e the percentile estimates in a Kp column vector, q, where.

Step 3. Generate B bootstrap samples from each of the original samples and estimate Q, the set of percentiles of interest, on each bootstrap sample. Let denote the estimate of the percentile profile Q for the hth bootstrap sample from the ith population (and). Further, let be the B × p matrix of bootstrap percentile estimates from the ith population. There is no restriction on the bootstrap sampling plan; we illustrate the method with standard n out of n sampling.

Step 4. Compute the p × p covariance matrix for where IB is

the identity matrix of dimension B and 1 is the column vector of ones of length B. Then, arrange the variance- covariance matrices in an overall block-diagonal covariance matrix, V, of dimension Kp × Kp where the off- diagonal matrices are 0:

Step 5. Consider the c × Kp contrast matrix A where 1 ≤ c ≤ (Kp − 1). To test the null hypothesis H0: Aq = 0, calculate the Wald test statistic

(1)

Step 6. The confidence interval for any linear function of a set of percentiles, say, aq, where a is a 1 × Kp vector of constants, is calculated by

(2)

where z1−a/2 is the 1 − a/2 critical value of the standard normal distribution. In general, A is chosen to jointly test the hypotheses of interest and each a, which could be an individual row of A, is taken separately to construct confidence intervals on the individual population percentiles corresponding to the linear combinations created by A. The level of α in Equation (2) may be adjusted for multiple tests using a method such as Bonferroni adjustment. The following numerical example illustrates some potential choices of a for generating confidence intervals of interest.

2.3. A Numerical Example for Two Populations

For illustrative purposes, consider the following example with simulated data with hypothetical values of q and V. We are given two independent samples (K = 2) and wish to test the equality of three percentiles (p = 3) between them, which we will refer to as the percentile profile Q = (Q1, Q2, Q3). Suppose that we are interested in the 25th, 50th, and 75th percentiles; i.e., u = (0.25, 0.5, 0.75). Let q1 = (5.04, 8.38, 11.21) and q2 = (4.00, 6.28, 9.95) be the estimates of the first and second populations’ percentile profiles, respectively. Arranging the sample estimates in a column vector q, we find

Next we obtain B bootstrap sample estimates of Q for each population and calculate the variance-covariance matrix of the bootstrap sample estimates. Then we calculate V, which is a Kp × Kp block-diagonal matrix with zeros (imposed by the assumption of independence) on the off diagonal blocks. In our example simulated data, we calculated V equal to

If we are interested in testing the equality of percentile profiles (H0: q1j = q2j for), we choose A = [Ip| −Ip]. With point estimates of the percentile profiles, q, and estimated covariance matrix V, we have everything necessary to jointly test the hypothesis H0: Q1 = Q2 as in Equation (1) and construct confidence intervals for Q1 - Q2 as in Equation (2). The Wald statistic is W = 4.97 with 3 degrees of freedom so we conclude the differences between the percentile profiles are not statistically significant. To construct confidence intervals on the differences of the percentiles between the groups, we follow Equation (2). Let ai be the necessary constant vector to construct the confidence interval for the difference between the ith percentiles of the two populations. We have, , and; i.e., the rows of A. Using a Bonferroni adjustment on each estimate, the 95% confidence intervals are [−0.99, 3.07] for Q11 − Q21, [−0.17, 4.37] for Q12 − Q22, and [−1.78, 4.30] for Q13 − Q23.

Testing the equality of percentiles q1j = q2j for is just one of many possible contrasts we could construct. For example, we could test the equality of the inter quartile range (IQR), the difference of the 75th and 25th percentiles, between the two populations. To test this, we simply specify a different contrast matrix A, which would be a 1 × Kp vector, where. Let and denote the estimate of the first and second population’s IQR and the difference as. The variance estimate of is equal to

This calculation is completed via matrix computations as in Equations (1) and (2). For the example above,

and. The calculation using Equation (1) yields a W of 0.03 so we would conclude that the IQR of the populations are not significantly different.

To test the equality of the IQR for K populations we generalize the above results. Let be the 1 × p vector containing the contrast of interest. For testing the equality of the IQR with Q = (25, 50, 75),. For 2 populations, only one comparison is necessary and A is of dimension 1 × Kp. However, for K > 1, we require K − 1 comparisons, with A in the form of

where 0 is the 1 × p row vector of zeros. In this case, A is a (K − 1) × Kp matrix where the non-zero 1 × p vectors are and for.

