The multisample Cucconi test

Abstract

The multisample version of the Cucconi rank test for the two-sample location-scale problem is proposed. Even though little known, the Cucconi test is of interest for several reasons. The test is compared with some Lepage-type tests. It is shown that the multisample Cucconi test is slightly more powerful than the multisample Lepage test. Moreover, its test statistic can be computed analytically whereas several others cannot. A practical application example in experimental nutrition is presented. An R function to perform the multisample Cucconi test is given.

Appendices

Appendix 1

Cucconi (1968) does not organize the formal results on $\sum _{i=1}^{n_{k}}R_{ki}^{2}$ and $\sum _{i=1}^{n_{k}}( n+1-R_{ki})^{2}$ in lemmas and theorems but gives an outline of some derivations. Here we reorganize the results on the sums of squared ranks and squared antiranks in a clearer manner by reporting the whole proof of all results. Well known results about sum of powers of the first $n$ natural numbers will be used in this section.

Theorem 1

Under the null hypothesis

$$\begin{aligned} E\left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2}\right) =n_{k}\left( n+1\right) \left( 2n+1\right) /6 \ \forall k. \end{aligned}$$

Proof

Consider the population of the first $n$ squared natural numbers $1,2^{2},...,n^{2}$. $\frac{1}{n_k} \sum _{i=1}^{n_{k}}R_{ki}^{2}$ may be seen as the random variable defined by the mean of a random sample of $n_{k}$ values drawn without replacement from this population. Since the mean of the sample means is equal to the population mean it follows that

$$\begin{aligned} E\left( \frac{1}{n_{k}} \sum _{i=1}^{n_{k}}R_{ki}^{2}\right) =\frac{1}{n}\sum _{i=1}^{n}i^{2}=\left( n+1\right) \left( 2n+1\right) /6 \end{aligned}$$

$\forall k$, and the thesis follows immediately. $\square$

Theorem 2

Under the null hypothesis

$$\begin{aligned} Var\left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2}\right) =n_{k}\left( n-n_{k}\right) \left( n+1\right) \left( 2n+1\right) \left( 8n+11\right) /180 \forall k. \end{aligned}$$

Proof

With simple algebra we first compute the variance $\tau ^2$ of the population of the first $n$ squared natural numbers

$$\begin{aligned} \tau ^{2}&= \frac{1}{n}\sum _{i=1}^{n}\left[ i^{2}-\left( 2n+1\right) \left( n+1\right) /6\right] ^{2}\\&= \left( n^{2}-1\right) \left( 2n+1\right) \left( 8n+11\right) /180. \end{aligned}$$

Now, since the sampling is without replacement

$$\begin{aligned} Var\left( \frac{1}{n_{k}}\sum _{i=1}^{n_{k}}R_{ki}^{2}\right) = \frac{n-n_{k}}{n-1}\frac{\tau ^{2}}{n_{k}} \end{aligned}$$

(8)

$\forall k$, and the thesis follows immediately. $\square$

The following lemmas will be used for proving Theorem 3.

Lemma 1

Under the null hypothesis

$$\begin{aligned} E\left[ \left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2}\right) ^{2}\right]&= n_{k}\left( n-n_{k}\right) \left( n+1\right) \left( 2n+1\right) \left( 8n+11\right) /180\\&\quad +\,n_{k}^{2}\left( n+1\right) ^{2}\left( 2n+1\right) ^{2}/36 \ \forall k. \end{aligned}$$

Proof

Straightforward (we deliberately not factor the result). $\square$

Lemma 2

Under the null hypothesis

$$\begin{aligned} E\left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2}\sum _{i=1}^{n_{k}}R_{ki}\right) =n_{k}n\left( n+1\right) ^{2}\left( 2n_{k}+1\right) /12 \forall k. \end{aligned}$$

Proof

$$\begin{aligned} E\left( \, \sum _{i=1}^{n_{k}}R_{ki}^{2}\sum _{i=1}^{n_{k}}R_{ki}\right) =E\left( \,\sum _{i=1}^{n_{k}} R_{ki}^{3}\right) +E\left( \,\sum _{i=1}^{n_{k}}\sum _{j\ne 1}^{n_{k}}R_{ki}^{2}R_{kj}\right) . \end{aligned}$$

