Confidence intervals for assessing equivalence of two treatments with combined unilateral and bilateral data

Abstract

Responses from the paired organs are generally highly correlated in bilateral studies, statistical procedures ignoring the correlation could lead to incorrect results. Note the intraclass correlation in the study of combined unilateral and bilateral outcomes; 11 confidence intervals (CIs) including 7 asymptotic CIs and 4 Bootstrap-resampling CIs for assessing the equivalence of 2 treatments are derived under Rosner's correlated binary data model. Performance is evaluated with respect to the empirical coverage probability (ECP), the empirical coverage width (ECW) and the ratio of the mesial non-coverage probability to the non-coverage probability (RMNCP) via simulation studies. Simulation results show that (i) all CIs except for the Wald CI and the bias-corrected Bootstrap percentile CI generally produce satisfactory ECPs and hence are recommended; (ii) all CIs except for the bias-corrected Bootstrap percentile CI provide preferred RMNCPs and are more symmetrical; (iii) as the measurement of the dependence increases, the ECWs of all CIs except for the score CI and the profile likelihood CI show increasing patterns that look like linear, while there is no obvious pattern on the ECPs of all CIs except for the profile likelihood CI. A data set from an otolaryngologic study is used to illustrate the proposed methods.

1. Introduction

In randomized clinical trials, it is frequently that we may collect data from the paired organs or body parts, and the outcomes may be either bilateral (e.g. two organs are sick) or unilateral response (only one organ or a body part is sick). Consequently, it causes the data generally highly correlated in bilateral studies, and statistical procedures ignoring the intraclass correlation of the bilateral observations may lead to incorrect inference [4,7,9,20,21]. Taking the intraclass correlation into consideration, Rosner [20] presented an intraclass correlation model for analyzing ophthalmologic data to which a person may have contributed two eyes worth of information. Le [12] considered the testing for linear trends in proportions using correlated otolaryngology or ophthalmology data. Note that asymptotic test procedures could also yield unacceptably high type I error rates for small sample studies and sparse data structures even if the intraclass correlation is taken into consideration, Tang et al. [26] proposed the exact unconditional and approximate unconditional procedures based on three test statistics (i.e. the Wald statistic, the statistics based on a dependence model and an independence model proposed by Rosner and Milton [21]). Followed by Tang et al. [26], eight test statistics for testing the equality of two treatments and the corresponding asymptotical and approximate unconditional test procedures in the bilateral study are developed by Tang et al. [25]. Tang et al. [23] investigated the goodness-of-fit and the model selection for a few popular statistical models for correlated paired binary outcomes. A variety of confidence intervals for estimating the difference between the proportions of responders in a randomized two-armed clinical trial are proposed in Pei et al. [18].

However, individuals may produce either unilateral data (e.g. data from only one organ) or bilateral data (e.g. data from two organs) in many medical comparative studies (e.g. otolaryngologic or ophthalmologic studies). For example, to evaluate the efficacy of two antibiotics (i.e. Cefaclor and Amoxicillin) for the treatment of otitis media with effusion (OME) in an otolaryngologic study, Mandel et al. [15] considered a randomized double-blinded clinical trial. In this trial, a total of 214 children (293 ears) underwent unilateral or bilateral tympanocentesis before they were randomly assigned to one of the two treatments. After a 14-day course of treatment with one of the antibiotics, the outcome of each child was recorded at the end of the treatment. In this study, only 203 evaluable children without repeat tympanocentesis, treatment change or tympanic membrane perforations have received one of the treatments. For the group with unilateral disease, two results were determined: cured and not-cured; and for the group with bilateral disease, three results, i.e. cured (both ears become OME-cured), partially cured (only one ear becomes OME-cured) and both ears are not cured were recorded. The data are reported in Table 1.

Table 1.

OME status after 14 days and 42 days of treatment (in terms of No. of children).

	Amoxicillin			Cefaclor
No. of OME-free ears	0	1	2	0	1	2
Unilateral	39	27	–	24	38	–
Bilateral	15	3	13	14	9	21

For this study, it is important to test whether the cure rates are identical between the Cefaclor and Amoxicillin groups. Under the equal correlation coefficient model, Pei et al.[17] considered the equivalence testing of two successful cure rates and developed several asymptotic test procedures under the independent and the dependent models, respectively. However, CI estimators for comparative studies with combined unilateral and bilateral binary data have not been developed. In this article, we consider the CI construction for proportion difference in comparative medical studies with combined unilateral and bilateral binary outcomes under Rosner's correlated binary data model.

This article is organized as follows. The data structure and the probability model are described in Section 2. Seven asymptotic confidence intervals and four Bootstrap-resampling CIs are proposed in Section 3. In Section 4, the performance of all CIs is evaluated via simulation studies in terms of the ECP, the ECW and the RMNCP. An illustration of our methodologies with otolaryngological data is presented in Section 5. The paper closes with a brief conclusion and discussion in Section 6.

2. Data structure

Let $m_{h^{\mathrm{\prime }}i}^{(1)}$ represent the number of individuals with $h^{\mathrm{\prime }}$ ear being cured in the ith treatment for the unilateral group, and $m_{hi}^{(2)}$ represent the number of individuals with h ear/ears being cured in the ith treatment for the bilateral group, $p_{h^{\mathrm{\prime }}i}^{(1)}$ and $p_{hi}^{(2)}$ be the corresponding probabilities ( $h^{\mathrm{\prime }}=0,1$, $h=0,1,2,i=0,1$). Let $m_{+i}^{(s)}=\sum _{h=0}^s{m}_{hi}^{(s)}$ (s = 1, 2) and $m_{++}=\sum _{s=1}^2\sum _{i=0}^1{m}_{+i}^{(s)}$.

Let $Z_{ij}^{(1)}=1$ if the jth individual received the ith treatment in the unilateral group is OME-cured and $Z_{ij}^{(1)}=0$ otherwise. In the bilateral group, $Z_{ijk}^{(2)}=1$ if the kth ear of the jth individual received the ith treatment is OME-cured and $Z_{ijk}^{(2)}=0$ otherwise ( $i=0,1,j=1,2,\dots ,m_{+i}^{(s)},s=1,2,k=1,2$). Following Rosner [20], it is eligible to assume that the correlation of the responses from the two ears should be the same for the two treatments, i.e.

$$\begin{array}{rl}Pr(Z_{ij}^{(1)}& =1)=Pr(Z_{ijk}^{(2)}=1)={\lambda }_i,\\ \text{and}Pr(Z_{ijk}^{(2)}& =1\mid Z_{ij,3-k}^{(2)}=1)=R{\lambda }_i\end{array}$$

for $i=0,1,j=1,2,\dots ,m_{+i}^{(s)},s=1,2,k=1,2$. As shown in [20], the correlation coefficient between $Z_{ij1}^{(2)}$ and $Z_{ij2}^{(2)}$ is $\rho ={\lambda }_i(R-1)/(1-{\lambda }_i)$ for i = 0, 1, then R represents a measurement of the dependence between two ears of the same individual in two treatments. Specially, R = 1 if two ears are completely independent, while $R{\lambda }_i=1(i=0,1)$ if two ears are completely dependent. By simple calculation, we can have the following probability model under Rosner's assumption:

$$\begin{array}{r}\begin{array}{rl}{p}_{0i}^{(1)}& =1-{\lambda }_i,p_{1i}^{(1)}={\lambda }_i,\\ p_{0i}^{(2)}& =1+R{\lambda }_i^2-2{\lambda }_i,p_{1i}^{(2)}=2{\lambda }_i(1-R{\lambda }_i),p_{2i}^{(2)}=R{\lambda }_i^2\end{array}\end{array}$$ (1)

for i = 0, 1. The data structure and the probability model for the combined unilateral and bilateral outcomes are given in Table 2.

Table 2.

The data structure and probability model for combined unilateral and bilateral data.

	Unilatral		Bilateral
No. of ears with OME-free	i = 0	i = 1	i = 0	i = 1	Total
0	$m_{00}^{(1)}(p_{00}^{(1)})$	$m_{01}^{(1)}(p_{01}^{(1)})$	$m_{00}^{(2)}(p_{00}^{(2)})$	$m_{01}^{(2)}(p_{01}^{(2)})$
1	$m_{10}^{(1)}(p_{10}^{(1)})$	$m_{11}^{(1)}(p_{11}^{(1)})$	$m_{10}^{(2)}(p_{10}^{(2)})$	$m_{11}^{(2)}(p_{11}^{(2)})$
2	–	–	$m_{20}^{(2)}(p_{20}^{(2)})$	$m_{21}^{(2)}(p_{21}^{(2)})$
Total	$m_{+0}^{(1)}(1.0)$	$m_{+1}^{(1)}(1.0)$	$m_{+0}^{(2)}(1.0)$	$m_{+1}^{(2)}(1.0)$	$m_{++}$

According to Rosner's assumption, the log-likelihood function of the parameters ${\lambda }_0$, ${\lambda }_1$ and R for the observation data $\mathbf{m}=\{(m_{0i}^{(1)}$, $m_{1i}^{(1)}$, $m_{0i}^{(2)}$, $m_{1i}^{(2)}$, $m_{2i}^{(2)})$: $i=0,1\}$ is given by

$$\begin{array}{rl}l(\mathbf{m};{\lambda }_0,{\lambda }_1,R)& =C+m_{00}^{(1)}\log (1-{\lambda }_0)+m_{10}^{(1)}\log {\lambda }_0+m_{01}^{(1)}\log (1-{\lambda }_1)+m_{11}^{(1)}\log {\lambda }_1\\ & +m_{00}^{(2)}\log (1+R{\lambda }_0^2-2{\lambda }_0)+m_{10}^{(2)}\log [2{\lambda }_0(1-R{\lambda }_0)]+m_{20}^{(2)}\log (R{\lambda }_0^2)\\ & +m_{01}^{(2)}\log (1+R{\lambda }_1^2-2{\lambda }_1)+m_{11}^{(2)}\log [2{\lambda }_1(1-R{\lambda }_1)]+m_{21}^{(2)}\log (R{\lambda }_1^2),\end{array}$$ (2)

where C is a constant that doesn't involve parameters.

