# A Two-Stage Unrelated Question Randomized Response Model for Estimating the Prevalence of Stigmatized Attribute

Authors: Ghulam Narjis; Javid Shabbir; Ronald Onyango
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

The randomized response technique (RRT) is an effective method designed to obtain the stigmatized information from respondents while assuring the privacy. In this study, we propose a new two-stage unrelated question RRT model to estimate the prevalence of sensitive attributeπ when the proportion of unrelated innocuous attribute is known and unknown. A simulation study is carried out to validate the theoretical results of proposed estimators. The utility of proposed two-stage unrelated question RRT model under stratification is also explored. An efficiency comparison between the proposed two-stage unrelated question RRT model and the Singh and Suman RRT model is carried out numerically under simple and stratified random sampling.


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## Body

## 1. Introduction

In social and behavioral surveys, often researchers may ask the sensitive questions from the respondents, i.e., the illegal use of drugs, tax evasion, homosexuality, cheating in examinations, extramarital affairs, illegal income, or domestic violence. It is highly possible that the respondents may distort the truth or refuse to answer if asked the stigmatized questions directly, and these are due to social desirability response bias. In brief, social desirability bias (SDB) is the idiosyncrasy created by an interviewee in answering the stigmatized questions unfaithfully in the desire of leaving good impression on the interviewer. For example, suppose the question of interest is “are you addicted to any drug?” If asked directly, a respondent may be reluctant to admit and respond with a “no” perhaps, embarrassed to reveal their true behavior in fear of social stigma or negative judgments. Under such circumstances, researchers end up dealing with fabricated or distorted responses, thus leading to inaccurate or invalid estimates of the sensitive behaviors they are studying [1].To overcome nonresponse and the falsified responses, Warner [2] suggested an ingenious method of randomized response technique (RRT) to estimate the unknown population proportion of sensitive attribute π. In the Warner [2] randomization device such as a deck having two types of cards, a proportion p of the cards have the statement “I have the sensitive attribute A,” and the remaining cards have the statement “I do not have the sensitive attribute Ac.” Each selected respondent is asked to choose a card at random from the well-shuffled deck and report “yes” (or “no”) if his/her actual status matches (unmatches) with the statement on the card. The whole process is unseen by the interviewer so that he does not know the respondent’s answers, whether it comes from the sensitive question or from its complement. Some other developments for the estimation of population proportion are due to Mangat et al. [3], Bhargava and Singh [4]; Gupta et al. [5]; Kim and Elam [6] and Singh and Tarray [7]; etc.Motivated by the works of Singh and Suman [1], we suggest a two-stage unrelated randomized response model for estimating the population proportion possessing the stigmatized attribute A when unrelated non-stigmatized attribute X is known and unknown. It differs from the Singh and Suman [1] RRT model in the sense that in the second stage of randomized device, we in-cooperate the element of truthful response about the sensitive attribute.


## 2. Proposed Two-Stage Unrelated Question RRT Model

We propose a two-stage unrelated question RRT model to estimate the proportionπ of persons who possess the stigmatized attribute A under the study when the proportion of unrelated innocuous attribute X is known and unknown, respectively.


### 2.1. When the Proportion of Unrelated Innocuous Attribute Is Known

In the proposed two-stage RRT model, each selected respondent in the sample is given two randomization devicesR1,R2. The first-stage randomized device R1 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Go to the randomized deviceR2The probabilities areT and 1−T, respectively. Second-stage randomized device R2 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Do you possess nonsensitive attributeX?(iii)
Draw one more cardThe probabilities arep1, p2, and p3, respectively, such that ∑i=13pi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute A. The whole process of randomization is unseen by the interviewer and does not know whether the respondent’s answers come from the first draw or from the second draw. So, the privacy of interviewee(s) is protected and they will respond without any fear.The probability of getting answer “yes” from the respondent is as follows:(1)θ1=Tπ+1−Tp1π+p2πx1+p3kk−1+p32kk−1π,where k is the total number of cards in the deck R2. Solving equation (1) for π, an estimator for π is given as follows:(2)π^PI=θ^1−1−Tp21+p3k/k−1πxT+1−Tp1+p3k/k−1p1+p3,where θ^1 is the proportion of “yes” answer obtained from n sampled respondents. The random variable θ^1=n′/n follows the binomial distribution with parameters n,θ1, and therefore, the properties of an estimator π^PI may be summarized in the following theorem.Theorem 1.
The estimatorπ^PI is an unbiased for π with variance and is given as follows:(3)Vπ^PI=π1−πn+1−T+1−Tp1+p3k/k−1p1+p3−21−Tp21+p3k/k−1πxnT+1−Tp1+p3k/k−1p1+p3+1−Tp21+p3k/k−1πx1−1−Tp21+p3k/k−1πxnT+1−Tp1+p3k/k−1p1+p32.Proof is simple so omitted.Lemma 1.
The unbiased estimate of the varianceVπ^PI is given as follows:(4)V^π^PI=θ^11−θ^1n−1T+1−Tp1+p3k/k−1p1+p32.


