Exact Sign-Test Intervals for a Median

From a Sign Test to a Confidence Interval

A sign test can assess a proposed population median, but the same calculation can do more: it can identify a range of median values that the sample does not reject at a chosen level. The resulting interval uses order statistics rather than a normal approximation. Its coverage is exact under the stated continuous-sampling assumptions, and may exceed the requested confidence level because the binomial count is discrete.

Let $X_1,\ldots,X_n$ be an independent, identically distributed sample from a continuous population with median $m$. Continuity ensures that an observation equals $m$ with probability zero. At the true median, each observation is below $m$ with probability $1/2$, so the number below $m$ has a binomial distribution. This is the key fact behind the interval.

Definition: Write $X_{(1)}\leq X_{(2)}\leq\cdots\leq X_{(n)}$ for the ordered sample values. For an integer $c$ satisfying $0\leq c<n/2$, the central sign-test interval is $$ \left(X_{(c+1)},\,X_{(n-c)}\right]. $$ Its exact coverage probability under the continuous model is $$ C_n(c)=1-2\sum_{j=0}^{c}{n\choose j}2^{-n}. $$

The endpoints are order statistics, not estimates of a standard error. A smaller $c$ gives a wider interval and higher coverage; a larger $c$ gives a narrower interval and lower coverage. To obtain confidence of at least $1-\alpha$, choose the largest permitted $c$ for which $C_n(c)\geq 1-\alpha$. If no $c$ beyond zero meets the target, use $c=0$, provided its coverage meets the target.

Why the Binomial Formula Gives the Coverage

For a fixed candidate median $m$, define $B$ as the number of sample observations strictly below $m$. At the true median, continuity gives $P(X_i<m)=1/2$. Independence makes the indicators of the events $X_i<m$ independent Bernoulli random variables, so

B\sim\operatorname{Bin}(n,1/2).

The interval contains $m$ exactly when $X_{(c+1)}<m\leq X_{(n-c)}$. The first inequality holds when at least $c+1$ observations are below $m$, or $B\geq c+1$. The second holds when fewer than $n-c$ observations are below $m$, or $B\leq n-c-1$. Thus the interval covers $m$ precisely when

c+1\leq B\leq n-c-1.

The complementary outcomes are $B\leq c$ and $B\geq n-c$. These tails have equal probabilities because a $\operatorname{Bin}(n,1/2)$ distribution is symmetric. They are disjoint when $c<n/2$, which is why their probabilities can be added:

P(B\leq c\text{ or }B\geq n-c) =2P(B\leq c) =2\sum_{j=0}^{c}{n\choose j}2^{-n}.

Subtracting this rejection probability from one gives $C_n(c)$. This also explains why the actual confidence level need not equal the requested level exactly: only certain binomial tail probabilities are available.

Set the model and confidence target.
Assume independent observations from a continuous population and specify the desired minimum confidence, such as 90% or 95%.

Evaluate candidate tail probabilities.
For each candidate $c$, calculate $C_n(c)=1-2P\{\operatorname{Bin}(n,1/2)\leq c\}$, considering only $0\leq c<n/2$.

Select the largest acceptable $c$.
Use the largest $c$ whose exact coverage is at least the target. A larger value would make the interval too narrow to retain the required coverage.

Read off the endpoints.
Order the observations and report $\left(X_{(c+1)},X_{(n-c)}\right]$, with the actual coverage stated separately from the requested level.

Worked Order-Statistic Intervals

Worked Example: A 90% Interval From Eight Measurements

A hypothetical laboratory samples $n=8$ independent measurements from a continuous population with median $m$. The ordered observed measurements are 4, 7, 9, 12, 15, 18, 21, and 26 units. Construct the narrowest central sign-test interval with coverage at least 90%, and calculate its exact coverage.

Method. Compute the binomial lower tail for candidate values of $c$. The denominator is $2^8=256$.

For $c=0$, the two rejection tails contain only $B=0$ and $B=8$. Therefore

C_8(0)=1-2\left(\frac{{8\choose0}}{256}\right) =1-\frac{2}{256} =\frac{254}{256} =0.9922.

For $c=1$, the lower-tail probability is

P(B\leq1)=\frac{{8\choose0}+{8\choose1}}{256} =\frac{1+8}{256} =\frac{9}{256}.

Thus

C_8(1)=1-\frac{18}{256} =\frac{238}{256} =0.9297.

For $c=2$, the cumulative lower tail is $(1+8+28)/256=37/256$, giving $C_8(2)=1-74/256=182/256=0.7109$. The largest $c$ meeting the 90% target is therefore $c=1$. Its endpoints have ranks $c+1=2$ and $n-c=7$, so the interval is

\left(X_{(2)},X_{(7)}\right]=(7,21].

The interval has exact coverage $238/256=0.9297$, or 92.97%, rather than exactly 90%. Common mistake avoided: using $c=2$ because it produces a narrower interval would reduce coverage to 71.09%, below the required level.

Worked Example: A Paired-Difference Median

A hypothetical study records before-and-after measurements for 14 independent subjects. Let each difference be the after measurement minus the before measurement. Assume the differences are independent observations from a continuous distribution with median $m_D$, and that none is zero. Their ordered values are $-12,-9,-7,-5,-3,-1,2,4,6,8,10,13,16,20$. Find the central interval with coverage at least 90%.

Method. Apply the sign-test interval to the differences themselves. The paired design defines each observation as one subject's difference; the target here is the population median difference, not the median of either measurement separately.

