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06-24-2017 12:59 PM

Hello,

I wish to study trends in age over time in 3 large samples (>1000 individuals each) from the same population and get p for linear trend.

Samples are independent and data are available for 4 time periods separated by equal time-intervals.

Would the following be correct? Any other suggestion?

proc glm;

class (time period);

model AGE=(time period);

contrast 'linear' (time period) -3 -1 1 3;

run;

Thanks!

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06-24-2017 01:48 PM

EMB wrote:

Hello,

I wish to study trends in age over time in 3 large samples (>1000 individuals each) from the same population and get p for linear trend.

Samples are independent and data are available for 4 time periods separated by equal time-intervals.

Would the following be correct? Any other suggestion?

proc glm;

class (time period);

model AGE=(time period);

contrast 'linear' (time period) -3 -1 1 3;

run;

There are a few things you haven't told us, such as:

- For each individual, are the time intervals exactly the same?
- How can age be a response variable? If you measure the idividual's age at four different time periods ... isn't the individual's age already a known function of the age at the first time interval, plus the delta between the other intervals? What is the source of randomness in this data?

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Posted in reply to PaigeMiller

06-24-2017 03:10 PM

The study includes incident cases of a specific disease, and data are collected at defined periods, each period includes different participants. As part of the descriptive analysis I would like to look at trends over time in age of the diagnosed individuals.

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06-27-2017 01:38 PM - edited 06-27-2017 01:39 PM

EMB wrote:

As part of the descriptive analysis I would like to look at trends over time in age of the diagnosed individuals.

I am totally not understanding this part.

Trends of what? Your sentence (and code) implies you want to find the trend over time of AGE. That cannot be right. Your age increases 1 year for every 1 year of change in time. No statistics or regression or p-values needed.