The hypothesis test for a population mean in Chapter 7 can be done using t distribution in the case of a small sample if the population is assumed as a normal distribution. As such, if we make some assumptions about a population distribution and test a population parameter using sample data, it is called a parametric test. The hypothesis tests for two population parameters in Chapter 8 and the analysis of variances in Chapter 9 are also parametric tests, because they assume that populations are normal distributions.

However, real world data may not be appropriate to assume that a population follows a normal distribution, or there may not be enough number of samples to assume a normal distribution. In some cases, data collected are not continuous or can be ordinal such as rank, then the parametric tests are not appropriate. In such cases, methods to test population parameters by converting the data into signs or ranks without assuming on population distributions are called the distribution-free or nonparametric tests.

Since the nonparametric test utilizes the converted data such as signs or ranks, there may be some loss of information about the data. Therefore, if a population can be assumed as a normal distribution, there is no reason to use the nonparametric tests. In fact, when a population follows a normal distribution, a nonparametric test has a higher probability of the type 2 error at the same significance level. However, a nonparametric test would be more appropriate if the data are from a population that do not follow a normal distribution.

The hypothesis test for a population mean in Chapter 7 is based on the theory of the central limit theorem for the sampling distribution of all possible sample means. However, the nonparametric test use signs by examining whether data values are small or large from the central location parameter of the population (the Sign Test of 10.1.1), or use ranks by calculating the ranking of the data (the Wilcoxon Signed Rank Test of Section 10.1.2). Here, the central location parameter can be the population mean or the population median, but usually referring to the population median that is not affected by an extreme point of the data.

Estimation of a population parameter can also be made by using a nonparametric method, but this chapter only introduces nonparametric hypothesis tests. Those interested in the nonparametric estimation should refer to the relevant literature.

Let's take a look at the sign test with the following examples.

When the population median is \(M\), the sign test is used to test whether \(M = M_0\) or \(M \gt M_0\) (or \(M \lt M_0\), or \(M \ne M_0\)). However, if the population distribution is symmetrical to the mean, the sign test is the same as the test of the population mean, because mean and median are the same in this case.

When there are \(n\) number of samples, the test statistic for the sign test uses the number of data which are greater than \(M_0\) which is denoted \(n_+\). The sign test uses the random variable of ‘the number of + signs (\(n_+\))’ which follows a binomial distribution with parameters \(n\) and \(p\)=0,5, i.e., \(B(n,0.5)\) when the null hypothesis is true. You can use the number of data which are less than \(M_0\), i.e., \(n_- = n - n_+\) and \(n_-\) also follows a binomial distribution \(B(n,0.5)\). Let us use \(n_+\) in this section. \(B(n,0.5)_{\alpha}\) represents the 100\(\times \alpha\) right tail percentile, but the accurate percentile value may not exist, because it is a discrete distribution. In this case, middle value of two nearest percentile is often used. Table 10.1.1 summarizes the decision rule for each type of hypothesis of the sign test.

As studied in Chapter 5, the binomial distribution \(B(n,0.5)\) can be approximated by the normal distribution \(N(0.5n,0.5^2 n)\) if \(n\) is sufficiently large. Therefore, if the sample size is large, the test statistic \(n_+\) = 'the number of plus sign data' can be tested using the normal distribution \(N(0.5n,0.5^2 n)\). Table 10.1.2 summarizes the decision rule for each hypothesis of the sign test in the case of large samples.

The sign test described in the previous section converted sample data to either + or - symbols by examining whether the data were larger or smaller than the medium \(M_0\). In this case, most of the information that the original sample data have is lost. In order to apply the Wilcoxon signed rank test, we subtract \(M_0\) first from the sample data and take the absolute value of this data. Assign ranks on these absolute values and calculate the sum of the larger ranks than \(M_0\) and the sum of the smaller ranks than \(M_0\). If two rank sums are similar, we conclude that the population median is equal to \(M_0\). This signed rank sum test is the most widely used nonparametric method for testing the central location parameter of a population. This test takes into account not only whether the sample data are larger or smaller than \(M_0\), but also the relative size of the sample data from \(M_0\).

