R 如何为lm()建立平衡的单向方差分析

R 如何为lm()建立平衡的单向方差分析,r,regression,linear-regression,lm,anova,R,Regression,Linear Regression,Lm,Anova,我有以下数据: 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), Mor

我有以下数据:

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
感谢您的帮助

我们首先使用
stack()

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
现在,列
y
是自变量(应答),列
group
是因变量(协变量)

在进行统计建模之前,我们可以使用
boxplot
可视化您的组数据:

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

我们看到,“NS”组和“EXSM”组在平均值上没有明显差异,但其他两个水平在平均值上有很大差异。让我们调用
lm()


很高兴知道您使用了
stack
anova(fit)
plot(fit)
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
DAT$group <- factor(DAT$group, levels = c("NS", "EXSM", "Less.5", "More.5"))
boxplot(y ~ group, DAT)  ## formula method for boxplot
fit <- lm(y ~ group, data = DAT)
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