4.4 - Normal approximations for rank tests

4.4 - Normal approximations for rank tests

For samples beyond the supplied Wilcoxon tables, a normal approximation turns a rank sum into a familiar standardised statistic. Learn the two different mean–variance formulae, apply continuity corrections, and keep the tail consistent with the rank sum you chose.

Large-sample signed-rank distribution

If a signed-rank test has 2424 observations, can you use a table that ends at 2020? Use the large-sample distribution of a specified rank sum instead.

The original assumptions remain: independent random observations from a continuous population symmetric about its median, with no zero deviations or tied absolute ranks. For paired data, these are assumptions about the population of differences. An approximation does not remove them.

Under the null, rank rr contributes either rr or 00 to the positive rank sum PP, with equal probabilities. Its mean contribution is r/2r/2, and its variance is r2/4r^2/4. Independent signs give

E(P)=12∑r=1nr,Var⁡(P)=14∑r=1nr2.\mathrm E(P)=\frac12\sum_{r=1}^n r, \qquad \operatorname{Var}(P)=\frac14\sum_{r=1}^n r^2.

Large-sample signed-rank model

For sufficiently large nn, either specified sum PP or QQ is approximated under the null by

N(n(n+1)4,n(n+1)(2n+1)24).\mathrm N\left(\frac{n(n+1)}4,\frac{n(n+1)(2n+1)}{24}\right).

These mean and variance formulae are in the formula booklet. They apply to PP or QQ, not to T=min⁡(P,Q)T=\min(P,Q).

The second parameter is a variance: take its square root when standardising. Rank sums are integers, so include the observed integer's full interval: P≥pP\ge p uses a normal boundary p−0.5p-0.5; P≤pP\le p uses p+0.5p+0.5.

Worked example 1: an upper-tail approximation

Question. An independent random sample of 2424 observations satisfies the signed-rank assumptions. Relative to the proposed population median m0m_0, the positive rank sum is P=219P=219. Use a normal approximation to test H0:m=m0H_0:m=m_0 against H1:m>m0H_1:m>m_0, at 5%5\%.

Plan. A larger median gives a large positive rank sum, so approximate P(P≥219)\mathrm P(P\ge219).

Under H0H_0,

μP=24(25)4=150σP2=24(25)(49)24=1225z=218.5−15035=1.957142857…\begin{aligned} \mu_P&=\frac{24(25)}4=150 \\ \sigma_P^2&=\frac{24(25)(49)}{24}=1225 \\ z&=\frac{218.5-150}{35}=1.957142857\ldots \end{aligned}

The approximate upper-tail probability is 0.02516530.0251653, which is below 0.050.05.

Answer. Reject H0H_0. There is sufficient evidence at the 5%5\% level that the population median exceeds m0m_0.

Check. The total rank sum is 300300, so Q=81Q=81. Its lower-tail boundary is 81.581.5, giving (81.5−150)/35=−1.957142857…(81.5-150)/35=-1.957142857\ldots, the reflected tail with the same probability.

The half-unit correction is small on the plot, but it determines which whole-number outcomes the shaded area approximates.

[DIAGRAM: asset_name: Continuity correction for a signed-rank sum; asset_slug: cam9231-signed-rank-normal-tails; caption: The upper tail from P equals 219 starts at 218.5, while the equivalent lower tail from Q equals 81 ends at 81.5; description: Two exact normal density plots with mean 150 and standard deviation 35. The left plot shades values above 218.5 to approximate P at least 219. The right plot shades values below 81.5 to approximate Q at most 81. Each shows the mean, the half-unit boundary and an inset integer interval showing the relevant observed rank sum. Both shaded areas are approximately 0.0252.]
The upper tail from P equals 219 starts at 218.5, while the equivalent lower tail from Q equals 81 ends at 81.5

Change only the alternative. Here the two-sided result is close enough to the 5%5\% boundary that retaining intermediate precision matters.

Worked example 2: doubling a near-boundary tail

Question. For the same 2424 observations and P=219P=219, test H0:m=m0H_0:m=m_0 against H1:m≠m0H_1:m\ne m_0, at 5%5\%, using a normal approximation.

Plan. Use the same corrected tail and include its equally extreme reflected tail.

The null model is still N(150,1225)\mathrm N(150,1225), and the corrected upper-tail standardised value is still 1.957142857…1.957142857\ldots. Therefore

p≈2P(Z>1.957142857…)=0.0503307>0.05.p\approx2\mathrm P(Z>1.957142857\ldots) =0.0503307>0.05.

Do not round the standardised value to 1.961.96 before making this comparison.

Answer. Do not reject H0H_0 using this normal approximation. There is insufficient evidence at the 5%5\% level that the population median differs from m0m_0.

Check. The two-sided normal cutoff is approximately 1.9599641.959964, slightly larger than 1.957142857…1.957142857\ldots. The critical-value comparison gives the same decision.

The table and the approximation concern the same null sign allocations. Use the supplied exact table when it covers the sample size; the large-sample calculation has approximation error, especially when its result lies near a decision boundary.

Watch out

The smaller sum TT is folded towards the lower end and is not normally distributed with the displayed mean. Choose PP or QQ, then select its tail.

Use the lower positive-rank tail in this fresh test.

Large-sample rank-sum distribution

For two independent samples, the null distribution depends on both sample sizes, rather than on the number of paired differences.

Pool the observations and rank them increasingly. Let the smaller sample have size m≤nm\le n, with sum RmR_m. Under the identical-populations null, each pooled rank has the same chance of belonging to this sample. Its mean rank is (m+n+1)/2(m+n+1)/2, so the expected sum is m(m+n+1)/2m(m+n+1)/2.

The ranks assigned to one sample are selected without replacement and are dependent. Do not calculate their variance as if they were independent uniform draws; use the supplied rank-sum variance.

Large-sample rank-sum model

For sufficiently large independent samples satisfying the rank-sum assumptions, under H0H_0 of identical populations,

Rm ∼˙ N(m(m+n+1)2,mn(m+n+1)12).R_m\ \dot\sim\ \mathrm N\left( \frac{m(m+n+1)}2, \frac{mn(m+n+1)}{12} \right).

Both formulae are in the formula booklet. This model approximates RmR_m, not WW.

The populations must still be continuous and symmetric about their medians, with independent random samples and no pooled ties. Use the actual m,nm,n, and a half-unit continuity correction for the integer rank sum.

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