[J_Holland]

Hi, I've used regression in excell to determin the 95% concentration interval of the slope and intercept of my lineair calibration. I got a questions about how to determine concentration using the formula you get containing the 95% CI's, a formula like this example: y= 1,5x(± 0,2)x + 1,2(±0,5).

Should I only use the 95% CI of the slope? or also the intercept? I notice I get a very wide range of  possible concentrations when I include both 95%CI's (of the slope and intercept). When I only use the 95% CI of the slope I get much "prettier" result which actually seem usefull.

Can anybody tell me how I should use a formule containing 95% CI limits, to determine the concentration my analysis, and its upper and lower expected concentration (95%)?

[mjc123]

Don't add the CIs (confidence intervals, by the way, not concentration intervals) of the slope and intercept, because the slope and intercept are strongly correlated, so you can't add their variances as if they were independent. That's why you get such wide limits. More detailed analysis of the regression gives (IIRC) the following formula for the confidence interval:
CI = ±t_95%*s/sqrt(n)*{1 + (x - x)²/(x² - x²)}^1/2
where t is Student's t-statistic, s is the standard error, and I have written means with underlines because I can't work out how to write overlines. (The means refer to the set of n data points used to calculate the regression line.) Note how this increases as x deviates from x.
Actually this refers to the confidence interval of the regression value of y for a particular value of x. More relevant for you is the prediction interval, which is the likely deviation of a single measurement from the regression line. This is given by
PI = ±t_95%*s/sqrt(n)*{n + 1 + (x - x)²/(x² - x²)}^1/2
However, all this refers to the deviation of a y measurement for a given x value. What you want is the deviation of the estimate of x from a measured y value. There is a formula for this, and I worked it out myself once long ago, but I was told by a real analytical chemist not to use it, because it gives a spurious precision to your result - actually, given all the other possible sources of error in a measurement, the uncertainty in your result is much greater than the regression confidence interval. So unfortunately I can't really answer your question!

[J_Holland]

I'm sorry, I meanth "confidence interval" ofcourse

And I've asked some old classmates, who told me there was a different formula to determine the possible spread withing 95% chance of a measured sample. So it wasn't related to the formula I mentioned before.

It's a very long formula with parameters in dutch, my native language, so I cant really translate it for you guys. But I found an old excell spreadsheet wich helps me calculate it fast. Thanks for your reply though!

[scwilson]

I have a bit to add to this discussion. Without getting too technical, the uncertainties in the slope and y-intercept are not sufficient to fully estimate the uncertainty in a concentration as determined form a calibration curve.

Please look here to see the formula that is necessary:

http://www.chemicalforums.com/index.php?topic=71468.msg258369#msg258369

That post describes the formula. S_r can be determined in Excel using the function "STEYX" and Sxx can be determined in excel using the function "DEVSQ" and inputting your x-values for the calibration curve--i.e. the concentrations of your standards. Once calculated, this formula represents the standard deviation of your unknown's concentration as determined from your calibration curve.

From there, you just need to determine your t-value for 95% confidence. You can do this by googline "table of critical t-values" and select the appropriate value (p = 0.05). Your degrees of freedom is N-2, since your linear regression depends on x AND y values. Most students mistakenly assume they have N-1 degrees of freedom for a linear regression; this is not true!

For a WONDERFUL text on this EXACT sort of question, please see the following:

http://acad.depauw.edu/harvey_web/eText%20Project/AnalyticalChemistry2.0.html

It's a link to a FREE and WELL-WRITTEN e-textbook called "Analytical Chemsitry 2.0" by David Harvey. Chapter 5 is relevant to your current studies. Formula 5.25 (page 178) describes the formula I linked to earlier and describes it beautifully. Example problem 5.11 (page 179) is an example of EXACTLY what you're asking with a complete solution provided.