Let $\delta ={\lambda }_1-{\lambda }_0$ be the difference between two proportions for two treatments, then ${\lambda }_1={\lambda }_0+\delta$. Therefore, the log-likelihood function given in Equation (2) becomes

$$\begin{array}{rl}\ell (\mathbf{m};\delta ,{\lambda }_0,R)& =C+m_{00}^{(1)}\log (1-{\lambda }_0)+(m_{10}^{(1)}+m_{10}^{(2)}+2m_{20}^{(2)})\log {\lambda }_0\\ & +m_{01}^{(1)}\log (1-{\lambda }_0-\delta )\\ & +(m_{11}^{(1)}+m_{11}^{(2)}+2m_{21}^{(2)})\log ({\lambda }_0+\delta )+m_{00}^{(2)}\log (1+R{\lambda }_0^2-2{\lambda }_0)\\ & +(m_{20}^{(2)}+m_{21}^{(2)})\log R+m_{10}^{(2)}\log (1-R{\lambda }_0)+m_{11}^{(2)}\log [1-R({\lambda }_0+\delta )]\\ & +m_{01}^{(2)}\log [1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta )].\end{array}$$ (3)

It is easily shown that the parameter vector $(\delta ,{\lambda }_0,R)$ satisfies the following conditions:

$$\begin{array}{l}\text{when}\delta \ge 0,0\le {\lambda }_0\le 1-\delta \text{and}max\{{\displaystyle \frac{2({\lambda }_0+\delta )-1}{({\lambda }_0+\delta {)}^2}},0.0\}\le R\le {\displaystyle \frac{1}{{\lambda }_0+\delta }},\\ \text{when}\delta <0,-\delta \le {\lambda }_0\le 1\text{and}max\{{\displaystyle \frac{2{\lambda }_0-1}{{\lambda }_0^2}},0.0\}\le R\le {\displaystyle \frac{1}{{\lambda }_0}}.\end{array}$$

In order to investigate whether there is a significant difference between the two treatments, we are interested in the confidence interval construction for the proportion difference (i.e. δ) in this article. Seven asymptotic confidence interval estimators and four Bootstrap-resampling confidence intervals are developed and evaluated as follows.

3. Confidence interval estimators

3.1. CIs based on Wald-type statistics

It is easily shown that the sample estimates of ${\lambda }_0$ and ${\lambda }_1$ are, respectively, given by

$${\hat{\lambda }}_0=\frac{m_{10}^{(1)}+m_{10}^{(2)}+2m_{20}^{(2)}}{m_{+0}^{(1)}+2m_{+0}^{(2)}},{\hat{\lambda }}_1=\frac{m_{11}^{(1)}+m_{11}^{(2)}+2m_{21}^{(2)}}{m_{+1}^{(1)}+2m_{+1}^{(2)}}.$$ (4)

Therefore, the sample estimate of δ can be given by

$$\hat{\delta }=\frac{m_{11}^{(1)}+m_{11}^{(2)}+2m_{21}^{(2)}}{m_{+1}^{(1)}+2m_{+1}^{(2)}}-\frac{m_{10}^{(1)}+m_{10}^{(2)}+2m_{20}^{(2)}}{m_{+0}^{(1)}+2m_{+0}^{(2)}},$$ (5)

and the maximum likelihood estimation $\hat{R}$ of the parameter R can be obtained by solving the following equation:

$$\frac{m_{00}^{(2)}{\hat{\lambda }}_0^2}{1+R{\hat{\lambda }}_0^2-2{\hat{\lambda }}_0}-\frac{m_{10}^{(2)}{\hat{\lambda }}_0}{1-R{\hat{\lambda }}_0}+\frac{m_{20}^{(2)}+m_{21}^{(2)}}{R}\frac{m_{01}^{(2)}{\hat{\lambda }}_1^2}{1+R{\hat{\lambda }}_1^2-2{\hat{\lambda }}_1}-\frac{m_{11}^{(2)}{\hat{\lambda }}_1}{1-R{\hat{\lambda }}_1}=0.$$ (6)

It is easily shown that the variance of ${\hat{\lambda }}_i$ (i = 0, 1) can be given by

$$\text{Var}({\hat{\lambda }}_i)=\frac{m_{+i}^{(1)}{\lambda }_i(1-{\lambda }_i)+2m_{+i}^{(2)}{\lambda }_i[1+(R-2){\lambda }_i]}{(m_{+i}^{(1)}+2m_{+i}^{(2)}{)}^2}.$$ (7)

Therefore, the variance of $\hat{\delta }$ is given by ${\sigma }^2={\sigma }^2(\delta ,{\lambda }_0,R)=\text{Var}({\hat{\lambda }}_0)+\text{Var}({\hat{\lambda }}_1),$ and it can be estimated by

$$\begin{array}{rl}{\hat{\sigma }}^2& ={\sigma }^2(\hat{\delta },{\hat{\lambda }}_0,\hat{R})=\frac{m_{+0}^{(1)}{\hat{\lambda }}_0(1-{\hat{\lambda }}_0)+2m_{+0}^{(2)}{\hat{\lambda }}_0[1+(\hat{R}-2){\hat{\lambda }}_0]}{(m_{+0}^{(1)}+2m_{+0}^{(2)}{)}^2}\\ & +\frac{m_{+1}^{(1)}({\hat{\lambda }}_0+\hat{\delta })(1-{\hat{\lambda }}_0-\hat{\delta })+2m_{+1}^{(2)}({\hat{\lambda }}_0+\hat{\delta })[1+(\hat{R}-2)({\hat{\lambda }}_0+\hat{\delta })]}{(m_{+1}^{(1)}+2m_{+1}^{(2)}{)}^2}\end{array}$$ (8)

and

$$\begin{array}{rl}{\tilde{\sigma }}^2(\delta )& ={\sigma }^2(\delta ,{\tilde{\lambda }}_0,\tilde{R})\\ & =\frac{m_{+0}^{(1)}{\tilde{\lambda }}_0(1-{\tilde{\lambda }}_0)+2m_{+0}^{(2)}{\tilde{\lambda }}_0[1+(\tilde{R}-2){\tilde{\lambda }}_0]}{(m_{+0}^{(1)}+2m_{+0}^{(2)}{)}^2}\\ & +\frac{m_{+1}^{(1)}({\tilde{\lambda }}_0+\delta )(1-{\tilde{\lambda }}_0-\delta )+2m_{+1}^{(2)}({\tilde{\lambda }}_0+\delta )[1+(\tilde{R}-2)({\tilde{\lambda }}_0+\delta )]}{(m_{+1}^{(1)}+2m_{+1}^{(2)}{)}^2},\end{array}$$ (9)

respectively, where ${\tilde{\lambda }}_0={\tilde{\lambda }}_0(\delta )$, $\tilde{R}=\tilde{R}(\delta )$ are the constrained maximum-likelihood estimates (CMLEs) of ${\lambda }_0$ and R given the value of δ, and they can be obtained by solving the following equations:

$$\frac{\mathrm{\partial }\ell (\mathbf{m};\delta ,{\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0}=0,\frac{\mathrm{\partial }\ell (\mathbf{m};\delta ,{\lambda }_0,R)}{\mathrm{\partial }R}=0.$$ (10)

No closed-form exists, an iterative algorithm (e.g. Fisher-score iterative algorithm, refer to Appendix 1 for details) can be used to find the solutions to the above equations. When the solutions are out of the parameter space, we can use the search algorithm to find the CMLEs. According to the central limit theorem, it is easily shown that $T=(\hat{\delta }-\delta )/\sqrt{\mathrm{Var}(\hat{\delta })}$ is asymptotically distributed as standard normal distribution as all $m_{+i}^{(1)}$ and $m_{+i}^{(2)}$ ( $i=0,1$) are large. Therefore, the $100(1-\alpha )\mathrm{\%}$ confidence interval for δ based on the Wald method is given by

$${\text{CI}}_{w_1}=[max\{-1,\hat{\delta }-z_{\alpha /2}\hat{\sigma }\},min\{1,\hat{\delta }+z_{\alpha /2}\hat{\sigma }\}],$$ (11)

where $z_{\alpha /2}$ is the upper $\alpha /2$ percentile of the standard normal distribution.

Similar to Wilson [27], the $100(1-\alpha )\mathrm{\%}$ confidence lower and upper limits for δ can be obtained by solving the equations with respect to $\delta :(\hat{\delta }-\delta )/\tilde{\sigma }(\delta )=z_{\alpha /2}$ and $(\hat{\delta }-\delta )/\tilde{\sigma }(\delta )=-z_{\alpha /2}$, respectively. Let $f(\delta )=(\hat{\delta }-\delta )/\tilde{\sigma }(\delta )-z_{\alpha /2}$ and $g(\delta )=(\hat{\delta }-\delta )/\tilde{\sigma }(\delta )+z_{\alpha /2}$, the following bisection method can be used to find the solutions of these equations:

Step 1: Select a suitable constant h , for example, 0.1, 0.05, etc., start from $-1$ and search for the minimum positive integer k such that $f(-1+kh)\cdot f(-1+(k+1)h)\le 0$, where $[-1+kh,-1+(k+1)h]\subseteq [-1,1]$.

