95% CI for an estimated X given Y in a simple linear regression model

Question

Considering a simple linear regression model Y= beta0+beta1 x X, with beta0 and beta1 computed, I have to estimate the expected X given a new Y and 95% confidence intervals. I used the formula X=(Y-beta0)/slope. How can I compute in R the 95% interval for the calculated value of ind, given a new value of the height?

head(df)

   ind height
1 4.27    174
2 3.60    159
3 3.61    175

summary(lm(df$ind~df$height))

Call:
lm(formula = df$ind ~ df$height)

Residuals:
 Min       1Q   Median       3Q      Max 

-0.56263 -0.27596  0.03866  0.26632  0.55440 

Coefficients:
         Estimate Std. Error t value Pr(>|t|)   
(Intercept) -0.968895   1.512371  -0.641  0.52903   

df$height    0.027871   0.008985   3.102  0.00562 **

Residual standard error: 0.3287 on 20 degrees of freedom
Multiple R-squared: 0.3248,     Adjusted R-squared: 0.2911 
F-statistic: 9.622 on 1 and 20 DF,  p-value: 0.005621 

#

I tried:

pred.frame <- data.frame(ind=seq(3,5,0.25)) 
predict(bclm,int="c",level=0.95,data=pred.frame)
    fit      lwr      upr
1  174.5780 169.3146 179.8414
2  166.7696 163.6419 169.8973
3  166.8862 163.7806 169.9917
...............

Answer 1

You are looking for the so called calibration interval. You have to be careful with the exact question: is $Y$ a new observation and you are trying to estimate $X$, or are you looking for the point where the regression line reaches $Y$? The difference will determine whether you need to use the confidence interval of the regression line or its prediction interval. Since you are talking about the confidence interval of the line, I assume you are interested in the latter question.

The 95% confidence interval for the estimated $X$ will be given by the points where the $y=Y$ horizontal line intersects the lower and upper confidence bands of the regression line. So go through the lwr values, and check at which x value they crossed the $Y$, and similarly for the upr values. Depending on the sophistication you want, this can be done by just looking at the printout, or having R find the crossing point, or even using the uniroot function to find the point with arbitrary precision (you are using steps of 0.25, so that's your current precision).