If we denote the population median as \(M\), the signed rank sum test is to test whether the population median is \(M_0\) or greater than (or less than or not equal to) \(M_0\). However, if the population distribution is symmetric about the mean, the signed rank sum test becomes to test about the population mean because the population median and mean are the same. The basic statistical model is as follows: $$ X_i = M_0 + \epsilon_{i}, \quad i=1,2,...,n $$ where \(\epsilon_i\)’s are independent, symmetric about the mean 0 and follow the same distribution.

If \(x_1 , x_2 , ... , x_n\) are sample data, ranks of \(|x_i - M_0|\) are calculated first and the sum of ranks for the data which are greater than \(M_0\) (+ sign data of \(x_1 , x_2 , ... , x_n\)), denoted as \(R_+\), is calculated. \(R_+\) is the test statistic for the signed rank sum test and the sampling distribution of \(R_+\), denoted as \(w_{+}(n)\), is calculated for testing hypothesis by considering all possible cases. 『eStatU』 provides \(w_{+}(n)\) until \(n\) = 22. \(w_{+}(n)_{α}\) denotes right tail 100\(\times α\) percentile of the \(w_{+}(n)\) distribution, but it is not easy to find the exact percentile, because \(w_{+}(n)\) is a discrete distribution and is usually used to approximate the two adjacent values. Table 10.1.5 summarizes the decision rule for the Wilcoxon signed rank sum test for each type of hypothesis.

If the sample size is large enough, the test statistic \(R_+\) is approximated to a normal distribution with the following mean \(E(R_{+})\) and variance \(V(R_{+})\) when the null hypothesis is true. $$ \begin{align} E(R_+ ) &= \frac {n(n+1)}{4} \\ V(R_+ ) &= \frac {n(n+1)(2n+1)} {24} \end{align} $$

Table 10.1.6 summarizes the decision rule of the signed rank sum test for each type of hypothesis.

The distribution of \(w_{+}(n)\) is independent of the population distribution. In other words, the Wilcoxon signed rank sum test is a distribution free test. For example, if \(n\) = 3, the distribution of \(w_{+}(3)\) can be obtained as follows:

Therefore, the distribution of \(w_{+}(3)\) can be calculated as follows which is independent of the population distribution.

If there is a tie on the value of \(|x_i - M_0|\), the average rank is calculated when the ranking is obtained. In this case, the variance of \(R_+\) in case of large sample is calculated using the following modified formula. $$ V(R_+ ) = \frac{1}{24 } [n(n+1)(2n+1) - \frac{1}{2} \sum_{j=1}^{g} t_{j}(t_{j}-1)({t}_{j}+1) ] $$ Here \(g\) = (number of groups of tie), \(t_j\) = (size of \(j^{th}\) tie group, i.e., number of observations in the tie group). if there is no tie, size of \(j^{th}\) tie group is 1 and \(t_j\) = 1.

The testing hypothesis for the two population means in Chapter 8 used the t-distribution in case of a small sample, if each population could be assumed to be a normal distribution. However, the assumption that the population follows a normal distribution may not be appropriate for real world data, or that there may not be enough sample data to assume a normal distribution. Alternatively, if collected data is ordinal such as ranking, then the parametric t-test is not appropriate. In such cases, a nonparametric method is used to test parameters by converting data to ranks without assuming the distribution of the population. This section introduces the Wilcoxon rank sum test.

Nonparametric tests convert data into ranks, so there may be some loss of information about the data. Therefore, if data are normally distributed, there is no reason to apply a nonparametric test. However, a nonparametric method would be a more appropriate method if the data do not follow a normal distribution.

As in Chapter 8, this section introduces nonparametric tests for testing location parameters of two populations for the samples drawn independently from each population and for the samples drawn as paired.

Let's take a look at the Wilcoxon rank sum test with the following example.

