Student Support Forum: '3d plot, confidence interval, and model fitting' topicStudent Support Forum > General > "3d plot, confidence interval, and model fitting" |
I am attaching my data and notebook.
I have data points of five variables(z,x,y,a,b), and I drew a 3d plot which represents the (z,x,y) coordinates which are conditioned by the values of a and b. You can see them in my notebook attached.
Now, I am facing some challenges.
(1) (Xavier and Peter helped me, but I could not realize it.) First, the variable of main interest in the 3d plot is "z".
I want to obtain the 95% confidence interval of z - which is calculated by its standard deviation. In other words, could somebody help me obtain the upper bound and lower bound of each value of "z". For example, when z=3000, the upper bound should be 3000 + 1.96 (t or z scroe) * sd (standard deviation). Then, I would like to record them as new variables in my data. This is challenging. How can I add additional variables which I create. (e.g. z_upper and z_lower)
(2) Then, I hope to make two new 3d plots based on the upper and lower bounds of each value of "z". Maybe the upper bound of "z" will create another 3d plot above the original 3d plot I have already created in the notebook attached. So will the lower bound of "z" below the original 3d plot. So, finally, I hope to have three 3d plots in one image space at the same time.
(3) Sorry for this lengthy question. Finally, I want to know the best unknown function(which means the least residual values or the best fitting model) among the variables.
z = any unknown function of (x,y,a,b)
That should be nonlinear. IS IT POSSIBLE to obtain the best unknown function without specifying any type of function. Because I do not have any idea about how the function might look like, I cannot specify any type of function(e.g. qudratic, polynomial, exponential, logorithm...or any combination of these).
Under this circumstance, what do you think would be the best option for me to find the best fitting unknown function.
Just out of my imagination that
z = 0.23exp(x+x^2)+73xyz^2-a^4xy+ln(ab)*xy .... ............
Thank you so much for reading and many thanks in advance as always.
Attachment: David_practice.zip, URL: ,