The Wilcoxon Signed Rank (One-Sample) Test for the Mean

PURPOSE

Given a random sample of size n (or composite sample size n, each composite consisting of k aliquots), the Wilcoxon signed rank test can be used to test hypotheses regarding the population mean or median of the population from which the sample was selected.

ASSUMPTIONS AND THEIR VERIFICATION

The Wilcoxon signed rank test assumes that the data constitute a random sample from a symmetric continuous population. (Symmetric means that the underlying population frequency curve is symmetric about its mean/median.) Symmetry is a less stringent assumption than normality since all normal distributions are symmetric, but some symmetric distributions are not normal. The mean and median are equal for a symmetric distribution, so the null hypothesis can be stated in terms of either parameter. Tests for symmetry can be devised which are based on the chi-squared distribution, or a test for normality may be used. If the data are not symmetric, it may be possible to transform the data so that this assumption is satisfied. See Chapter 4 for more information on transformations and tests for symmetry.

LIMITATIONS AND ROBUSTNESS

Although symmetry is a weaker assumption than normality, it is nonetheless a strong assumption. If the data are not approximately symmetric, this test should not be used. For large sample sizes (n > 50), the t-test is more robust to violations of its assumptions than the Wilcoxon signed rank test. For small sample sizes, if the data are not approximately symmetric and are not normally distributed, this guidance recommends consulting a statistician before selecting a statistical test or changing the population parameter to the median and applying a different statistical test (Section 3.2.3).

The Wilcoxon signed rank test may produce misleading results if many data values are the same. When values are the same, their relative ranks are the same, and this has the effect of diluting the statistical power of the Wilcoxon test. Box 3-5 demonstrates the method used to break tied ranks. If possible, results should be recorded with sufficient accuracy so that a large number of equal values do not occur. Estimated concentrations should be reported for data below the detection limit, even if these estimates are negative, as their relative magnitude to the rest of the data is of importance.

SEQUENCE OF STEPS

Directions for the Wilcoxon signed rank test for a simple random sample and a systematic simple random sample are given in Box 3-5 and an example is given in Box 3-6 for samples sizes smaller than 20. For sample sizes greater than 20, the large sample approximation to the

Box 3-5: Directions for the Wilcoxon Signed Rank Test for Simple and Systematic Random Samples

Let X1, X2, . . . , Xn represent the n data points. The following describes the steps for applying the Wilcoxon signed rank test for a sample size (n) less than 20 for Case 1 (H0: µ # C); modifications for Case 2 (H0: µ $ C) are given in braces {}. If the sample size is greater than or equal to 20, use Box 3-7. STEP 1: If possible, assign values to any measurements below the detection limit. If

this is not possible, assign the value "Detection Limit divided by 2" to each value. Then subtract each observation Xi from C to obtain the deviations di = C - Xi. If any of the deviations are zero delete them and correspondingly reduce the sample size n.

STEP 2: Assign ranks from 1 to n based on ordering the absolute deviations |di| (i.e., magnitude of differences ignoring the sign) from smallest to largest. The rank 1 is assigned to the smallest value, the rank 2 to the second smallest value, and so forth. If there are ties, assign the average of the ranks which would otherwise have been assigned to the tied observations.

STEP 3: Assign the sign for each observation to create the signed rank. The sign is positive if the deviation di is positive; the sign is negative if the deviation di is negative.

STEP 4: Calculate the sum R of the ranks with a positive sign. STEP 5: Use Table A-6 of Appendix A to find the critical value w".

If R < w", {R > n(n+1)/2 - w"}, the null hypothesis may be rejected; proceed to Step 7.

Otherwise, there is not enough evidence to reject the null hypothesis, and the false acceptance error rate will need to be verified; proceed to Step 6.

STEP 6: If the null hypothesis (H0) was not rejected, calculate either the power of the test or the sample size necessary to achieve the false rejection and false acceptance error rates using a software package like the DEFT software (EPA , 1994). For large sample sizes, calculate,

m ' s 2_(z 1&"%z1&$)2 (µ₁&_C)2 % _(0.5)z2 1&"

where zp is the pth percentile of the standard normal distribution (Table A-1 of Appendix A). If 1.16m # n, the false acceptance error rate has been satisfied.

STEP 7: The results of the test may be:

1) the null hypothesis was rejected and it seems that the true mean is greater than C {less than C};

2) the null hypothesis was not rejected and the false acceptance error rate was satisfied and it seems that the true mean is less than C {greater than C}; or

3) the null hypothesis was not rejected and the false acceptance error rate was not satisfied and it seems that the true mean is greater than C {less than C} but conclusions are uncertain since the sample size was too small.