### 2.2. When the Proportion of Unrelated Innocuous Attribute Is Unknown

To estimate the population proportion of sensitive attribute,π, and unrelated innocuous attribute, πx, simultaneously, we applied a split-sample approach. In this method, a sample size n is split into two subsamples of sizes n1 and n2n1+n2=n, and in each subsample, different randomization devices are applied.The description of randomization devices in two subsample sizesn1 and n2 is given as follows.The layout of the two-stage randomized response model forn1 is given as follows.First-stage randomized deviceR11 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Go to the randomized deviceR12The probabilities areT1 and 1−T1, respectively. Second-stage randomized device R12 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Do you possess nonsensitive attributeX?(iii)
Draw one more cardThe probabilities arep1, p2, and p3, respectively, such that ∑i=13pi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute A.The layout of two-stage randomized response model for n2 is given as follows.First-stage randomized deviceR21 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Go to the randomized deviceR22The probabilities areT2 and 1−T2, respectively. Second-stage randomized device R22 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Do you possess nonsensitive attributeX?(iii)
Draw one more cardThe probabilities areq1, q2, and q3, respectively, such that ∑i=13qi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute A.The probability of getting answer “yes” from the respondent is as follows:(5)θ2=T1π+1−T1p1π+p2πx1+p3kk−1+p32kk−1π,and(6)θ3=T2π+1−T2q1π+q2πx1+q3kk−1+q32kk−1π.Solving equations (5) and (6) for π and πx, we have estimators for π and πx, respectively, as follows:(7)π^PII=1−T2q21+q3k/k−1θ^2−1−T1p21+p3k/k−1θ^3C1,where(8)C1=T11−T2q21+q3kk−1−T21−T1p21+p3kk−1+1−T11−T2p1q2−p2q11+p3kk−11+q3kk−1+1−T11−T2kk−1p32q21+q3kk−1−q32p21+p3kk−1≠0,and(9)π^Px=1D1T2+1−T2q11+q3kk−1+q32kk−1θ^2−T1+1−T1p11+p3kk−1+p32kk−1θ^3,where(10)D1=T21−T1p21+p3kk−1−T11−T2q21+q3kk−1+1−T11−T2p2q1−p1q21+p3kk−11+q3kk−1+1−T11−T2kk−1q32p21+p3kk−1−p32q21+q3kk−1≠0,where θ^2 and θ^3 are the proportion of “yes” answer obtained from n1 and n2 sampled respondents, respectively. Since the random variables θ^2 and θ^3 follow the binomial distribution with parameters n1,θ2 and n2,θ3, respectively, therefore, the properties of estimators π^PII and π^Px may be summarized in the following theorems.Theorem 2.
The estimatorsπ^PII and π^Px are unbiased for π and πx with variances, respectively, and are given as follows:(11)Vπ^PII=d22c32π1−π+c31−c3−2c2πxπ+c2πx1−c2πxn1C12+c22d32π1−π+d31−d3−2d2πxπ+d2πx1−d2πxn2C12,and(12)Vπ^Px=d32c32π1−π+c31−c3−2c2πxπ+c2πx1−c2πxn1D12+c32d32π1−π+d31−d3−2d2πxπ+d2πx1−d2πxn2D12,where(13)c2=1−T1p21+p3kk−1,c3=T1+1−T1p11+p3kk−1+p32kk−1,d2=1−T2q21+q3kk−1,d3=T2+1−T2q11+q3kk−1+q32kk−1.Proof is simple so omitted.The ratio of optimal sample sizesn1 and n2 for two independent samples to minimize Vπ^PII is given as follows:(14)n1n2π^PIIopt=d22c32π1−π+c31−c3−2c2πxπ+c2πx1−c2πxc22d32π1−π+d31−d3−2d2πxπ+d2πx1−d2πx.The minimum variance ofπ^PII is given as follows:(15)Vπ^PIImin=1nC12d22c32π1−π+c31−c3−2c2πxπ+c2πx1−c2πx+c22d32π1−π+d31−d3−2d2πxπ+d2πx1−d2πx2.Lemma 2.
The unbiased estimates of the variancesVπ^PII and Vπ^Px are given as follows:(16)V^π^PII=1−T2q21+q3k/k−12θ^21−θ^2n1−1C12+1−T1p21+p3k/k−12θ^31−θ^3n2−1C12,and(17)V^π^Px=T2+1−T2q11+q3k/k−1+q32k/k−12θ^21−θ^2n1−1D12+T1+1−T1p11+p3k/k−1+p32k/k−12θ^31−θ^3n2−1D12.


