Hypothesis Testing > How to Calculate the Least significant Difference

Watch the video or read the steps below:

## How to Calculate the Least Significant Difference (LSD): Overview

When you run an ANOVA (Analysis of Variance) test and get a significant result, that means at least one of the groups tested differs from the other groups. However, you can’t tell from the ANOVA test *which* group differs. In order to address this, Fisher developed the least significant difference test in 1935, which is only used when you reject the null hypothesis as a result of your hypothesis test results. The LSD calculates the smallest significant between two means as if a test had been run on those two means (as opposed to all of the groups together). This enables you to make direct comparisons between two means from two individual groups. Any difference larger than the LSD is considered a significant result.

## Least Significant Difference (LSD): Formula

The **formula **for the least significant difference is:

Where:

t = critical value from the t-distribution table

MSw = mean square within, obtained from the results of your ANOVA test

n = number of scores used to calculate the means

## How to Calculate the Least Significant Difference (LSD): Sample Problem

**Sample problem:** Calculate the Least Significant Difference for the difference between two means on Group 1 and Group 2 with the following test results:

Step 1: **Find the t-critical value.** The t-critical value for α=0.05, dfw = 36 is 2.028 (I used the TI-83 calculator to find a t-distribution value). *Make sure you are using the Within groups DF from your results!*

Step 2: **Insert the given values, the MSE from your results (I used 26.65 from Within groups on the above table) the t-distribution value from Step 2 into the least significant difference formula:**

LSD = 2.028 √ (26.65 * (2/10)) = 4.68.

Put this number aside for a moment,

Step 3: **Calculate ybar1-ybar2 from the results. **For this example, we get -8.7.

Step 4: **Compare Step 2 and Step 3.** If |ybar1-ybar2| ≥ LSD_{1,2} then you can reject the null hypothesis that the means are the same (H_{0}:μ_{1} = μ_{2}). Our value of 8.7 is larger than 4.68 so we can reject the null hypothesis.

*That’s it!*

**Tip**: The LSD will only make sense if you have a significant result from ANOVA (i.e. if you reject the null hypothesis). Therefore, you shouldn’t run the test if you do not get a significant result from ANOVA.

thank you T_T

Thanks for your explanations

Thanks so much.

thanks so much, it really helped

I need more solved exapmle

Good

Dis grt

I need more examples for clarification if you could simply write it in

other way around.

Thankyou!

Hello, Serah,

Could you clarify what you mean by “write it in

other way around.”? An example would help.

Thanks!

atleast i’ve got an idea

Help with steps to run fisher’s unprotected test using genstat 17

Sorry, Mduh…I don’t know anything about genstat.

I need some clarity and more solved examples

I’d be happy to add another example when I update the article :)

What did you need clarified?

Thanks for the enlightenment

THANKS SO MUCH I NOW HAVE A BETTER UNDERSTANDING OF WHEN TO USE LSD

It is ok and well explained for me