### Document

```Running Scheffe’s
Multiple Comparison Test
in Excel
For founding Inter-Groups
Differences after getting significant
results in overall ANOVA test
Scheffe’s Test
• Scheffe’s Test is a very popular and the most
conservative Post Hoc Test
(Post Hoc = unplanned before experiments).
• It is most useful in cases of UNequal sample
size, when the number of groups increase or
when planned comparison of contrast being
required (Not a Post Hoc test then)!
Part One
One-Way Anova
e.g. Completely Randomized Design
Unequal Sample Size
• An experiment with Completely Randomized Design
has been started with 10 equal weight chickens in each
group of Treatment A (Control), Treatment B,
Treatment C and Treatment D, with increasing dosage
of a new drug that might increase growing rate.
• However, some chickens have died during the
experiment, especially in groups with higher dosage.
• Please find any significant different increase of weight
among the 4 groups.
In Excel with ‘Analysis ToolPak’ Add-In activated, click Data, Data Analysis :-
Choose ‘Anova: Single Factor’ = One-way Anova
Select Data Area including Labels:-
Overall Anova result:The overall Anova result reject the null hypothesis that all group means are equal!
For finding exactly where the differences exist, we proceed to run Scheffe’s Test!!
Critical F value
to be used later
N
Group Means
to be used later
Within Group Variance MSw to be used later
Step One – Creating a
‘Critical F value for Scheffe’ - F
• This ‘Critical F Value for Scheffe’ is calculated by:
F = ‘F Crit’ in Anova X (Num. of Groups – 1)
= 3.0088 X 3
= 9.0264
Step Two – Calculation of
‘F for Scheffe’ for all combinations
Equation for calculation of Fs (F for Scheffe) :-
Any F > 9.0264 would indicate significant
differences
F for A vs B = (10.0456-10.3474)2/1.2408(1/10 +1/8)
= 0.0911/0.2792 = 0.3262 < 9.0264
F for A vs C = (10.0456-12.0956)2/1.2408(1/10 +1/6)
= 4.2025/0.3310 = 12.6964 > 9.0264
F for A vs D = (10.0456-11.8789)2/1.2408(1/10 +1/4)
= 3.3610/0.4343 = 7.7389 > 9.0264
F for B vs C = (10.3474-12.0956)2/1.2408(1/8 +1/6)
= 3.0562/0.3619 = 8.4449 > 9.0264
F for B vs D = (10.3474-11.8789)2/1.2408(1/8 +1/4)
= 2.3455/0.4653 = 5.0408 > 9.0264
F for C vs D = (12.0956-11.8789)2/1.2408(1/6 +1/4)
= 0.0470/0.5170 = 0.0910 > 9.0264
Significant Difference
between A and C !!
Counter Checking with SPSS
Using the same Data Set
Choose Post Hoc test e.g. Scheffe :-
Overall Anova result similiar to that in Excel:-
The Result in SPSS for the Scheffe Test well matched the result in Excel that only
Group A and Group C are found to be significantly different in their Group Mean.
Proving that the Excel results are
exactly equal to that in SPSS!!
• Although the Excel result for the Scheffe test well
matched that in SPSS, this might not be enough to
prove the figures they got are absolutely the same!
. Unlike in Tukey’s Test that ‘Critical Differences’ are
can be used for counter checking to 95% Conf. Int.
of the SPSS output.
But we can simply use another method by checking
the ‘F-Table’ backward.
For Proving that
the Excel result is consistent with SPSS!!
For Example, for Group A vs Group C :F for A vs C = (10.0456-12.0956)2/1.2408(1/10 +1/6)
= 4.2025/0.3310 = 12.6964 > 9.0264
12.6964/(k-1) = 12.6964/3 = 4.2321
Num of Groups - 1
Let’s Check the F Table on df (3,24)
0.016
SPSS output
4.2321
Excel
Result
Conclusion
• After activating the ‘Analysis TookPak’ Add-in in
Excel, we can have useful statistical tests to use
including different Anova tests.
• We find that, if overall Anova result is significant,
we can work further to run Post Hoc Test e.g.
Scheffe’s Test to find where the mean differences
exist, not too difficultly!
• For One-way Anova, the Excel result has been
proved to be consistent with SPSS, even with
Unequal Sample Size! Let’s go to Part 2 for Twoway Anova now!!
