Nonparametric tolerance intervals controlling percentages in both tails and simultaneous testing of quantiles

Abstract

The problem of constructing nonparametric equal-tailed tolerance intervals is considered. A method for finding minimum sample size required so that the rth order statistic and the largest rth order statistic form a p content– $(1-\alpha )$ coverage equal-tailed tolerance interval is proposed. A simultaneous test for quantiles is also provided. Minimum sample sizes that are required to construct equal-tailed TIs and those to carry out hypothesis tests at a given nominal level are presented. The methods are illustrated using data collected from a new secure messaging system.

1. Introduction

A tolerance interval (TI) is a statistical interval based on a random sample that would include a specified percentage of the population with high confidence. For example, a statistical interval that includes at least a proportion p of the sampled population with confidence $1-\alpha$ is usually called p-content– $(1-\alpha )$ coverage two-sided TI or simply $(p,1-\alpha )$ two-sided TI. Owen (1964) has introduced another two-sided TI with some constraints that have applications in sampling inspection plans. This TI controls the percentages of population in both tails, and is commonly referred to as the “equal-tailed” TI. Formally, the equal-tailed TI is constructed so that not only it captures at least proportion p of the population, but also at most 100 $(1-p)/2$ percentage of the population lies in the left tail and at most 100 $(1-p)/2$ percentage of the population lies in the right tail. Both two-sided TIs and equal-tailed TIs are available for most of the commonly used continuous populations such as the normal, exponential, Weibull and gamma. See the book by Krishnamoorthy and Mathew [6] and Hoang–Nguyen–Thuy and Krishnamoorthy [3].

Construction of tolerance intervals for parametric continuous distributions highly depends on the model assumption. Tolerance intervals that are appropriate for any continuous distribution can be constructed using a nonparametric approach. In general, if a sample does not fit a parametric model, then one may seek nonparametric statistical intervals for intended applications. For a given $(p,1-\alpha ,n)$, a nonparametric tolerance interval that satisfies specified content-coverage requirement may not exist. Let $\mathit{X}=(X_1,\dots ,X_n)$ be a random sample from a continuous distribution $F_X(x)$, and let $X_{(1)}<\cdots <X_{(n)}$ be the order statistics for the sample. A nonparametric tolerance interval is formed by a pair of order statistics that are determined by the content and coverage requirements. A $(p,1-\alpha )$ nonparametric TI of the form $(X_{(r)},X_{(s)})$ is constructed so that

$$P_{X_{(r)},X_{(s)}}\{F_X\left(X_{(s)}\right)-F_X\left(X_{(r)}\right)\ge p\}=1-\alpha ,$$

or equivalently,

$$P_{X_{(r)},X_{(s)}}\{P_X(X_{(r)}\le X\le X_{(s)}|X_{(r)},X_{(s)})\ge p\}=1-\alpha .$$ (1)

For simplicity and convenience, s is set as n−r + 1, so that $(X_{(r)},X_{(n-r+1)})$ is a $(p,1-\alpha )$ tolerance interval (see Wilks [8]). Note that the interval is formed by the rth smallest order statistic and the rth largest order statistic so that it would include at least a proportion p of the population with confidence $1-\alpha$. For a given $(p,1-\alpha ,r)$, minimum sample size needed to compute TI of the form $(X_{(r)},X_{(n-r+1)})$ can be obtained using some standard software packages, R-package ‘tolerance’ by Young [10] and the PC calculator that accompanies the book by Krishnamoorthy [4]. This nonparametric TIs are further improved by Young and Mathew [11]. For multivariate nonparametric tolerance region, refer to Young and Mathew [12] and for construction of tolerance intervals for nonparametric regression refer to Guo and Young [2].

The tolerance interval determined by (1) includes at least 100p percentage of the population with confidence $1-\alpha$. This means that at most $100(1-p)$% of the population lies outside the interval. However, it is not clear what percentage of population lies in the left tail and what percentage lies in the right-tail. Owen [7] has noted that there are practical situations where one wants to assess the percentages of data fall in both tails. He has proposed a sampling plan which is designed to accept a lot if the samples provide enough evidence to indicate that at most 100 $(\frac{1-p}{2})$% of items' characteristics falls below a lower specification limit $L_l$ and at most 100 $(\frac{1-p}{2})$% of items' characteristics falls above an upper specification limit $L_u$. Assuming normality, Owen (1964) has proposed an exact test. Also, see Chapter 2 of Krishnamoorthy and Mathew [6].

