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.

解决方案

1. <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.

2. 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.

解决方案

1. 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. Try as.factor() function before aov.

2. Is m the dependent variable? If yes, what is the visbility and soundvolume? 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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