- Thread starter Kari Gunson
- Start date

Or that you have had 99 different paths and you observed 16 turtles?

I was thinking of a Poisson model where most observed values were 0 and sometimes 1 and maybe a few 2, so that sum is 16. Then the Poisson parameter would be estimated as 16/99.

A likelihood interval could be formed, or a confidence interval by rejecting values that are not consistent with the observed sum of 16.

Or that you have had 99 different paths and you observed 16 turtles?

I was thinking of a Poisson model where most observed values were 0 and sometimes 1 and maybe a few 2, so that sum is 16. Then the Poisson parameter would be estimated as 16/99.

A likelihood interval could be formed, or a confidence interval by rejecting values that are not consistent with the observed sum of 16.

You can imagine these 16 structures as 16 columns and one row. Then you can do a chi-squared test with 6.1 as expected value and the observations as the number of crossings per structure. (Just do the calculations by hand with a pocket calculator or just excel.)

Next you could have the 16 structures as columns and the hours as rows; from 6 am to 10 pm is 16 hours. So you would have a table with 16x16 (rows x columns) and in each cell the number of crossings; 0, 1, 2…. Most cells would have 0 crossings, many have 1 and possibly a few with 2.

Then you can do a chi-squarred test on that or preferably a Fishers exact test. If you sum these values to the margin you would have the popularity of each structure and the popularity of the hour.

(You could do a logit model with structure and hour as explanatory factors and number of crossings as the dependent, But that maybe seems too complicated.)