In this article, we provide nonparametric methods of finding equal-tailed TIs and simultaneous testing of lower and upper quantiles of a continuous population. In the following section, we describe a method of determining TIs of the form $(X_{(r)},X_{(n-r+1)})$ so that at most 100 $(\frac{1-p}{2})$% of the population less than $X_{(r)}$ and at most 100 $(\frac{1-p}{2})$% of the population greater than $X_{(n-r+1)}$ with confidence $1-\alpha$. Notice that $(X_{(r)},X_{(n-r+1)})$ includes at least 100p% of the center of the population. In Section 3, we address the problem of simultaneous testing on lower $\frac{1-p}{2}$ and upper $\frac{1-p}{2}$ quantiles. This simultaneous test also involves determining a pair of statistics so that the type I error rates are within the nominal significance level. In Section 4, we provide an example to illustrate the methods. Some concluding remarks are given in Section 5.

2. Equal-tailed tolerance intervals

Let $X_{(1)}<\cdots <X_{(n)}$ be order statistics for a sample of size n from a continuous population with a distribution function $F_X(x)$. We like to construct a $(p,1-\alpha )$ equal-tailed TI of the form $(X_{(r)},X_{(s)})$, $1\le r<s\le n$, so that

$$P_{X_{(r)},X_{(s)}}(P(X<X_{(r)})<\frac{1-p}{2}\mathrm{and}P(X>X_{(s)})<\frac{1-p}{2}|X_{(r)},X_{(s)})=1-\alpha .$$

For convenience, write $p_t=\frac{1}{2}(1-p)$. The above probability can be expressed as

$$P_{X_{(r)},X_{(s)}}(F_X(X_{(r)})<p_t\mathrm{and}1-F_X(X_{(s)})<p_t|X_{(r)},X_{(s)})=1-\alpha .$$ (2)

It is well-known that $F(X_{(k)})$ is distributed like $U_{(k)}$, where $U_{(k)}$ is the kth order statistic for a sample of size n from a uniform $(0,1)$ distribution. Using this result, we can write (2) as

$$P_{U_{(r)},U_{(s)}}(U_{(r)}<p_t\mathrm{and}{U}_{(s)}>1-p_t)=1-\alpha .$$

The joint probability density function of $U_{(r)}$ and $U_{(s)}$, $1\le r<s\le n$, is given by

$$\begin{array}{rl}& f_{U_{(r)},U_{(s)}}(x,y)\\ & =\frac{n!}{(r-1)!(s-r-1)!(n-s)!}{x}^{r-1}(y-x{)}^{s-r-1}(1-y{)}^{n-s},0<x<y<1.\end{array}$$

For example, see Section 2.6 of Gibbons and Chakraborty [1]. Letting $c_{r,s;n}=\frac{n!}{(r-1)!(s-r-1)!(n-s)!}$, the probability in (2) can be written as

$$c(r,s;n){\int }_{1-p_t}^1{\int }_0^{p_t}{x}^{r-1}(y-x{)}^{s-r-1}(1-y{)}^{n-s}\mathrm{d}x\mathrm{d}y.$$ (3)

Let $b(x|a,b)$ and $B(x|a,b)$ denote the beta density function and the distribution function with shape parameters a>0 and b>0. As shown in the appendix, letting s = n−r + 1, the above integral can be simplified and expressed as

$$\mathrm{CP}(n|r,p)={\int }_{1-p_t}^{1}b(y|n-r+1,r)B(p_t/y|r,n-2r+1)\mathrm{d}y.$$ (4)

For a given $(r,p,1-\alpha )$, the minimum sample size n must be determined so that

$$\mathrm{CP}(n|r,p)\ge 1-\alpha ,$$ (5)

and $(X_{(r)},X_{(n-r+1)})$ is a $(p,1-\alpha )$ equal-tailed TI.

For a given $(r,p,1-\alpha )$, we computed the minimum sample size that satisfies the probability requirement (5) and reported them in Table 1 for values of $r=1(1)20$, p = 0.80, 0.90, 0.95 and 0.99 and $1-\alpha =0.90,0.95$ and 0.99. As an example, the minimum sample size required so that the 4th smallest order statistic and the 4th largest order statistic form a 0.90 content–0.95 coverage equal-tailed TI is 173. That is, if we have a sample of size 173, then $(X_{(4)},X_{(170)})$ is a (0.90, 0.95) equal-tailed TI. Furthermore, at most 5% of the population lies below $X_{(4)}$ and at most 5% of the population exceeds $X_{(170)}$ with confidence 0.95. For any value of $(p,1-\alpha ,r)$ not reported in Table 1, we can find the sample size n using the R function ssize.np.eq.TI(r, p, cl) given in the appendix.