Let's generalize the Wilcoxon rank sum test described in [Example 10.2.1]. Denote random samples selected independently from each of the two populations as follows. The sample sizes are \(n_1\) and \(n_2\) respectively, and \(n = n_1 + n_2\).

For convenience, assume \(n_1 \ge n_2\). If \(n_1 < n_2\), you can swap between \(X\) and \(Y\) .

The statistical model of the Wilcoxon rank sum test is as follows: $$ \begin{align} X_i &= M_1 + \epsilon_i & i=1,2, ... , n_i\\ Y_j &= M_1 + \Delta + \epsilon_j & j=1,2, ... , n_j \\ \end{align} $$ \(Y_j\) can also be written as \(Y_j = M_2 + \epsilon_j\) where \(M_2 = M_1 +\Delta \). Here \(\Delta\) is the difference between location parameters. \(\epsilon_i\)’s are independent and follow the same continuous distribution which is symmetric around 0.

The test statistic for the Wilcoxon rank sum test is the sum of ranks, \(R_2\), for \(Y_1 , Y_2, ... , Y_{n_2}\) based on the combined sample of \(X_1 , X_2, ... , X_{n_1}\),\(Y_1 , Y_2, ... , Y_{n_2}\). The distribution of the random variable \(R_2\) = ‘Sum of the ranks for \(Y\) sample’ can be obtained by investigating all possible cases of ranks for \(Y\) which is \({}_{n}P_{n_2}\) and is denoted as \(w_{2}(n_{1},n_{2})\). 『eStatU』 provides the Wilcoxon rank sum distribution \(w_{2}(n_{1},n_{2})\) and its table up to \(n\) = 25. \(w_{2}(n_{1},n_{2})_{α}\) denotes the right tail 100\(\times α\) percentile, but it might not be able to find the accurate percentile, because \(w_{2}(n_{1},n_{2})\) is a discrete distribution. In this case, middle value of two percentiles near \(w_{2}(n_{1},n_{2})_{α}\) is often used as an approximation. Table 10.2.4 summarizes the decision rule for each type of hypothesis.

When the null hypothesis is true, if the sample is large enough, the test statistic is approximated to the normal distribution with the following mean \(E(R_2 )\) and variance \(V(R_2 )\): $$ \begin{align} E(R_2 ) &= \frac {n_2 (n_1 + n_2 +1 ) } {2} \\ V(R_2 ) &= \frac {n_1 n_2 (n_1 + n_2 +1)} {12} \end{align} $$ Table 10.2.5 summarizes the decision rule for each hypothesis type of the Wilcoxon rank sum test if the sample is large enough.

The distribution of rank sum statistic, \(w_{2}(n_{1},n_{2})\), is not dependent on the population distribution. That is, the rank sum test is a distribution free test. For example, if \(n_1\) = 3 and \(n_2\) = 2, the distribution \(w_{2}(3, 2)\) can be found as follows. All possible cases of ranks for \(R_2\) is \({}_{5}C_{2}\) = 10.

Therefore, the distribution \(w_{2}(3, 2)\) is given regardless of the population distribution as follows:

If there is a tie in the combined sample, the average rank is assigned to each data. In this case, the variance of \(R_2\) should be modified in case of large sample as follows: $$ V(R_2 ) = \frac{n_1 n_2} {12} \left[n_1 + n_2 + 1 - \frac{ \sum_{j=1}^{g} {t}_{j} ({t}_{j}-1)({t}_{j}+1) } {(n_1 + n_2 ) (n_1 + n_2 - 1) } \right] $$ Here \(g\) = (number of tied groups), \(t_j\) = (size of \(j^{th}\) tie group, i.e., number of observations in the tie group). If there is no tie, size of \(j^{th}\) tie group is 1 and \(t_j\) = 1.

Section 8.1.2 discussed the testing hypothesis for two population means using paired samples. Paired samples are used when it is difficult to extract samples independently from two populations, or if independently extracted, the characteristics of each sample object are so different that the resulting analysis is meaningless. If two populations are normally distributed, the t-test was applied for the difference data of the paired samples as described in Section 8.1.2. However, if the normality assumption of two populations can not be satisfied, the Wilcoxon signed rank sum test in Section 10.1.2, which is a nonparametric test, can be applied to the difference data of the paired samples.

