4.4 - Non-parametric tests and the sign test

4.4 - Non-parametric tests and the sign test

How can you test a typical value when the population cannot be assumed normal? Learn how a sign test uses a population median, why its probability calculation is binomial, and how to carry out exact and approximate tests.

When a distribution-free method helps

Can an unusually long waiting time change the conclusion of a test about the typical wait? That depends on which information the test uses.

A small-sample test for a population mean usually needs a specified population model. For example, the one-sample tt-test assumes a normal population. Waiting times may instead have an unknown, strongly skewed distribution. A non-parametric test can answer a suitable question without assuming that particular distributional form.

Non-parametric test

A non-parametric test uses information such as signs or ranks without requiring a specified parametric population distribution such as a normal distribution.

It still has assumptions. Random sampling and independence matter; some rank tests also require symmetry.

The question matters as much as the method. A sign test concerns a population median, not a population mean. For a continuous distribution with median mm, half the probability lies on each side of mm. The two sides need not have the same shape.

Worked example 1: using a median without normality

Question. Independent random waiting times, in minutes, are 1.2,1.7,2.4,3.1,4.8,8.6,15.21.2,1.7,2.4,3.1,4.8,8.6,15.2. The population shape is unknown and may be right-skewed. Explain how signs could help investigate whether its median is 44 minutes.

Plan. Compare each wait with the proposed population median, rather than fitting a normal model.

The first four observations are below 44; the last three are above it. Thus the sign information is

−,−,−,−,+,+,+.-,-,-,-,+,+,+.

Under the proposed median, each independent observation has probability 1/21/2 of being above 44. We can investigate whether the observed imbalance is unusually large using that probability model.

Answer. A sign test can use the four below and three above without assuming a normal population. These counts alone are not yet a test conclusion.

Check. Sorting the seven observations gives sample median 3.13.1, which is below 44, consistent with having more observations below than above.

Change only the largest wait. This reveals both a strength and a limitation of signs.

Worked example 2: discarding magnitude

Question. Replace 15.215.2 by 152152 in Worked example 1. How does the information used by a sign test change?

Plan. Check whether that observation crosses the proposed median.

Both 15.215.2 and 152152 exceed 44. The replacement therefore leaves the sign row unchanged, with four negative and three positive signs. A much longer wait does not count as several positive observations.

Answer. The sign-test calculation is unchanged. This protects it from the size of an extreme value, but also discards potentially useful information about magnitudes.

Check. The new sample median is still 3.13.1, whereas the sample mean rises by (152−15.2)/7(152-15.2)/7. The mean and the sign information respond differently.

Distribution-free does not mean assumption-free, and a test about a median cannot be reported as a test about a mean.

Key idea

Choose the population question first. Signs can test a median without assuming normality or symmetry, provided observations are independent and the null gives equal probabilities on the two sides.

Use that distinction in a new setting.

Signs relative to a median

Each observation above a proposed median is one success in a binomial experiment under the null hypothesis.

Write di=xi−m0d_i=x_i-m_0, where m0m_0 is the proposed population median. Assign a positive sign when di>0d_i>0 and a negative sign when di<0d_i<0. Count signs, not the sizes of the deviations.

The sign-test null distribution

For nn independent observations from a continuous population, under H0:m=m0H_0:m=m_0, the number SS above m0m_0 satisfies

S∼B(n,1/2).S\sim\mathrm{B}(n,1/2).

The binomial probability formula is in the formula booklet; identifying this sign-test model is part of the method.

Continuity gives P(X=m0)=0\mathrm P(X=m_0)=0, so the two sign probabilities add to one. The calculations here have no observations equal to the tested median. A value equal to it has neither sign; do not label it positive. Procedures for such ties are outside the cases considered here.

Worked example 3: an exact number of positive signs

Question. Eight observations are sampled independently from a continuous population whose median is 77. Find the probability that exactly two exceed 77.

Plan. Count observations above the median as binomial successes.

Under the stated population median, S∼B(8,1/2)S\sim\mathrm B(8,1/2). Therefore

P(S=2)=(82)(12)8=28256=0.109375.\mathrm P(S=2)=\binom82\left(\frac12\right)^8 =\frac{28}{256}=0.109375.

Answer. The probability is 0.1093750.109375.

Check. There are 28=2562^8=256 equally likely sign sequences, and choosing the two positive positions gives 2828 sequences.

Change “exactly two” to “at most two”. A hypothesis test uses a tail, not just the probability of its observed count.

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