3. Simulation Studies

3.1. Accuracy of Bootstrap Variance Estimates

The convergence of bootstrap variance estimators of order statistics to the true variance has been demonstrated by Ghosh et al. [10] and Babu [11] , with the exact rate of convergence and coverage error of confidence intervals shown by Hall and Martin in [5] and [6] . Empirical studies in the present paper demonstrate this behavior for the normal and gamma distributions. Two million replicate samples were generated and the (⌊nu⌋ + 1)th order statistics corresponding to the values of u listed in Table 1 (for comparing N(0,1) distributions) and Table 2 (for comparing gamma(shape = 2, scale = 1)) were found as the empirical estimate of the variance/covariance of the particular order statistics, denoted as. The normal distribution and gamma distributions were chosen to investigate differences in the bootstrap covariance estimation between symmetric and highly-skewed distributions. The particular parameters chosen for the gamma distribution yields relatively large values of skewness, which is desired. These empirical estimates were compared to bootstrap variance/covariance estimates (denoted as) of the same order statistics following steps 1 - 4 from Section 2 using B = 1000. All simulations were programmed and run in R 3.1.2.

In general, the relative error of the bootstrap variance/covariance estimates decreases as sample size increases. In some instances (in small sample sizes, such as 20) the variance increases as sample size increases. This occurs in the extreme percentiles in both the normal and gamma samples. This likely occurs because the order statistics used for Q(0.05) and Q(0.95) are close to 1 and n. In bootstrap resampling, this causes artificially low estimates of variance when sample sizes are small due to the limited number of unique values that the particular

Table 1. Error relative to empirical expected value of bootstrap variance-covariance estimates from N(0,1).

Table 2. Error relative to empirical expected value of bootstrap variance-covariance estimates from gamma (2,1).

bootstrap order statistic may take. In these situations (Q(0.05) in normal samples, Q(0.05) and Q(0.95) in gamma samples) the expected behavior resumes at around n = 50. Also, it is important to note that all estimates of the variance of a single order statistic (excluding the 95th percentile estimate of gamma with n = 20) have positive bias while all estimates of covariance of two order statistics (excluding those that are arbitrarily small) have negative bias. It can be seen from Table 1 and Table 2 that the covariance estimates are considerably closer to being accurate than the variance estimates in terms of relative error. It would seem that because of the inflated bootstrap variance estimates that this would cause issues with the Wald test. In the next section further studies are shown to investigate the effects of the error of bootstrap variance and covariance estimates.

3.2. Performance of the Wald Test

Simulation studies were performed in which the Wald statistic was calculated for several percentile profiles that ranged from one up to nine equally-spaced percentiles. Other profiles were tested that included unevenly spaced percentiles with extreme percentiles in the tails of the distribution, referred to as Q1 = (0.05, 0.95), Q2 = (0.05, 0.25, 0.5, 0.75, 0.95), and Q3 = (0.05, 0.1, 0.25, 0.5, 0.75, 0.9, 0.95), which specify the values of u in q(u). For example, testing one equally-spaced percentile would equate to the 50th percentile, or u = 0.5. Testing four percentiles would equate to testing the 20th, 40th, 60th, and 80th percentiles, or u = 0.2, 0.4, 0.6, 0.8. These percentile profiles, particularly ones with more than one percentile, are chosen to compare the populations across the entire sample space. Although the Wald statistic will vary depending on the profile chosen for a given sample, selecting a profile with similar values of u will cause only slight changes in the results of the test (more details later).

Again, the standard normal and gamma (2,1) distributions were used for empirical Type I error estimates to examine differences between symmetric (normal) and highly-skewed (gamma) distributions. As can be seen from the empirical type I error in Table 3 and Table 4, the shape of the distribution has little effect on the asymptotic behavior of the test as empirical Type I error is roughly the same under both scenarios. Also, the error of the bootstrap estimate of percentile variance appears to be substantially moderated when used in the Wald test. For example, the normalized error of the bootstrap variance estimate of the 5th percentile (the sixth order statistic for n = 100) in N(0,1) is 0.226 (from Table 1). It is assumed that the 95th percentile is roughly the same due to symmetry of the distribution though not exact due to the particular order statistics used. However, the empirical type I error of the Wald test comparing the 5th and 95th percentiles between the populations is 0.0483, which is unexpected considering the magnitude of the error of the bootstrap variance estimates. In another example, the normalized error of the bootstrap variance estimates of Q1 = (0.05, 0.95) for n =100 from gamma (shape = 2, scale = 1) are 0.185 and 0.310, respectively (from Table 2). However, the Wald test yields an empirical type I error of just 0.0475.

