Math Help - Sine regression on calculator

I'm trying to find a model for months vs. inches of precipitation. I've entered those #s in lists 1 and 2 on my ti-84 plus. Then I followed directions, which said to select SinReg and then type L1, L2 and hit enter. I did, but I got ERR: SINGULAR MAT. Please help? Idk what's wrong, I have it in radian mode.

Any suggestions are appreciated :]

Originally Posted by janedoe

I'm trying to find a model for months vs. inches of precipitation. I've entered those #s in lists 1 and 2 on my ti-84 plus. Then I followed directions, which said to select SinReg and then type L1, L2 and hit enter. I did, but I got ERR: SINGULAR MAT. Please help? Idk what's wrong, I have it in radian mode.

Any suggestions are appreciated :]

There is a lot going on behind the scenes in the calculator that you don't know about. If you don't know anything about matrices, this will sound like gobbledygook.

The calculator uses matrix algebra to create regression curves.

It can only do this if your original matrix (i.e. your lists) is square and non-singular. In other words, you must have the same number of rows as columns, and there exists another matrix that can be multiplied by your original matrix to get the identity matrix. This "other matrix" is known as the inverse matrix.

In your case, your original matrix is singular, so no inverse matrix exists.

Originally Posted by Prove It

There is a lot going on behind the scenes in the calculator that you don't know about. If you don't know anything about matrices, this will sound like gobbledygook.

The calculator uses matrix algebra to create regression curves.

It can only do this if your original matrix (i.e. your lists) is square and non-singular. In other words, you must have the same number of rows as columns, and there exists another matrix that can be multiplied by your original matrix to get the identity matrix. This "other matrix" is known as the inverse matrix.

I doubt that, as long as you have enough data (more data points than parameters say) and it's not degenerate the regression should work. In a sense the more data the better.

In this case I would guess the data is degenerate in some way.

CB

Originally Posted by CaptainBlack

I doubt that, as long as you have enough data (more data points than parameters say) and it's not degenerate the regression should work. In a sense the more data the better.

In this case I would guess the data is degenerate in some way.

CB

Hmm, what would make it degenerate? (sorry, I know nothing about this lol and neither does my teacher!)