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Means equivalence test (TOST / Schuirmann) - statistical subject
Good morning,
My subject is mostly a request about a statistical test and not about the use of JMP, although I will have a small question about this at the end of the discussion.
I am completely confused about the use of two different methods of the TOST / Schuirmann average equivalence test (I am talking about equivalence and not average comparison!).
Here is the data:
Average 1: 101.0
Average 2: 101.9
Lower accepted bias (B-): -1
Upper bias accepted (B+): +2
σ1: 0.9
σ2: 1.5
n1:6
n2: 6
α: 0.05
tcritical: 1.81
Varpool: 1.52
σpool: 1.23
Here are the two methods used to assess equivalence.
1st method:
Calculation of the confidence interval:
CI = (X1 - X2) +/- t(1-2α,n1 + n2 -2)*σpool*(1/n1 + 1/n2)
Which gives [ 0.17 - 1.66 ], the interval being included between the limits of acceptance of the equivalence [ -1 and 2 ], the test concludes with the equivalence of the two averages.
2nd method:
Calculation of TLower and TUpper statistics.
TL = [(X1 - X2) - B-] / σpool * Root (1/n1 + 1/n2).
TU = [(X1 - X2) - B+] / σpool * Root (1/n1 + 1/n2).
I find for TLower: 2.69 and TUpper: -1.52.
T critical with α: 0.05 and dof 10: 1.81.
With the two t values, I cannot find the equivalence (t value -1.52 > -1.81).
So I would like to understand why this difference? Knowing that the confidence interval approach remains extremely rare (see link: https://www.ncbi.nlm.nih.gov/pmc/art...MMC-25-123.pdf )
Otherwise for the use of JMP (version 17.0 in my case), when I launch an equivalence test, JMP asks for a difference value to certify from when the non-equivalence is considered, but I cannot have a off-center bias (my example: -1; +2), since only one value must be entered.
Thank you in advance for your feedback on this subject ^^!
Cordially !
This post originally written in French and has been translated for your convenience. When you reply, it will also be translated back to French .
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Re: Test d'équivalence de moyennes (TOST / Schuirmann) - sujet statistique
I think there is an error in your CI formula. It should be: Confidence interval = (x1–x2) +/- t*√((sp2/n1) + (sp2/n2)) where sp2 = ((n1-1)s12 + (n2-1)s22) / (n1+n2-2).
I used this JMP script with the correct formula:
Names Default to Here( 1 );
xBar1 = 101.0;
xBar2 = 101.9;
t = Students t Quantile( 1-0.05/2, 6 + 6 - 2 );
n1 = n2 = 6;
s1 = 0.9;
s2 = 1.5;
sp = Sqrt( ((n1-1)*(s1^2)+(n2-1)*(s2^2)) / (n1+n2-2) );
ci = (xBar1-xBar2) + {-1,1}*t * Sqrt( (sp^2/n1)+(sp^2/n2) );
I get the result [-2.49120941403157, 0.691209414031555], which exceeds the stated equivalence interval of [-1,2].
JMP uses the second method you illustrate. JMP calculates from the data, not summary sample statistics, so I can't verify the conclusion by the second method.
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Re: Test d'équivalence de moyennes (TOST / Schuirmann) - sujet statistique
Hi @Mark_Bailey : Your t needs to be adjusted.
Should be
t = Students t Quantile( 1-0.05, 6 + 6 - 2 );
This is because, for equivalence at alpha=0.05,the 90% CI (1-2alpha CI) confidence interval is evaluated.
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Re: Test d'équivalence de moyennes (TOST / Schuirmann) - sujet statistique
Good catch! Amended script and new result.
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Re: Test d'équivalence de moyennes (TOST / Schuirmann) - sujet statistique
Hi @Arkawa : You've made some mathematical errors;
Corrections are: :
CI = (X1 - X2) +/- t(1-2α,n1 + n2 -2)*σpool*sqrt(1/n1 + 1/n2)
TL = [(X1 - X2) - B-] / (σpool * Root (1/n1 + 1/n2)).
TU = [(X1 - X2) - B+] / (σpool * Root (1/n1 + 1/n2)).
TL and TU can be derived from the CL formula, so the two methods are equivalent.
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Re: Test d'équivalence de moyennes (TOST / Schuirmann) - sujet statistique
Bonjour,
@Mark_Bailey et @MRB3855 merci pour vos retours très insctructifs !!
Je viens de tester à la main les deux méthodes, pour l'intervalle de confiance je trouve : -0.37 - 2.21 indiquant une absence d'équivalence.
Pour la statistique je trouve : TL = 2.69 et TU = -1.52 pour des Tcritique de -1.81 / + 1.81, non équivalence prouvée, est-ce que quelqu'un peut me vérifier svp ?
Cordialement !