### 2.3. Simulation Study

We perform a simulation study to validate the theoretical results of proposed two-stage unrelated question RRT model. The simulated results of proposed estimatorsπ^PI and π^PII are presented in Tables 1 and 2, respectively. When the proportion of unrelated innocuous attribute is known, the design parameters T and p1, p2=1−p1/3, and p3=1−p1−p2 were allowed to vary while k=45 and sample size n=1000 were fixed. Similarly, when the proportion of unrelated innocuous attribute is unknown, the design parameters T1=0.3, T2=0.6, sample size n=1000, and k=45 were fixed, while p1, p2=1−p1/3,p3=1−p1−p2, q1, q2=1−q1/3, and q3=1−q1−q2 were allowed to vary. The optimum sample sizes n1 and n2 were obtained using equation (14) for each case. The number of trials for each table is set to be 10,000.Table 1
Simulation results of proposed estimatorπ^PI, when trials=10000,n=1000 at various choices of π, πx, p1,p2=1−p1/3, and p3=1−p1−p2, respectively.
πp1πxT=0.3T=0.5T=0.7Eπ^PIVπ^PIEπ^pVπ^PIEπ^PIVπ^PIEmpiricalTheoreticalEmpiricalTheoreticalEmpiricalTheoretical0.30.20.10.328260.000390.000360.318130.000310.000300.309680.000260.000260.30.299940.000410.000410.309190.000320.000320.305180.000260.000270.50.271590.000440.000440.300070.000330.000330.299840.000280.000270.70.243030.000460.000470.290500.000340.000340.295140.000280.000280.90.214810.000480.000480.281640.000360.000350.289940.000280.000280.40.10.314260.000310.000300.272760.000350.000360.285250.000290.000290.30.300310.000330.000320.263660.000350.000370.280370.000280.000290.50.285920.000330.000340.254360.000360.000380.275380.000290.000300.70.271220.000360.000360.245550.000370.000380.270480.000300.000300.90.257370.000360.000370.309420.000260.000270.305400.000250.000240.60.10.305840.000260.000260.304460.000280.000270.302610.000250.000250.30.300370.000280.000270.299950.000280.000280.300220.000260.000250.50.294230.000280.000280.292250.000270.000260.295180.000240.000240.70.288520.000290.000290.290070.000260.000270.294220.000240.000240.90.282850.000290.000300.288030.000270.000270.293210.000250.000240.80.10.301480.000240.000230.301050.000220.000220.300650.000220.000220.30.299750.000240.000230.300480.000230.000220.299980.000220.000220.50.298540.000240.000240.300070.000220.000230.300070.000220.000220.70.297390.000240.000240.299350.000230.000230.299650.000230.000220.90.295850.000250.000250.299280.000230.000230.299480.000220.000220.60.20.10.670940.000500.000480.609050.000380.000380.605030.000320.000310.30.642480.000480.000480.599800.000380.000370.600460.000310.000310.50.614190.000480.000470.591060.000380.000370.595290.000300.000310.70.585730.000460.000450.581880.000360.000360.590300.000310.000300.90.557430.000430.000410.572460.000370.000350.585100.000310.000300.40.10.635450.000380.000380.623800.000340.000330.613190.000290.000290.30.620860.000380.000380.619090.000330.000330.610740.000290.000290.50.606820.000380.000370.614540.000330.000330.607830.000290.000290.70.592680.000370.000360.597850.000290.000280.599080.000260.000260.90.578840.000360.000340.595990.000270.000280.597540.000270.000260.60.10.614020.000320.000320.594300.000280.000280.596490.000260.000260.30.608630.000320.000310.602480.000260.000260.601300.000250.000250.50.602800.000310.000310.602070.000270.000260.600890.000250.000250.70.597400.000310.000300.601550.000260.000260.600850.000260.000250.90.591420.000300.000290.601040.000260.000260.600350.000260.000250.80.10.603190.000270.000270.600500.000260.000260.600260.000250.000250.30.601650.000280.000270.600150.000260.000260.599940.000250.000250.50.600760.000270.000270.599600.000250.000260.599780.000250.000250.70.599180.000270.000260.599240.000260.000260.599670.000250.000250.90.597880.000270.000260.598800.000260.000260.599090.000250.00025Table 2
Simulation results of proposed estimatorπ^PII, when trials=10000,n=1000, T1=0.3, T2=0.6, and various choices of π, πx, p1,p2=1−p1/3, p3=1−p1−p2, q1,q2=1−q1/3, and q3=1−q1−q2, respectively.
πp1πxq1=0.3q1=0.5q1=0.7Eπ^PIIVπ^PIIn1Eπ^PIIVπ^PIIn1Eπ^PIIVπ^PIIn1EmpiricalTheoreticalEmpiricalTheoreticalEmpiricalTheoretical0.300.20.10.289620.001860.001563160.286110.000680.000712280.290260.000380.000401370.30.291100.002060.001693240.288780.000750.000762370.292420.000390.000411430.50.292470.002290.001793280.291160.000790.000792420.294520.000400.000421480.70.293170.002390.001873300.293530.000820.000822440.296440.000430.000431500.90.294650.002530.001933290.296590.000840.000842440.298410.000420.000441510.40.10.333420.017310.006274080.292030.001620.001413070.292340.000540.000551910.30.330800.018630.006724120.293170.001720.001493120.293780.000540.000571960.50.327980.020230.007094140.293890.001820.001553160.295390.000580.000592000.70.323860.020760.007404150.296180.001890.001603180.297080.000580.000602030.90.323100.021200.007634150.296510.002000.001653190.298460.000600.000612040.60.10.616290.467200.035795380.336350.032080.010144280.295000.001200.001142860.30.645730.489680.037735370.339130.034490.010574290.295990.001200.001182890.50.671940.509520.039435360.333200.036000.010964310.297300.001270.001202910.70.689720.518940.040905360.330540.036860.011304310.297440.001310.001232930.90.714720.525290.042135360.324710.037020.011604320.297740.001330.001262940.80.10.302920.001240.001037260.311810.005210.003146290.490631.227240.090034760.30.306070.001290.001067220.316840.005250.003226280.481621.254980.091834760.50.309660.001370.001097190.318450.005420.003306260.460591.239650.093554760.70.313290.001370.001127170.322690.005340.003376250.476991.319460.095174760.90.317140.001310.001147160.325840.005560.003446240.452831.302950.096714770.60.20.10.577900.003500.001993230.570700.001060.000892370.579400.000510.000481440.30.577870.003590.001983250.573590.001110.000882390.581650.000520.000481450.50.580000.003730.001943240.575390.001120.000872380.583260.000520.000471440.70.583330.003700.001883210.578870.001150.000852340.585690.000510.000471410.90.584400.003680.001793160.581870.001140.000822290.587800.000510.000461370.40.10.667380.036620.007954120.583500.002690.001753130.584140.000750.000661980.30.669270.036720.007874120.584030.002720.001733130.585270.000730.000661980.50.660500.036050.007734120.585970.002760.001713130.586890.000750.000651970.70.656260.035700.007524110.587550.002760.001673100.588390.000760.000641950.90.652260.035320.007234080.588900.002790.001633070.589980.000760.000631920.60.11.259820.861430.044535370.676500.066580.012424300.590690.001790.001372900.31.263510.887320.044075370.672960.065250.012304300.590760.001830.001362900.51.287050.844120.043365370.673350.067140.012134300.592670.001810.001352890.71.289050.817260.042425370.667150.066470.011924290.593170.001830.001332880.91.280310.771760.041255380.666130.065080.011674280.594400.001790.001312860.80.10.604240.001590.001247200.621650.007040.003766270.964052.457960.106684760.30.607770.001550.001237210.625410.006950.003736270.975772.462690.105974760.50.610780.001530.001227220.628140.006800.003696270.952242.488180.105174760.70.613810.001500.001207230.629630.006760.003656280.943092.428280.104274760.90.617430.001480.001187250.635470.006690.003606290.928642.468690.10329476The simulated results of proposed estimatorsπ^PI and π^PII provide strong support to our earlier findings that π^PI and π^PII are unbiased estimators of π, respectively. The results of proposed estimators for both known and unknown cases are discussed as follows:(i)
Table1 presents the simulated results of proposed estimator π^PI, which indicates that Eπ^PI and empirical variances Vπ^PI are very close to the corresponding true parameter values of π and theoretical values of Vπ^PI, respectively. For example, when p1=0.6, T=0.3, π=0.3, and πx=0.5, Table 1 showed that the simulated mean is Eπ^PI=0.29423 with empirical variance Vπ^PI=0.00028 and theoretical variance Vπ^PI=0.00028, respectively.(ii)
The results in Table2, for example, show that when π=0.30, πx=0.5, T1=0.30, T2=0.60, p1=0.2, and q1=0.5, that is, the value of simulated mean Eπ^PII=0.29353 is close to the value of parameter π and empirical variance Vπ^PII=0.00082 is close to corresponding theoretical variance Vπ^PII=0.00082, respectively, these findings indicate that the proposed estimator π^PII is an unbiased estimator of π.Note that we also perform a simulation study on other combinations of design parameters and it came to our notice that the simulated values ofEπ^PII produce infinite or negative point estimates and variances that are unusually higher at some combinations. These are due to algebraic limitations (division by zero or value denominator is negative). From time to time, that can be entirely avoidable because these parameters can be assigned at our will and can be changed without difficulties.So, we conclude that the proposed estimators,π^PI and π^PII, are in good agreement with the corresponding theoretical values.


