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For the mean comparison among variables, Least Significant difference (LSD) test is the most common method. Today I’ll introduce LSD test using R Studio.

Here is one data.

Cultivar=rep(c("CV1"), each=12)
Nitrogen=rep(rep(c("N0","N1","N2","N3"), each=3))
Block=rep(c("I","II","III"),4)
Yield=c (99, 109, 89, 115, 142, 133, 121, 157, 142, 125, 150, 139)
dataA=data.frame(Cultivar, Nitrogen, Block, Yield)
dataA

   Cultivar  Nitrogen  Block  Yield
1  CV1       N0        I      99
2  CV1       N0        II     109
3  CV1       N0        III    89
4  CV1       N1        I      115
5  CV1       N1        II     142
6  CV1       N1        III    133
7  CV1       N2        I      121
8  CV1       N2        II     157
9  CV1       N2        III    142
10 CV1       N3        I      125
11 CV1       N3        II     150
12 CV1       N3        III    139

This data is about the yield difference of CV1 in response to 4 different nitrogen fertilizer (N0 ,N1, N2, N3). First of all, let’s check the mean per each nitrogen fertilizer.

mean(subset(dataA, Nitrogen=="N0")$ Yield) #99
mean(subset(dataA, Nitrogen=="N1")$ Yield) #130
mean(subset(dataA, Nitrogen=="N2")$ Yield) #140
mean(subset(dataA, Nitrogen=="N3")$ Yield) #138

It seems that yield is different from nitrogen fertilizers, but we need to confirm it statistically. First, I’ll run One-Way ANOVA to check there is a difference in final yield among nitrogen fertilizer through the below codes.

Fert_ANOVA= aov(Yield ~ Nitrogen + factor(Block), data=dataA)
summary(Fert_ANOVA)

               Df   Sum Sq  Mean Sq  F value  Pr(>F)   
Nitrogen       3    3248    1082.7   19.14    0.00179 **
factor(Block)  2    1207    603.3    10.66    0.01059 * 
Residuals      6    339     56.6                   
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Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

The result shows that different nitrogen fertilizer affected final yield (p-value: 0.00179), indicating that yield is different from N0 to N3. Now I’d like to know which nitrogen fertilizer brings about the greatest yield.

To compare the mean of yield per each nitrogen fertilizer, let’s install agricolae package and run LSD test through the below codes.

install.packages("agricolae")
library(agricolae)
LSD_Test= LSD.test(Fert_ANOVA,"Nitrogen")
LSD_Test

$statistics
   MSerror   Df   Mean    CV        t.value    LSD
   56.58333  6    126.75  5.934666  2.446912   15.02855

$parameters
        test  p.ajusted   name.t     ntr  alpha
  Fisher-LSD      none    Nitrogen   4    0.05

$means
   Yield      std r       LCL      UCL Min Max   Q25 Q50   Q75
N0    99 10.00000 3  88.37321 109.6268  89 109  94.0  99 104.0
N1   130 13.74773 3 119.37321 140.6268 115 142 124.0 133 137.5
N2   140 18.08314 3 129.37321 150.6268 121 157 131.5 142 149.5
N3   138 12.52996 3 127.37321 148.6268 125 150 132.0 139 144.5

$comparison
NULL

$groups
     Yield    groups
N2   140      a
N3   138      a
N1   130      a
N0    99      b

The LSD Test indicates that yield was not different from N1, N2, N3 fertilizer applications, while N0 application resulted in the lowest yield (If the letter is the same, it means the mean is the same). If N0 is 0kg nitrogen application, it makes sense that yield is lower than under nitrogen applications.

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What if Two-Way ANOVA? If we want to compare the mean about interaction between two variables, how can we do that?

Here is one data.

