R 中的一种方式方差分析和 TUKEY 有条件 [英] one way ANOVA and TUKEY in R with conditions
问题描述
我试图找出包含以下 6 个因素的变量 stim_ending_t 之间的平均差异:1, 1.5, 2, 2.5, 3, 3.5
您可以访问 df
问:如何在比较平均值的条件下进行方差分析,其中包含现在关注图像"和现在关注声音".
问:我也想通过事后测试来跟进.
这是我尝试过的方法,但显然不是正确的方法!
# 计算单向方差分析测试res.aov <- aov(m ~ stim_ending_t, 数据 = clean_test_master2)摘要(res.aov)Df Sum Sq Mean Sq F 值 Pr(>F)stim_ending_t 1 7.589 7.589 418.8 <2e-16 ***残差 34 0.616 0.018---表示.代码:0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
我认为 aov 的结果有问题!stim_ending_t 有 6 个因子,因此自由度 (Df) 应该 = 5,而不是上表中的 != 1.
# 事后测试TukeyHSD(res.aov, conf.level = 0.99)这是我收到的消息TukeyHSD.aov(res.aov, conf.level = 0.99) 中的错误:拟合模型中没有因子另外: 警告信息:在复制中(粘贴(〜",xx),数据= mf):忽略非因素:stim_ending_t
注意:参与者在一个会话中完成实验,从任一条件-Opening_text 开始,随机并完成另一个.
- <块引用>
以下6个因素:1, 1.5, 2, 2.5, 3, 3.5
不是!如果您将其视为因素,它将是 一个 具有 六个级别 的因素.您将其用作定量变量,请参阅方差分析表中的
Df
.它应该是 5 而不是 1.在aov
之前尝试as.factor()
函数. m
是因变量吗?如果是,visbility
和soundvolume
是多少?如果它们也是因素,那么独立性假设是错误的.在这种情况下,您应该将这些因素引入到模型中.
I am trying to find the mean differences between my variable stim_ending_t which contains the following 6 factors: 1, 1.5, 2, 2.5, 3, 3.5
You can access the df Here
stim_ending_t visbility soundvolume Opening_text m sd coefVar
<dbl> <dbl> <dbl> <chr> <dbl> <dbl> <dbl>
1 1 0 0 Now focus on the Image 1.70 1.14 0.670
2 1 0 0 Now focus on the Sound 1.57 0.794 0.504
3 1 0 1 Now focus on the Image 1.55 1.09 0.701
4 1 0 1 Now focus on the Sound 1.77 0.953 0.540
5 1 1 0 Now focus on the Image 1.38 0.859 0.621
6 1 1 0 Now focus on the Sound 1.59 0.706 0.444
7 1.5 0 0 Now focus on the Image 1.86 0.718 0.387
8 1.5 0 0 Now focus on the Sound 2.04 0.713 0.350
9 1.5 0 1 Now focus on the Image 1.93 1.00 0.520
10 1.5 0 1 Now focus on the Sound 2.14 0.901 0.422
Here is a visual representation of my data
Q: How I can do ANOVA with the condition of comparing the mean by "Opening_test" which contains "Now focus on the Image", and "Now focus on the Sound."
Q: Also I want to follow that with post hoc test.
Here is what I have tried but apparently is not the right way!
# Compute one-way ANOVA test
res.aov <- aov(m ~ stim_ending_t, data = clean_test_master2)
summary(res.aov)
Df Sum Sq Mean Sq F value Pr(>F)
stim_ending_t 1 7.589 7.589 418.8 <2e-16 ***
Residuals 34 0.616 0.018
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
I think there is something wrong with result from aov! stim_ending_t has 6 factors, so Dgree of fredom (Df) should = 5 not != 1 from the above table.
# post hoc test
TukeyHSD(res.aov, conf.level = 0.99)
Here is the message I got
Error in TukeyHSD.aov(res.aov, conf.level = 0.99) :
no factors in the fitted model
In addition: Warning message:
In replications(paste("~", xx), data = mf) :
non-factors ignored: stim_ending_t
Note: the participants completed the experiment at one session by starting with either condition-Opening_text, randomly and completing the other one.
-
the following 6 factors: 1, 1.5, 2, 2.5, 3, 3.5
Not! It will be one factor with six levels, if you will treat it as factor. You are using it as quantitative variable, see
Df
in ANOVA table. It should be 5 instead 1. Tryas.factor()
function beforeaov
. Is
m
the dependent variable? If yes, what is thevisbility
andsoundvolume
? If they are also factors, the assumption of independence is faulty. In this case you should introduce those factors to model.
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