如何將 95% 的置信區間添加到 ggplot 中因子水平的比例圖中？

Question

我想建立在我對之前提出的問題得到的很好答案的基礎上：

ggplot2 中因子水平內的圖表比例而不是計數

我希望以代碼為基礎：

var1 <- c("Left", "Right", NA, "Left", "Right", "Right", "Right", "Left", "Left", "Right", "Left", "Left","Left", "Right", "Left", "Right", "Right", "Right", "Left", "Left", "Right", NA, "Left", "Left","Left", "Right", NA, "Left", "Right", "Right", "Right", "Left", "Left", "Right", "Left", "Left","Left", "Right", "Left", "Right", "Right", "Right", "Left", "Left", "Right", NA, "Left", "Left")
var2 <- c("Higher", "Lower", NA, "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", "Slightly higher","Higher", "Lower", "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", NA, "Slightly lower","Higher", "Lower", NA, "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", "Slightly higher","Higher", "Lower", "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly lower", "Higher", "Higher", "Higher", NA, "Slightly lower")
df <- as.data.frame(cbind(var1, var2))

library(dplyr)
library(ggplot2)

df %>%
  na.omit() %>%
  group_by(var1, var2) %>%
  summarise(n = n()) %>%
  mutate(n = n/sum(n)) %>%
  ungroup() %>%
  ggplot() + aes(var2, n, fill = var1) + 
  geom_bar(position = "dodge", stat = "identity") + 
  labs(x="Left or Right",y="Count")+
  scale_y_continuous() +
  scale_fill_discrete(name = "Answer:")+ theme_classic()+ 
  theme(legend.position="top")  +
  scale_fill_manual(values = c("black", "red"))

以 95% 置信區間的形式向圖表上的每個條添加誤差條。 我試圖在術語中添加

upperE=(1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n), lowerE=(-1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n).

但是，唉，我不斷收到錯誤...

我還嘗試為圖表制作一個全新的 dataframe，因此：

var1 <- c("Left", "Right", NA, "Left", "Right", "Right", "Right", "Left", "Left", "Right", "Left", "Left","Left", "Right", "Left", "Right", "Right", "Right", "Left", "Left", "Right", NA, "Left", "Left","Left", "Right", NA, "Left", "Right", "Right", "Right", "Left", "Left", "Right", "Left", "Left","Left", "Right", "Left", "Right", "Right", "Right", "Left", "Left", "Right", NA, "Left", "Left")
var2 <- c("Higher", "Lower", NA, "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", "Slightly higher","Higher", "Lower", "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", NA, "Slightly lower","Higher", "Lower", NA, "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly higher", "Higher", "Higher", "Higher", "Slightly higher","Higher", "Lower", "Slightly higher", "Slightly higher", "Slightly higher", "Lower", "Slightly lower", "Higher", "Higher", "Higher", NA, "Slightly lower")
df <- as.data.frame(cbind(var1, var2))



dat <- df %>%
  na.omit() %>%
  group_by(var1, var2) %>%
  summarise(n = n()) %>%
  mutate(prop = n/sum(n),upperE=1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n, lowerE=-1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n)

test <- ggplot(dat, aes(x=var2, y = prop, fill = var1))+ 
  geom_bar(position = "dodge", stat = "identity") + geom_errorbar(aes(ymin = lowerE, ymax = upperE),position="dodge")+
  labs(x="Answer",y="Proportion")+
  scale_fill_discrete(name = "Condition:")+ theme_classic()+ 
  theme(legend.position="top") 

這給了我錯誤條，但在 Y 軸上位於 0，而不是在每個條的頂部......

有沒有人有什么建議？ 謝謝！

Answer 1

我現在已經弄清楚了如何讓誤差條位於每個條上適當的 position - 我需要將誤差條的 ymin 和 ymax 規范與正在繪制的值相關聯，因此：

dat <- df %>%
  na.omit() %>%
  group_by(var1, var2) %>%
  summarise(n = n()) %>%
  mutate(prop = n/sum(n),upperE=1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n, lowerE=-1.96*sqrt(n/sum(n))*(1-(n/sum(n)))/n)

test <- ggplot(dat, aes(x=var2, y = prop, fill = var1))+ 
  geom_bar(position = "dodge", stat = "identity") + geom_errorbar(aes(ymin = prop+lowerE, ymax = prop+upperE),width = .2, position=position_dodge(.9))+
  labs(x="Answer",y="Proportion")+
  scale_fill_discrete(name = "Condition:")+ theme_classic()+ 
  theme(legend.position="top") 

這給了：

Answer 2

95%CI 中 SE 的比例公式為： se = sqrt((p * (1-p))/n 。所以我認為在上面的解決方案中說明了： sqrt(n/sum(n) * 1-(n/sum(n))/n) 。但是， n只有成功的計數。完整的樣本是sum(n) 。所以它實際上應該是sqrt(n/sum(n) * (1-(n/sum(n))/**sum**(n)) 。

Answer 3

超級舊線程，但以防萬一有人仍然偶然發現這一點：已投票答案中的置信區間公式不正確。

它應該是：

mutate(prop = n/sum(n),
         upperE=1.96*sqrt(n/sum(n)*(1-(n/sum(n)))/sum(n)), 
         lowerE=-1.96*sqrt(n/sum(n)*(1-(n/sum(n)))/sum(n)))

. 使用用於置信區間的公式，您只需對公式的第一位求平方根。 但是，您需要對整個公式取平方根（Z 分數除外）。