For $n=14$, $2^{14}=16384$. At $c=3$, the cumulative lower-tail numerator is

{14\choose0}+{14\choose1}+{14\choose2}+{14\choose3} =1+14+91+364=470.

The two tails together have probability $2(470)/16384=940/16384$, so the coverage is

C_{14}(3)=1-\frac{940}{16384} =\frac{15444}{16384} =0.9426.

For the next candidate, $c=4$, add ${14\choose4}=1001$ to the lower-tail numerator. The coverage becomes

C_{14}(4)=1-\frac{2(470+1001)}{16384} =\frac{13442}{16384} =0.8204.

Therefore $c=3$ is the largest value that achieves at least 90% coverage. The endpoint ranks are $c+1=4$ and $n-c=11$. Hence the interval for the population median difference is

\left(D_{(4)},D_{(11)}\right]=(-5,10].

The exact coverage is $15444/16384=0.9426$, to four decimals. Common mistake avoided: the upper rank is $n-c=11$, not $n-c+1=12$; using rank 12 would not invert the stated sign-test acceptance rule.

Worked Example: Selecting Endpoints for a 95% Target

A hypothetical actuary studies 12 independent, continuous claim-processing times, measured in days. Treat the observations as draws from a population with median $m$. In increasing order, the values are 3.1, 4.0, 4.8, 5.2, 5.9, 6.3, 7.1, 7.8, 8.4, 9.6, 11.2, and 13.0. Find the narrowest central interval with coverage at least 95%.

Method. Compare successive values of $c$ using the $\operatorname{Bin}(12,1/2)$ tails. The denominator is $2^{12}=4096$.

At $c=2$, the cumulative tail numerator is $1+12+66=79$. Thus

C_{12}(2)=1-\frac{2(79)}{4096} =\frac{3938}{4096} =0.9614.

At $c=3$, add ${12\choose3}=220$, giving a cumulative numerator of $299$. Then

C_{12}(3)=1-\frac{2(299)}{4096} =\frac{3498}{4096} =0.8540.

Since $c=2$ meets the 95% target and $c=3$ does not, $c=2$ is the largest acceptable value. The interval endpoints are the third and tenth order statistics:

\left(X_{(3)},X_{(10)}\right]=(4.8,9.6]\text{ days}.

Its exact coverage is $3938/4096=0.9614$, to four decimals. The procedure gives 96.14% coverage, which is at least the requested 95%. Common mistake avoided: reporting 95% as the interval's exact coverage would confuse the target with the coverage actually attained by the discrete binomial rule.

Conditions and Exam Traps

The exact formula is not a universal interval for every distribution or every data set. Its binomial derivation requires independent observations from a continuous population and a median $m$ for which each observation has probability $1/2$ of lying below it. For paired data, apply the method to the differences: the differences from distinct pairs must be independent and drawn from a continuous distribution with the target median. A symmetry assumption is not needed for this sign-based interval.

Continuity matters. If the population is discrete, observations can equal the median with positive probability, and the below-median sign probability need not be $1/2$. Likewise, ties created by rounding or repeated observed values may make the continuous-model calculation questionable. Do not silently treat a data set with relevant ties as if the continuous, no-tie assumptions had automatically been satisfied; the appropriate treatment depends on the data model and how values were recorded.

There is also an important restriction on doubling the tails. The identity $P(B\leq c\text{ or }B\geq n-c)=2P(B\leq c)$ requires disjoint tails, which holds for $0\leq c<n/2$. For example, if $n=2$ and $c=1$, the events $B\leq1$ and $B\geq1$ overlap at $B=1$. Doubling the lower tail would give $2(1+2)/4=1.5$, an impossible probability. The rejection event is actually certain in that case, with probability 1; the correct remedy is to enforce the disjoint-tail restriction before using the doubled-tail formula.

Key takeaway: For a continuous population median, choose the largest $c$ with $0\leq c<n/2$ whose exact binomial coverage meets the target. The resulting interval is $\left(X_{(c+1)},X_{(n-c)}\right]$; for paired data, replace $X$ by the paired differences.

Check Your Understanding

For each problem, state the selected value of $c$, the interval ranks, and the exact coverage when requested. Assume independent observations from a continuous population unless the question concerns paired differences.

A sample has $n=7$ observations and the desired coverage is at least 90%. What is the narrowest central sign-test interval rule, and what is its exact coverage? (Answer: $c=0$; $(X_{(1)},X_{(7)}]$; $126/128=0.9844$.)
For $n=10$, construct the narrowest central interval with coverage at least 95%. (Answer: $c=1$; $(X_{(2)},X_{(9)}]$; coverage $1002/1024=0.9785$.)
For $n=16$, find the narrowest central interval with coverage at least 90%. (Answer: $c=4$; $(X_{(5)},X_{(12)}]$; coverage $60502/65536=0.9232$.)
A paired study has 11 independent, continuous, nonzero differences. Find the central interval rule with coverage at least 90%, including its exact coverage. (Answer: $c=2$; $(D_{(3)},D_{(9)}]$; coverage $1914/2048=0.9346$.)
For $n=8$, a proposed interval is $(X_{(2)},X_{(7)}]$. Determine its $c$ and exact coverage, and explain why doubling the tails is valid. (Answer: $c=1$; coverage $238/256=0.9297$; the rejection tails $B\leq1$ and $B\geq7$ are disjoint.)

What You'll Learn