The results in Table 1 through Table 4 indicate that the Wald test is robust to large error of the bootstrap variance estimates of the percentiles, at least in the sense that type I error is less than the target value, 0.05 in this case. Also, profiles with more percentiles require greater sample size before convergence to the theoretical alpha

Table 3. Empirical Type I error (10,000 replicates) of Wald Test from N(0,1).

Table 4. Empirical Type I error (10,000 replicates) of Wald Test with from gamma (shape = 2, scale = 1).

level. For the parameters chosen in Table 3 and Table 4, the empirical type I error is adequate for samples as few as 20 in most cases, and for profiles with extreme percentiles (i.e. 5th and 95th) the type I error is adequate between sample size 40 and 60. As a general rule of thumb, sample sizes of at least 50 will give reliable results for profiles of any size and any percentile.

As mentioned previously, the primary purpose of the test is to compare the distributions of random variables among populations along the variable’s entire domain but have the added precision of testing the equality of specific parameters, i.e., the percentiles. With certain percentile profiles, the test has similarities of both: (1) tests that are designed to detect any difference between distributions, such as the KS test, and (2) tests on specific population parameters, such as the t-test for means. For comparative purposes, there is no procedure to the authors’ knowledge that explicitly tests a set of percentiles among populations. Rank or location tests as well as global tests of equality are not appropriate comparisons; nevertheless, we used the KS test as a benchmark of the power under various distributions.

For each set of parameters for the distributions listed in Table 5 and Table 6, the sample size necessary for the KS test to achieve a power of 0.8 was found (based on 100,000 replicate samples). With the same sample size for the particular distributions, the proposed test was performed for various percentile profiles and the empirical power calculated based on 10,000 replicate samples. Results of the simulations that test equally-space percentile profiles are listed in Table 5 and unequally-spaced profiles are in Table 6. The test is competitive with the KS in most of the scenarios tested. With some distributions and percentile profiles, the test has greater power than the KS test. For equally-spaced percentiles, the profiles with 1, 2, 3, and 4 percentiles perform similarly, with the 2 and 3 percentile profiles having slightly greater average power across all scenarios. For unequally-spaced percentile profiles, profile Q2 performed the best while the profile containing only the upper and lower extremes, Q1, performed the worst. These simulation results outline a general strategy for using this procedure.

In general, there is not likely interest in a specific value of u, but there may be such interest in arbitrary values in the small neighborhood around u. For instance, in a test involving the median, we may not be really interested in the actual value at which half the observations are above and half below, but rather some measure of the general “center” of the distribution by which we could compare the two populations. Similarly, we could choose to test the equality of 25th or 20th percentile, not necessarily because this is a specific parameter of interest, but merely some measure of the lower half of the distribution. If we were to construct confidence intervals at each value of q(u), we could examine the extent of the differences at these exact locations, but likely interpret them as just a sample of the relationship between the distributions in the segment near q(u). Essentially, we are systematically sampling along the domain of the populations to detect differences between the distributions. Therefore, if there is not a specific profile of interest to be tested, the percentile profile should be selected to adequately cover the entire domain, particularly in the tails of the distributions where discrepancies frequently exist.

4. Illustrative Example: C-Reactive Protein and Diabetes

We demonstrate the use of the procedure with an illustrative example involving C-reactive protein (CRP) levels

Table 5. Simulation results compared to Kolmogorov-Smirnov test at 80% power with equally-spaced percentiles.

Table 6. Simulation results compared to Kolmogorov-Smirnov test at 80% power with unequally-spaced percentiles.

and diabetes using data from the 2009-2010 National Health and Nutrition Examination Surveys (NHANES). CRP is found in blood plasma and serves as a biomarker for inflammation. Physicians find it informative to monitor patients who are ill because CRP levels tend to rise with injury or disease progression and decline with healing or disease regression. Suppose we wished to test for differences in the distribution of CRP between groups based on diabetes status determined by plasma fasting glucose (PFG). Individuals with PFG (mg/dL) of 99 or below are classified as “normal”, between 100 and 125 are classified as “prediabetes”, and 126 or above as “diabetes”, as designated by the American Diabetes Association (http://www.diabetes.org/diabetes-basics/diagnosis/). We used the procedure to test that the 5th, 10th, 25th, 50th, 75th, 90th, and 95th percentiles were equal between (1) normal males and males with diabetes and (2) normal females and females with diabetes (Only data on adults between the ages of 18 and 79 were analyzed).

