Linear Least Squares Excel Template

[danielu13]

Homework Statement

I had to make a Linear Least Squares Excel template for my first week's Physics lab to analyze linear data for the rest of the semester. All of my values come out correctly, except the standard deviation values. I was wondering if someone could take a look at my template and see what the problem is.

Homework Equations

The Attempt at a Solution

See attached file.

 

[Mark44]

Homework Statement

I had to make a Linear Least Squares Excel template for my first week's Physics lab to analyze linear data for the rest of the semester. All of my values come out correctly, except the standard deviation values. I was wondering if someone could take a look at my template and see what the problem is.

Homework Equations

The Attempt at a Solution

See attached file.

You are using D (E26) in formulas before you calculate it. Your formulas for m and b need to come after (i.e., below or to the right of) values that are used to compute them.

 

[danielu13]

Okay, I didn't realize that Excel worked in order like that. So it would work correctly if I put m and b below the D calculation, or could I put a new cell above these that is =E26, and change E26 for the new cell in the calculation formulas for m and b?

 

[Mark44]

Okay, I didn't realize that Excel worked in order like that. So it would work correctly if I put m and b below the D calculation,

Yes.

or could I put a new cell above these that is =E26, and change E26 for the new cell in the calculation formulas for m and b?

I don't know what you're saying here. At any rate, you don't want to use a value in a formula before you have computed it. That's why you had all those #DIV/0! errors.

 

[Ray Vickson]

Yes.
I don't know what you're saying here. At any rate, you don't want to use a value in a formula before you have computed it. That's why you had all those #DIV/0! errors.

Another way to prevent this is to add a small positive quantity to the denominator, so (for example) instead of dividing by some quantity D you divide by something like D + 0.000001. That will make no practical difference in almost every real case. Alternatively, you could divide by max(D, .000001); so as long as D >= 0.000001 the denominator is not changed: it is still D.

RGV