bayesian fish length-weight relationship

narcilili

New Member
hello,
I am working on fish length-weight relationship: log W = log a + b log L with a sample size of n = 112. i wonder if i can get some feedback on using jasp summary stats bayesian one sample t test as a substitute for the frequentist t statistic = (b-3)/SE = 3.691, where b is the observed regression coefficient, 3 is the isometric value, SE is the standard error of the regression coefficient. The null is there is no difference between the b and 3.
This is what I got using the summary stats Bayesian one sample t test where the alternative is 56 times more likely than the null. cheers

Attachments

• 80.6 KB Views: 6

narcilili

New Member
Hello hlsmith,
Thank you for your interest. log W = log a + b log L is the linear form of W = aL^b
and is used to estimate a, the intercept, and b, the slope, W is weight, L is length. In fisheries science, the b is then compared to 3 which is called isometric growth. Using the frequentist t = (b-3)/SE = 3.691, where b is the observed regression coefficient (2.642) , 3 is the isometric value, SE is the standard error of the regression coefficient (0.097). The null is there is no difference between the b and 3. using the frequentist approach, df = n-2 = 110 two sided at 0.05 the t tabular is 1.98 and the null is rejected and 2.642 is significantly different from 3. However, I would like to use bayesian analysis and I used JASP summary stats bayesian one sample t test.
I need a second opinion on what I did. I am grateful you are interested.

hlsmith

Less is more. Stay pure. Stay poor.
Hey, I posted then fig that is what you meant. I will reply tomorrow when I have more time.

hlsmith

Less is more. Stay pure. Stay poor.
All of those logs is what was throwing my off. They may be necessary in the biological sciences, I would not now. However the logging of in the raw data of the DV and IV would result in the percentage change in the weight given a percentage change in in the length. Not sure if the intercept needs logging?

I am guessing you are also using flat priors?

If you fit this model and the above attachment was your output, how is three coming into play? Also, I get it is your comparative constant, but can you describe its underlying rationale? Are you trying to say the mean has to be greater than three to be significant? Not sure if the logging is messing with this, perhaps 3 was already logged. Otherwise, the way I would typically do this is look at the posterior for your beta and show its mass is beyond this threshold. Also, since you have a threshold, this would likely be a one-sided test and all of your alpha (level of significance) should be on one-side. Sorry for the quick reply - I have a busy morning.

narcilili

New Member
hello hlsmith,
thank you for your continuing interest. let me share with you what Somy Kuriakose of Fishery Resources Assessment Division ICAR-Central Marine Fisheries Research Institute wrote (https://core.ac.uk/download/pdf/95776221.pdf):
Because the volume of a 3-dimensional object is proportional to the cube of length for a regularly shaped solid, it is expected that the exponent b would have a value of more or less b = 3. Thus, the weight of a fish is proportional to cube of the length of the fish. For an ideal fish which maintains the same shape b=3. Most species of fish do change their shape as they grow and so a cube relationship between length and weight would hardly be expected. In practice, fish that have thin elongated bodies will tend to have values of b that are less than 3 while fish that have thicker bodies will tend to have values of b that are greater than 3. Thus this also help to determine whether somatic growth is isometric (b=3) or allometric. Values of b smaller, equal and larger than 3 indicate isometry, negative allometry and positive allometry respectively.
the t test is explained in the photo i have attached.

cheers

Attachments

• 68.5 KB Views: 0