Date:         Fri, 30 May 2008 19:19:21 -0500
Reply-To:     paul@WUBIOS.WUSTL.EDU
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Paul Thompson <paul@WUBIOS.WUSTL.EDU>
Subject:      Re: modelling lactation curves
Comments: To: nchapinal@YAHOO.COM
In-Reply-To:  <0fdf7a2c-f2f5-4422-80d9-c5a27386acd3@s21g2000prm.googlegroups.com>
Content-Type: text/plain; charset=ISO-8859-1; DelSp="Yes"; format="flowed"

Quoting nchapinal@YAHOO.COM:


Make it a quadratic model.

1) Center day.  I assume that "day" is 0 at calving, and goes for 300
days or so.  To center it, subtract the mean day.  Call that

cday=day-150

Of course, the intercept is now the middle of the measurement period.

2) Compute cday-squared. Call it cday2 or cdaysq.

Model production=cday cdaysq treatment cday*treatment cdaysq*treatment;

If you don't center day, day and daysq are highly collinear, and the
estimates have high standard errors.

> Hi,
>
> milk production follows a curve, as some of you may already now. When
> a high producer cow calves, the milk production per day is aroung 10
> kg a day, but increases very quickly getting soon a peak (even more
> than 40 kg/day) and then it starts decreasing slowly.
> I have seen paper that compare the milk production between two
> treatments by using PROC MIXED and REPEATED.
>
> For instance:
>
> proc mixed;
> class cow treatment;
> model production = treatment day treatment*day;
> repeated /subject=cow type=cs; *they usually try different types to
> see which one fits better;
> run;
>
> However, to me, that is not totally sound, since milk production is
> not linear.
> Is there any way to overcome this problem? Is it going to make any
> difference in the results?
>
> Thanks!
>