如何为 lm() 设置平衡单向方差分析 [英] How to set up balanced one-way ANOVA for lm()

问题描述

我有数据:

dat <- data.frame(NS = c(8.56, 8.47, 6.39, 9.26, 7.98, 6.84, 9.2, 7.5),
                  EXSM = c(7.39, 8.64, 8.54, 5.37, 9.21, 7.8, 8.2, 8),
                  Less.5 = c(5.97, 6.77, 7.26, 5.74, 8.74, 6.3, 6.8, 7.1),
                  More.5 = c(7.03, 5.24, 6.14, 6.74, 6.62, 7.37, 4.94, 6.34))

#     NS EXSM Less.5 More.5
# 1 8.56 7.39   5.97   7.03
# 2 8.47 8.64   6.77   5.24
# 3 6.39 8.54   7.26   6.14
# 4 9.26 5.37   5.74   6.74
# 5 7.98 9.21   8.74   6.62
# 6 6.84 7.80   6.30   7.37
# 7 9.20 8.20   6.80   4.94
# 8 7.50 8.00   7.10   6.34

每列提供一组数据.我使用组索引变量:

Each column gives data from a group. I use group index variable:

group <- c(rep("NS",8), rep("EXSM",8), rep("More.5",8), rep("Less.5",8))

当我尝试命令时出现错误

My error occurs when I try the command

fit <- lm(NS ~ group, data = dat)
Error in model.frame.default(formula = NS ~ group, data = dat, drop.unused.levels = TRUE) : 
  variable lengths differ (found for 'group')

我是 lm() 函数的新手，我哪里做错了?我知道在这之后我只需要打电话

I am new to lm() function and where am I doing wrong? I know that after this I just have to call

anova(fit)
plot(fit)

感谢任何帮助！

推荐答案

我们首先使用 stack() 来重塑你的数据:

We first use stack() to reshape your data:

DAT <- setNames(stack(dat), c("y", "group"))
#       y  group
# 1  8.56     NS
# 2  8.47     NS
# 3  6.39     NS
# 4  9.26     NS
# 5  7.98     NS
# 6  6.84     NS
# 7  9.20     NS
# 8  7.50     NS
# 9  7.39   EXSM
# 10 8.64   EXSM
# 11 8.54   EXSM
# 12 5.37   EXSM
# 13 9.21   EXSM
# 14 7.80   EXSM
# 15 8.20   EXSM
# 16 8.00   EXSM
# 17 5.97 Less.5
# 18 6.77 Less.5
# 19 7.26 Less.5
# 20 5.74 Less.5
# 21 8.74 Less.5
# 22 6.30 Less.5
# 23 6.80 Less.5
# 24 7.10 Less.5
# 25 7.03 More.5
# 26 5.24 More.5
# 27 6.14 More.5
# 28 6.74 More.5
# 29 6.62 More.5
# 30 7.37 More.5
# 31 4.94 More.5
# 32 6.34 More.5

分类变量应编码为因子.我们使用 factor 进行编码.使用 levels 参数指定因子水平.

Categorical variable should be coded as factor. We use factor for coding. Use the levels argument to specify factor levels.

DAT$group <- factor(DAT$group, levels = c("NS", "EXSM", "Less.5", "More.5"))

现在，列 y 是自变量(响应)，而列 group 是因变量(协变量)

Now, column y is the independent variable (response), while column group is the dependent variable (covariate)

在统计建模之前，我们可以使用 boxplot 来可视化您的组数据:

Before statistical modelling, we can use boxplot to visualize your group data:

boxplot(y ~ group, DAT)  ## formula method for boxplot

我们看到组NS"和EXSM"的均值似乎没有明显差异，但其他两个级别的均值差异很大.让我们调用 lm():

We see that group "NS" and "EXSM" do not appear to have noticeable difference in mean, but other two levels are quite different in mean. Let's call lm():

fit <- lm(y ~ group, data = DAT)

要分析您的模型，请使用 summary() 和 anova():

For analysis of your model, use summary() and anova():

summary(fit)

# Call:
# lm(formula = y ~ group)

# Residuals:
#      Min       1Q   Median       3Q      Max 
# -2.52375 -0.52750  0.07187  0.56281  1.90500 

# Coefficients:
#             Estimate Std. Error t value Pr(>|t|)    
# (Intercept)   8.0250     0.3553  22.585   <2e-16 ***
# groupEXSM    -0.1312     0.5025  -0.261   0.7959    
# groupLess.5  -1.7225     0.5025  -3.428   0.0019 ** 
# groupMore.5  -1.1900     0.5025  -2.368   0.0250 *  
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

# Residual standard error: 1.005 on 28 degrees of freedom
# Multiple R-squared:  0.3709,  Adjusted R-squared:  0.3035 
# F-statistic: 5.502 on 3 and 28 DF,  p-value: 0.004231

anova(fit)
# Analysis of Variance Table

# Response: y
#           Df Sum Sq Mean Sq F value   Pr(>F)   
# group      3 16.674  5.5579  5.5025 0.004231 **
# Residuals 28 28.282  1.0101                    
# ---
# Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

这篇关于如何为 lm() 设置平衡单向方差分析的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！