Question

所以想象一下这两组女性和男性的年龄：

 femalesage<-c(30,52,59,25,26,72,46,32,64,45)
 malesage<-c(40,56,31,63,63,78,42,45,67)

我可以很容易地做一个t.test（女性年龄，malesage）来达到以下结果：

 t.test(femalesage,malesage)

Welch Two Sample t-test

 data:  femalesage and malesage
 t = -1.2013, df = 16.99, p-value = 0.2461
 alternative hypothesis: true difference in means is not equal to 0
 95 percent confidence interval:
 -24.224797   6.647019
 sample estimates:
 mean of x mean of y 
 45.10000  53.88889

现在，假设我有相同的数据组织，所以这样的事情：

ages<-c(30,52,59,25,26,72,46,32,64,45,40,56,31,63,63,78,42,45,67)
genders<-c("F","F","F","F","F","F","F","F","F","F","M","M","M","M","M","M","M","M","M","M")
df<-data.frame(ages, genders)

我希望使用某种回归测试获得与威尔士双样本t检验类似的结果，测试Beta1 = 0与Beta1的斜率不等于0，其中B1是性别系数和反应是年龄。知道我怎么能得到相同的结果？

Answer 1

t检验和线性回归都是一般线性模型的特例。在单个预测器的情况下，对回归系数的显着性的测试等同于t检验的显着性。

R的t.test函数允许以两种不同的方式指定输入数据：如您所做的那样作为两个单独的向量，或者像我在这里一样使用公式接口。同样，执行简单线性回归的lm函数需要公式接口。在这种情况下，这使得两个函数调用相同，我们只需要更改函数的名称。

您的数据：

ages <- c(30,52,59,25,26,72,46,32,64,45,40,56,31,63,63,78,42,45,67)
genders <- c("F","F","F","F","F","F","F","F","F","F","M","M","M","M","M","M","M","M","M","M")
df <- data.frame(ages, genders)

t检验：

t.test(ages ~ genders, data = df)

    Welch Two Sample t-test

data:  ages by genders
t = -1.2013, df = 16.99, p-value = 0.2461
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -24.224797   6.647019
sample estimates:
mean in group F mean in group M 
       45.10000        53.88889

（几乎）相同的回归：

summary(lm(ages ~ genders, data = df))

Call:
lm(formula = ages ~ genders, data = df)

Residuals:
   Min     1Q Median     3Q    Max 
-22.89 -13.49   0.90  11.11  26.90 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   45.100      5.060   8.914 8.12e-08 ***
gendersM       8.789      7.351   1.196    0.248    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 16 on 17 degrees of freedom
Multiple R-squared:  0.07756,   Adjusted R-squared:  0.0233 
F-statistic: 1.429 on 1 and 17 DF,  p-value: 0.2483

请注意，性别的t和beta与p值几乎相同。

找到平行t检验的平均值和检验斜率= 0 t检验

1 个答案: