5.07a–5.07c - Single-sample non-parametric tests

Choosing signs and ranks

Nine of ten independently sampled battery lifetimes exceed a claimed median. Can that be convincing evidence even when nobody knows the shape of the lifetime distribution? A normal curve is unnecessary: under the claim, each lifetime is equally likely to fall on either side of the median.

For a continuous population with median $m$, half the probability lies below $m$ and half above it. The median need not equal the mean. In a strongly right-skewed distribution, a few large values can pull the mean well above the median.

A sign records only which side of a reference value an observation occupies. A rank records relative position. For example, the differences $-20,1,2,3$ have one negative sign and three positive signs; the negative difference also has the largest magnitude. A sign test discards that last fact. A signed-rank test retains it.

The last two rows are developed in the next lesson. Their basis is already visible: paired tests keep each genuine match; a rank-sum test instead asks whether one independent sample tends to occupy the small or large ranks in the pooled data. Under identical continuous populations, each allocation of sample labels to the pooled ranks is equally likely. A claim specifically about different medians needs populations of the same shape and spread, differing only in location.

These tests are useful for small samples whose population shape is not known to be normal. “Non-parametric” does not mean “no hypotheses”, “no parameters” or “no assumptions”. The single-sample tests here concern the population median. For the signed-rank test, symmetry is an additional assumption; the sign test does not require it.

The distinction matters before any calculation. Extra information is useful only when the assumptions needed to use it are credible.

The single-sample sign test

Count observations above the hypothetical median and use an exact binomial tail. This converts an unknown population shape into a familiar probability calculation.

Let $m$ be the population median and suppose $\mathrm{H}_0:m=m_0$. For independent observations from a continuous population, the indicators of exceeding $m_0$ are independent successes with probability $1/2$. In this lesson no observation equals $m_0$.

The observed count is not itself a probability. Include outcomes at least as extreme as the one observed, in the direction of the alternative.

Worked example 2 changes only the alternative: a change in either direction matters. The two-sided probability therefore includes the equally extreme lower tail.

Here is a useful small-sample surprise: if all $6$ observations exceed a proposed median, the one-sided probability is $(1/2)^6=1/64<0.05$. You can obtain evidence with six observations without claiming that six measurements reveal the population's shape.

The next sample changes the count by one. A modest-looking change can move a discrete tail across the significance level.

Constructing signed ranks

A signed-rank test distinguishes a tiny negative difference from a large negative difference. Subtract the hypothetical median first, rank the magnitudes, then return the signs.

For data $12,18,19,23,24,25,26,27$, compare with $20$. The differences are $-8,-2,-1,3,4,5,6,7$. Ranking the original data would give the negative differences ranks $1,2,3$; that loses the fact that the value $12$ is the furthest from $20$.

[DIAGRAM: asset_name: Signs become signed ranks; asset_slug: h245_5_07_signed_rank_weights; caption: Rank absolute differences from the proposed median, then restore their signs; description: Eight differences from 20 shown in order of increasing magnitude, with negative differences -1, -2 and -8 coloured clay and positive differences 3, 4, 5, 6 and 7 coloured blue. Bars have heights equal to absolute ranks 1 through 8, labels show the signed differences, and the two sums are W minus equals 11 and W plus equals 25.]