For males, there were 533 within the normal group and 160 in the diabetes group. The difference in percentile profiles between normal males and diabetes males is highly significant with a Wald test statistic of 42.9 with seven degrees of freedom (p < 0.0001). For females, there were 867 within the normal group and 123 in the diabetes group with a Wald statistic of 14.0 (p = 0.0517), indicating there is likely a difference between the groups’ percentile profiles. It is important to remember that alternate profiles will result in different values of Wald statistics from the differences in the sample order statistics as well as the distribution of the order statistics within the sample. Due to this uncertainty of estimates, a more informative analysis likely involves confidence intervals as seen in Table 7.

Let d represent the vector of differences between population percentile estimates, equal to qꞌAꞌ where q is the combined vector of population percentile estimates and A = [Ip| − Ip]. Let di = qiD − qiN be the estimated difference between the ith percentile of the diabetes and normal groups. Recall from Equation (2) that the confidence interval for the difference is where ai is the ith row of A. In this case, with four comparisons, it would be appropriate to adjust z1−α/2 for each di to ensure the family−wise error rate does not exceed 0.05; thus a Bonferroni correction was made. Estimates and confidence intervals are presented Table 7. By examining the differences and calculating confidence intervals at specified percentiles that span the entire domain, better insight into the differences between populations is possible. For both males and females, it is not a simple shift in location of CRP between normal and diabetes groups; there appears to be increasing differences as the percentile increases.

5. Conclusions

The Wald test provides a natural framework for testing equality of linear combinations of percentiles among multiple groups. Although estimating percentiles and obtaining estimates of variance via bootstrap samples has been previously investigated, the use of the Wald test for hypothesis testing has not been proposed. As shown with simulation studies, the procedure has excellent asymptotic properties and is readily implemented in practical applications as illustrated in this paper. The selection of a percentile profile, coupled with the ability to construct various contrasts, allows us to compare populations across the entire domain in detail and detect important differences at multiple locations.

The test may have limited use in some circumstances as the test relies on large-sample assumptions. Some profile/distribution combinations may yield inaccurate bootstrap estimates of percentile variance for small sample sizes. As shown in simulation studies, variance estimates of extreme percentiles (order statistics) near 1 and n are inaccurate in small sample sizes. However, these errors seem to have limited effect on the performance of the Wald test. With reasonable sample sizes, it is an excellent nonparametric procedure for population comparison and testing and is competitive with standard nonparametric tests for testing equality of distributions, specifically the two sample Kolmogorov-Smirnov test.

Another important aspect of the procedure is the choice of percentiles to be tested. If the exact percentiles are not important, a profile (with equally- or unequally-spaced percentiles) along the entire domain of the distribution is effective as an overall nonparametric test of global inequality, but can also be used to test the equality of linear combinations of the percentile profiles to obtain specific estimates at these locations. More research can be done to determine the optimal allocation of percentiles within a profile, perhaps based on features of the data.

Table 7. Estimates of C-reactive protein differences in percentiles between diabetes and normal groups for males and females.

Further work can investigate the sample size limitations and distribution effect upon the bootstrap estimates of variance how this in turn affects the Wald test.

Acknowledgements

This research was supported by 1 U54 GM104940 from the National Institute of General Medical Sciences of the NIH, which funds the Louisiana Clinical and Translational Science Center. William D. Johnson also receives support from the National Center For Complementary & Integrative Health and the Office of Dietary Supplements of the National Institutes of Health under Award Number 3 P50AT002776 which funds the Botanical Research Center of Pennington Biomedical Research Center and the Department of Plant Biology and Pathology in the School of Environmental and Biological Sciences (SEBS) of Rutgers University. The content of this manuscript is solely the responsibility of the authors and does not necessarily represent the official views of the National Institutes of Health.

NOTES

*Corresponding author.

Cite this paper
Johnson, W. and Romer, J. (2016) Hypothesis Testing of Population Percentiles via the Wald Test with Bootstrap Variance Estimates. Open Journal of Statistics, 6, 14-24. doi: 10.4236/ojs.2016.61003.
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