In case of the paired samples, first calculate the differences (\(d_i = x_{i1} -x_{i2}\)) for each paired sample as shown in Table 10.2.6. For the data of differences, we examine the normality to check whether the parametric test can be applicable or not. If it is not applicable, we apply the Wilcoxon signed rank sum test on the differences.

Let's take a look at the next example.

The Wilcoxon signed rank test for the paired samples is to test whether the population median of the differences between two populations, \(M_d\), is zero or not. If we denote the paired samples as \((x_1 , y_1 ), (x_2 , y_2 ) , ... , (x_n , y_n )\), the Wilcoxon signed rank sum test calculate the difference \(d_i = x_i - y_i\) first and assign ranks on \(|d_i|\). The sum of ranks of \(|d_i|\) which has + sign of \(d_i\), \(R_+\), is used as the test statistic. 『eStatU』 provides the distribution of \(R_+\), denoted as \(w_{+}(n)\), up to \(n\) = 22. \(w_{+}(n)_{α}\) refers to the right tail 100 \(\times α\) percentile of this distribution which may not have an accurate percentile value, because it is a discrete distribution. In this case the average of two values near \(w_{+}(n)_{α}\) is used approximately. Table 10.2.9 summarizes the decision rule of the Wilcoxon signed rank sum test for paired samples by the type of hypothesis.

If the sample size of the paired sample is large, use the normal distribution approximation formula shown in Table 10.1.6.

The testing hypothesis for several population means in Chapter 9 was possible if each population could be assumed to be a normal distribution and has the same population variance. However, the assumption that the population follows a normal distribution may not be true for real-world data, or that there may not be enough data to assume a normal distribution. Alternatively, if data are ordinal such as ranks, then the parametric test is not appropriate. In this case, a nonparametric test is used by converting data into ranks without making assumptions about the population distribution. This section introduces the Kruskal-Wallis test corresponding to the completely randomized design of experiments and the Friedman test corresponding to the randomized block design of experiments in Chapter 9.

Since nonparametric tests are done by using the converted data such as ranks, there may be some loss of information about the data. Therefore, if data are normally distributed, there is no reason to apply a nonparametric test. However, a nonparametric test would be a more appropriate method if data were selected from a population that did not follow a normal distribution.

The Kruskal–Wallis test extends the Wilcoxon rank sum test for two populations. Consider the following example.

Let us generalize the Kruskal–Wallis \(H\) test described so far with an example. Denote random samples collected independently from the \(k\) populations (at each level of one factor) when their sample sizes are \(n_1 , n_2 , ... , n_k\) as follows: (\(n = n_1 + n_2 + \cdots + n_k\)).

The statistical model of the Kruskal-Wallis test is as follows: $$ X_{ij} = \mu + \tau_i + \epsilon_{ij}, \quad i=1,2,... k; j=1,2,...,n_i \; \text{where} \sum_{i=1}^{k} \tau_i = 0. $$ Here \(\tau_i\) represents the effect of the level \(i\) and \(\epsilon_{ij}\)’s are independent and follow the same continuous distribution. The hypothesis of the Kruskal-Wallis test is as follows: $$ \begin{align} H_0 &: \tau_1 = \tau_2 = \cdots = \tau_k \\ H_1 &: \text{At least one pair of } \tau_i \text{ is not equal.} \end{align} $$ For the Kruskal–Wallis test, ranking data for the combined sample must be created. Table 10.3.4 is a notation of ranking data for each level.