Step 2: Use the bisection method to find the root of $f(\delta )=0$ on the interval $[-1+kh,-1+(k+1)h]$, then we obtain the lower limit of the interval, denoted as ${\delta }_{wl}$.

Step 3: Similarly, start from 1 and search for the minimum positive integer k such that $g(1-kh)\cdot g(1-(k+1)h)\le 0$, where $[1-kh,1-(k+1)h]\subseteq [-1,1]$. Use the bisection method to find the root of $g(\delta )=0$ on the interval $[1-kh,1-(k+1)h]$, then we obtain the upper limit of the interval, denoted as ${\delta }_{wu}$.

Therefore, the $100(1-\alpha )\mathrm{\%}$ confidence interval for δ based on the Wilson method is given by ${\text{CI}}_{w_2}=[{\delta }_{wl},{\delta }_{wu}]$.

3.2. CI based on Agresti–Coull method

As shown in many literatures, the Wald CI given by Equation (11) usually performs not well when the sample size is small in the sense that it usually provides empirical coverage probabilities lower than the pre-specified confidence level. Adding a small count to every cell before computing the interval limits is a common strategy. For example, Agresti and Coull [1] suggested a Wald CI by adding 2 for one-sample binomial problems. Similar to Agresti and Coull [1], we use the simulation approach to find the small count to be added. The simulation study shows that the Wald CI given in Equation (11) by adding 0.25 to all cell counts (i.e. $m_{h^{\mathrm{\prime }}i}^{(1)}$ and $m_{hi}^{(2)}$, $h^{\mathrm{\prime }}=0,1$, h = 0, 1, 2, i = 0, 1) performs well. We denote this CI as ${\text{CI}}_{mw}$.

3.3. CI based on the inverse hyperbolic tangent transformation

Note that the Wald CI and the CI based on the Wilson method are derived from the normal approximation of $\hat{\delta }$. However, when the sample size is small or the data have a sparse structure, the asymptotical distribution of $\hat{\delta }$ is usually highly skewed. In this case, the inverse hyperbolic tangent transformation (i.e. Fisher's Z transformation [11]) can be used to improve the normal approximation of $\hat{\delta }$. By using this transformation for $\hat{\delta }$, we have ${\text{tanh}}^{-1}(\hat{\delta })=\frac{1}{2}\log [(1+\hat{\delta })/(1-\hat{\delta })]$. It is easily shown that the expectation of ${\text{tanh}}^{-1}(\hat{\delta })$ is given by $E[{\text{tanh}}^{-1}(\hat{\delta })]={\text{tanh}}^{-1}(\delta )$, and the variance of ${\text{tanh}}^{-1}(\hat{\delta })$ is given by ${\sigma }_z^2=\text{Var}({\text{tanh}}^{-1}(\hat{\delta }))=\text{Var}(\hat{\delta })/(1-{\delta }^2{)}^2$ by using the delta method. The corresponding variance estimation of ${\text{tanh}}^{-1}(\hat{\delta })$ is given by ${\hat{\sigma }}_z^2={\hat{\sigma }}^2/(1-{\hat{\delta }}^2{)}^2$. When all $m_{+i}^{(1)}$ and $m_{+i}^{(2)}$ (i = 0, 1) are large, the test statistic $[{\text{tanh}}^{-1}(\hat{\delta })-{\text{tanh}}^{-1}(\delta )]/{\hat{\sigma }}_z$ is asymptotically distributed as a standard normal distribution, so the $100(1-\alpha )\mathrm{\%}$ confidence interval for ${\text{tanh}}^{-1}(\delta )$ can be given by $[{\delta }_l^{(z)},{\delta }_u^{(z)}]$, where

$${\delta }_l^{(z)}={\text{tanh}}^{-1}(\hat{\delta })-z_{\alpha /2}{\hat{\sigma }}_z,{\delta }_u^{(z)}={\text{tanh}}^{-1}(\hat{\delta })+z_{\alpha /2}{\hat{\sigma }}_z.$$

Therefore, the $100(1-\alpha )\mathrm{\%}$ confidence interval for δ can be obtained via the inverse transformation of ${\text{tanh}}^{-1}(\delta )$, which is given by

$${\text{CI}}_z=[(\exp (2{\delta }_l^{(z)})-1)/(\exp (2{\delta }_l^{(z)})+1),(\exp (2{\delta }_u^{(z)})-1)/(\exp (2{\delta }_u^{(z)})+1)].$$ (12)

If $\hat{\delta }$ equals $-1$ or 1, the confidence limits for ${\text{tanh}}^{-1}(\delta )$ are not defined. In this case, we just take the CI for δ to be $[-1,1]$.

3.4. CI based on the Haldane method

According to the central limit theorem, since $(\hat{\delta }-\delta )/\sqrt{\mathrm{Var}(\hat{\delta })}$ is asymptotically distributed as the standard normal distribution when the sample size is large enough, then we have

$$\mathrm{P}(|(\hat{\delta }-\delta )/\sqrt{\mathrm{Var}(\hat{\delta })}|\le z_{\alpha /2})\approx 1-\alpha .$$

Let $\eta ={\lambda }_1-{\lambda }_0$, then ${\lambda }_0=(\eta -\delta )/2$ and ${\lambda }_1=(\eta +\delta )/2$. Thus, the variance $\mathrm{Var}(\hat{\delta })$ can be expressed in terms of parameters δ and η. Similar to Beal [3], we can consider the following quadratic equation of δ:

$$a{\delta }^2-b\delta +c\le 0,$$

where

$$\begin{array}{l}a=AB-z_{\alpha /2}^2\{2(R-2)(B{m}_{+0}^{(2)}+A{m}_{+1}^{(2)})-(m_{+0}^{(1)}B+m_{+1}^{(1)}A)\}/4,\\ b=2AB\hat{\delta }+z_{\alpha /2}^2\{(1-\eta )(m_{+1}^{(1)}A-m_{+0}^{(1)}B)+2((R-2)\eta +1)(m_{+1}^{(2)}A-m_{+0}^{(2)}B)\}/2,\\ c=AB{\hat{\delta }}^2-z_{\alpha /2}^2\eta \{(2(R-2)\eta +4)(m_{+0}^{(2)}B+m_{+1}^{(2)}A)-((\eta -2)(m_{+0}^{(1)}B+m_{+1}^{(1)}A)\}/4\end{array}$$

with $A=(m_{+0}^{(1)}+2m_{+0}^{(2)}{)}^2$ and $B=(m_{+1}^{(1)}+2m_{+1}^{(2)}{)}^2$. If a>0 and $b^2-4ac\ge 0$, the asymptotic $100(1-\alpha )\mathrm{\%}$ confidence limits of δ are given by $[{\delta }_l(\eta ),{\delta }_u(\eta )]$, where ${\delta }_l(\eta ,R)=max\{-1.0,(b-\sqrt{b^2-4ac})/2a\}$, ${\delta }_u(\eta ,R)=min\{1.0,(b+\sqrt{b^2-4ac})/2a\}$. The unknown parameters η and R can be estimated by $\hat{\eta }={\hat{\lambda }}_1+{\hat{\lambda }}_0$ and $\hat{R}$, respectively, therefore, the asymptotic $100(1-\alpha )\mathrm{\%}$ confidence interval for δ is given by

$${\text{CI}}_h=[{\delta }_l(\hat{\eta },\hat{R}),{\delta }_u(\hat{\eta },\hat{R})].$$ (13)

As noted by Beal [3], this confidence interval can be regarded as the extension of ‘Haldane Interval’ to account for the intraclass correlation of the combined unilateral and bilateral data. Note that the probability that the estimation of ${\lambda }_i$ is 0 cannot be negligible when ${\lambda }_i$ are small, especially when $m_{+i}^{(1)}$ and $m_{+i}^{(2)}$ (i = 0, 1) are small. In this case, we can adopt the simple adjustment by adding 0.5 to each cell.

3.5. CI based on the score test statistic

Using the general theory of efficient scores proposed by Rao [19], the score statistic for testing the hypothesis $H_0:\delta ={\delta }_0$ can be given by

$$T_{sc}({\delta }_0)=\frac{\mathrm{\partial }\ell (\mathbf{m};\delta ,{\lambda }_0,R)}{\mathrm{\partial }\delta }\sqrt{I^{11}}{|}_{\delta ={\delta }_0,{\lambda }_0={\tilde{\lambda }}_0({\delta }_0),R=\tilde{R}({\delta }_0)},$$

where

$$\begin{array}{rl}\frac{\mathrm{\partial }\ell (\mathbf{m};\delta ,{\lambda }_0,R)}{\mathrm{\partial }\delta }& =\frac{m_{11}^{(1)}+m_{11}^{(2)}+2m_{21}^{(2)}}{{\lambda }_0+\delta }-\frac{m_{01}^{(1)}}{1-{\lambda }_0-\delta }\\ & -\frac{2m_{01}^{(2)}[1-R({\lambda }_0+\delta )]}{1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta )}-\frac{m_{11}^{(2)}R}{1-R({\lambda }_0+\delta )}\end{array}$$

is the score function, and $I^{11}$ is the $(1,1)$th element of the inverse of the Fisher information matrix, which is given by

$$\begin{array}{r}\begin{array}{rl}{I}^{11}& ={[I_{11}-\left(\begin{array}{cc}{I}_{12}& I_{13}\end{array}\right){\left(\begin{array}{cc}{I}_{22}& I_{23}\\ I_{23}& I_{33}\end{array}\right)}^{-1}\left(\begin{array}{c}{I}_{12}\\ I_{13}\end{array}\right)]}^{-1}\\ & =\frac{I_{22}{I}_{33}-(I_{23}{)}^2}{2I_{12}{I}_{13}{I}_{23}+I_{11}{I}_{22}{I}_{33}-I_{11}(I_{23}{)}^2-I_{22}(I_{13}{)}^2-I_{33}(I_{12}{)}^2},\end{array}\end{array}$$

where $I_{sl}$ ( $1\le s\le l\le 3$) is the element of the Fisher information matrix (refer to Appendix for details).

