The one-sample CI for a normal mean and PI for a single observation from a normal distribution were both based on the central t distribution. A CI for a particular percentile (e.g., the 1st percentile or the 95th percentile) of a normal population distribution is based on the noncentral t distribution. A particular distribution of this type is specified by both df and the value of the noncentrality parameter δ (δ = 0 gives the central t distribution). The key result is that the variable 

has a noncentral t distribution with df =  n – 1 and δ = – (z percentile) √n.

Let t._025,ν,δ and t_.975,ν,δ denote the critical values that capture upper-tail area .025 and lower-tail area .025, respectively, under the noncentral t curve with m df and non centrality parameter δ (when δ = 0, t_.975 = −t_.025, since central t distributions are symmetric about 0).

a. Use the given information to obtain a formula for a 95% confidence interval for the (100p)th percentile of a normal population distribution.

b. For δ = 6.58 and df = 15, t_.975 and t_.025 are (from software) 4.1690 and 10.9684, respectively. Use this information to obtain a 95% CI for the 5th percentile of the beer alcohol distribution considered in Exercise 17.

Data From Exercise 17

Transcribed Image Text:

T: X-μ o/√n - (z percentile) √n S/o