双样本比例检验

R 中的假设检验

Richie Cotton

Data Evangelist at DataCamp

比较两个比例

$H_{0}$：30岁及以上与30岁以下的SO用户作为业余爱好者的比例相同。

$H_{0}$：$p_{\geq30} - p_{<30} = 0$

$H_{A}$：30岁及以上与30岁以下的SO用户作为业余爱好者的比例不同。

$H_{A}$：$p_{\geq30} - p_{<30} \neq 0$

alpha <- 0.05

计算 z 分数

$$ z = \frac{(\hat{p}_{\geq30} - \hat{p}_{<30}) - 0}{\text{SE}(\hat{p}_{\geq30} - \hat{p}_{<30})} $$

$$ \text{SE}(\hat{p}_{\geq30} - \hat{p}_{<30}) = \sqrt{\dfrac{\hat{p} \times (1 - \hat{p})}{n_{\geq30}} + \dfrac{\hat{p} \times (1 - \hat{p})}{n_{<30}}} $$

$\hat{p}$ 是 $p$ 的合并估计（共同且未知的成功比例）。

$$ \hat{p} = \frac{n_{\geq30} \times \hat{p}_{\geq30} + n_{<30} \times \hat{p}_{<30}}{n_{\geq30} + n_{<30} } $$

只需计算4个数：$\hat{p}_{\geq30}$、$\hat{p}_{<30}$、$n_{\geq30}$、$n_{<30}$。

获取 z 分数所需的数值

stack_overflow %>%
  group_by(age_cat) %>%
  summarize(
    p_hat = mean(hobbyist == "Yes"),
    n = n()
  )

# A tibble: 2 x 3
  age_cat     p_hat     n
  <chr>       <dbl> <int>
1 At least 30 0.773  1050
2 Under 30    0.843  1216

z_score

-4.217

使用 prop_test() 进行比例检验

library(infer)
stack_overflow %>% 
  prop_test(

    hobbyist ~ age_cat,                   # proportions ~ categories

    order = c("At least 30", "Under 30"), # which p-hat to subtract

    success = "Yes",                      # which response value to count proportions of

    alternative = "two-sided",            # type of alternative hypothesis

    correct = FALSE                       # should Yates' continuity correction be applied?

  )

# A tibble: 1 x 6
  statistic chisq_df   p_value alternative lower_ci upper_ci
      <dbl>    <dbl>     <dbl> <chr>          <dbl>    <dbl>
1      17.8        1 0.0000248 two.sided     0.0605    0.165

让我们来练习！

R 中的假设检验