### 2.4. Relative Efficiency

To show the dominance of the proposed estimators, the empirical comparisons are made over the Singh and Suman [1] estimators. The relative efficiency of proposed estimators with respect to the Singh and Suman [1] estimators when the proportion of unrelated innocuous attribute is known and unknown is defined as follows:(18)RE1=Vπ^SIVπ^PIandRE2=Vπ^SIIVπ^PII.The variance of the Singh and Suman [1] estimators π^SI and π^SII, respectively, is given as follows:(19)Vπ^SI=π1−πn+1−T+1−Tp11+p3k/k−1−21−Tp21+p3k/k−1πxnT+1−Tp11+p3k/k−1+1−Tp21+p3k/k−1πx1−1−Tp21+p3k/k−1πxnT+1−Tp11+p3k/k−12,and(20)Vπ^SII=d22c12π1−π+c11−c1−2c2πxπ+c2πx1−c2πxn1C2+c22d12π1−π+d11−d1−2d2πxπ+d2πx1−d2πxn2C2,where(21)C=T11−T2q21+q3kk−1−T21−T1p21+p3kk−1+1−T11−T2p1q2−p2q11+p3kk−11+q3kk−1≠0,and(22)c1=T1+1−T1p11+p3kk−1,d1=T2+1−T2q11+q3kk−1.In Table3, we observed that for all possible combinations of design parameters, the values of RE1 are substantially greater than 1, which shows the domination of proposed estimator π^PI over the Singh and Suman [1] estimator π^SI. Further, we observed a decreasing trend in relative efficiency when values of T and p1 increase. The empirical results of RE2 are presented in Table 4 for different choices of design parameters. At the assumed values of design parameters, we observed the dominance of proposed estimator π^PII over the Singh and Suman [1] estimator π^SII. Moreover, we also observed that the efficiency can either be increased or decreased by changing the combinations of design parameters. So, we conclude that the proposed two-stage RRT model is relatively efficient than the Singh and Suman [1] RRT model when the proportion of unrelated innocuous attribute is known and unknown.Table 3
Relative efficiencyRE1 of proposed estimator π^PI with respect to the Singh and Suman [1] estimator π^PI, when k=45 and at various choices of T, p1, p2=1−p1/3, and p3=1−p1−p2, respectively.
p1πxπ=0.3π=0.6TT0.30.50.70.30.50.70.20.11.581.321.161.751.451.240.21.641.341.161.811.471.250.31.691.361.171.861.491.250.41.731.371.171.921.511.260.51.771.391.181.971.541.270.61.801.401.182.031.561.270.71.831.421.192.091.581.280.81.861.431.192.161.611.290.91.891.441.192.241.641.300.40.11.231.151.081.341.221.130.21.251.151.081.351.231.130.31.261.161.081.361.241.130.41.271.171.091.381.241.140.51.281.171.091.401.251.140.61.291.181.091.411.261.140.71.301.181.091.431.271.140.81.311.181.091.451.271.150.91.321.191.091.461.281.150.60.11.091.061.031.131.091.060.21.091.061.031.141.101.060.31.091.061.031.141.101.060.41.091.061.031.141.101.060.51.101.061.041.151.101.060.61.101.061.041.151.101.060.71.101.071.041.151.101.060.81.101.071.041.161.111.060.91.111.071.041.161.111.060.80.11.021.011.011.031.021.010.21.021.011.011.031.021.010.31.021.011.011.031.021.010.41.021.011.011.031.021.010.51.021.011.011.031.021.010.61.021.011.011.031.021.010.71.021.011.011.031.021.010.81.021.011.011.031.021.010.91.021.011.011.031.021.01Table 4
Relative efficiencyRE2 of proposed estimator π^PII with respect to the Singh and Suman [1] estimator π^SII, when k=45, q1=0.25, at various choices of π, πx, T1, T2, p1, p2, q2, p3=1−p1−p2, and q3=1−q1−q2, respectively.
T20.30.5p10.250.250.350.350.450.450.250.250.350.350.450.45p20.150.150.20.20.150.150.150.150.20.20.150.15q20.450.650.450.650.450.650.450.650.450.650.450.65πT1πx0.30.20.33.2814.441.856.201.371.643.006.041.893.181.441.640.53.4215.041.906.371.401.683.126.261.933.241.461.660.73.5315.611.946.551.421.713.216.461.973.311.471.690.40.31.822.381.351.621.201.281.832.211.421.621.251.320.51.872.451.371.661.221.301.872.261.441.651.261.330.71.912.521.391.691.231.321.912.311.461.671.271.340.60.31.341.431.151.201.101.121.371.441.201.241.131.150.51.361.461.161.221.111.141.381.461.211.251.141.160.71.371.481.171.241.121.151.401.481.211.261.141.170.80.31.121.131.051.061.041.041.131.151.071.081.051.060.51.121.141.061.071.041.051.141.151.081.091.061.060.71.131.151.061.081.041.061.141.161.081.091.061.060.60.20.33.7016.231.996.681.481.773.386.742.023.381.541.750.53.8817.522.067.131.511.873.527.092.083.511.571.800.74.0721.712.138.791.552.273.687.682.143.761.601.900.40.32.052.681.451.751.281.372.062.471.521.731.331.400.52.112.841.481.841.301.442.112.561.541.781.351.430.72.183.431.522.231.331.702.172.721.571.881.361.500.60.31.481.591.221.281.161.191.511.591.271.311.191.210.51.511.651.241.341.171.231.531.621.281.341.201.230.71.531.931.261.571.181.411.551.691.291.391.201.270.80.31.191.211.091.111.071.081.211.221.111.121.091.090.51.201.241.101.141.081.101.211.231.121.131.091.100.71.211.371.111.271.081.201.221.261.121.161.091.12


### 2.5. Privacy Protection

Lanke [8] proposed a measure of privacy protection based on the idea that a person belongs to the sensitive group A might be reluctant to reveal the actual group, whereas a member of complementary group Ac is expected to be quite willing to announce so. Thus, a membership in A may be embarrassing but not that in Ac. Thus, the higher the conditional probability PA|R, R∈"Yes",″No" belongs to A given a certain answer R, the greater is the embarrassment caused by giving that response. The Lanke measure (hereafter L measure) is as follows.In the proposed two-stage RRT model, the conditional probability that a respondent reporting “yes” also belongs to the sensitive groupA is given as follows:(23)PA|Yes=T+1−Tp1+p2πx1+p3k/k−1+p32k/k−1πTπ+1−Tp1π+p2πx1+p3k/k−1+p32k/k−1π.The conditional probability that a respondent reporting “No” also belongs to the sensitive groupA is given as follows:(24)PA|No=1−T−1−Tp1+p2πx1+p3k/k−1−1−Tp32k/k−1π1−Tπ+1−Tp1π+p2πx1+p3k/k−1+p32k/k−1π.Comparing the two RRT models, the model with the smaller value of Lanke [8] privacy measure will be more protective than the other. From Figure 1, we observed that the Singh and Suman [1] model and the proposed two-stage RRT model are equally protected.Figure 1
Measure of privacy protection, whenk=45, πx=0.35, T=0.3, p2=1−p1/3, and p3=1−p1−p2, respectively.(25)Li=maxPA|Yes,PA|No,i=P,S.


