|---|---|---|---|
| classtype | 1,274 | |||
| regular | 689 (54%) | 689 (100%) | 0 (0%) | |
| small | 585 (46%) | 0 (0%) | 585 (100%) | |
| reading | 1,274 | 628.80 (36.73) | 625.49 (35.88) | 632.70 (37.37) |
| math | 1,274 | 631.59 (38.84) | 628.84 (37.94) | 634.83 (39.66) |
| graduated | 1,274 | 1,108 / 1,274 (87%) | 597 / 689 (87%) | 511 / 585 (87%) |
| 1 n (%); Mean (SD); n / N (%) | ||||
mean(star$reading[star$D=='small']) - mean(star$reading[star$D=='regular'])## [1] 7.2115 需要注意的議題
5.1 選樣偏誤(selection bias)
5.2 比對兩組樣本的差異(Balance Check)
5.3 平均值差異的標準誤
\[ \sqrt{var(\hat{Y}_{treated}+\hat{Y}_{control})} =\sqrt{var(\hat{Y}_{treated}) + var(\hat{Y}_{control})} =\sqrt{\frac{\sigma^2_{T}}{N_{T}} + \frac{\sigma^2_{C}}{N_{C}}} \]
v.small <- var(star$reading[star$D == 'small']); v.small## [1] 1396v.regular <- var(star$reading[star$D == 'regular']); v.regular## [1] 1287n.group<- star %>% group_by (D) %>%
summarise(n = n())
n.small<-as.numeric(n.group[2,2]); n.regular<-as.numeric(n.group[1,2])
PoolSE<-sqrt((v.small/n.small)+(v.regular/n.regular)); PoolSE## [1] 2.063library(gtsummary)
star %>% tbl_summary(
by = D,
statistic = list(
all_continuous() ~ "{mean} ({sd})",
all_categorical() ~ "{n} / {N} ({p}%)"
),
digits = all_continuous() ~ 2) %>%
add_p(pvalue_fun = ~ style_pvalue(.x, digits = 2)) %>%
add_overall() %>%
add_n() %>%
modify_header(label ~ "**Variable**") | Variable | N | Overall, N = 1,2741 | regular, N = 6891 | small, N = 5851 | p-value2 |
|---|---|---|---|---|---|
| classtype | 1,274 | <0.001 | |||
| regular | 689 / 1,274 (54%) | 689 / 689 (100%) | 0 / 585 (0%) | ||
| small | 585 / 1,274 (46%) | 0 / 689 (0%) | 585 / 585 (100%) | ||
| reading | 1,274 | 628.80 (36.73) | 625.49 (35.88) | 632.70 (37.37) | <0.001 |
| math | 1,274 | 631.59 (38.84) | 628.84 (37.94) | 634.83 (39.66) | 0.013 |
| graduated | 1,274 | 1,108 / 1,274 (87%) | 597 / 689 (87%) | 511 / 585 (87%) | 0.71 |
| gender | 1,274 | 572 / 1,274 (45%) | 300 / 689 (44%) | 272 / 585 (46%) | 0.29 |
| interest | 1,274 | 0.42 | |||
| 1 | 257 / 1,274 (20%) | 142 / 689 (21%) | 115 / 585 (20%) | ||
| 2 | 235 / 1,274 (18%) | 129 / 689 (19%) | 106 / 585 (18%) | ||
| 3 | 128 / 1,274 (10%) | 68 / 689 (9.9%) | 60 / 585 (10%) | ||
| 4 | 280 / 1,274 (22%) | 138 / 689 (20%) | 142 / 585 (24%) | ||
| 5 | 374 / 1,274 (29%) | 212 / 689 (31%) | 162 / 585 (28%) | ||
| 1 n / N (%); Mean (SD) | |||||
| 2 Pearson’s Chi-squared test; Wilcoxon rank sum test | |||||
R計算如下：
v.small <- var(star$reading[star$D == 'small']); v.small## [1] 1396v.regular <- var(star$reading[star$D == 'regular']); v.regular## [1] 1287n.group<- star %>% group_by (D) %>%
summarise(n = n())
n.small<-n.group[2,2]; n.regular<-n.group[1,2]
p1 <- (((n.small-1)*v.small)+((n.regular-1)*v.regular))
PoolSD<-sqrt(p1/(n.small+n.regular-2));
PoolSE <- PoolSD*sqrt((1/n.small)+(1/n.regular)); PoolSE## n
## 1 2.056- 兩個聯合標準誤的公式的計算結果一致，都是2.0左右。
- 因為2倍的標準誤仍然小於實驗效果(2*2.05<7.2)，可以結論實驗效果並不是隨機發生。換句話說，小班制真的會提高成績。
- 套用以上的標準誤公式，可以確定平均數的差異是否大到不是隨機發生。
6 作業
- 請以圖形表示受試組與實驗組的次數分佈。
- 請問這兩組受訪者的贊成分數的平均數分別是多少？
- 請問這兩組受訪者的贊成分數的平均數差異為何？
- 請問聯合標準誤為何？
7 更新內容日期
最後更新日期 02/24/2024