[Rasch] Mean diferences and measurement error

Adams, Ray adams at acer.edu.au
Sun Jun 3 11:27:27 EST 2007

it depends upon the estimator you are using.  For simplicity let's assume you are using MLEs or WLEs for estimators of location on the latent variable.  The with a t-test you luck out and will automatically be taking measurement error into account.  You are in luck because the population variance estimate that you get from the MLEs is biased upwards due to the presence of measurement error.  When you use this (biased) variance estimate in a t-test you are (unwittlingly) reducing the t value.  A little bit of mathematics shows that the resulting reduction is equivalent to multiplying the t value by the reliability.  In other words the estimated t is reduced due to the addition of the measurement.  Of of course this does not hold if you use an unbiased estimate of the variance, then you'd need to add the measurement error.  Also, note that while this holds for a t-test it does not hold for other statistics.


From: rasch-bounces at acer.edu.au on behalf of Andrés Burga León
Sent: Sun 6/3/2007 1:58 AM
To: rasch
Subject: [Rasch] Mean diferences and measurement error

Hello to everybody on the list:


I was thinking about this subject and I wan't to know your opinions:


In every statistical test, you found that if you wan't to assess mean differences you could use a simple t test: (mean1 - mean 2) / standard error. But this formula only considers the sampling error. It assumes perfect reliability.


What about the measurement error? Why didn't consider it in the assessment of mean differences. I'm not expert in this subject, but could it be possible to made a linear composite of sampling error and measurement error? I mean something like sqrt(sampling error^2 + measurement error^2) and so assess better mean groups differences?


Many greetings




Mg. Andrés Burga León

Profesor Asociado

Facultad de Psicología


Universidad Peruana Cayetano Heredia

Amendariz 497

Lima 18



Teléfono: 242 - 8191



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