The name is absent

Formul∣

Default case. I basically use the method of Mood and Graybill (1963, 408). Let x₁ ≤ x₂ ≤ ∙ ∙ ∙ ≤ x_n be a sample of
size n arranged in ascending order. Denote the estimated gth centile of the æ’s as c_q. We require that 0 < q < 100. Let
R = (n + 1)g∕100 have integer part r and fractional part f, that is, r = int(R) and f = R—r. (If R is itself an integer, then
r = R and f = 0.) Note that 0 ≤ r ≤ n. For convenience, define x∣> = x₁ and x_n+∣ = x_n. Then C_q is estimated by

c_q = x_r + f × (x_r+ι - x_r),

that is, c_q is a weighted average of x_r and a⅛₊ι. Loosely speaking, a (conservative) p% confidence interval for C_q involves
finding the observations ranked t and и which correspond respectively to the a = (100 — p)∕200 and 1 — a quantiles of a
binomial distribution with parameters n and g∕100, i.e., B(n,g∕100). More precisely, define the )th value (i = 0,... ,n) of
the cumulative binomial distribution function to be F_i = P(X ≤ г), where X has distribution B(n,g∕100). For convenience,
let Γ-∣ = 0 and F_n+1 = 1. Then i is found such that F_t ≤ a and F_t+1 > a, and и is found such that I--F¹_u ≤ a and
1 - P∙_u-ι > a.

With the cci option in force, the (conservative) confidence interval is (2⅛₊1,2⅛+1) and its actual coverage is F_u — F_t.

The default case uses linear interpolation on the F_i as follows. Let

g = (a - F_t)/(F_t+1 - F_t),

h = [α - (1 - F_u)]/[(1 - F_u) -(1- .F¹_a.₁)]

= (ɑ — 1 + F_u)/( F¹_u-1 — F_u).

Then the interpolated lower and upper confidence limits (c_qL,c_qu) for C_q are

CqL = Xt+1 +g× (≈⅛₊2 - ≈⅛₊₁)

CqU — a'_u ∣ ∣ h × (j'_u∣∣ Xu)∙

For example, suppose we want a 95% confidence interval for the median of a sample of size 13. So n = 13, q = 50, p = 95,
α = .025, .R = 14 × 50/100 = 7, f = 0. The median is therefore the 7th observation. Some example data x_i and the values of
F_i are as follows:

(c50L,c50σ) = (x3,x11) = (10,77),

with actual coverage P₁₀ — F₂ = .9888 — .0112 = .9776 (97.8% confidence). For the interpolation calculation, we have

g = (.025 - .0112)/(.0461 - .0112) = .395,
h = (.025 - 1 + .9888)/(.0998 - .9539) = .395.

So

⅛ol = X3 + ∙395 × (a?4 — Ж3) = 10 + .395 × 5 = 11.98,

<⅛(jγ = xιι — ∙395 × (жц — жю) = 77 — .395 × 18 = 69.89.

normal case. The value of c_q is as above. Its s.e. is given by the formula

s_q = ʌ/g(lθθ — q)! [100nZ(c_q; x, s)j

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