Cultivar=rep(c("CV1","CV2"),each=12)
Nitrogen=rep(rep(c("N0","N1","N2","N3"), each=3),2)
Block=rep(c("I","II","III"),8)
Yield=c (99, 109, 89, 115, 142, 133, 121, 157, 142, 125, 150, 139, 82, 104, 99, 117, 125, 127, 145, 154, 154, 151, 166, 175)
dataA=data.frame(Cultivar,Nitrogen,Block,Yield)
dataA

   Cultivar Nitrogen  Block  Yield
1  CV1       N0       I      99
2  CV1       N0       II     109
3  CV1       N0       III    89
4  CV1       N1       I      115
5  CV1       N1       II     142
6  CV1       N1       III    133
7  CV1       N2       I      121
8  CV1       N2       II     157
9  CV1       N2       III    142
10 CV1       N3       I      125
11 CV1       N3       II     150
12 CV1       N3       III    139
13 CV2       N0       I      82
14 CV2       N0       II     104
15 CV2       N0       III    99
16 CV2       N1       I      117
17 CV2       N1       II     125
18 CV2       N1       III    127
19 CV2       N2       I      145
20 CV2       N2       II     154
21 CV2       N2       III    154
22 CV2       N3       I      151
23 CV2       N3       II     166
24 CV2       N3       III    175

This data is a factorial experiment between two genotype (CV1, CV2) and 4 different nitrogen fertilizer (N0, N1, N2, N3), and final yield was measured. First, I’ll run Two-Way ANOVA to check there is an interaction between genotype and nitrogen fertilizer about yield through the below codes.

Fert_ANOVA= aov(Yield~Cultivar+Nitrogen+Cultivar:Nitrogen+factor(Block), data=dataA)
summary(Fert_ANOVA)

                  Df   Sum Sq  Mean Sq  F value   Pr(>F)    
Cultivar           1   253     253      4.99      0.042327 *  
Nitrogen           3   10695   3565     70.17     1.12e-08 ***
factor(Block)      2   1505    752      14.81     0.000351 ***
Cultivar:Nitrogen  3   1040    347       6.82     0.004604 ** 
Residuals         14   711     51                     
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Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

There is an interaction between genotype and nitrogen fertilizer about final yield (p-value: 0.004604). Now, I’ll compare the means among the interactions through the below codes.

LSD_Test= LSD.test(Fert_ANOVA, c("Cultivar", "Nitrogen"))
LSD_Test

       test   p.ajusted  name.              t ntr  alpha
  Fisher-LSD  none       Cultivar:Nitrogen  8      0.05

$means
       Yield       std r       LCL      UCL Min Max   Q25 Q50   Q75
CV1:N0    99 10.000000 3  90.17386 107.8261  89 109  94.0  99 104.0
CV1:N1   130 13.747727 3 121.17386 138.8261 115 142 124.0 133 137.5
CV1:N2   140 18.083141 3 131.17386 148.8261 121 157 131.5 142 149.5
CV1:N3   138 12.529964 3 129.17386 146.8261 125 150 132.0 139 144.5
CV2:N0    95 11.532563 3  86.17386 103.8261  82 104  90.5  99 101.5
CV2:N1   123  5.291503 3 114.17386 131.8261 117 127 121.0 125 126.0
CV2:N2   151  5.196152 3 142.17386 159.8261 145 154 149.5 154 154.0
CV2:N3   164 12.124356 3 155.17386 172.8261 151 175 158.5 166 170.5

$comparison
NULL

$groups
         Yield    groups
CV2:N3   164      a
CV2:N2   151      b
CV1:N2   140     bc
CV1:N3   138      c
CV1:N1   130     cd
CV2:N1   123      d
CV1:N0    99      e
CV2:N0    95      e

Under the N0, yield was the lowest in both CV1 and CV2, and yield was the greatest in CV2 when N3 was applied.

full code: https://github.com/agronomy4future/r_code/blob/main/How_to_conduct_Least_Significant_difference_(LSD)_test_using_R_STUDIO.ipynb

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