## 2.1. When the Proportion of Unrelated Innocuous Attribute Is Known

In the proposed two-stage RRT model, each selected respondent in the sample is given two randomization devicesR1,R2. The first-stage randomized device R1 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Go to the randomized deviceR2The probabilities areT and 1−T, respectively. Second-stage randomized device R2 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Do you possess nonsensitive attributeX?(iii)
Draw one more cardThe probabilities arep1, p2, and p3, respectively, such that ∑i=13pi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute A. The whole process of randomization is unseen by the interviewer and does not know whether the respondent’s answers come from the first draw or from the second draw. So, the privacy of interviewee(s) is protected and they will respond without any fear.The probability of getting answer “yes” from the respondent is as follows:(1)θ1=Tπ+1−Tp1π+p2πx1+p3kk−1+p32kk−1π,where k is the total number of cards in the deck R2. Solving equation (1) for π, an estimator for π is given as follows:(2)π^PI=θ^1−1−Tp21+p3k/k−1πxT+1−Tp1+p3k/k−1p1+p3,where θ^1 is the proportion of “yes” answer obtained from n sampled respondents. The random variable θ^1=n′/n follows the binomial distribution with parameters n,θ1, and therefore, the properties of an estimator π^PI may be summarized in the following theorem.Theorem 1.
The estimatorπ^PI is an unbiased for π with variance and is given as follows:(3)Vπ^PI=π1−πn+1−T+1−Tp1+p3k/k−1p1+p3−21−Tp21+p3k/k−1πxnT+1−Tp1+p3k/k−1p1+p3+1−Tp21+p3k/k−1πx1−1−Tp21+p3k/k−1πxnT+1−Tp1+p3k/k−1p1+p32.Proof is simple so omitted.Lemma 1.
The unbiased estimate of the varianceVπ^PI is given as follows:(4)V^π^PI=θ^11−θ^1n−1T+1−Tp1+p3k/k−1p1+p32.


## 2.2. When the Proportion of Unrelated Innocuous Attribute Is Unknown

To estimate the population proportion of sensitive attribute,π, and unrelated innocuous attribute, πx, simultaneously, we applied a split-sample approach. In this method, a sample size n is split into two subsamples of sizes n1 and n2n1+n2=n, and in each subsample, different randomization devices are applied.The description of randomization devices in two subsample sizesn1 and n2 is given as follows.The layout of the two-stage randomized response model forn1 is given as follows.First-stage randomized deviceR11 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Go to the randomized deviceR12The probabilities areT1 and 1−T1, respectively. Second-stage randomized device R12 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Do you possess nonsensitive attributeX?(iii)
Draw one more cardThe probabilities arep1, p2, and p3, respectively, such that ∑i=13pi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute A.The layout of two-stage randomized response model for n2 is given as follows.First-stage randomized deviceR21 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Go to the randomized deviceR22The probabilities areT2 and 1−T2, respectively. Second-stage randomized device R22 consists of the following statements:(i)
Do you possess sensitive attributeA?(ii)
Do you possess nonsensitive attributeX?(iii)
Draw one more cardThe probabilities areq1, q2, and q3, respectively, such that ∑i=13qi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute A.The probability of getting answer “yes” from the respondent is as follows:(5)θ2=T1π+1−T1p1π+p2πx1+p3kk−1+p32kk−1π,and(6)θ3=T2π+1−T2q1π+q2πx1+q3kk−1+q32kk−1π.Solving equations (5) and (6) for π and πx, we have estimators for π and πx, respectively, as follows:(7)π^PII=1−T2q21+q3k/k−1θ^2−1−T1p21+p3k/k−1θ^3C1,where(8)C1=T11−T2q21+q3kk−1−T21−T1p21+p3kk−1+1−T11−T2p1q2−p2q11+p3kk−11+q3kk−1+1−T11−T2kk−1p32q21+q3kk−1−q32p21+p3kk−1≠0,and(9)π^Px=1D1T2+1−T2q11+q3kk−1+q32kk−1θ^2−T1+1−T1p11+p3kk−1+p32kk−1θ^3,where(10)D1=T21−T1p21+p3kk−1−T11−T2q21+q3kk−1+1−T11−T2p2q1−p1q21+p3kk−11+q3kk−1+1−T11−T2kk−1q32p21+p3kk−1−p32q21+q3kk−1≠0,where θ^2 and θ^3 are the proportion of “yes” answer obtained from n1 and n2 sampled respondents, respectively. Since the random variables θ^2 and θ^3 follow the binomial distribution with parameters n1,θ2 and n2,θ3, respectively, therefore, the properties of estimators π^PII and π^Px may be summarized in the following theorems.Theorem 2.
The estimatorsπ^PII and π^Px are unbiased for π and πx with variances, respectively, and are given as follows:(11)Vπ^PII=d22c32π1−π+c31−c3−2c2πxπ+c2πx1−c2πxn1C12+c22d32π1−π+d31−d3−2d2πxπ+d2πx1−d2πxn2C12,and(12)Vπ^Px=d32c32π1−π+c31−c3−2c2πxπ+c2πx1−c2πxn1D12+c32d32π1−π+d31−d3−2d2πxπ+d2πx1−d2πxn2D12,where(13)c2=1−T1p21+p3kk−1,c3=T1+1−T1p11+p3kk−1+p32kk−1,d2=1−T2q21+q3kk−1,d3=T2+1−T2q11+q3kk−1+q32kk−1.Proof is simple so omitted.The ratio of optimal sample sizesn1 and n2 for two independent samples to minimize Vπ^PII is given as follows:(14)n1n2π^PIIopt=d22c32π1−π+c31−c3−2c2πxπ+c2πx1−c2πxc22d32π1−π+d31−d3−2d2πxπ+d2πx1−d2πx.The minimum variance ofπ^PII is given as follows:(15)Vπ^PIImin=1nC12d22c32π1−π+c31−c3−2c2πxπ+c2πx1−c2πx+c22d32π1−π+d31−d3−2d2πxπ+d2πx1−d2πx2.Lemma 2.
The unbiased estimates of the variancesVπ^PII and Vπ^Px are given as follows:(16)V^π^PII=1−T2q21+q3k/k−12θ^21−θ^2n1−1C12+1−T1p21+p3k/k−12θ^31−θ^3n2−1C12,and(17)V^π^Px=T2+1−T2q11+q3k/k−1+q32k/k−12θ^21−θ^2n1−1D12+T1+1−T1p11+p3k/k−1+p32k/k−12θ^31−θ^3n2−1D12.