It is $E\left( \,\sum _{i=1}^{n_{k}}R_{ki}^{3}\right) =\frac{n_k}{n}\sum _{j=1}^{n}j^{3}=n_kn\left( n+1\right) ^{2}/4$ and it is

$$\begin{aligned} E\left( \sum _{i=1}^{n_{k}}\sum _{j\ne 1}^{n_{k}}R_{ki}^{2}R_{kj}\right)&= \frac{n_k(n_k-1)}{n\left( n-1\right) }\sum _{j=1}^{n}j^{2}\sum _{l\ne j}^{n}l\\&= \frac{n_k(n_k-1)}{n\left( n-1\right) }\left( \,\sum _{j=1}^{n}j^{2}\sum _{l=1}^{n}l-\sum _{j=1}^{n}j^{3}\right) \\&= n_k(n_k-1)n\left( n+1\right) ^{2}/6. \end{aligned}$$

Finally it follows with simple algebra that

$$\begin{aligned} E\left( \sum _{i=1}^{n_{k}}R_{ki}^{2}\sum _{i=1}^{n_{k}}R_{ki}\right) =n_{k}n\left( n+1\right) ^{2}\left( 2n_{k}+1\right) /12 \end{aligned}$$

$\forall k$. $\square$

Theorem 3

Under the null hypothesis

$$\begin{aligned} Cor\left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2},\sum _{i=1}^{n_{k}}\left( n+1-R_{ki}\right) ^{2}\right) =-\frac{30n+14n^{2}+19}{\left( 8n+11\right) \left( 2n+1\right) } \ \forall k. \end{aligned}$$

Proof

It is

$$\begin{aligned}&Cor\left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2},\sum _{i=1}^{n_{k}}\left( n+1-R_{ki}\right) ^{2}\right) \nonumber \\&\quad =\frac{E\left[ \left( \sum _{i=1}^{n_{k}}R_{ki}^{2}\right) \left( \, \sum _{i=1}^{n_{k}}\left( n+1-R_{ki}\right) ^{2}\right) \right] -n_{k}^{2}\left( 2n+1\right) ^{2}\left( n+1\right) ^{2}/36}{n_{k}\left( n-n_{k}\right) \left( n+1\right) \left( 2n+1\right) \left( 8n+11\right) /180}. \end{aligned}$$

It remains to compute

$$\begin{aligned}&E\left[ \left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2}\right) \left( \, \sum _{i=1}^{n_{k}}\left( n+1-R_{ki}\right) ^{2}\right) \right] \nonumber \\&\quad \quad =n_{k}\left( n+1\right) ^{2}E\left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2}\right) +E \left[ \!\left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2}\right) ^{2}\right] -2\left( n+1\right) E\!\left( \, \sum _{i=1}^{n_{k}}R_{ki}^{2}\sum _{i=1}^{n_{k}}R_{ki}\right) \!. \end{aligned}$$

Using Lemma 1 it follows that

$$\begin{aligned}&Cov\left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2},\sum _{i=1}^{n_{k}}\left( n+1-R_{ki}\right) ^{2}\right) \nonumber \\&\quad \quad =n_{k}\left( n+1\right) ^{2}E\left( \, \sum _{i=1}^{n_{k}}R_{ki}^{2}\right) +Var\left( \, \sum _{i=1}^{n_{k}}R_{ki}^{2}\right) -2\left( n+1\right) E\left( \, \sum _{i=1}^{n_{k}}R_{ki}^{2}\sum _{i=1}^{n_{k}}R_{ki}\right) . \end{aligned}$$

Using Theorem 1, Theorem 2 and Lemma 2 it follows with simple algebra that

$$\begin{aligned}&Cov\left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2},\sum _{i=1}^{n_{k}}\left( n+1-R_{ki}\right) ^{2}\right) \nonumber \\&\quad =-n_{k}\left( n+1\right) \left( 30n+14n^{2}+19\right) \left( n-n_{k}\right) /180 \end{aligned}$$

and finally that

$$\begin{aligned} Cor\left( \,\sum _{i=1}^{n_{k}}R_{ki}^{2},\sum _{i=1}^{n_{k}}\left( n+1-R_{ki}\right) ^{2}\right) =-\frac{30n+14n^{2}+19}{\left( 8n+11\right) \left( 2n+1\right) } \end{aligned}$$

$\forall k$.$\square$

Appendix 2

An R function for performing the multisample Cucconi test follows.

To analyze the data considered in Sect. 6 run the following code.