The sum of squares for the one-way analysis of variance studied in Chapter 9 by using the ranking data in Table 10.3.4 are as follows: $$\small \begin{align} SST & = \sum_{i=1}^k \sum_{j=1}^{n_i} ( R_{ij} - {\overline R}_{\cdot \cdot} )^2 = \sum_{k=1}^n (k -{\overline R}_{\cdot \cdot} )^2 = n(n+1)(n-1) \\ SSTr & = \sum_{i=1}^k \sum_{j=1}^{n_i} ( {\overline R}_{i \cdot} - {\overline R}_{\cdot \cdot} )^2 \\ SSE & = SST - SSTr \\ \end{align} $$ Also, the statistic for the \(F\)-test is as follows: $$ F = \frac {MSTr}{MSE} = \frac { \frac{SSTr}{k-1}} {\frac{SSE}{n-k}} = \frac {\frac{SSTr}{k-1}} {\frac{SST-SSTr}{n-k}} = \frac{\frac{n-k}{k-1}} {\frac{SST}{SSTr} -1} $$ Since \(SST\) is a constant, the statistic for the \(F\)-test is proportional to \(SSTr\).

The statistic for the Kruskal-Wallis test \(H\) is proportional to \(SSTr\) as follows: $$\small \begin{align} H & = \frac {12}{n(n+1)} \sum_{i=1}^{k} n_{i} ( {\overline {R}}_{i \cdot} - {\overline {R}}_{\cdot \cdot} )^{2} \\ & = \frac{12}{n(n+1)} \sum_{i=1}^{k} \frac{{R}_{i \cdot}^2}{n_i} - 3(n+1) \\ \end{align} $$

The multiplication constant \( \frac{12}{n(n+1)}\) in the definition of \(H\) statistics is intended to ensure that the statistic follows approximately the chi-square distribution with \(k-1\) degrees of freedom.

The distribution of the Kruskal-Wallis test statistic \(H\), denoted as \(h(n_1 ,n_2 , ... , n_k )\), can be obtained by considering all possible cases of ranks {\(1, 2, ... ,n\)} which is \(n!\). 『eStatU』 provides the table of \(h(n_1 ,n_2 , ... , n_k )\) up to \(n\) = 10. \(h(n_1 ,n_2 , ... , n_k )_α\) denotes the right tail 100\(\times α\) percentile, but it might not have the exact value of this percentile, because \(h(n_1 ,n_2 , ... , n_k )\) is a discrete distribution. In this case, the middle of two adjacent values of 100\(\times α\) percentile is often used. The decision rule of the Kruskal-Wallis test is as Table 10.3.5.

The distribution of the Kruskal-Wallis \(H\) statistic is independent of a population distribution. In other words, the Kruskal-Wallis test is a distribution-free test.

If the null hypothesis is true and the sample size is large enough, the test statistic \(H\) is approximated by the chi-square distribution with \(k-1\) degrees of freedom. Table 10.3.6 summarizes the decision rule for the Kruskal-Wallis test in case of large samples.

If there is a tie in the combined sample, the average rank is assigned to each data. In this case, the statistic \(H\) shall be modified as follows: $$ H' = \frac{H} {1 - \sum_{j=1}^{g} \frac{T_j}{n^3 - n} } $$ Here \(g\)= (number of tied groups), \(T_j = \sum_{j=1}^{g} {t}_{j} ({t}_{j}-1)({t}_{j}+1) \) where \(t_j\) = (the size of the \(j^{th}\) tie group, i.e., the number of observations in the tie group). If there is no tie, the size of the \(j^{th}\) tie group is 1 and \(t_j\) = 1.

In Section 9.2, we studied the randomized block design to measure the fuel mileage of three types of cars which reduce the impact of the block factor, i.e., driver. If each population follows a normal distribution, sample data are analyzed using the F-test based on the two-way analysis of variance without the interaction. However, the assumption that a population follows a normal distribution may not be appropriate for real-world data, or that there may not be enough data to assume a normal distribution. Alternatively, if the data collected might not be continuous and are ordinal such as ranks, then the parametric test is not appropriate. In such cases, nonparametric tests are used to test parameters by converting data to ranks without assuming the distribution of the population. This section introduces the Friedman test corresponding to the randomized block design experiments in Section 9.2.2.

Let us take a look at the Friedman test using [Example 9.2.1] which was the car fuel mileage measurement problem.

