11-43. Confidence Intervals for Quantiles:

Let X1, . . . ,Xn represent a set of sample random variables taken from a population of size N. For the quantile P, let γp be its sample realization, where p depicts the proportion of the population below γp (e.g., for quartiles, p = 0.25j, j = 1, 2, 3; for deciles, p = 0.10j, j = 1, . . . , 9; and for percentiles, p = 0.01j, j = 1, . . . , 99) and σp = 

1 − nN

 p(1−p)

n .

Then we may form (L1,L2) = (p − kσp, p + kσp) and thus, from the sample observations (which have been placed in an increasing sequence x(1), . . . , x(n)), let us determine the point 1 below which the proportion L1 of the sample frequencies fall; and the point 2 below which the proportion L2 of the sample frequencies lie. Then if k = 2, an approximate 95%

confidence interval for P is (1, 2).

For sufficiently large n (> 50) ,±k may be replaced by the N(0, 1) upper and lower percentage points ±zα/2 to give (1, 2) = (p−zα/2σp, p+zα/2σp)

so that now a 100(1 − α)% confidence interval for P is (1, 2). To determine an interval estimate for the median, an alternative approach is to first order the realizations of the sample random variables by increasing magnitude as x(1), . . . , x(n). Then for n > 50, a 100(1−α)% confidence interval for the population median is 1 = x(h) ≤ median ≤ x(n−h+1) = 2, where h is approximated as h = n−zα/2

√

n−1 2 . Suppose a set of n = 60 observations taken from a population of size N = 2000 have been arranged by order of magnitude as:

5, 5, 7, 7, 7, 8, 9, 9, 10, 10, 10, 10, 11, 14, 14, 20, 20, 22, 22, 25, 27, 27, 30, 30, 30, 35, 38, 38, 39, 40, 41, 45, 45, 50, 50, 50, 50, 53, 54, 55, 55, 58, 59, 61, 61, 61, 62, 64, 64, 64, 67, 71, 71, 80, 85, 85, 89, 90, 90, 90.

Find a 95% confidence interval for the median. Also, find a 99%
confidence interval for the first quartile Q1.