As shown in [19], $T_{sc}$ is asymptotically distributed as a standard normal distribution under $H_0:\delta ={\delta }_0$ as all $m_{+i}^{(1)}$ and $m_{+i}^{(2)}$ (i = 0, 1) are large. Therefore, the $100(1-\alpha )\mathrm{\%}$ confidence interval for δ based on the score test can be given by

$${\text{CI}}_{sc}=[{\delta }_{sl},{\delta }_{su}],$$ (14)

where the lower limit ${\delta }_{sl}$ is the solution of the equation $T_{sc}({\delta }_0)=z_{\alpha /2}$ with respect to ${\delta }_0$, and the upper limit ${\delta }_{su}$ is the solution of the equation $T_{sc}({\delta }_0)=-z_{\alpha /2}$, respectively. Similarly, no closed form exists; the above bisection method given in Section 3.1 can be used to obtain the solutions.

3.6. CI based on the profile-likelihood-ratio test

For testing the hypothesis $H_0:\delta ={\delta }_0$, the likelihood ratio test statistic is given by

$$T_l({\delta }_0)=2[\ell (\mathbf{m};\hat{\delta },{\hat{\lambda }}_0,\hat{R})-\ell (\mathbf{m};{\delta }_0,{\tilde{\lambda }}_0({\delta }_0),\tilde{R}({\delta }_0))].$$

Similarly, $T_l({\delta }_0)$ is asymptotically distributed as the Chi-square distribution with one degree of freedom under $H_0:\delta ={\delta }_0$ when all $m_{+i}^{(1)}$ and $m_{+i}^{(2)}\to \mathrm{\infty }$ (i = 0, 1). Therefore, the asymptotic profile likelihood CI for δ can be obtained by inverting the likelihood ratio test $T_l({\delta }_0)$, i.e. the $100(1-\alpha )\mathrm{\%}$ confidence lower and upper limits for δ satisfy

$$2[\ell (\mathbf{m};\hat{\delta },{\hat{\lambda }}_0,\hat{R})-\ell (\mathbf{m};{\delta }_0,{\tilde{\lambda }}_0({\delta }_0),\tilde{R}({\delta }_0))]\le {\chi }_{1,\alpha }^2,$$

where ${\chi }_{1,\alpha }^2$ is the upper α percentile of the chi-square distribution with one degree of freedom. Let $f({\delta }_0)=2[\ell (\mathbf{m};\hat{\delta },{\hat{\lambda }}_0,\hat{R})-\ell (\mathbf{m};{\delta }_0,{\tilde{\lambda }}_0({\delta }_0),\tilde{R}({\delta }_0))]-{\chi }_{1,\alpha }^2$. Similarly, the following bisection method can be used to find the upper and lower limits:

Step 1: Select a suitable constant h, for example, 0.1, 0.05, etc., starting from $-1$ and search for the minimum positive integer k such that $f(-1+kh)\cdot f(-1+(k+1)h)\le 0$, where $[-1+kh,-1+(k+1)h]\subseteq [-1,1]$.

Step 2: Use the bisection method to find the root of $f({\delta }_0)=0$ on the interval $[-1+kh,-1+(k+1)h]$, then we obtain the lower limit of the interval, denoted as ${\delta }_{ll}$.

Step 3: Similarly, start from 1 and search for the minimum positive integer k such that $f(1-kh)\cdot f(1-(k+1)h)\le 0$, where $[1-kh,1-(k+1)h]\subseteq [-1,1]$. Use the bisection method to find the root of $f({\delta }_0)=0$ on the interval $[1-kh,1-(k+1)h]$, then we obtain the upper limit of the interval, denoted as ${\delta }_{lu}$.

Therefore, the $100(1-\alpha )\mathrm{\%}$ confidence interval for δ based on the likelihood ratio test is given by

$${\text{CI}}_l=[{\delta }_{ll},{\delta }_{lu}].$$ (15)

3.7. Bootstrap-resampling CIs

It is well known that the Bootstrap-resampling method has been applied extensively in many fields. For example, Efron and Tibshirani [10] and Shao and Tu [22] used the Bootstrap-resampling method to estimate the variability of complicated statistics, and Li [13] suggested the use of a Bootstrap procedure to generate the empirical distribution of the test statistic in ROC analysis. In particular, when the sample size is small, confidence intervals for proportion difference based on the large sample approximation may not be reliable. In this case, the Bootstrap-resampling method is usually recommended to construct CIs [10,22]. Therefore, we consider the following parametric Bootstrap-resampling procedure to construct the CI for δ:

Step 1: Given the observed data $\mathbf{m}=\{(m_{0i}^{(1)}$, $m_{1i}^{(1)}$, $m_{0i}^{(2)}$, $m_{1i}^{(2)}$, $m_{2i}^{(2)})$: $i=0,1\}$, we can obtain the parameter estimators ${\hat{\lambda }}_0$, ${\hat{\lambda }}_1$, $\hat{\delta }$ and $\hat{R}$ of the parameters ${\lambda }_0$, ${\lambda }_1$, δ and R via Equations (4)–(6). Let ${\hat{\sigma }}^2$ be the estimate of ${\sigma }^2$ that is calculated from the observed data.

Step 2: Generate the Bootstrap sample ${\mathbf{m}}^{\ast }=\{(m_{0i}^{\ast (1)}$, $m_{1i}^{\ast (1)}$, $m_{0i}^{\ast (2)}$, $m_{1i}^{\ast (2)}$, $m_{2i}^{\ast (2)})$: $i=0,1\}$ from the product of binomial and trinomial distributions based on the estimated parameters, i.e. $m_{1i}^{\ast (1)}$ follows Binomial distribution $(m_{+i}^{(1)};{\hat{p}}_{1i}^{(1)})$ and $(m_{0i}^{\ast (2)},m_{1i}^{\ast (2)},m_{2i}^{\ast (2)})$ follows Trinomial distribution $(m_{+i}^{(2)};{\hat{p}}_{0i}^{(2)},{\hat{p}}_{1i}^{(2)},{\hat{p}}_{2i}^{(2)})$ (i = 0, 1), where $p_{h^{\mathrm{\prime }}i}^{(1)}$ and $p_{hi}^{(2)}$ ( $h^{\mathrm{\prime }}=0,1$, h = 0, 1, 2) are given by Equation (1), in which parameters ${\lambda }_0$, ${\lambda }_1$ and R are substituted by their estimators ${\hat{\lambda }}_0$, ${\hat{\lambda }}_1$ and $\hat{R}$, respectively.

Step 3: For each generated sample ${\mathbf{m}}^{\ast }$, calculate the estimations ${\hat{\lambda }}_0^{\ast }$, ${\hat{\delta }}^{\ast }$ and ${\hat{R}}^{\ast }$ of ${\lambda }_0$, δ and R via Equations (4)–(6).

Step 4: Independently repeating the above process (i.e. Steps 2–3) B times, we can obtain B Bootstrap estimators ${\hat{\delta }}^{\ast (b)}$, ${\hat{\lambda }}_0^{\ast (b)}$ and ${\hat{R}}^{\ast (b)}$ of δ, ${\lambda }_0$ and R ( $b=1,2,\dots ,B$), respectively. The B Bootstrap estimators $\{{\hat{\delta }}^{\ast (b)}{\}}_{b=1}^B$ are then ordered from the smallest to the largest, and let ${\hat{\delta }}_{(1)}^{\ast }$, ${\hat{\delta }}_{(2)}^{\ast }$, ··· , ${\hat{\delta }}_{(B)}^{\ast }$ be the ordered values.

(i) Bootstrap percentile CI

Following Efron and Tibshirani [10], the $100(1-\alpha )\mathrm{\%}$ Bootstrap percentile CI for δ is given by

$${\text{CI}}_{b_1}=[{\hat{\delta }}_{([B\alpha /2])},{\hat{\delta }}_{([B(1-\alpha /2)])}],$$ (16)

where $[a]$ denotes the maximum integer not greater than a.