## 2.3. Simulation Study

We perform a simulation study to validate the theoretical results of proposed two-stage unrelated question RRT model. The simulated results of proposed estimatorsπ^PI and π^PII are presented in Tables 1 and 2, respectively. When the proportion of unrelated innocuous attribute is known, the design parameters T and p1, p2=1−p1/3, and p3=1−p1−p2 were allowed to vary while k=45 and sample size n=1000 were fixed. Similarly, when the proportion of unrelated innocuous attribute is unknown, the design parameters T1=0.3, T2=0.6, sample size n=1000, and k=45 were fixed, while p1, p2=1−p1/3,p3=1−p1−p2, q1, q2=1−q1/3, and q3=1−q1−q2 were allowed to vary. The optimum sample sizes n1 and n2 were obtained using equation (14) for each case. The number of trials for each table is set to be 10,000.Table 1
Simulation results of proposed estimatorπ^PI, when trials=10000,n=1000 at various choices of π, πx, p1,p2=1−p1/3, and p3=1−p1−p2, respectively.
πp1πxT=0.3T=0.5T=0.7Eπ^PIVπ^PIEπ^pVπ^PIEπ^PIVπ^PIEmpiricalTheoreticalEmpiricalTheoreticalEmpiricalTheoretical0.30.20.10.328260.000390.000360.318130.000310.000300.309680.000260.000260.30.299940.000410.000410.309190.000320.000320.305180.000260.000270.50.271590.000440.000440.300070.000330.000330.299840.000280.000270.70.243030.000460.000470.290500.000340.000340.295140.000280.000280.90.214810.000480.000480.281640.000360.000350.289940.000280.000280.40.10.314260.000310.000300.272760.000350.000360.285250.000290.000290.30.300310.000330.000320.263660.000350.000370.280370.000280.000290.50.285920.000330.000340.254360.000360.000380.275380.000290.000300.70.271220.000360.000360.245550.000370.000380.270480.000300.000300.90.257370.000360.000370.309420.000260.000270.305400.000250.000240.60.10.305840.000260.000260.304460.000280.000270.302610.000250.000250.30.300370.000280.000270.299950.000280.000280.300220.000260.000250.50.294230.000280.000280.292250.000270.000260.295180.000240.000240.70.288520.000290.000290.290070.000260.000270.294220.000240.000240.90.282850.000290.000300.288030.000270.000270.293210.000250.000240.80.10.301480.000240.000230.301050.000220.000220.300650.000220.000220.30.299750.000240.000230.300480.000230.000220.299980.000220.000220.50.298540.000240.000240.300070.000220.000230.300070.000220.000220.70.297390.000240.000240.299350.000230.000230.299650.000230.000220.90.295850.000250.000250.299280.000230.000230.299480.000220.000220.60.20.10.670940.000500.000480.609050.000380.000380.605030.000320.000310.30.642480.000480.000480.599800.000380.000370.600460.000310.000310.50.614190.000480.000470.591060.000380.000370.595290.000300.000310.70.585730.000460.000450.581880.000360.000360.590300.000310.000300.90.557430.000430.000410.572460.000370.000350.585100.000310.000300.40.10.635450.000380.000380.623800.000340.000330.613190.000290.000290.30.620860.000380.000380.619090.000330.000330.610740.000290.000290.50.606820.000380.000370.614540.000330.000330.607830.000290.000290.70.592680.000370.000360.597850.000290.000280.599080.000260.000260.90.578840.000360.000340.595990.000270.000280.597540.000270.000260.60.10.614020.000320.000320.594300.000280.000280.596490.000260.000260.30.608630.000320.000310.602480.000260.000260.601300.000250.000250.50.602800.000310.000310.602070.000270.000260.600890.000250.000250.70.597400.000310.000300.601550.000260.000260.600850.000260.000250.90.591420.000300.000290.601040.000260.000260.600350.000260.000250.80.10.603190.000270.000270.600500.000260.000260.600260.000250.000250.30.601650.000280.000270.600150.000260.000260.599940.000250.000250.50.600760.000270.000270.599600.000250.000260.599780.000250.000250.70.599180.000270.000260.599240.000260.000260.599670.000250.000250.90.597880.000270.000260.598800.000260.000260.599090.000250.00025Table 2
Simulation results of proposed estimatorπ^PII, when trials=10000,n=1000, T1=0.3, T2=0.6, and various choices of π, πx, p1,p2=1−p1/3, p3=1−p1−p2, q1,q2=1−q1/3, and q3=1−q1−q2, respectively.
πp1πxq1=0.3q1=0.5q1=0.7Eπ^PIIVπ^PIIn1Eπ^PIIVπ^PIIn1Eπ^PIIVπ^PIIn1EmpiricalTheoreticalEmpiricalTheoreticalEmpiricalTheoretical0.300.20.10.289620.001860.001563160.286110.000680.000712280.290260.000380.000401370.30.291100.002060.001693240.288780.000750.000762370.292420.000390.000411430.50.292470.002290.001793280.291160.000790.000792420.294520.000400.000421480.70.293170.002390.001873300.293530.000820.000822440.296440.000430.000431500.90.294650.002530.001933290.296590.000840.000842440.298410.000420.000441510.40.10.333420.017310.006274080.292030.001620.001413070.292340.000540.000551910.30.330800.018630.006724120.293170.001720.001493120.293780.000540.000571960.50.327980.020230.007094140.293890.001820.001553160.295390.000580.000592000.70.323860.020760.007404150.296180.001890.001603180.297080.000580.000602030.90.323100.021200.007634150.296510.002000.001653190.298460.000600.000612040.60.10.616290.467200.035795380.336350.032080.010144280.295000.001200.001142860.30.645730.489680.037735370.339130.034490.010574290.295990.001200.001182890.50.671940.509520.039435360.333200.036000.010964310.297300.001270.001202910.70.689720.518940.040905360.330540.036860.011304310.297440.001310.001232930.90.714720.525290.042135360.324710.037020.011604320.297740.001330.001262940.80.10.302920.001240.001037260.311810.005210.003146290.490631.227240.090034760.30.306070.001290.001067220.316840.005250.003226280.481621.254980.091834760.50.309660.001370.001097190.318450.005420.003306260.460591.239650.093554760.70.313290.001370.001127170.322690.005340.003376250.476991.319460.095174760.90.317140.001310.001147160.325840.005560.003446240.452831.302950.096714770.60.20.10.577900.003500.001993230.570700.001060.000892370.579400.000510.000481440.30.577870.003590.001983250.573590.001110.000882390.581650.000520.000481450.50.580000.003730.001943240.575390.001120.000872380.583260.000520.000471440.70.583330.003700.001883210.578870.001150.000852340.585690.000510.000471410.90.584400.003680.001793160.581870.001140.000822290.587800.000510.000461370.40.10.667380.036620.007954120.583500.002690.001753130.584140.000750.000661980.30.669270.036720.007874120.584030.002720.001733130.585270.000730.000661980.50.660500.036050.007734120.585970.002760.001713130.586890.000750.000651970.70.656260.035700.007524110.587550.002760.001673100.588390.000760.000641950.90.652260.035320.007234080.588900.002790.001633070.589980.000760.000631920.60.11.259820.861430.044535370.676500.066580.012424300.590690.001790.001372900.31.263510.887320.044075370.672960.065250.012304300.590760.001830.001362900.51.287050.844120.043365370.673350.067140.012134300.592670.001810.001352890.71.289050.817260.042425370.667150.066470.011924290.593170.001830.001332880.91.280310.771760.041255380.666130.065080.011674280.594400.001790.001312860.80.10.604240.001590.001247200.621650.007040.003766270.964052.457960.106684760.30.607770.001550.001237210.625410.006950.003736270.975772.462690.105974760.50.610780.001530.001227220.628140.006800.003696270.952242.488180.105174760.70.613810.001500.001207230.629630.006760.003656280.943092.428280.104274760.90.617430.001480.001187250.635470.006690.003606290.928642.468690.10329476The simulated results of proposed estimatorsπ^PI and π^PII provide strong support to our earlier findings that π^PI and π^PII are unbiased estimators of π, respectively. The results of proposed estimators for both known and unknown cases are discussed as follows:(i)
Table1 presents the simulated results of proposed estimator π^PI, which indicates that Eπ^PI and empirical variances Vπ^PI are very close to the corresponding true parameter values of π and theoretical values of Vπ^PI, respectively. For example, when p1=0.6, T=0.3, π=0.3, and πx=0.5, Table 1 showed that the simulated mean is Eπ^PI=0.29423 with empirical variance Vπ^PI=0.00028 and theoretical variance Vπ^PI=0.00028, respectively.(ii)
The results in Table2, for example, show that when π=0.30, πx=0.5, T1=0.30, T2=0.60, p1=0.2, and q1=0.5, that is, the value of simulated mean Eπ^PII=0.29353 is close to the value of parameter π and empirical variance Vπ^PII=0.00082 is close to corresponding theoretical variance Vπ^PII=0.00082, respectively, these findings indicate that the proposed estimator π^PII is an unbiased estimator of π.Note that we also perform a simulation study on other combinations of design parameters and it came to our notice that the simulated values ofEπ^PII produce infinite or negative point estimates and variances that are unusually higher at some combinations. These are due to algebraic limitations (division by zero or value denominator is negative). From time to time, that can be entirely avoidable because these parameters can be assigned at our will and can be changed without difficulties.So, we conclude that the proposed estimators,π^PI and π^PII, are in good agreement with the corresponding theoretical values.


