【发布时间】:2012-11-15 11:03:47
【问题描述】:
我了解与之前帖子的对比,我认为我在做正确的事情,但它并没有达到我的预期。
x <- c(11.80856, 11.89269, 11.42944, 12.03155, 10.40744,
12.48229, 12.1188, 11.76914, 0, 0,
13.65773, 13.83269, 13.2401, 14.54421, 13.40312)
type <- factor(c(rep("g",5),rep("i",5),rep("t",5)))
type
[1] g g g g g i i i i i t t t t t
Levels: g i t
当我运行这个时:
> summary.lm(aov(x ~ type))
Call:
aov(formula = x ~ type)
Residuals:
Min 1Q Median 3Q Max
-7.2740 -0.4140 0.0971 0.6631 5.2082
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 11.514 1.729 6.659 2.33e-05 ***
typei -4.240 2.445 -1.734 0.109
typet 2.222 2.445 0.909 0.381
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 3.866 on 12 degrees of freedom
Multiple R-squared: 0.3753, Adjusted R-squared: 0.2712
F-statistic: 3.605 on 2 and 12 DF, p-value: 0.05943
这里我的参考是我的类型“g”,所以我的typei是类型“g”和类型“i”的区别,我的typet是类型“g”和类型“t”的区别.
我想在这里再看两个对比,typei+typeg 和类型“t”的区别以及类型“i”和类型“t”的区别
对比
> contrasts(type) <- cbind( c(-1,-1,2),c(0,-1,1))
> summary.lm(aov(x~type))
Call:
aov(formula = x ~ type)
Residuals:
Min 1Q Median 3Q Max
-7.2740 -0.4140 0.0971 0.6631 5.2082
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 10.8412 0.9983 10.860 1.46e-07 ***
type1 -0.6728 1.4118 -0.477 0.642
type2 4.2399 2.4453 1.734 0.109
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 3.866 on 12 degrees of freedom
Multiple R-squared: 0.3753, Adjusted R-squared: 0.2712
F-statistic: 3.605 on 2 and 12 DF, p-value: 0.05943
当我尝试通过更改参考来进行第二次对比时,我得到了不同的结果。我不明白我的对比有什么问题。
【问题讨论】:
-
统计问题的正确位置是stats.stackexchange.com。
标签: r statistics anova