EPA QA/G-9 Final

QA00 Version 3 - 13 July 2000

Box 3-6: An Example of the Wilcoxon Signed Rank Test for a Simple Random Sample

Consider the following 10 data points: 974 ppb, 1044 ppb, 1093 ppb, 897 ppb, 879 ppb, 1161 ppb, 839 ppb, 824 ppb, 796 ppb, and one observation below the detection limit of 750 ppb. This data will be used to test the hypothesis: H0: µ $ 1000 ppb vs. HA: µ < 1000 ppb (Case 2). The decision maker has specified a 10% false rejection decision error limit (") at 1000 ppb (C), and a 20% false acceptance decision error limit ($) at 900 ppb (µ1).

STEP 1: Assign the value 375 ppb (750 divided by 2) to the data point below the detection limit. Subtract C (1000) from each of the n observations Xi to obtain the deviations di = 1000 - Xi. This is shown in row 2 of the table below.

Xi 974 1044 1093 897 879 1161 839 824 796 375 di 26 -44 -93 103 121 -161 161 176 204 625 |di| 26 44 93 103 121 161 161 176 204 625 rank 12 2 3 4 5 6.5 6.5 8 9 10 s-rank 1 -2 -3 4 5 -6.5 6.5 8 9 10 STEP 2: Assign ranks from 1 to n based on ordering the absolute deviations |di| (magnitude ignoring

any negative sign) from smallest to largest. The absolute deviations are listed in row 3 of the table above. Note that the data have been sorted (rearranged) for clarity so that the absolute deviations are ordered from smallest to largest.

The rank 1 is assigned to the smallest value, the rank 2 to the second smallest value, and so forth. Observations 6 and 7 are ties, therefore, the average (6+7)/2 = 6.5 will be assigned to the two observations. The ranks are shown in row 4.

STEP 3: Assign the sign for each observation to create the signed rank. The sign is positive if the deviation di is positive; the sign is negative if the deviation di is negative. The signed rank is shown in row 5.

STEP 4: R = 1 + 4 + 5 + 6.5 + 8 + 9 + 10 = 43.5.

STEP 5: Table A-6 of Appendix A was used to find the critical value w" where " = 0.10. For this

example, w0.10 = 15. Since 43.5 > (10x11)/2 - 15 = 40, the null hypothesis may be rejected. STEP 7: The null hypothesis was rejected with a 10% significance level using the Wilcoxon signed rank

Box 3-7: Directions for the Large Sample Approximation to the Wilcoxon Signed Rank Test for Simple and Systematic Random Samples

Let X1, X2, . . . , Xn represent the n data points where n is greater than or equal to 20. The following describes the steps for applying the large sample approximation for the Wilcoxon signed rank test for Case 1 (H0: µ # C); modifications for Case 2 (H0: µ $ C) are given in braces {}.

STEP 1: If possible, assign values to any measurements below the detection limit. If this is not possible, assign the value "Detection Limit divided by 2" to each value. Then subtract each observation Xi from C to obtain the deviations di = C - Xi. If any of the deviations are zero delete them and correspondingly reduce the sample size n.

STEP 2: Assign ranks from 1 to n based on ordering the absolute deviations |di| (i.e., magnitude of differences ignoring the sign) from smallest to largest. The rank 1 is assigned to the smallest value, the rank 2 to the second smallest value, and so forth. If there are ties, assign the average of the ranks which would otherwise have been assigned to the tied observations. STEP 3: Assign the sign for each observation to create the signed rank. The sign is positive if the

deviation di is positive; the sign is negative if the deviation di is negative. STEP 4: Calculate the sum R of the ranks with a positive sign.

STEP 5: Calculate w ' where p = 1 - " {p = "} and zp n(n % ₁₎

4

% _x

p n(n % 1) (2n % 1)/24

is the pth _{percentile of the standard normal distribution (Table A-1 of Appendix A).} STEP 6: If R < w {R > w}, the null hypothesis may be rejected. Go to Step 8.

Otherwise, there is not enough evidence to reject the null hypothesis, and the false acceptance error rate will need to be verified. Go to Step 7.

STEP 7: If the null hypothesis (H0) was not rejected, calculate either the power of the test or the sample size necessary to achieve the false rejection and false acceptance error rates using a software package like the DEFT software (EPA , 1994). For large sample sizes, calculate,

m ' s 2_(z 1&"%z1&$)2 (µ₁&_C)2 % _(0.5)z2 1&"

where zp is the pth percentile of the standard normal distribution (Table A-1 of Appendix A). If 1.16m # n, the false acceptance error rate has been satisfied.

STEP 8: The results of the test may be:

1) the null hypothesis was rejected and it seems that the true mean is greater {less} than C; 2) the null hypothesis was not rejected and the false acceptance error rate was satisfied and it seems that the true mean is less than C {greater than C}; or

3) the null hypothesis was not rejected and the false acceptance error rate was not satisfied and it seems that the true mean is greater than C {less than C} but conclusions are uncertain since the sample size was too small.

EPA QA/G-9 Final

QA00 Version 3 - 15 July 2000