How Do You Calculate the Standard Deviation of Total Average in MRI ROIs?

AI Thread Summary
To evaluate the standard deviation of the average value across multiple regions of interest (ROIs) in magnetic resonance imaging, the inverse variance method is recommended. The total variance is calculated as the sum of the inverse variances of each ROI, with weights assigned based on their respective standard deviations. This approach ensures that ROIs with smaller standard deviations contribute more significantly to the overall average. The discussion emphasizes the importance of proper weighting in calculating TotalAvg. This method provides a more accurate representation of variability across the selected ROIs.
xfshi2000
Messages
30
Reaction score
0
Hi all:
In magnetic resonance imaging such as human brain head image, I select several regions of interest (ROI) with the size of 8 pixel. Thus the average value and standard deviation are computed for each ROI. Now I want to evaluate the average value over different ROIs. For example, TotalAvg=[AvgVal(ROI1)+AvgVal(ROI2)+AvgVal(ROI3)]/3. How do I calculate the standard deviation of TotalAvg? It seems that If I use error propagation formula, it make no sense. Would anyone like to give one solution? Thanks

xf
 
Physics news on Phys.org
The inverse variance (square of standard deviation) of the average value is the sum of the inverse variances of the individual ROIs. The variance is the square of the standard deviation. The ROI values you add should be weighted using their inverse variances. An ROI value with a small standard deviation has more weight than an ROI with a large one.

Se eqn(2) in
http://controls.engin.umich.edu/wiki/index.php/Basic_statistics:_mean,_median,_average,_standard_deviation,_z-scores,_and_p-value

Bob S
 
Last edited by a moderator:
Thank you Bob. I got it.

xf
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top