How to Use Fisher’s Least Significant Difference (LSD) in R

A one-way ANOVA is used to determine whether or not there is a statistically significant difference between the means of three or more independent groups.

The hypotheses used in a one-way ANOVA are as follows:

  • H0: The means are equal for each group.
  • HA: At least one of the means is different from the others.

If the p-value from the ANOVA is less than some significance level (like α = .05), we can reject the null hypothesis and conclude that at least one of the group means is different from the others.

But in order to find out exactly which groups are different from each other, we must conduct a post-hoc test.

One commonly used post-hoc test is Fisher’s least significant difference (LSD) test.

You can use the LSD.test() function from the agricolae package to perform this test in R.

The following example shows how to use this function in practice.

Example: Fisher’s LSD Test in R

Suppose a professor wants to know whether or not three different studying techniques lead to different exam scores among students.

To test this, she randomly assigns 10 students to use each studying technique and records their exam scores.

The following table shows the exam scores for each student based on the studying technique they used:

We can use the following code to create this dataset and perform a one-way ANOVA on it in R:

#create data frame
df <- data.frame(technique = rep(c("tech1", "tech2", "tech3"), each = 10),
                   score = c(72, 73, 73, 77, 82, 82, 83, 84, 85, 89,
                             81, 82, 83, 83, 83, 84, 87, 90, 92, 93,
                             77, 78, 79, 88, 89, 90, 91, 95, 95, 98))

#view first six rows of data frame

  technique score
1     tech1    72
2     tech1    73
3     tech1    73
4     tech1    77
5     tech1    82
6     tech1    82

#fit one-way ANOVA
model <- aov(score ~ technique, data = df)

#view summary of one-way ANOVA

            Df Sum Sq Mean Sq F value Pr(>F)  
technique    2  341.6  170.80   4.623 0.0188 *
Residuals   27  997.6   36.95                 
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Since the p-value in the ANOVA table (.0188) is less than .05, we can conclude that not all of the mean exam scores between the three groups are equal.

Thus, we can proceed to perform Fisher’s LSD test to determine which group means are different.

The following code shows how to do so:


#perform Fisher's LSD
   MSerror Df Mean       CV  t.value     LSD
  36.94815 27 84.6 7.184987 2.051831 5.57767

        test p.ajusted    name.t ntr alpha
  Fisher-LSD      none technique   3  0.05

      score      std  r      LCL      UCL Min Max   Q25  Q50   Q75
tech1  80.0 5.868939 10 76.05599 83.94401  72  89 74.00 82.0 83.75
tech2  85.8 4.391912 10 81.85599 89.74401  81  93 83.00 83.5 89.25
tech3  88.0 7.557189 10 84.05599 91.94401  77  98 81.25 89.5 94.00


      score groups
tech3  88.0      a
tech2  85.8      a
tech1  80.0      b

[1] "group"

The portion of the output that we’re most interested in is the section titled $groups. The techniques that have different characters in the groups column are significantly different.

From the output we can see:

  • Technique 1 and Technique 3 have significantly different mean exam scores (since tech1 has a value of “b” and tech3 has a value of “a”)
  • Technique 1 and Technique 2 have significantly different mean exam scores (since tech1 has a value of “b” and tech2 has a value of “a”)
  • Technique 2 and Technique 3 do not have significantly different mean exam scores (since they both have a value of “a”)

Additional Resources

The following tutorials explain how to perform other common tasks in R:

How to Perform a One-Way ANOVA in R
How to Perform a Bonferroni Post-Hoc Test in R
How to Perform Scheffe’s Post-Hoc Test in R

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