## 2.4. Relative Efficiency

To show the dominance of the proposed estimators, the empirical comparisons are made over the Singh and Suman [1] estimators. The relative efficiency of proposed estimators with respect to the Singh and Suman [1] estimators when the proportion of unrelated innocuous attribute is known and unknown is defined as follows:(18)RE1=Vπ^SIVπ^PIandRE2=Vπ^SIIVπ^PII.The variance of the Singh and Suman [1] estimators π^SI and π^SII, respectively, is given as follows:(19)Vπ^SI=π1−πn+1−T+1−Tp11+p3k/k−1−21−Tp21+p3k/k−1πxnT+1−Tp11+p3k/k−1+1−Tp21+p3k/k−1πx1−1−Tp21+p3k/k−1πxnT+1−Tp11+p3k/k−12,and(20)Vπ^SII=d22c12π1−π+c11−c1−2c2πxπ+c2πx1−c2πxn1C2+c22d12π1−π+d11−d1−2d2πxπ+d2πx1−d2πxn2C2,where(21)C=T11−T2q21+q3kk−1−T21−T1p21+p3kk−1+1−T11−T2p1q2−p2q11+p3kk−11+q3kk−1≠0,and(22)c1=T1+1−T1p11+p3kk−1,d1=T2+1−T2q11+q3kk−1.In Table3, we observed that for all possible combinations of design parameters, the values of RE1 are substantially greater than 1, which shows the domination of proposed estimator π^PI over the Singh and Suman [1] estimator π^SI. Further, we observed a decreasing trend in relative efficiency when values of T and p1 increase. The empirical results of RE2 are presented in Table 4 for different choices of design parameters. At the assumed values of design parameters, we observed the dominance of proposed estimator π^PII over the Singh and Suman [1] estimator π^SII. Moreover, we also observed that the efficiency can either be increased or decreased by changing the combinations of design parameters. So, we conclude that the proposed two-stage RRT model is relatively efficient than the Singh and Suman [1] RRT model when the proportion of unrelated innocuous attribute is known and unknown.Table 3
Relative efficiencyRE1 of proposed estimator π^PI with respect to the Singh and Suman [1] estimator π^PI, when k=45 and at various choices of T, p1, p2=1−p1/3, and p3=1−p1−p2, respectively.
p1πxπ=0.3π=0.6TT0.30.50.70.30.50.70.20.11.581.321.161.751.451.240.21.641.341.161.811.471.250.31.691.361.171.861.491.250.41.731.371.171.921.511.260.51.771.391.181.971.541.270.61.801.401.182.031.561.270.71.831.421.192.091.581.280.81.861.431.192.161.611.290.91.891.441.192.241.641.300.40.11.231.151.081.341.221.130.21.251.151.081.351.231.130.31.261.161.081.361.241.130.41.271.171.091.381.241.140.51.281.171.091.401.251.140.61.291.181.091.411.261.140.71.301.181.091.431.271.140.81.311.181.091.451.271.150.91.321.191.091.461.281.150.60.11.091.061.031.131.091.060.21.091.061.031.141.101.060.31.091.061.031.141.101.060.41.091.061.031.141.101.060.51.101.061.041.151.101.060.61.101.061.041.151.101.060.71.101.071.041.151.101.060.81.101.071.041.161.111.060.91.111.071.041.161.111.060.80.11.021.011.011.031.021.010.21.021.011.011.031.021.010.31.021.011.011.031.021.010.41.021.011.011.031.021.010.51.021.011.011.031.021.010.61.021.011.011.031.021.010.71.021.011.011.031.021.010.81.021.011.011.031.021.010.91.021.011.011.031.021.01Table 4
Relative efficiencyRE2 of proposed estimator π^PII with respect to the Singh and Suman [1] estimator π^SII, when k=45, q1=0.25, at various choices of π, πx, T1, T2, p1, p2, q2, p3=1−p1−p2, and q3=1−q1−q2, respectively.
T20.30.5p10.250.250.350.350.450.450.250.250.350.350.450.45p20.150.150.20.20.150.150.150.150.20.20.150.15q20.450.650.450.650.450.650.450.650.450.650.450.65πT1πx0.30.20.33.2814.441.856.201.371.643.006.041.893.181.441.640.53.4215.041.906.371.401.683.126.261.933.241.461.660.73.5315.611.946.551.421.713.216.461.973.311.471.690.40.31.822.381.351.621.201.281.832.211.421.621.251.320.51.872.451.371.661.221.301.872.261.441.651.261.330.71.912.521.391.691.231.321.912.311.461.671.271.340.60.31.341.431.151.201.101.121.371.441.201.241.131.150.51.361.461.161.221.111.141.381.461.211.251.141.160.71.371.481.171.241.121.151.401.481.211.261.141.170.80.31.121.131.051.061.041.041.131.151.071.081.051.060.51.121.141.061.071.041.051.141.151.081.091.061.060.71.131.151.061.081.041.061.141.161.081.091.061.060.60.20.33.7016.231.996.681.481.773.386.742.023.381.541.750.53.8817.522.067.131.511.873.527.092.083.511.571.800.74.0721.712.138.791.552.273.687.682.143.761.601.900.40.32.052.681.451.751.281.372.062.471.521.731.331.400.52.112.841.481.841.301.442.112.561.541.781.351.430.72.183.431.522.231.331.702.172.721.571.881.361.500.60.31.481.591.221.281.161.191.511.591.271.311.191.210.51.511.651.241.341.171.231.531.621.281.341.201.230.71.531.931.261.571.181.411.551.691.291.391.201.270.80.31.191.211.091.111.071.081.211.221.111.121.091.090.51.201.241.101.141.081.101.211.231.121.131.091.100.71.211.371.111.271.081.201.221.261.121.161.091.12