Part 2
Two-Way Anova
e.g. Scheffe’s HDS Test using Excel
in
aXb factorial Design
With Replication
- 6 cages each with 4 rats have been used for a Completely
Randomized Two-Factors (a x b factorial) With Replication
Design Experiment.
- The 24 rats had been assigned randomly to be subjects for
the ‘combinations’ of factor one (Diet A, B, C, D) with factor
two (Lighting 1, 2, 3-2 times each). The response is a ‘score’
after the 12 ‘treatments’ e.g. a growing rate in body weight
within a certain period of time.
Please find any Significant Differences caused by the two
factors.
Running the Scheffe’s Test
in Excel
e.g.Two-way Anova
aXb Factorial Design
With Replication
In Excel with ‘Analysis ToolPak’ Add-In activated, click Data, Data Analysis :-
Choose ‘Two-Factor With Replication’:-
Select Data Area including all Labels :-
A closer look:-
Range Including Labels
Number of rows
of Replication
Output :-
Overall Anova Results
Pair of degree of freedom
To be used for checking F!!
Overall Anova Result
Lighting
(Significant)
Diet
(Significant)
Interaction
(Not Significant)
‘Critical F’ value for Scheffe’s Test
‘MSE’ for
Scheffe’s Test
For the factor ‘Diet’
the Group Means are:-
Step One – Creating a
‘Critical F value for Scheffe’ - F
• This ‘Critical F Value for Scheffe’ is calculated by:
F = ‘Critical F in Anova’ X (Num. of Groups – 1)
= 3.4903 X 3
= 10.4709
Step Two – Calculation of
‘F for Scheffe’ for all combinations
Equation for Group Differences in Scheffe’s Test:-
Any F > 10.479 would indicate significant
differences for ‘Diet’
F for A vs B = (13.9167-22.3333)2/10.417(1/6+1/6)
= 70.8392/3.4726 = 20.3977> 10.4709
F for A vs C = (13.9167-13.8333)2/10.417(1/6 +1/6)
= 0.0070/3.4726 = 2.0158 < 10.479
F for A vs D = (13.9167-21.4167)2/10.4167(1/6+1/6)
= 56.25/3.4726 = 16.198 > 10.479
F for B vs C = (22.3333-13.8333)2/10.4167(1/6 +1/6)
= 72.25/3.4726 = 20.8057 > 10.479
F for B vs D = (10.3474-11.8789)2/10.4167(1/6 +1/6)
= 0.8402/3.4726= 0.2420 < 10.479
F for C vs D = (13.8333-21.4167)2/10.4767(1/6 +1/6)
= 57.5080/3.4726 = 16.5605 > 10.479
Significant Difference
between A and B !!
Significant Difference
between A and D !!
Significant Difference
between B and C !!
Significant Difference
between C and D !!
Counter Checking with SPSS
Using the same Data Set
The overall results are identical with that in Excel output previously:-
Proving that the Excel result are
exactly equal to that in SPSS!!
• Although the Excel result for the Scheffe test well
matched that in SPSS, this might not be enough to
prove the figures they got are absolutely the same!
. Unlike in Tukey’s Test that ‘Critical Differences’ are
can be used for counter checking to 95% Conf. Int.
of the SPSS output.
But we can simply use another method by checking
the ‘F-Table’ backward.
For Proving that
the Excel result is consistent with SPSS!!
F for A vs B = (13.9167-22.3333)2/10.4167(1/10 +1/10) SPSS sign.
20.3977/ 3 = 6.7992
0.006
(Group Num. -1)
F for A vs D = (13.9167-21.4167)2/10.4167(1/10 +1/0)
16.1980/3 = 5.3993
0.014
F for B vs C = (22.3333-13.8333)2/10.4167(1/10 +1/10)
20.8057/ 3 = 6.3952
0.006
F for C vs D = (13.8333-21.4167)2/10.4767(1/10 +1/10)
16.5605/3 = 5.5202
0.013
The Significance checked from F value from Excel well matched that in the SPSS output!!
Conclusion
• After activating the ‘Analysis TookPak’ Add-in in
Excel, we can have useful statistical tests to use
including different Anova tests.
• We find that, if overall Anova result is significant,
we can work further in Excel to run Post Hoc Test
e.g. Scheffe’s Test to find where the mean
differences exist, not too difficultly!
• We find that this is not only possible in One-way
Anova, but even in Two-way Anova, such as aXb
factorial tests!!
Thank You very much!
```