配对 t 检验

R 中的假设检验

Richie Cotton

Data Evangelist at DataCamp

美国共和党总统投票数据集

state	county	repub_percent_08	repub_percent_12
Alabama	Bullock	25.69	23.51
Alabama	Chilton	78.49	79.78
Alabama	Clay	73.09	72.31
Alabama	Cullman	81.85	84.16
Alabama	Escambia	63.89	62.46
Alabama	Fayette	73.93	76.19
Alabama	Franklin	68.83	69.68
...	...	...	...

500 行；每行代表一次总统选举的县级投票结果。

¹ https://dataverse.harvard.edu/dataset.xhtml?persistentId=doi:10.7910/DVN/VOQCHQ

假设

问题：2008 年相较 2012 年，共和党候选人的得票百分比是否更低？

$H_{0}$: $\mu_{2008} - \mu_{2012} = 0$

$H_{A}$: $\mu_{2008} - \mu_{2012} < 0$

设显著性水平 $\alpha = 0.05$。

数据为配对数据，因为每个百分比对应同一县。

从两个样本到一个样本

sample_data <- repub_votes_potus_08_12 %>% 
  mutate(diff = repub_percent_08 - repub_percent_12)

ggplot(sample_data, aes(x = diff)) +
  geom_histogram(binwidth = 1)

diff 变量的直方图——大多数值在 -10 到 10 之间，存在一些离群值。

计算差值的样本统计量

sample_data %>% 
  summarize(xbar_diff = mean(diff))

  xbar_diff
1 -2.643027

修订后的假设

旧假设

$H_{0}$: $\mu_{2008} - \mu_{2012} = 0$

$H_{A}$: $\mu_{2008} - \mu_{2012} < 0$

新假设

$H_{0}$: $\mu_{\text{diff}} = 0$

$H_{A}$: $ \mu_{\text{diff}} < 0$

$t = \dfrac{\bar{x}_{\text{diff}} - \mu_{\text{diff}}}{\sqrt{\dfrac{s_{diff}^2}{n_{\text{diff}}}}}$

$df = n_{diff} - 1$

计算 p 值

n_diff <- nrow(sample_data)

s_diff <- sample_data %>% 
  summarize(sd_diff = sd(diff)) %>%
  pull(sd_diff)

t_stat <- (xbar_diff - 0) / sqrt(s_diff ^ 2 / n_diff)

-16.06374

degrees_of_freedom <- n_diff - 1

$t = \dfrac{\bar{x}_{\text{diff}} - \mu_{\text{diff}}}{\sqrt{\dfrac{s_{\text{diff}}^2}{n_{\text{diff}}}}}$

$df = n_{\text{diff}} - 1$

p_value <- pt(t_stat, df = degrees_of_freedom)

2.084965e-47

用 t.test() 检验两个均值差异

t.test(

  # Vector of differences
  sample_data$diff,

  # Choose between "two.sided", "less", "greater"
  alternative = "less",

  # Null hypothesis population parameter
  mu = 0

)

    One Sample t-test

data:  sample_data$diff
t = -16.064, df = 499, p-value < 2.2e-16
alternative hypothesis: true mean is less than 0
95 percent confidence interval:
     -Inf -2.37189
sample estimates:
mean of x 
-2.643027

t.test() 与 paired = TRUE

t.test(
  sample_data$repub_percent_08,
  sample_data$repub_percent_12,
  alternative = "less",
  mu = 0,
  paired = TRUE
)

    Paired t-test

data:  sample_data$repub_percent_08 and 
       sample_data$repub_percent_12
t = -16.064, df = 499, p-value < 2.2e-16
alternative hypothesis: true difference in means 
                        is less than 0
95 percent confidence interval:
     -Inf -2.37189
sample estimates:
mean of the differences 
              -2.643027

未配对 t.test()

t.test(
  x = sample_data$repub_percent_08,
  y = sample_data$repub_percent_12,
  alternative = "less",
  mu = 0
)

未配对 t 检验更易出现假阴性（统计功效更低）。

    Welch Two Sample t-test

data:  sample_data$repub_percent_08 and 
       sample_data$repub_percent_12
t = -2.8788, df = 992.76, p-value = 0.002039
alternative hypothesis: true difference in means
                        is less than 0
95 percent confidence interval:
      -Inf -1.131469
sample estimates:
mean of x mean of y 
 56.52034  59.16337

Vamos praticar!

R 中的假设检验