mean_x=mean([x{:,2}]); % convert cell array to matrix, mean by row
Your x is a cell array, with entries that might be different number of rows because the ode*() routines can output a variable number of rows when you provide a time span that is exactly two elements.
The syntax you are using [x{:,2}] is going to access the second element of the cell array only, and extra the one matrix there. Your initial conditions are length 18, so that is going to be a something-by-18 matrix, Then the [] around the single something-by-18 matrix is going to leave it as a something-by-18 matrix. Your x{:,2} looks on the surface like it might be selecting multiple entries in x, but x is a row vector of cell arrays because of the way you build it with [x,{x0}] so there is only one row in x and so x{:,2} is going to be x{1,2}
You then take the mean of that something-by-18 matrix along the first non-scalar dimension, so you will get a 1 x 18 vector.
std_x=mean_x/size(x,2); % std dev of mean(y) is mean(y)/nObs
size(x,2) is going to be the number of cells in the cell array, which will be n. But why would you divide the mean created from a single cell entry by the number of iterations you did?
SEM=(std_x/sqrt(length(x)))
why are you mixxing using length(x) and size(x,2) to get the number of entries in x ? Be consistent or people will get stuck trying to figure out if there are any circumstances under which the two could differ.
ts = tinv([0.025 0.975],length(x)-1); % T-Score
That is going to be a vector of length 2.
CI95 = mean_x + ts.*(std_x/sqrt(length(x)));
mean_x is 1 x 18, std_x is 1 x 18. sqrt(length(x)) is scalar, so you have 1 x 18 + 1 x 2 .* 1 x 18 . But 1 x 2 .* 1 x 18 is going to fail.
If you made
ts = tinv([0.025; 0.975],length(x)-1); % T-Score
then it would be 2 x 1 and that can be .* with 1 x 18 giving 2 x 18, which can be added to 1 x 18, giving 2 x 18.
