3
$\begingroup$

I've conducted a Welch Anova from summary statistics with the following valid results:

# Welch's ANOVA (One-way analysis of means) 

data: group and scores
F value  df1  df2     p-value  
5.35     2    16.83   0.015912  *

---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

summary statistics

group   size   mean   sd
<chr>   <int>  <dbl>  <dbl>
A       10     77.3   11.324016 
B       10     89.3   4.808557  
C       10     84.7   5.292552  

underlying data

df <- data.frame(group = rep(c('A', 'B', 'C'), each = 10),
                 score = c(64, 66, 68, 75, 78, 94, 98, 79, 71, 80,
                           91, 92, 93, 85, 87, 84, 82, 88, 95, 96,
                           79, 78, 88, 94, 92, 85, 83, 85, 82, 81))
#summary
group   size   mean   sd
<chr>   <int>  <dbl>  <dbl>
A       10     77.3   11.324016 
B       10     89.3   4.808557  
C       10     84.7   5.292552

I found some references on how to calculate omega squared: https://peterstatistics.com/Packages/python-docs/effect_sizes/eff_size_omega_sq.html

Since i dont have calculated any sum of squares, which formula is appropriate to retrieve omega squared? The reference states, all formulas give the same result. But somehow the results using the formulas from Kirk 1996 and Caroll and Nordholm 1975 differ from the formula provided by Hays 1973 (and Albers and Lakens 2018). Any advice?

Improve this question
$\endgroup$
1
  • $\begingroup$ Thank you for asking a new question. $\endgroup$
    – gung - Reinstate Monica
    Commented Aug 28 at 20:31

2 Answers 2

Reset to default
3
$\begingroup$

The first formula (Kirk) does not require any sums of squares, only df1, F, and N. You have all of that.

# F value  df1  df2     p-value  
# 5.35     2    16.83   0.015912
((5.3492-1)*2) / (2*(5.3492-1) + 30)  # [1] 0.2247742
Improve this answer
$\endgroup$
2
$\begingroup$

There is a direct relationship between s (standard deviation) and SS. $s=\sqrt {\frac {SS} {n-1}}$ as explained in more details in this CV post.
You also should look at this link; it shows how to conduct ANOVA from just summary statistics, step by step, with formulas and examples. This will let you compute all the SS's, F, etc., and use whichever formula you desire.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you very much. Thats straight forward. I'm just wondering, because i used the welch anova to account for unequal variances, but only the approach from hays relies on the weighted means square, whereas the formulas given by kirk and nordholm don't. So my gut tells me the hays formula is more appropriate. Else I could calculate all the SS's and get the (non estimated) omega squared as proposed by you. But again should i use the degrees of freedom calculated by the sum of adjusted weights or not? $\endgroup$
    – whoo
    Commented Aug 29 at 17:14
  • 1
    $\begingroup$ If you used a Welch ANOVA (which one should use all the time, btw), then you should use the d.f.'s for the Welch ANOVA; indeed hays p.486 seems to be more appropriate. I honestly would calculate $\omega^2$ a few different ways, just to see how different (or, hopefully, not) the answers are. $\endgroup$
    – jginestet
    Commented Aug 29 at 18:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.