(ii) Bootstrap percentile-t CI

Let ${\hat{\sigma }}^{\ast (b)}=\sigma ({\hat{\delta }}^{\ast (b)},{\hat{\lambda }}_0^{\ast (b)},{\hat{R}}^{\ast (b)})$ be the estimated standard deviation of the bth Bootstrap estimator ${\hat{\delta }}^{\ast (b)}$. For each of the B bootstrap samples, we can obtain $\{t^{\ast (b)}=({\hat{\delta }}^{\ast (b)}-\hat{\delta })/{\hat{\sigma }}^{\ast (b)}:b=1,2,\dots ,B\}$. Following Efron and Tibshirani [10], the $100(1-\alpha )\mathrm{\%}$ bootstrap percentile-t confidence interval for δ can be obtained by

$${\text{CI}}_{b_2}=[max\{-1,\hat{\delta }-t_{([(1-\alpha /2)B])}^{\ast }\hat{\sigma }\},min\{1,\hat{\delta }+t_{([(1-\alpha /2)B])}^{\ast }\hat{\sigma }\}],$$ (17)

where $t_{(b)}^{\ast }$'s denote the ordered values of $t^{\ast (b)}$'s from the smallest to the largest and $\hat{\sigma }$ is the positive squared root of ${\hat{\sigma }}^2$ given in Step 1.

(iii) Bias-corrected Bootstrap percentile CI

Following DiCiccio and Efron [8], the $100(1-\alpha )\mathrm{\%}$ confidence interval for δ can be given by

$${\text{CI}}_{b_3}=[{\hat{\delta }}_{([B{\alpha }_1])},{\hat{\delta }}_{([B{\alpha }_2])}],$$ (18)

where ${\alpha }_1=\mathrm{\Phi }(2{\hat{z}}_0-z_{1-\alpha /2})$, ${\alpha }_2=\mathrm{\Phi }(2{\hat{z}}_0+z_{1-\alpha /2})$ with $z_0={\mathrm{\Phi }}^{-1}(\frac{1}{B}\sum _{b=1}^{B}I({\hat{\delta }}^{\ast (b)}<\hat{\delta }))$. Here, $\mathrm{\Phi }(\cdot )$ is the standard normal distribution function, and ${\mathrm{\Phi }}^{-1}(\cdot )$ is its inverse.

(iv) Bootstrap percentile-t CI combining with the inverse hyperbolic tangent transformation

Since that a variance stabilizing transformation much like the inverse hyperbolic tangent transformation can give good results when this transformation is used to construct CI for numbers lying between −1 and +1, then we consider a Bootstrap percentile-t CI combining with the inverse hyperbolic tangent transformation. First, for each of the B bootstrap samples, we can obtain $\{t_z^{\ast (b)}=[{\text{tanh}}^{-1}({\hat{\delta }}^{\ast (b)})-{\text{tanh}}^{-1}(\hat{\delta })]\cdot [1-({\hat{\delta }}^{\ast (b)}{)}^2]/{\hat{\sigma }}^{\ast (b)}:b=1,2,\dots ,B\}$, then the $100(1-\alpha )\mathrm{\%}$ bootstrap percentile-t confidence interval for ${\text{tanh}}^{-1}(\delta )$ can be obtained by $[{\delta }_{bzl},{\delta }_{bzu}]=[{\text{tanh}}^{-1}(\hat{\delta })-t_{z([(1-\alpha /2)B])}^{\ast }\hat{\sigma }/[1-({\hat{\delta }}^{\ast (b)}{)}^2]$, ${\text{tanh}}^{-1}(\hat{\delta })+t_{z([(1-\alpha /2)B])}^{\ast }\hat{\sigma }/[1-({\hat{\delta }}^{\ast (b)}{)}^2]]$, where $t_{z(b)}^{\ast }$'s denote the ordered values of $t_z^{\ast (b)}$'s from the smallest to the largest. Therefore, the $100(1-\alpha )\mathrm{\%}$ bootstrap confidence interval for δ is given by

$${\text{CI}}_{b_4}=[(\exp (2{\delta }_{bzl})-1)/(\exp (2{\delta }_{bzl})+1),(\exp (2{\delta }_{bzu})-1)/(\exp (2{\delta }_{bzu})+1)].$$ (19)

Note that the asymptotical methods based on the large sample assumption do not necessarily control their actual coverage probabilities at the pre-specified confidence level for small sample sizes. In this case, some adjusted methods can be used to construct the CI for proportion difference, for example, the approximate unconditional methods proposed in Tang et al. [24], the exact binomial method and mid-P method modified from the exact binomial method proposed in Li et al. [14]. However, for two groups of combined unilateral and bilateral data, a severe computing burden will be encountered for our problem even for the very small sample size $m_{+0}^{(1)}=m_{+1}^{(1)}=m_{+0}^{(2)}=m_{+1}^{(2)}=10$. Therefore, we have not adopted the exact and approximate unconditional methods in this article.

4. Simulation study

In this section, we investigate the performance of the proposed confidence intervals via simulation studies in terms of the empirical coverage probability (ECP) and the empirical coverage width (ECW). In general, a method to construct a CI is better if the ECP is closer to the pre-specified confidence level and the ECW is smaller. Following Newcombe [16], the location of a CI can be characterized in terms of the balance between the mesial non-coverage probability (MNCP) and the distal non-coverage probability (DNCP). A simple index, i.e. the ratio of the MNCP to the non-coverage probability (RMNCP=MNCP/( $1-$ECP )) can be used to evaluate the location of a CI, where DNCP and MNCP are defined with respect to the true value of δ. These evaluation indices are defined as follows.

(i) Empirical coverage probability

$$\text{ECP}=\frac{1}{K}\sum _{k=1}^{K}I(\delta \in [{\delta }_l({\mathbf{m}}^{(k)}),{\delta }_u({\mathbf{m}}^{(k)})]),$$

where K is the number of replications and $I(\cdot )$ is the indicator function, ${\mathbf{m}}^{(k)}=\{(m_{0i}^{(1)}$, $m_{1i}^{(1)}$, $m_{0i}^{(2)}$, $m_{1i}^{(2)}$, $m_{2i}^{(2)}{)}^{(k)}$: $i=0,1\}$ is the kth replication, and $[{\delta }_l({\mathbf{m}}^{(k)}),{\delta }_u({\mathbf{m}}^{(k)})]$ is the CI that is constructed from the set of ${\mathbf{m}}^{(k)}$ by any of the 11 methods under evaluation.

(ii) Empirical coverage width

$$\text{ECW}=\frac{1}{K}\sum _{k=1}^K({\delta }_u({\mathbf{m}}^{(k)})-{\delta }_l({\mathbf{m}}^{(k)})).$$

(iii) Ratio of the mesial non-coverage probability to the non-coverage probability

According to Newcombe [16], the mesial non-coverage probability (MNCP) is defined as

$$\text{MNCP}=\frac{1}{K}\sum _{k=1}^{K}I(\delta \in A({\mathbf{m}}^{(k)})),$$

where $A({\mathbf{m}}^{(k)})$ is given by

$$A({\mathbf{m}}^{(k)})=\{\begin{array}{ll}[-1.0,{\delta }_l({\mathbf{m}}^{(k)})),& \text{if}\delta >0.0,\\ ({\delta }_u({\mathbf{m}}^{(k)}),1.0]\cup [-1.0,{\delta }_l({\mathbf{m}}^{(k)})),& \text{if}\delta =0.0,\\ ({\delta }_u({\mathbf{m}}^{(k)}),1.0],& \text{if}\delta <0.0,\end{array}$$

and distal non-coverage probability (DNCP) is defined as

$$\text{DNCP}=\frac{1}{K}\sum _{k=1}^{K}I(\delta \in B({\mathbf{m}}^{(k)})),$$

where $B({\mathbf{m}}^{(k)})$ is given by

$$B({\mathbf{m}}^{(k)})=\{\begin{array}{ll}[-1.0,{\delta }_l({\mathbf{m}}^{(k)})),& \text{if}\delta <0.0,\\ ({\delta }_u({\mathbf{m}}^{(k)}),1.0],& \text{if}\delta >0.0.\end{array}$$

The ratio of the mesial non-coverage probability to the non-coverage probability (RMNCP) is then defined as

$$\text{RMNCP}=\text{MNCP}/\text{NCP}=\text{MNCP}/(\text{MNCP}+\text{DNCP}).$$

As shown in [16], a CI is classified as satisfactory if the RMNCP lies in [0.4, 0.6], and too mesially located if the ratio is smaller than 0.4, and too distal if it is greater than 0.6.

To evaluate the proposed methods, we consider three kinds of sample size designs of $(m_{+0}^{(1)}$, $m_{+1}^{(1)})$, $m_{+0}^{(2)}$, $m_{+1}^{(2)})$: (i) small sample size $(20$, 20, 20, $20)$; (ii) moderate sample size $(30$, 30, 30, $30)$; and (iii) large sample size $(50$, 50, 50, $50)$. We only report the simulation results for balanced sample sizes due to no substantial difference for the performance between balanced and unbalanced sample sizes. For parameter settings of ( ${\lambda }_0$, δ, R), a total of $54=3\times 3\times 6$ combinations with $\delta =0.0$, 0.10, 0.15, ${\lambda }_0=0.1$, 0.3, 0.5 and $R=1.0(0.1)1.5$, i.e. from 1.0 to 1.5 with step size 0.1 are considered in the simulation studies. A total of 5000 replications are conducted to the simulation studies and the bootstrap CI of δ is based on 2000 replications. According to the definition of MNCP, when $\delta =0.0$, the RMNCP is always equal to 1.0. Therefore, we only report RMNCPs of all CIs for the situations with $\delta \ne 0$.

