Thread: Confidence Intervals when Using a Linear Regression

Hi,

I am doing data analysis for an Atomic Absorption Spectroscopy experiment, and the question I have is pretty basic.

I have done a linear regression on my data, and used the LINEST function in Excel to get s_y, s_m, and s_b. Since I am interested in the value of the x-intercept, it was easy to propagate the values from the LINEST function to find the s value for that value. My problem lies in finding the 95% CI for the x-intercept.

I took replicate data for each of the points I plotted, so each one is actually the average of 3 measurements under the same conditions. Since I had 5 sets of conditions, there were 15 points total going into this analysis. Does this mean I must divide the s value I propagated by sqrt(15) and multiply it by the appropriate t-value for the 95% CI with 13 degrees of freedom? Or have I gotten lost somewhere?

Thanks.

Hey zorz.

Does the LINEST function return the standard errors of the estimated parameters? If not do you know what function does?

Okay, I lave looked up the literature for the LINEST function, and it seems the s_m, s_y, and s_b statistics it gives are all standard errors of those values. LINEST itself performs a linear regression on a given set of datapoints (in this case, 4), and the output is the formula for the best fit line, as well as the standard error values I mentioned above (and R^2 and a few others, too).

I think this means I have answered my own question now. The fact I did a linear regression means the uncertainty in my x-values should be much less than that in my y-values. This is the case, and in fact I am meant to neglect the x-value uncertainty. This means that even though each datapoint is the average of three separate measurements, I can treat each one as though it had no uncertainty. Also, if I am trying to find the 95% CI based upon the standard errors given by LINEST, I just have to find the appropriate t-value and multiply it by the propagated uncertainty. Since I have 4 datapoints, would that mean I should use the t-value for 2 degrees of freedom (4.303) in my calculations?