# Confidence (Prediction) Interval after transformation

1. Dec 8, 2007

### mattmns

Hi, I am a stuck on figuring out how to get a 95% prediction confidence interval after I do a transformation. Here are some details.

So we have some data on the brain weight and body weight of different mammals, and we are to perform linear regression on the data, predicting brain weight from body weight. One of the transformations we have to do is ln(Br Wt) vs ln(Bo Wt). Using this transformation we are to predict the brain weight of the cat and get a 95% CI for this cat.

I got the following equation from the above transformation: ln(Br Wt) = 2 + .849*ln(Bo Wt)

The Cat's body weight is 3.3 (kg, but this should not be relevant).

So the predicted brain weight of the cat under this transformation is e^(2 + .849*ln(3.3)) = 20.37 which is reasonable given our data (the actual brain weight of this cat is 25.6 according to our data).

So now I am to find a 95% CI for the cat and this is where I am stuck at.

We have the following formula:

CI: $$\hat{Y_p} \pm t_{crit}SE_{pred}(\hat{Y_p})$$

Where our critical t is going to be 2.09302 (DF_res = 19)

And $$SE_{pred}(\hat{Y_p}) = s_{y|x}\sqrt{1 + \frac{1}{n} + \frac{(X_p - \bar{X})^2}{s_x^2(n-1)}}$$

We have $s_{y|x} = 0.590996, \bar{X} = 1.25, s_x = 0.94, n = 21, X_p = 3.3$

So $SE_{pred}(\hat{Y_p}) = .6049550631$

Now we should have everything we need to find this CI.

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I am sure that a 95% CI for ln(Cat Br Wt) would be (1.74, 4.28) [The center of this CI being ~3.01 = 2 + .849*ln(3.3)]. However, if I then raise each of these to the e power, my CI is no longer centered around my predicted value of 20.365. Have any ideas about how to fix this problem? Am I missing something silly here? Thanks!