### Author Topic: constraints=none  (Read 661 times)

#### amikka

• Newbie
• Posts: 1
##### constraints=none
« on: October 29, 2013, 09:02:30 AM »
Hi,

we are analyzing position-effects with a multifaceted model.
In the first step we only used the term model=item and constraints=cases.
In the second step we used the term model= item+position and constraints=cases. In this case it seems that the position parameters are constrained by setting the mean of the postion parameter to zero. Because we wanted a free estimate for all position parameters we tried to analyze the same model (model=item+position) with constraints=none. We fixed no parameters and get an free estimate for all item parameters and all position parameters. If we compare the item parameters of constraints=cases and constraints=none the differ only in a constant. The Correlation between the different estimates of the parameter is 1.0. So everything seems fine.

But we wonder how the item response model in the case constraints=none is identified because neither the mean of the latent variable nor the mean of the item parameter is set equal zero?
In the conquest manual only three of the four possible types (constraints=cases, items, default) are described.

Thanks,
Anna

#### Eveline Gebhardt

• Full Member
• Posts: 103
##### Re: constraints=none
« Reply #1 on: October 30, 2013, 01:26:50 AM »
Hi Anna

The model is not identified when using constraints=none and no parameters are anchored to fixed values. (Doesn't CQ give a warning?) The scale could be placed anywhere between minus and plus infinity. However, the differences between the parameter estimates are the same as when using contraints=cases/items and so a scatter plot of item parameters would show a straight line.

Facet parameters are always fixed so that they add up to zero. Their values are relative to each other and to the average item difficulty.

Best wishes
Eveline