Figures 1(i), 2(i) and 3(i) report the change of ECPs of two-sided $95\mathrm{\%}$ CIs for δ as the change of R. It is observed that ${\text{CI}}_{w_1}$ and ${\text{CI}}_{b_3}$ have slightly deflated ECPs when the sample size is small (e.g. $m_{+0}^{(1)}=m_{+1}^{(1)}=m_{+0}^{(2)}=m_{+1}^{(2)}=20$), and ${\text{CI}}_{mw}$ is slightly conservative for $\delta =0.0$, ${\lambda }_0=0.1$ under the small sample size (i.e. $m_{+0}^{(1)}=m_{+1}^{(1)}=m_{+0}^{(2)}=m_{+1}^{(2)}=20$), other CIs generally perform well in the sense that they can well control their ECPs around the pre-specified confidence level. Although the intra-class correlation is not ignorable in bilateral data analysis, it seems that the CIs except for ${\text{CI}}_l$ are independent of data structure, i.e. there is no obvious pattern on ECPs as R changes as shown in Figures 1–3, while the ECP of ${\text{CI}}_l$ increases as the increase of R for ${\lambda }_0=0.5$ with $\delta =0.1$ and 0.15.

Figure 1.

(i) ECPs of two-sided 95% CIs of δ versus R; (ii) ECWs of two-sided 95% CIs of δ versus R under sample size $(m_{+0}^{(1)},m_{+1}^{(1)},m_{+0}^{(2)},m_{+1}^{(2)})=(20,20,20,20)$.

Figure 2.

(i) ECPs of two-sided 95% CIs of δ versus R; (ii) ECWs of two-sided 95% CIs of δ versus R under sample size $(m_{+0}^{(1)},m_{+1}^{(1)},m_{+0}^{(2)},m_{+1}^{(2)})=(30,30,30,30)$.

Figure 3.

(i) ECPs of two-sided 95% CIs of δ versus R; (ii) ECWs of two-sided 95% CIs of δ versus R under sample size $(m_{+0}^{(1)},m_{+1}^{(1)},m_{+0}^{(2)},m_{+1}^{(2)})=(50,50,50,50)$.

Figures 1(ii), 2(ii) and 3(ii) report the change of ECWs of two-sided $95\mathrm{\%}$ CIs for δ as the change of R. It is interesting to find that ECWs of the CIs except for ${\text{CI}}_{sc}$ and ${\text{CI}}_l$ tend to be wider as R increases, and can be seen increasing patterns that look like linear, while ECWs of ${\text{CI}}_{sc}$ and ${\text{CI}}_l$ show the patterns that look like nonlinear. As expected, ECWs of all CIs are shorter with the increase of the sample size.

Figure 4 reports the change of RMNCPs of two-sided $95\mathrm{\%}$ CIs for δ as the change of R. It is observed that all CIs except for ${\text{CI}}_{b_3}$ usually have satisfactory interval locations, as their ratios of the mesial non-coverage probability to total non-coverage probability are close to 0.5 even for small sample sizes, while ${\text{CI}}_{b_3}$ has too mesially located interval in some cases. The larger the sample sizes, the closer to 0.5 the RMNCPs of all CIs. As shown in Figure 4, it seems that there is no obvious pattern on RMNCPs as R changes.

Figure 4.

RMNCPs of two-sided 95% CIs of δ versus R under different sample sizes of $(m_{+0}^{(1)},m_{+1}^{(1)},m_{+0}^{(2)},m_{+1}^{(2)})$: (i) $(20,20,20,20)$; (ii) $(30,30,30,30)$; (iii) $(50,50,50,50)$.

To further investigate the overall performance of the CIs across a range of values for the nuisance parameters ${\lambda }_0$ and R for given values of δ: i.e. $\delta =-0.2(0.1)0.2$ , let $c_1=max\{(2{\lambda }_0-1)/{\lambda }_0^2,0.0\}$ and $c_2=max\{[2({\lambda }_0+\delta )-1]/({\lambda }_0+\delta {)}^2,0.0\}$; if $\delta \le 0.0$, ${\lambda }_0=(-\delta +\frac{1+\delta }{10})(\frac{1+\delta }{5})(-\delta +\frac{9(1+\delta )}{10})$, $R=(c_1+\frac{1/{\lambda }_0-c_1}{10})(\frac{1/{\lambda }_0-c_1}{5})(c_1+\frac{9(1/{\lambda }_0-c_1)}{10})$; if $\delta >0$, ${\lambda }_0=\frac{1-\delta }{10}(\frac{1-\delta }{5})\frac{9(1-\delta )}{10}$, $R=(c_2+\frac{1/({\lambda }_0+\delta )-c_2}{10})(\frac{1/({\lambda }_0+\delta )-c_2}{5})(c_2+\frac{9(1/({\lambda }_0+\delta )-c_2)}{10})$, where $a(b)c$ means that the value is from a to c with step size b. Four sample size designs of $(m_{+0}^{(1)}$, $m_{+1}^{(1)})$, $m_{+0}^{(2)}$, $m_{+1}^{(2)})$ are considered in this study, i.e. (i) $(20,20,20,20)$ ; (ii) $(30,30,30,30)$ ; (iii) $(50,50,50,50)$ ; and (iv) $(30,50,30,50)$ . Boxplots of ECPs, ECWs and RMNCPs of various CIs are reported in Figure 5.

Figure 5.

Boxplots of ECPs, ECWs, RMNCPs of two-sided 95% CIs for δ under different sample sizes of $(m_{+0}^{(1)},m_{+1}^{(1)},m_{+0}^{(2)},m_{+1}^{(2)})$: (i) $(20,20,20,20)$; (ii) $(30,30,30,30)$; (iii) $(50,50,50,50)$ and (iv) $(30,50,30,50)$.

In terms of the coverage probability, the boxplots of ECP in Figure 5 show that the results of various methods are generally satisfactory. Furthermore, ${\text{CI}}_{mw}$, ${\text{CI}}_{w2}$ and ${\text{CI}}_{sc}$ perform better than the others, as their median empirical coverage probabilities are closer to the preassigned confidence level even under small sample sizes. When the sample size increases, the performance of other CIs show improvements. The results given in Figures 1–3 also support these findings.

With regard to the width of the various CIs, the boxplots of ECW in Figure 5 show that ${\text{CI}}_{sc}$ has wider interval width than the others under small and moderate sample sizes (e.g. $(m_{+0}^{(1)}$, $m_{+1}^{(1)}$, $m_{+0}^{(2)}$, $m_{+1}^{(2)})$ = $(20,20,20,20)$ and $(30,30,30,30)$), as its median empirical coverage widths are larger than the others. However, when the sample size is large, all CIs have similar median ECWs.

For the location of the various CIs, the boxplots of RMNCP in Figure 5 show that the RMNCPs of ${\text{CI}}_{sc}$ are almost in $[0.4,0.6]$, suggesting that the CI has a satisfactory location. Except for ${\text{CI}}_{b3}$, the other CIs can also produce good results, as their median RMNCPs are closer to 0.5, while for ${\text{CI}}_{b3}$, the values of RMNCP suggest that the CI tends to be slightly too mesial when the sample size is not large; the results given in Figures 1–3 also support these findings.

In summary, with the increase of the sample size, ECPs of all CIs are closer to the pre-specified confidence level, and all CIs have better interval locations and smaller interval widths. Generally, all CIs except for ${\text{CI}}_{w_1}$ and ${\text{CI}}_{b_3}$ can usually achieve the nominal coverage and exhibit shorter interval widths from small to large sample sizes, and hence are recommended to practical applications.

5. Real example

To illustrate the proposed methods in this article, we re-visit the otolaryngological study in Section 1. Let i = 0 be the treatment of Amoxicillin and i = 1 be the treatment of Cefaclor, respectively. According to Equations (4)–(6), $\hat{\delta }=0.1558$, ${\hat{\lambda }}_0=0.4375$ and $\hat{R}=1.4679$. The corresponding $95\mathrm{\%}$ confidence intervals for δ based on various methods are reported in Table 3.

Table 3.

Various $95\mathrm{\%}$ confidence intervals of δ for real data.

Methods	CI	Width	Methods	CI	Width
${\text{CI}}_{w1}$	$(0.0241,0.2875)$	0.2634	${\text{CI}}_l$	$(-0.0780,0.2247)$	0.3027
${\text{CI}}_{mw}$	$(0.0228,0.2845)$	0.2617	${\text{CI}}_{b1}$	$(0.0191,0.2839)$	0.2648
${\text{CI}}_z$	$(0.0221,0.2840)$	0.2619	${\text{CI}}_{b2}$	$(0.0348,0.2768)$	0.2420
${\text{CI}}_h$	$(0.0233,0.2853)$	0.2620	${\text{CI}}_{b3}$	$(0.0255,0.2770)$	0.2515
${\text{CI}}_{w2}$	$(0.0193,0.2813)$	0.2620	${\text{CI}}_{b4}$	$(0.0320,0.2750)$	0.2430
${\text{CI}}_{sc}$	$(-0.0563,0.2127)$	0.2690

It is noteworthy that the lower limits of all CIs except for ${\text{CI}}_{sc}$ and ${\text{CI}}_l$ are all greater than 0.0. The results indicate that there is no evidence to support rejecting the hypothesis $\delta =0$ at the $95\mathrm{\%}$ confidence level using ${\text{CI}}_{sc}$ and ${\text{CI}}_l$, although we reject it at this confidence level using the other nine methods. In this case, which result is reliable? The results of our simulation study show that ${\text{CI}}_{sc}$ usually have a little wider interval widths than other CIs, although it can well control it's ECPs around the pre-specified confidence level. Since the upper limits of all CIs are greater than 0.0. Moreover, the lower limits of ${\text{CI}}_{sc}$ and ${\text{CI}}_l$ are very close to 0.0. Therefore, we are more inclined to the conclusion that the treatment of Amoxicillin is more effective than the treatment of Cefaclor at the $5\mathrm{\%}$ nominal level.