Table 1.

Sample size required so that $(X_{(r)},X_{(n-r+1)})$ is a $(p,\gamma )$ equal-tailed TI.

	$1-\alpha =0.90$				$1-\alpha =0.95$				$1-\alpha =0.99$
r	p = 0.80	0.90	0.95	0.99	0.80	0.90	0.95	0.99	0.8	0.90	0.95	0.99
1	29	58	118	593	35	72	146	734	51	104	210	1057
2	46	93	187	941	54	109	221	1110	72	146	294	1483
3	61	124	249	1251	70	142	286	1440	90	182	368	1851
4	75	153	307	1541	85	173	348	1748	106	216	435	2191
5	89	180	363	1820	100	202	406	2042	122	248	500	2514
6	103	207	417	2092	114	230	463	2327	137	279	562	2825
7	116	234	470	2357	127	258	519	2605	152	309	622	3127
8	129	260	522	2617	141	285	573	2877	167	338	681	3421
9	141	285	573	2874	154	311	626	3145	181	367	738	3710
10	154	310	623	3128	167	338	679	3409	195	395	795	3994
11	166	335	673	3379	180	364	731	3669	209	423	851	4274
12	179	360	723	3627	193	389	783	3927	222	450	906	4550
13	191	385	772	3874	205	415	834	4183	236	477	960	4823
14	203	409	821	4118	218	440	884	4437	249	504	1014	5093
15	215	433	870	4362	230	465	934	4688	262	531	1067	5360
16	227	457	918	4603	243	490	984	4938	275	557	1120	5626
17	239	481	966	4844	255	515	1034	5186	288	583	1173	5889
18	251	505	1014	5083	267	539	1083	5433	301	609	1225	6151
19	263	529	1061	5321	279	564	1132	5679	314	635	1277	6411
20	274	552	1109	5558	292	588	1181	5923	327	661	1328	6669

These nonparametric TIs are exact in the sense that the coverage probabilities are always greater than or equal to the nominal level. In general, nonparametric methods are known to be conservative in the sense that the coverage probabilities could be appreciably larger than the nominal level for some cases. To judge the degree of conservatism, we estimated the coverage probabilities of the TIs for some values of $(p,1-\alpha ,r,n)$ given in Table 1. The coverage probabilities were estimated for a linear combination of two Weibull distributions (i) 0.3 Weibull(3, 2) + 0.7 Weibull(2,1) and a combination of two gamma distributions (ii) 0.4 gamma(3,1) + 0.6 gamma(1,2). Here, Weibull $(c,b)$ denotes the Weibull distribution with the shape parameter c and the scale parameter b and gamma $(a,b)$ denotes the gamma distribution with the shape parameter a and the scale parameter b. Monte Carlo estimates of the coverage probabilities are reported in Table 2. We see in this table that for each entry in the first column, the corresponding estimated coverage probability $\widehat{1-\alpha }$ is very close to $1-\alpha$. Closeness of these probabilities indicates that in the nonparametric setup, these TIs are not conservative. However, they could be wider than those constructed using parametric models.

Table 2.

Coverage probabilities of equal-tailed TIs.

	(i)	(ii)
$(p,1-\alpha ,r,n)$	$\widehat{1-\alpha }$	$\widehat{1-\alpha }$
(.90, .90, 1, 58)	.901	.900
(.90, .90, 3, 124)	.903	.895
(.90, .95, 1, 72)	.950	.951
(.80, .90, 8, 128)	.897	.901
(.90, .99, 3, 182)	.990	.990
(.90, .95, 12, 389)	.952	.947

Note: (i) 0.3 Weibull(3,2) + .7 Weibull(2,1); (ii) 0.4 gamma(3,1) + 0.6 gamma(1,2)

3. Simultaneous hypothesis testing about population quantiles

Consider the acceptance sampling plan where a lot will be accepted if collected samples provide sufficient evidence to indicate that at most a proportion $\frac{1-p}{2}$ of items' characteristics are less than a lower specification limit $L_l$ and at most a proportion $\frac{1-p}{2}$ of items' characteristics are greater than an upper specification limit $U_l$. If $Q_p$ denotes the pth quantile of the population, then a solution to the acceptance sampling plan can be obtained by testing if $L_l<Q_{\frac{1-p}{2}}$ and $Q_{\frac{1+p}{2}}<U_l$ based on a sample. The hypotheses of interest here are

$$H_0:L_l\ge Q_{\frac{1-p}{2}}\mathrm{or}{Q}_{\frac{1+p}{2}}\ge U_l\mathrm{vs}.H_a:L_l<Q_{\frac{1-p}{2}}\mathrm{and}{Q}_{\frac{1+p}{2}}<U_l.$$ (6)

A nonparametric test for the above hypotheses is to be developed so that the type I error rate is within the nominal level α.

