一、同：
1.原理：
自变量不同产生的差异/随机因素产生的差异，来检验这个自变量不同产生的差异是否足够大到，可以结论说这个自变量对因变量有显著影响。
2.检验显著性
t^2 = F
二、异：
| t.test | F.test | |
|---|---|---|
| 1.indication | 2组的均值比较 | 2组及以上的比较 |
| 方差齐（组内差异相等） | 组内差异&组间差异 | |
| 小样本（n<30) | ||
| 当比较方差齐的两组均值时，P(t.test)=P(F.test) | ||
| 2. methods | (组1均值-组2均值)/ 方差 | 求和（每组均值-总均值）/求和（每个值-组内均值） |
| 1.独立样本T检验，2.配对样本T检验,3.单样本T检验 | 1.单因素（一组多变量），2.多因素 | |
三、一个例子
当比较方差齐的两组时，P(t.test)=P(F.test) CSDN
#1.两组数据，方差齐
weight<-scan()
16.68 20.67 18.42 18 17.44 15.95 18.68 23.22 21.42 19 18.92 NA
V<-rep(c('LY1','DXY'),rep(6,2))
df<-data.frame(V,weight)
a<-subset(df$weight,V=='LY1')
b<-subset(df$weight,V=='DXY')
var.test(a,b) #检验是否方差齐
#p-value =0.6653，，接受H0，方差齐
#{
F test to compare two variances
data: a and b
F = 0.6729, num df = 5, denom df = 4, p-value =
0.6653
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.07185621 4.97127448
sample estimates:
ratio of variances
0.6728954
}
t.test(a,b,var.equal=T,paired = F)
#p-value = 0.0571
#{ Two Sample t-test
data: a and b
t = -2.1808, df = 9, p-value = 0.0571
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-4.86513222 0.08913222
sample estimates:
mean of x mean of y
17.860 20.248
}
fit<-aov(weight~V,data=df)
summary(fit)
#p-value = 0.0571
#t^2=(-2.1808)^2 = F=4.756
#{
Df Sum Sq Mean Sq F value Pr(>F)
V 1 15.55 15.55 4.756 0.0571 .
Residuals 9 29.43 3.27
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
1 observation deleted due to missingness
}
#2.两组数据，方差不齐
weight<-scan()
16.68 20.67 18.42 18 17.44 30 18.68 23.22 21.42 19 18.92 82
V<-rep(c('LY1','DXY'),rep(6,2))
df<-data.frame(V,weight)
a<-subset(df$weight,V=='LY1')
b<-subset(df$weight,V=='DXY')
var.test(a,b)
#p-value= 0.002832,方差不齐
#{
F test to compare two variances
data: a and b
F = 0.038913, num df = 5, denom df = 5, p-value
= 0.002832
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.005445095 0.278085194
sample estimates:
ratio of variances
0.03891273
}
t.test(a,b,var.equal=T,paired = F)
#p-value = 0.3488
#{
Two Sample t-test
data: a and b
t = -0.98304, df = 10, p-value = 0.3488
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-33.77097 13.09431
sample estimates:
mean of x mean of y
20.20167 30.54000
}
t.test(a,b,var.equal=F,paired = F) #Welch法，校正方差不齐
#p-value = 0.3676
#{
Welch Two Sample t-test
data: a and b
t = -0.98304, df = 5.3885, p-value = 0.3676
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-36.79643 16.11976
sample estimates:
mean of x mean of y
20.20167 30.54000
}
fit<-aov(weight~V,data=df)
summary(fit)
#p-value = 0.349
#{
Df Sum Sq Mean Sq F value Pr(>F)
V 1 321 320.6 0.966 0.349
Residuals 10 3318 331.8
}
