+ Reply to Thread
Results 1 to 2 of 2

Thread: Error from calculation of X from Y response in linear regression

  1. #1
    Points: 4, Level: 1
    Level completed: 7%, Points required for next Level: 46

    Thanked 0 Times in 0 Posts

    Error from calculation of X from Y response in linear regression

    Hello, I have what I am pretty sure is a simple problem that I can't get a hold on and could use some help!

    I am sampling air and using a gas chromatograph (GC) quantify my results. I want to calculate a total combined error for my final number, but I am stuck on the standard error of the mass that I get back from my GC's calibration's linear regression.

    Here's the whole story: I have an instrument that I have calibrated with known concentrations, x, to get instrument responses, y. I perform a linear regression to get an equation, y = mx + b. I take a sample for a measured time at a measured flow rate to get my total volume of air collected. This number has simple errors associated with them that I have not problem combining to get my volume and associated error. I then run the sample on my calibrated GC, get an instrument response, y, then use the equation for the line I got from my linear regression to calculate the mass in the sample, x = (y-b)/m.

    So how do I calculate the error associated with that calculated x! The closest I have come is the equation at the bottom of the table in this link:


    When I use that, I get reasonable numbers, I was just hoping for a better reference to use. Any ideas?

    Thanks in advance!

  2. #2
    Points: 12,676, Level: 73
    Level completed: 57%, Points required for next Level: 174
    Master Tagger
    GretaGarbo's Avatar
    Thanked 462 Times in 402 Posts

    Re: Error from calculation of X from Y response in linear regression

    Search for the delta method or Gauss approximation.

    In gas chromatograph (GC) data, the data is often skewed. It is common to take the log on such variable (to do "a normalising transformation"). If you have y1 = mx+b, then taking log so y2 =log(y1), then with the mean of y2, then exp(mean(y2)) will estimate the median in y1, not the mean in y1. This might have relevance for your 'analysis'.

+ Reply to Thread


Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts

Advertise on Talk Stats