Remark 3.1

Assuming normality, Owen (1964) has proposed a test that rejects the null hypothesis at the level of significance α if $L_l\le \overline{X}-k_{h}S$ and $\overline{X}+k_{h}S\le U_l,$ where the critical value $k_h$ is to be determined so that $P(\overline{X}-k_{h}S\ge L_l\mathrm{and}\overline{X}+k_{h}S\le U_l|H_0)\le \alpha .$ Krishnamoorthy and Mathew [5] considered the above hypothesis testing problem in an application of establishing equivalency of a sampling device to the OSHA standard.

A way to develop a nonparametric test for the hypotheses in (6) is to determine the minimum sample size that is required so that

$$P_{X_{(r)},X_{(n-r+1)}}(P(X\le X_{(r)})>p_t\mathrm{and}P(X\le X_{(n-r+1)})<1-p_t)\le \alpha ,$$ (7)

where $p_t=(1-p)/2$. Proceeding as in the earlier section, the above requirement can be written as

$$P_{U_{(r)},U_{(n-r+1)}}(U_{(r)}>p_t\mathrm{and}{U}_{(n-r+1)}<1-p_t)\le \alpha ,$$

where $U_{(r)}$ is the rth order statistic for a sample of size n. By considering the probability of the complimentary event, we see that the minimum sample size that satisfies the above requirement also satisfies

$$P_{U_{(r)},U_{(n-r+1)}}(U_{(r)}<p_t\mathrm{or}{U}_{(n-r+1)}>1-p_t)\ge 1-\alpha .$$ (8)

Using the De Morgan law and the result that $U_{(r)}$ has the beta $(r,n-r+1)$ distribution, the above requirement can be expressed as

$$B(p_t|r,n-r+1)+\overline{B}(1-p_t|n-r+1,r)-\mathrm{CP}(n|r,p)\ge 1-\alpha ,$$ (9)

where $B(x|a,b)$ denotes the beta distribution function, $\overline{B}(x|a,b)=1-B(x|a,b)$ and $\mathrm{CP}(n|r,p)$ is given in (4).

We have computed sample sizes required to satisfy (9) for values of $\alpha =0.01,0.05,0.10$, and p = 0.80, 0.90, 0.95, 0.99, and presented them in Table 3. As an example, suppose a lot will be accepted at the level of significance 0.05, if the specification limits $L_l\le Q_{0.05}$ and $U_l\ge Q_{0.95}$. Referring to $(\alpha ,p,r)=(0.05,0.90,1)$ in Table 3, we find n = 28. The lot will be accepted if $L_l\le X_{(1)}$ and $X_{(28)}\le U_l$. For any value of $(p,\alpha ,r)$ not reported in Table 1, we can find the sample size n using the R function ssize.np.quantile.test(r, p, alpha) in the appendix.

Table 3.

Sample size required so that $P(Q_{\frac{1-p}{2}}\le X_{(r)}\mathrm{and}{X}_{(n-r+1)}\le Q_{\frac{1+p}{2}}|H_0)=\alpha$.

	$\alpha =0.10$				$\alpha =0.05$				$\alpha =0.01$
r	p = 0.80	0.90	0.95	0.99	0.80	0.90	0.95	0.99	0.8	0.90	0.95	0.99
1	11	22	45	230	14	29	59	299	21	44	90	459
2	23	46	94	472	27	55	112	567	36	75	153	775
3	34	69	140	704	39	80	162	818	50	103	210	1061
4	45	92	185	930	51	104	210	1061	63	130	264	1333
5	56	114	229	1153	63	128	258	1298	76	156	316	1595
6	67	136	273	1374	74	151	304	1531	88	181	367	1851
7	78	157	317	1593	85	174	350	1761	101	206	417	2102
8	88	179	360	1811	96	196	395	1990	113	231	466	2349
9	99	201	404	2028	108	219	441	2216	125	255	515	2594
10	110	222	447	2243	119	241	485	2441	137	279	563	2836
11	120	243	490	2458	130	263	530	2665	148	303	611	3076
12	131	265	532	2672	141	285	574	2887	160	326	658	3314
13	141	286	575	2886	152	307	619	3109	172	350	705	3550
14	152	307	617	3099	163	329	663	3330	183	373	752	3785
15	163	328	660	3312	173	351	707	3550	195	396	799	4019
16	173	349	702	3524	184	373	750	3769	206	419	845	4252
17	184	371	744	3736	195	395	794	3988	218	442	891	4484
18	194	392	787	3947	206	416	838	4206	229	465	937	4714
19	204	413	829	4158	217	438	881	4424	240	488	983	4944
20	215	434	871	4369	227	460	924	4641	251	511	1029	5173

