您没有指定wtd.t.test函数来自哪个包，因此我将假设使用来自“weights”包的函数。根据文档，前两个参数是来自两个组的数据，第3个和第4个参数是两个组中观察值的权重。如果未提供第4个参数，则给定的权重将用于两个组。这意味着您编写的代码正在测试Ya1的加权平均值是否不同于sec1的加权平均值。这似乎不是你想做的。我认为lm更适合您的用例：

# Make some example data
sec1 <- factor(sample(0:1, replace=TRUE, size=700))
Ya1 <- rnorm(700) + as.numeric(sec1)
weights1T <- 1.4^(rnorm(700))
# Use lm() to perform a weighted t-test
summary(lm(Ya1 ~ sec1, weights=weights1T))

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其给出：

> summary(lm(Ya1 ~ sec1, weights=weights1T))

Call:
lm(formula = Ya1 ~ sec1, weights = weights1T)

Weighted Residuals:
    Min      1Q  Median      3Q     Max 
-3.1921 -0.6672 -0.0374  0.7025  4.4411 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)  0.92035    0.05376   17.12   <2e-16 ***
sec11        1.11120    0.07874   14.11   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 1.061 on 698 degrees of freedom
Multiple R-squared:  0.222, Adjusted R-squared:  0.2209 
F-statistic: 199.1 on 1 and 698 DF,  p-value: < 2.2e-16

型
如果你真的想使用wtd.t.test，你可以这样做：

library(weights)
ysplit <- split(Ya1, sec1)
wsplit <- split(weights1T, sec1)
wtd.t.test(y1split[[1]], y1split[[2]], w1split[[1]], w1split[[2]])

型
它给出了与lm()几乎相同的答案：

> wtd.t.test(x=ysplit[[1]], y=ysplit[[2]],
+            weight=wsplit[[1]], weighty=wsplit[[2]])
$test
[1] "Two Sample Weighted T-Test (Welch)"

$coefficients
  t.value        df   p.value 
-13.50571 697.25403   0.00000 

$additional
 Difference      Mean.x      Mean.y    Std. Err 
-1.00357229  1.04628894  2.04986124  0.07430724 

Warning message:
In wtd.t.test(y1split[[1]], y1split[[2]], w1split[[1]], w1split[[2]]) :
  Treating data for x and y separately because they are of different lengths

型

在我看来结果就在你面前。

$additional
       Difference       Mean.x       Mean.y     Std. Err 
     -0.496466247  0.003533753  0.500000000

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Mean.x和Mean.y给予了第一组和第二组的平均值（在代码中称为组0和1，或Ya1和sec1）。
Difference显然是Mean.x减去Mean.y