Hello, Sorry to ask this question, I have been searching quite a lot but I could not find an answer. I have seen one post about this subject but I did not really understand the answer. I have a linear regression with 95% confidence interval. I would like to find the x-absciss of the intersection point with the upper confidence interval curve and a straight line (y=y0). For this I apply a formula of inverse regression from Draper and Smith, (Applied Regression analysis,1981): Xu=mean(X)+(b1(Yo-mean(Y))+t*s*(((Yo-mean(Y))?/Sxx)+(b1?/n)-(t?*s?/n*Sxx))^(1/2))/(b1?-t?*s?/Sxx) with: n=number of points in the regression Yo:ordinate of the intersection point b1:slope of linear regression s:sample standard deviation Sxx=sum(Xi-mean(X))? t=t(v,1-a/2) with a=0.05 and v=number of degrees of freedom of s? To check my code, I plot the regression line with its confidence interval (using predict.lm) and y=Yo, and it was clear that results were not as expected. It seems that the formula for confidence interval used by Draper and Smith, is the following: Y=bo+b1*X +/- t*s*((1/n)+(X-mean(X))?/Sxx)^(1/2)) with the same parameters as above and bo being the second parameter from regression. I also plotted this equation and I did not obtain the same confidence interval as the one predicted by predict.lm.... I would like to know which formula is used in predict.lm to calculate correctly the confidence interval in order to calculate correctly the intersection point... I try to read the code but I did not get a clue... Any help would be really welcome, thanks a lot for your time !! Have a good day!! Chlo? -- View this message in context: http://www.nabble.com/equation-of-confidence-interval-calculated-by-predict.lm-tp21377924p21377924.html Sent from the R help mailing list archive at Nabble.com.