The nonparametric tests are exact. However, like in the preceding section, these tests could be too conservative in some cases. To judge the error rates of the tests, we estimated the type I error rates for some values of $(p,\alpha ,r,n)$ given in Table 3. As in the preceding section, we used the same combination of two Weibull distributions, namely, (i) 0.3 Weibull(3, 2) + 0.7 Weibull(2,1) and the same combination of gamma distributions (ii) 0.4 gamma(3,1) + 0.6 gamma(1,2). Monte Carlo estimates of the error rates are given in Table 4. We note in this table that for each entry in the first column, the corresponding estimated error rates $\hat{\alpha }$ is very close to α. Comparison of these values indicates that in the nonparametric setup, these tests are not conservative.

Table 4.

Type I error rates of the quantile tests.

	(i)	(ii)
$(p,\alpha ,r,n)$	$\hat{\alpha }$	$\hat{\alpha }$
(.90, .10, 1, 22)	.098	.098
(.90, .10, 3, 69)	.097	.099
(.90, .05, 1, 29)	.048	.049
(.80, .10, 8, 88)	.097	.102
(.90, .01, 3, 103)	.011	.011
(.90, .05, 12, 285)	.056	.051

Note: (i) 0.3 Weibull(3,2) + .7 Weibull(2,1); (ii) 0.4 gamma(3,1) + 0.6 gamma(1,2)

4. Example

This example and the data are taken from Wurscher et al. [9]. It is concerned of efficacy and security of a New Secure Messaging System (NSMS). The system can send messages with a desired level of security, but the amount of time between sending and receiving the message is not well understood for this system. It is desirable to send and receive messages within 20 seconds for the system to be useable in the field. To understand the distribution of the delivery times, a sample of n = 52 runs was selected and the delivery times were recorded. The data are given in Table 5 of Wurscher et al. [9], and they are reproduced here in Table 5. In order to assess the speed of the delivery of messages, Wurscher et al. [9] wanted to place an upper bound below which 90% of delivery times lies. These authors have noted that the histogram of the data is not symmetric indicating that normality is not tenable. They have used a (.90, .90) nonparametric upper tolerance limit as an upper bound. We also checked if the data fit other parametric models such as the lognormal, Weibull, two-parameter exponential and gamma, and found none of them fits the data.

Table 5.

Delivery Times (in seconds) for high load low file size messages.

5.126429	8.212724	9.106340	9.253784	10.423324
12.393966	12.609172	12.670580	12.972754	13.195848
13.204790	13.447015	13.603801	13.629823	13.735626
13.738818	13.827530	14.171431	14.184233	14.445580
14.912384	15.412042	15.435410	15.464379	15.777407
16.333406	16.359170	16.364353	16.420189	16.667285
16.967149	17.018127	17.118655	17.160076	17.164883
17.688404	17.768533	18.238697	18.257817	18.405642
18.580804	18.763560	19.348932	19.500355	19.745636
19.748467	19.881802	20.735652	20.819050	21.452909
21.660328	21.717389

Using the R function ssize.np.eq.TI(r, p, cl) given in the appendix, we found that the required sample size of 52 so that $(X_{(1)},X_{(52)})=(5.126429,21.452909)$ is a (.861, .95) equal-tailed TI. Noting that $(1-.861)/2\approx .07$, we can say that at most 7% of the delivery times is less than 5.126429 seconds and at most 7% of the delivery times is greater than 21.717389 seconds with confidence 95%. Furthermore, using the R function ssize.np.eq.TI(2, .795, .95), we found n = 52. That is, $(X_{(2)},X_{(n-1)})=(8.212724,21.660328)$ is a (.795, .95) equal-tailed TI. This means that at most 10% of the delivery times is less than 8.212724 seconds and at most 10% of the delivery times is greater than 21.660328 seconds.