Let's generalize the Friedman test described so far using the above example. Assume that there are \(k\) number of levels and denote the rank of \(n\) number of data as follows:

A statistical model of the Friedman test is as follows: $$ X_{ij} = \mu + \tau_{i} + \beta_{j} + \epsilon_{ij}, \quad i=1,2,...,k; \;\; j=1,2,...,n $$ Here \(\tau_i\) is the effect of level \(i\) which satisfies \(\sum_{i=1}^{k} \tau_{i} = 0\) and \(\beta_{j}\) is the effect of block \(j\) which satisfies \(\sum_{j=1}^{n} \beta_{j} = 0\). \(\epsilon_{ij}\) ’s are independent and follows the same continuous distribution.

The hypothesis of the Friedman test is as follows: $$ \begin{align} H_0 &: \tau_1 = \tau_2 = \cdots = \tau_k \\ H_1 &: \text{At least one pair of } \tau_i \text{ is not equal.} \end{align} $$ For the Friedman test, ranking data for each block must be created. Table 10.3.11 is the notation of ranking data for each level.

If we apply the analysis of variance for the rank data of Table 10.3.11 instead of the observation data in Section 9.2, the total sum of squares, \(SST\), and the block sum of squares \(SSB\) are constants. The treatment sum of squares \(SSTr\) is as follows: $$\small \begin{align} SSTr &= \sum_{i=1}^{k} n( {\overline R}_{i \cdot} - {\overline R}_{\cdot \cdot} )^2 \\ SST &= SSTr + SSE \end{align} $$ Therefore, the \(F\)-test statistic can be written as follows: $$ F = \frac{MSTr}{MSE} = c \frac{ {SSTr } } { {SST - SSTr - SSE } } $$ Here \(c\) is a constant. That is, since \(SST\) is a constant, \(F\)-test statistic is proportional to \(SSTr\).

The Friedman test statistic \(S\) is proportional to \(SSTr\) as follows: $$\small \begin{align} S &= \frac{12}{k(k+1)} SSTr = \frac{12 n}{k(k+1)} \sum_{i=1}^{k} ({\overline R}_{i \cdot} - {\overline R}_{\cdot \cdot} )^2 \\ &= \frac{12}{nk(k+1)} \sum_{i=1}^{k} {R}_{i \cdot} ^2 - 3n(k+1) \end{align} $$

The reason \(S\) statistic has the constant multiplication of \(\frac{12}{k(k+1)}\) is to make \(S\) which follows a chi-square distribution with \(k-1\) degrees of freedom.

The distribution of the Friedman test statistic \(S\) is denoted as \(S(k,n)\). 『eStatU』 provides the distribution of \(S(k,n)\) up to \(n \le 8\) if \(k = 3\) and up to \(n \le 6\) if \(k = 4\). \(S(k,n)_{α}\) denotes the right tail 100\(\times α\) percentile, but there might not be the exact percentile, because it is a discrete distribution. In this case, the middle value of two nearest \(S(k,n)_{α}\) is often used approximately. Table 10.3.12 is the summary of decision rule of the Friedman test.

The distribution of the Friedman statistic \(S\) is independent of the population distribution. In other words, the Friedman test is a distribution-free test.

If the null hypothesis is true and if the sample is large enough, the test statistic \(S\) is approximated by the chi-square distribution with \(k-1\) degrees of freedom. Table 10.3.13 summarizes the decision rule for the Friedman test in case of large sample.

If there is a tie in the block, the average rank is assigned to each data. In this case, the statistic \(S\) shall be modified as follows: $$ S' = \frac{S}{1 - \sum_{j=1}^{g} \frac{T_j}{np(p^2 -1)} } $$ Here \(g\) = (number of tied groups), \(T_{j} = \sum_{j=1}^{g} {t}_{j} ({t}_{j}-1)({t}_{j}+1) \) where \(t_j\) = (the size of the \(j^{th}\) tie group, i.e., the number of observations in the tie group). If there is no tie, the size of the \(j^{th}\) tie group is 1 and \(t_j\) = 1.