Here's my situation.

I have a multiple linear regression which I've used to come up with a prediction interval to predict a value y for a given (x1,x2,x3,x4,x5,x6). It reads something like lower: 30, upper:48.

I also have the same exact thing to predict a value y* at another given (x1*,x2*,x3*,x4*,x5*,x6*). It reads something like lower:35, upper:51.

I want to answer this question:
What is the probability that the value y* is greater than the value y?

I think it's a basic question, but I'm not sure. I could likely come up with this probability if I knew the formula for how the prediction interval is calculated in a multi-variable situation.
Here's what I think should be done, but I wanted to run it by you guys first.

Prediction Intervals are based on a t-distribution with (n-6) degrees of freedom (I have a forced 0 y-int). So I believe the margin of error calculated is then some constant multiplied by the corresponding value from the t-distribution (t_.05/2 with n-6 degrees of freedom). The "some constant" would be the standard error of this particular estimate.

I then just do a basic 2 sample t-test using the point estimate prediction as the means and these constants as the standard errors with my n-6 degrees of freedom. Is this accurate?

Is there a better way?


  • 2
    $\begingroup$ Although it is a basic question, it likely needs an original analysis to be answered. Computations with independent $t$ distributions will not do the trick because the two prediction intervals will be strongly correlated. So, although the formula is explicitly given at stats.stackexchange.com/questions/9131, it's only a start: I don't think you can use that directly to solve your problem. A Bayesian analysis might be more tractable and perhaps would conform better to how you are thinking about this problem too, assuming you want a probability conditional on the existing data. $\endgroup$ – whuber May 23 '14 at 19:15
  • $\begingroup$ I didn't think about taking a Bayesian approach. I think that would indeed fit what I need more precisely. I'll have to brush up on Bayesian analysis (finish reading Kruschke) $\endgroup$ – Anthony Tyler May 23 '14 at 20:55

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