6. Conclusion and discussion

We consider the problem of CI construction for the proportion difference based on combined unilateral and bilateral data, which are commonly observed in the paired organs or two body parts studies. Seven asymptotic CIs, i.e. ${\text{CI}}_{w1}$, ${\text{CI}}_{w2}$, ${\text{CI}}_{mw}$, ${\text{CI}}_z$, ${\text{CI}}_h$, ${\text{CI}}_{sc}$ and ${\text{CI}}_l$ based on Rosner's dependence model, are constructed. Together with 4 bootstrap resampling methods, i.e. ${\text{CI}}_{b1}$, ${\text{CI}}_{b2}$, ${\text{CI}}_{b3}$ and ${\text{CI}}_{b4}$, we have developed 11 methods to construct CIs. A large-scale empirical study has been conducted to evaluate the performance of these 11 methods from different aspects. Overall, the empirical results suggest that all CIs except for ${\text{CI}}_{w_1}$ and ${\text{CI}}_{b_3}$ generally perform well from small to large sample size designs, while ${\text{CI}}_{w_1}$ and ${\text{CI}}_{b_3}$ produce a little deflated ECPs when the sample size is small (e.g. $(m_{+0}^{(1)}$, $m_{+1}^{(1)}$, $m_{+0}^{(2)}$, $m_{+1}^{(2)})=(20,20,20,20)$), and ${\text{CI}}_{aw}$ is slightly conservative for $\delta =0.0$, ${\lambda }_0=0.1$ under small sample sizes (e.g. $(m_{+0}^{(1)}$, $m_{+1}^{(1)}$, $m_{+0}^{(2)}$, $m_{+1}^{(2)})=(20,20,20,20)$). All CIs can well control their ECPs close to the pre-specified confidence level with preferred RMNCPs and shorter interval widths when sample sizes are large. Therefore, all CIs except for ${\text{CI}}_{w_1}$ and ${\text{CI}}_{b_3}$ are recommended for practical applications for small to large sample size designs, and when the sample size is large, ${\text{CI}}_{w_1}$ and ${\text{CI}}_{b_3}$ can also be recommended for applications.

The intra-class correlation is not ignorable in bilateral data analysis. Therefore, simulation studies are conducted to investigate the effect of the measurement of dependence (i.e. R) on ECP, ECW and RMNCP of various CIs. It is interesting to find that the ECWs of all CIs except for ${\text{CI}}_{sc}$ and ${\text{CI}}_l$ show like a linear increasing trend as R increases, and ${\text{CI}}_{sc}$ and ${\text{CI}}_l$ show like a nonlinear trend. However, except for ${\text{CI}}_l$, there is no obvious pattern on ECPs of other CIs, and for all CIs, there is no obvious pattern on RMNCPs.

It is well known that confidence intervals for proportion difference based on the large sample approximation may not be reliable when the sample size is small. In this case, other CI construction methods are available in the literature, such as exact unconditional (e.g. [2,5,6]) and approximate unconditional methods [24]. However, for two groups of combined unilateral and bilateral data, a severe computing burden will be encountered for our problem even under very small sample sizes. In addition, several methods proposed in this article perform well even under small sample sizes. Therefore, we do not consider these methods in this article.

The equivalence testing of two cure rates in the paired organs or two body parts studies has been investigated in other literature, for example, [18,20,21,25,26]; however, they considered the situation in which only bilateral binary data are available. When unilateral data are also available, although Pei et al. [17] have investigated these combined data, they just considered the hypothesis testing under the equal correlation coefficient model. In this article, we investigate a number of methods to construct CIs in the presence of both bilateral and unilateral data under Rosner's correlated binary data model, and several effective methods are recommended for practical applications.

Appendix 1. The restrained MLEs ${\tilde{\lambda }}_0(\delta )$ and $\tilde{R}(\delta )$ of ${\lambda }_0$ and R given the value of δ

Given the value of δ, the log-likelihood function of $\mathbf{m}$ is given by

$$\begin{array}{rl}\ell ({\lambda }_0,R)& =C+m_{00}^{(1)}\log (1-{\lambda }_0)+(m_{10}^{(1)}+m_{10}^{(2)}+2m_{20}^{(2)})\log {\lambda }_0+m_{01}^{(1)}\log (1-{\lambda }_0-\delta )\\ & +(m_{11}^{(1)}+m_{11}^{(2)}+2m_{21}^{(2)})\log ({\lambda }_0+\delta )+m_{00}^{(2)}\log (1+R{\lambda }_0^2-2{\lambda }_0)\\ & +(m_{20}^{(2)}+m_{21}^{(2)})\log R+m_{10}^{(2)}\log (1-R{\lambda }_0)+m_{11}^{(2)}\log [1-R({\lambda }_0+\delta )]\\ & +m_{01}^{(2)}\log [1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta )].\end{array}$$

Differentiating $\ell ({\lambda }_0,R)$ with respect to ${\lambda }_0$ and R yields

$$\begin{array}{rl}\frac{\mathrm{\partial }\ell ({\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0}& =\frac{m_{10}^{(1)}+m_{10}^{(2)}+2m_{20}^{(2)}}{{\lambda }_0}-\frac{m_{00}^{(1)}}{1-{\lambda }_0}-\frac{m_{10}^{(2)}R}{1-R{\lambda }_0}+\frac{2m_{00}^{(2)}(R{\lambda }_0-1)}{1+R{\lambda }_0^2-2{\lambda }_0}\\ & +\frac{m_{11}^{(1)}+m_{11}^{(2)}+2m_{21}^{(2)}}{{\lambda }_0+\delta }-\frac{m_{01}^{(1)}}{1-({\lambda }_0+\delta )}-\frac{m_{11}^{(2)}R}{1-R({\lambda }_0+\delta )}\\ & +\frac{2m_{01}^{(2)}[R({\lambda }_0+\delta )-1]}{1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta )}=0,\\ \frac{\mathrm{\partial }\ell ({\lambda }_0,R)}{\mathrm{\partial }R}& =\frac{m_{20}^{(2)}+m_{21}^{(2)}}{R}-\frac{m_{10}^{(2)}{\lambda }_0}{1-R{\lambda }_0}+\frac{m_{00}^{(2)}{\lambda }_0^2}{1+R{\lambda }_0^2-2{\lambda }_0}\\ & -\frac{m_{11}^{(2)}({\lambda }_0+\delta )}{1-R({\lambda }_0+\delta )}+\frac{m_{01}^{(2)}({\lambda }_0+\delta {)}^2}{1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta )}=0.\end{array}$$

Then, the restrained MLEs ${\tilde{\lambda }}_0(\delta )$ and $\tilde{R}(\delta )$ of ${\lambda }_0$ and R satisfy the following equations:

$$\frac{\mathrm{\partial }\ell ({\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0}=0,\frac{\mathrm{\partial }\ell ({\lambda }_0,R)}{\mathrm{\partial }R}=0.$$

Differentiating $\frac{\mathrm{\partial }\ell ({\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0}$ and $\frac{\mathrm{\partial }\ell ({\lambda }_0,R)}{\mathrm{\partial }R}$ with respect to ${\lambda }_0$ and R leads to

$$\begin{array}{rl}\frac{{\mathrm{\partial }}^2\ell ({\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0^2}& =-\frac{m_{10}^{(1)}+m_{10}^{(2)}+2m_{20}^{(2)}}{{\lambda }_0^2}-\frac{m_{00}^{(1)}}{(1-{\lambda }_0{)}^2}-\frac{m_{10}^{(2)}{R}^2}{(1-R{\lambda }_0{)}^2}\\ & -\frac{2m_{00}^{(2)}(R^2{\lambda }_0^2-2R{\lambda }_0-R+2)}{(1+R{\lambda }_0^2-2{\lambda }_0{)}^2}-\frac{m_{11}^{(1)}+m_{11}^{(2)}+2m_{21}^{(2)}}{({\lambda }_0+\delta {)}^2}\\ & -\frac{m_{01}^{(1)}}{[1-({\lambda }_0+\delta ){]}^2}-\frac{m_{11}^{(2)}{R}^2}{[1-R({\lambda }_0+\delta ){]}^2}\\ & -\frac{2m_{01}^{(2)}[R^2({\lambda }_0+\delta {)}^2-2R({\lambda }_0+\delta )-R+2]}{[1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta ){]}^2}\end{array}$$

$$\begin{array}{rl}\frac{{\mathrm{\partial }}^2\ell ({\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0\mathrm{\partial }R}& =-\frac{m_{10}^{(2)}}{(1-R{\lambda }_0{)}^2}+\frac{2m_{00}^{(2)}{\lambda }_0(1-{\lambda }_0)}{(1+R{\lambda }_0^2-2{\lambda }_0{)}^2}-\frac{m_{11}^{(2)}}{[1-R({\lambda }_0+\delta ){]}^2}\\ & +\frac{2m_{01}^{(2)}({\lambda }_0+\delta )[1-({\lambda }_0+\delta )]}{[1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta ){]}^2}\\ \frac{{\mathrm{\partial }}^2\ell ({\lambda }_0,R)}{\mathrm{\partial }{R}^2}& =-\frac{m_{20}^{(2)}+m_{21}^{(2)}}{R^2}-\frac{m_{10}^{(2)}{\lambda }_0^2}{(1-R{\lambda }_0{)}^2}-\frac{m_{00}^{(2)}{\lambda }_0^4}{(1+R{\lambda }_0^2-2{\lambda }_0{)}^2}\\ & -\frac{m_{11}^{(2)}({\lambda }_0+\delta {)}^2}{[1-R({\lambda }_0+\delta ){]}^2}-\frac{m_{01}^{(2)}({\lambda }_0+\delta {)}^4}{[1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta ){]}^2}.\end{array}$$