Suppose we like to test $H_0:L_l\ge Q_{.075}\mathrm{or}{Q}_{.925}\ge U_l\mathrm{vs}.H_a:L_l<Q_{.075}\mathrm{and}{Q}_{.925}<U_l,$ where $L_l=8.5$ and $U_l=21.5$ are specified values, at the level 0.05. Noting that p = 0.85, we find the required sample size n = 52 at $\alpha =0.052$ and r = 3. This sample size is not available in Table 3. We computed it using the R function ssn.quantile.test(r, p, alpha) given in the appendix. Note that there are 52 observations in Table 5. Also, $L_l=8.5\le 9.106340=X_{(3)}$ and $X_{(50)}=21.452909\le 21.5$, which provides evidence to support that at most 7.5% of the delivery times is less than 8.5 seconds and at most 7.5% of the delivery times is greater than 21.5 seconds. Furthermore, at least 85% of the delivery times falls in the interval $(8.5,21.5)$.

5. Concluding remarks

We provided a nonparametric approach to compute an equal-tailed TI, which is also known as the TI controlling percentages in both tails or TI capturing center of the population. For a given $(p,1-\alpha )$, this equal-tailed TIs are expected to be wider than the corresponding two-sided nonparametric TIs (see Section 8.4 of Krishnamoorthy and Mathew [6]). Furthermore, an equal-tailed TI requires larger sample size than the one requires to compute the corresponding two-sided TI, because the former TI not only captures at least a specified proportion of the population, but also controls the percentages of the population lie in both tails. We also provided a nonparametric test for quantiles that can be used in an accepting sampling plan if the sample data do not fit any continuous parametric distribution. It should be noted that the nonparametric methods proposed here are applicable to find TIs for any continuous distribution. However, large samples are needed to find a TI with specified values of content level p and confidence level $1-\alpha$. For some values of $(p,1-\alpha )$, we can not compute nonparametric TI using our method. For example, to construct a $(.90,.95)$ equal-tailed TI using the extreme order statistics, a sample of at least 72 observations is required (See Table 1). For any sample size smaller than 72, we can not find a $(.90,.95)$ equal-tailed nonparametric TI using the proposed method.

To help practitioners of statistics, we provide R function ssize.np.eq.TI(r, p, alpha) to determine the sample size to compute an equal-tailed TI at a desired value of $(r,p,1-\alpha )$, and R function ssize.np.quantile.test(r, p, alpha) to determine the sample size for simultaneous testing of quantiles.

Appendix 1.

To simplify the integral

$$\mathrm{CP}(r,s|n)=c(r,s;n){\int }_{1-p_t}^1{\int }_0^{p_t}{x}^{r-1}(y-x{)}^{s-r-1}(1-y{)}^{n-s}\mathrm{d}x\mathrm{d}y,$$ (A1)

we first use the transformation $u=x/y$ so that $\mathrm{d}x=y\mathrm{d}u$ and $u:0\to p_t/y$ and write the above integral as

$$c(r,s;n)\mathrm{Beta}(r,s-r){\int }_{1-p_t}^1{y}^{s-1}(1-y{)}^{n-s}B(p_t/y|r,s-r)\mathrm{d}y,$$

where Beta $(a,b)$ is the usual beta function. Recognizing the beta kernel in the above integral, we can further simplify the above expression as

$$c(r,s;n)\mathrm{Beta}(r,s-r)\mathrm{Beta}(s,n-s+1){\int }_{1-p_t}^{1}b(y|s,n-s+1)B(p_t/y|r,s-r)\mathrm{d}y.$$

It can be readily verified that the constant terms on the left hand side of the integral simplifies to 1, and thus we get

$$\mathrm{CP}(r,s|n)={\int }_{1-p_t}^{1}b(y|s,n-s+1)B(p_t/y|r,s-r)\mathrm{d}y.$$ (A2)

Replacing s with n−r + 1, we get (4).

Appendix 2. R code

The following function computes the sample size n so that the interval $(X_{(r)},X_{(n-r+1)})$ is a (p, cl) equal-tailed TI. This function also returns the value of the actual coverage probability.

The following function computes the sample size n so that $P(X_{(r)}\ge Q_{\frac{1-p}{2}}\mathrm{and}{X}_{(n-r+1)}\le Q_{\frac{1+p}{2}}|H_0)=\alpha$. This function also returns the value of the attained level.

Disclosure statement

No potential conflict of interest was reported by the author(s).