Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

262 Some Elementary Statistical Inferences

that is,P[S≤cα/ 2 ]=α/2, whereSis distributedb(n, 1 /2). Then note also that
P[S ≥n−cα/ 2 ]=α/2. (Because of the discreteness of the binomial distribu-
tion, either take a value ofαfor which these probabilities are correct or change the
equalities to approximations.) Thus it follows from expression (4.4.8) that

P[Ycα/ 2 +1<ξ 1 / 2 <Yn−cα/ 2 ]=1−α. (4.4.9)

Hence, when the sample is drawn, ifycα/ 2 +1andyn−cα/ 2 are the realized values of

the order statisticsYcα/ 2 +1andYn−cα/ 2 , then the interval

(ycα/ 2 +1,yn−cα/ 2 ) (4.4.10)

is a (1−α)100% confidence interval forξ 1 / 2.
To illustrate this confidence interval, consider the data of Example 4.4.4. Sup-
pose we want an 88% confidence interval forξ 1 / 2 .Thenα/2=0.060. Thencα/ 2 =4
becauseP[S≤4] =pbinom(4,15,.5)=0.059, where the distribution ofSis bino-
mial withn=15andp=0.5. Therefore, an 88% confidence interval forξ 1 / 2 is
(y 5 ,y 11 )=(96,106).
The R functiononesampsgn(x)computes a confidence interval for the median.
For the data in Example 4.4.4, the codeonesampsgn(x,alpha=.12)computes the
confidence interval (96,106) for the median.

Note that because of the discreteness of the binomial distribution, only certain
confidence levels are possible for this confidence interval for the median. If we further
assume thatf(x) is symmetric aboutξ, Chapter 10 presents other distribution free
confidence intervals where this discreteness is much less of a problem.

EXERCISES

4.4.1. Obtain closed-form expressions for the distribution quantiles based on the
exponential and Laplace distributions as discussed in Example 4.4.6.

4.4.2.Suppose the pdff(x) is symmetric about 0 with cdfF(x). Show that the
probability of a potential outlier from this distribution is 2F(4q 1 ), whereF−^1 (0.25) =
q 1 Use this to obtain the probability that an observation is a potential outlier for
the following distributions.

(a)The underlying distribution is normal. Use theN(0,1) distribution.

(b)The underlying distribution islogistic; that is, the pdf is given by

f(x)=

e−x

(1 +e−x)^2

, −∞<x<∞. (4.4.11)

(c)The underlying distribution is Laplace, with the pdf

f(x)=

1

2

e−|x|, −∞<x<∞. (4.4.12)