From $E(m_{hi}^{(k)}/m_{+i})=p_{hi}^{(k)}$ (h = 0, 1, 2, i = 0, 1 and k = 1, 2), we have

$$\begin{array}{rl}{I}_{11}& =E(-\frac{{\mathrm{\partial }}^2\ell ({\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0^2})=2m_{+0}^{(2)}[\frac{R^2{\lambda }_0^2-2R{\lambda }_0-R+2}{1+R{\lambda }_0^2-2{\lambda }_0}+\frac{1}{{\lambda }_0}+\frac{R^2{\lambda }_0}{1-R{\lambda }_0}]+\frac{m_{+0}^{(1)}}{{\lambda }_0(1-{\lambda }_0)}\\ & +2m_{+1}^{(2)}[\frac{R^2({\lambda }_0+\delta {)}^2-2R({\lambda }_0+\delta )-R+2}{1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta )}+\frac{1}{{\lambda }_0+\delta }+\frac{R^2({\lambda }_0+\delta )}{1-R({\lambda }_0+\delta )}],\\ I_{12}& =E(-\frac{{\mathrm{\partial }}^2\ell ({\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0\mathrm{\partial }R})=2m_{+0}^{(2)}{\lambda }_0[\frac{{\lambda }_0-1}{1+R{\lambda }_0^2-2{\lambda }_0}+\frac{1}{1-R{\lambda }_0}]\\ & +2m_{+1}^{(2)}({\lambda }_0+\delta )[\frac{{\lambda }_0+\delta -1}{1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta )}+\frac{1}{1-R({\lambda }_0+\delta )}],\\ I_{22}& =E(-\frac{{\mathrm{\partial }}^2\ell ({\lambda }_0,R)}{\mathrm{\partial }{R}^2})=m_{+0}^{(2)}[\frac{{\lambda }_0^4}{1+R{\lambda }_0^2-2{\lambda }_0}+\frac{2{\lambda }_0^3}{1-R{\lambda }_0}+\frac{{\lambda }_0^2}{R}]\\ & +m_{+1}^{(2)}[\frac{({\lambda }_0+\delta {)}^4}{1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta )}+\frac{2({\lambda }_0+\delta {)}^3}{1-R({\lambda }_0+\delta )}+\frac{({\lambda }_0+\delta {)}^2}{R}].\end{array}$$

Then, the Fisher information matrix is given by

$$I({\lambda }_0,R)=\left(\begin{array}{cc}{I}_{11}& I_{12}\\ I_{12}& I_{22}\end{array}\right).$$

Then the restrained MLEs ${\tilde{\lambda }}_0(\delta )$ and $\tilde{R}(\delta )$ by iteratively solving the following equation

$$\left(\begin{array}{c}{\lambda }_0^{(t+1)}\\ R^{(t+1)}\end{array}\right)=\left(\begin{array}{c}{\lambda }_0^{(t)}\\ R^{(t)}\end{array}\right)+(I({\lambda }_0^{(t)},R^{(t)}){)}^{-1}{\left(\begin{array}{c}{\displaystyle \frac{\mathrm{\partial }\ell ({\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0}}\\ {\displaystyle \frac{\mathrm{\partial }\ell ({\lambda }_0,R)}{\mathrm{\partial }R}}\end{array}\right)|}_{{\lambda }_0={\lambda }_0^{(t)},R=R^{(t)}},t=0,1,2,\dots .$$

Appendix 2. Derivation of score statistic $T_{sc}$

Differentiating $\frac{\mathrm{\partial }\ell (\delta ,{\lambda }_0,R)}{\mathrm{\partial }\delta }$, $\frac{\mathrm{\partial }\ell (\delta ,{\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0}$, $\frac{\mathrm{\partial }\ell (\delta ,{\lambda }_0,R)}{\mathrm{\partial }R}$ with respect to δ, ${\lambda }_0$ and R, respectively. Thus, it is easily shown that the Fisher information matrix is given by

$$\begin{array}{r}I(\delta ,{\lambda }_0,R)=\left(\begin{array}{ccc}{I}_{11}& I_{12}& I_{13}\\ I_{12}& I_{22}& I_{23}\\ I_{13}& I_{23}& I_{33},\end{array}\right),\end{array}$$

where

$$\begin{array}{rl}{I}_{11}& =E(-\frac{{\mathrm{\partial }}^2\ell (\delta ,{\lambda }_0,R)}{\mathrm{\partial }{\delta }^2})\\ & =2m_{+1}^{(2)}[\frac{R^2({\lambda }_0+\delta {)}^2-2R({\lambda }_0+\delta )-R+2}{1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta )}+\frac{1}{{\lambda }_0+\delta }+\frac{R^2({\lambda }_0+\delta )}{1-R({\lambda }_0+\delta )}]\\ & +\frac{m_{+1}^{(1)}}{({\lambda }_0+\delta )[1-({\lambda }_0+\delta )]},I_{12}=E(-\frac{{\mathrm{\partial }}^2\ell (\delta ,{\lambda }_0,R)}{\mathrm{\partial }\delta \mathrm{\partial }{\lambda }_0})=I_{11},\\ I_{13}& =E(-\frac{{\mathrm{\partial }}^2\ell (\delta ,{\lambda }_0,R)}{\mathrm{\partial }\delta \mathrm{\partial }R})=2m_{+1}^{(2)}({\lambda }_0+\delta )[\frac{{\lambda }_0+\delta -1}{1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta )}+\frac{1}{1-R({\lambda }_0+\delta )}],\\ I_{22}& =E(-\frac{{\mathrm{\partial }}^2\ell (\delta ,{\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0^2})\\ & =I_{11}+2m_{+0}^{(2)}[\frac{R^2{\lambda }_0^2-2R{\lambda }_0-R+2}{1+R{\lambda }_0^2-2{\lambda }_0}+\frac{1}{{\lambda }_0}+\frac{R^2{\lambda }_0}{1-R{\lambda }_0}]+\frac{m_{+0}^{(1)}}{{\lambda }_0(1-{\lambda }_0)},\\ I_{23}& =E(-\frac{{\mathrm{\partial }}^2\ell (\delta ,{\lambda }_0,R)}{\mathrm{\partial }{\lambda }_0\mathrm{\partial }R})=I_{13}+2m_{+0}^{(2)}{\lambda }_0[\frac{{\lambda }_0-1}{1+R{\lambda }_0^2-2{\lambda }_0}+\frac{1}{1-R{\lambda }_0}],\\ I_{33}& =E(-\frac{{\mathrm{\partial }}^2\ell (\delta ,{\lambda }_0,R)}{\mathrm{\partial }{R}^2})=m_{+0}^{(2)}[\frac{{\lambda }_0^4}{1+R{\lambda }_0^2-2{\lambda }_0}+\frac{2{\lambda }_0^3}{1-R{\lambda }_0}+\frac{{\lambda }_0^2}{R}]\\ & +m_{+1}^{(2)}[\frac{({\lambda }_0+\delta {)}^4}{1+R({\lambda }_0+\delta {)}^2-2({\lambda }_0+\delta )}+\frac{2({\lambda }_0+\delta {)}^3}{1-R({\lambda }_0+\delta )}+\frac{({\lambda }_0+\delta {)}^2}{R}],\end{array}$$

and the first main-diagonal element of the inverse of the Fisher information matrix $I(\delta ,{\lambda }_0,R)$ is given by

$$\begin{array}{rl}{I}^{11}& =[I_{11}-\left(\begin{array}{cc}{I}_{12}& I_{13}\end{array}\right){\left(\begin{array}{cc}{I}_{22}& I_{23}\\ I_{23}& I_{33}\end{array}\right)}^{-1}\left(\begin{array}{c}{I}_{12}\\ I_{13}\end{array}\right){]}^{-1}\\ & =\frac{I_{22}{I}_{33}-(I_{23}{)}^2}{2I_{12}{I}_{13}{I}_{23}+I_{11}{I}_{22}{I}_{33}-I_{11}(I_{23}{)}^2-I_{22}(I_{13}{)}^2-I_{33}(I_{12}{)}^2}.\end{array}$$

Therefore, the score statistic for testing $H_0:\delta ={\delta }_0$ can be given by

$$T_{sc}={\left[\frac{\mathrm{\partial }\ell (\delta ,{\lambda }_0,R)}{\mathrm{\partial }\delta }\sqrt{I^{11}}\right]|}_{\delta ={\delta }_0,{\lambda }_0={\tilde{\lambda }}_0({\delta }_0),R=\tilde{R}({\delta }_0)},$$

where ${\tilde{\lambda }}_0({\delta }_0)$ and $\tilde{R}({\delta }_0)$ are the constrained MLEs of ${\lambda }_0$ and R under $H_0$: $\delta ={\delta }_0$, respectively.

Funding Statement

The work of Qiu was supported by the National Natural Science Foundation of China [Grant Nos. 11871124, 11471060] and the Natural Science Foundation of Chongqing [Grant No. cstc2018jcyjAX0241, cstc2020jcyj-msxmX0232].

Disclosure statement

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