Consider a regression model Y = X(J +

e, e '"

N(O, a2 J) and suppose that we want to predict the value of a future observation, say Yo, that will be independent of Y and be distributed N(x~(J, ( 2).

(a) Find the distribution of Yo - x~{3

(b) Find a 95% symmetric two-sided prediction interval for Yo.

Hint: A prediction interval is similar to a confidence interval except that, rather than having a random but observable interval around an unobservable parameter, one has a random but observable interval around an unobserved random variable.

(c) Let 'f/ E (0, .5]. The 100'f/th percentile of the distribution of Yo is, say, 'Y('f/) = x~(J + z('f/)a. (Note that z('f/) is a negative number.)

Find a (1 - 0:)100% lower confidence bound for 'Y('f/). In reference to the distribution of Yo, this lower confidence bound is referred to as a lower 'f/ tolerance point with confidence coefficient (1 - 0:)100%.

For example, if'f/ = 0.1, 0: = 0.05, and Yo is the octane value of a batch of gasoline manufactured under conditions xo, then we are 95%

confident that no more than 10% of all batches produced under Xo will have an octane value below the tolerance point.

Hint: Use a noncentral t distribution based on x~{3 - 'Y('f/).

Comment: For more detailed discussions of prediction and tolerance (and we all know that tolerance is a great virtue), see Geisser (1993), Aitchison and Dunsmore (1975), and Guttman (1970).