## 2.5. Privacy Protection

Lanke [8] proposed a measure of privacy protection based on the idea that a person belongs to the sensitive group A might be reluctant to reveal the actual group, whereas a member of complementary group Ac is expected to be quite willing to announce so. Thus, a membership in A may be embarrassing but not that in Ac. Thus, the higher the conditional probability PA|R, R∈"Yes",″No" belongs to A given a certain answer R, the greater is the embarrassment caused by giving that response. The Lanke measure (hereafter L measure) is as follows.In the proposed two-stage RRT model, the conditional probability that a respondent reporting “yes” also belongs to the sensitive groupA is given as follows:(23)PA|Yes=T+1−Tp1+p2πx1+p3k/k−1+p32k/k−1πTπ+1−Tp1π+p2πx1+p3k/k−1+p32k/k−1π.The conditional probability that a respondent reporting “No” also belongs to the sensitive groupA is given as follows:(24)PA|No=1−T−1−Tp1+p2πx1+p3k/k−1−1−Tp32k/k−1π1−Tπ+1−Tp1π+p2πx1+p3k/k−1+p32k/k−1π.Comparing the two RRT models, the model with the smaller value of Lanke [8] privacy measure will be more protective than the other. From Figure 1, we observed that the Singh and Suman [1] model and the proposed two-stage RRT model are equally protected.Figure 1
Measure of privacy protection, whenk=45, πx=0.35, T=0.3, p2=1−p1/3, and p3=1−p1−p2, respectively.(25)Li=maxPA|Yes,PA|No,i=P,S.


## 3. Use of Proposed Two-Stage RRT Model in Stratified Random Sampling

We now present the proposed two-stage RRT model using stratification and then using a SRSWR sampling in each stratum to draw a subsample.


### 3.1. When the Proportion of Unrelated Innocuous Attribute Is Known

The response from each respondent in thehth stratum is taken through two randomized devices Rh1,Rh2. The layout of two-stage randomized response model in the hth stratum is given as follows: first-stage randomized device Rh1 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Go to the randomized deviceRh2The probabilities areTh and 1−Th, respectively. Second-stage randomized device Rh2 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Do you possess nonsensitive attributeXh?(iii)
Draw one more cardThe probabilities areph1, ph2, and ph3, respectively, such that ∑i=13phi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute Ah. The probability of “yes” answer in stratum h is as follows:(26)θ1h=Thπh+1−Thph1πh+ph2πxh1+ph3khkh−1+ph32khkh−1πh,where kh be the total number of cards in the proposed deck, Rh2, in stratum h. Solving equation (26) for πh, we have an estimator for πh as follows:(27)π^PIh=θ^1h−1−Thph21+ph3kh/kh−1πxhTh+1−Thph1+ph3kh/kh−1ph1+ph3.An unbiased estimator ofπ is given as follows:(28)π^PIst=∑h=1LWhπ^PIh,where Wh=Nh/N for h=1,2,..,L; Nh is the total number of individuals in the hth stratum; and N is total number of individuals in the population.The variance ofπ^P1st is given as follows:(29)Vπ^PIst=∑h=1LWh2nhVPIh,where(30)VPIh=πh1−πh+1−Th+1−Thph1+ph3kh/kh−1ph1+ph3−21−Thph21+ph3kh/kh−1πxhTh+1−Thph1+ph3kh/kh−1ph1+ph3+1−Thph21+ph3kh/kh−1πxh1−1−Thph21+ph3kh/kh−1πxhTh+1−Thph1+ph3kh/kh−1ph1+ph32.The variances of the Singh and Suman [1] RRT model when the proportion of unrelated innocuous attribute is known are given as follows:(31)Vπ^SIst=∑h=1LWh2nhVSIh,where(32)VSI=πh1−πh+1−Th+1−Thph11+ph3kh/kh−1−21−Thph21+ph3kh/kh−1πxhTh+1−Thph11+ph3kh/kh−1+1−Thph21+ph3kh/kh−1πxh1−1−Thph21+ph3kh/kh−1πxhTh+1−Thph11+ph3kh/kh−12.Using Theorem 5.7 (see Cochran [9]), the variance, i.e., Vπ^ist, is minimized under the Neyman allocation after ignoring FPC as follows:(33)Vπ^istmin=1n∑h=1LWhVih2,i=SI,PI.


### 3.2. When the Proportion of Unrelated Innocuous Attribute Is Unknown

To estimate the proportion of sensitive attribute and unrelated innocuous attribute simultaneously, the population in each stratumh is split into two sub-sample sizes such that mh=mh1+mh2, and in each subsample, two randomized response devices are provided to respondent. The layout of proposed two-stage RRT model in the hth stratum for two subsamples is described as follows.First-stage randomized deviceRh11 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Go to the randomized deviceRh12The probabilities areTh1 and 1−Th1, respectively. Second-stage randomized device Rh12 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Do you possess nonsensitive attributeXh?(iii)
Draw one more cardThe probabilities areph1, ph2, and ph3, respectively, such that ∑i=13phi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute Ah.The description of two-stage randomized response model for the second sample of sizemh2 in stratum h is given as follows.First-stage randomized deviceRh21 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Go to the randomized deviceRh22The probabilities areTh2 and 1−Th2, respectively. Second-stage randomized device Rh22 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Do you possess nonsensitive attributeXh?(iii)
Draw one more cardThe probabilities areqh1, qh2, and qh3, respectively, such that ∑i=13qhi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute Ah.The probability of getting answer “yes” from the respondent in thehth stratum is as follows:(34)θ2h=Th1πh+1−Th1ph1πh+ph2πxh1+ph3khkh−1+ph32khkh−1πh,and(35)θ3h=Th2πh+1−Th2qh1πh+qh2πxh1+qh3khkh−1+qh32khkh−1πh.Solving equations (35) and (36) for πh and πxh, we have estimators for πh and πxh, respectively, as follows:(36)π^PIIh=1−Th2qh21+qh3kh/kh−1θ^2h−1−Th1ph21+ph3kh/kh−1θ^3hCh1,where(37)Ch1=Th11−Th2qh21+qh3khkh−1−Th21−Th1ph21+ph3khkh−1+1−Th11−Th2ph1qh2−ph2qh11+ph3khkh−11+qh3khkh−1+1−Th11−Th2khkh−1ph32qh21+qh3khkh−1−qh32ph21+ph3khkh−1≠0,and(38)π^Pxh=Th2+1−Th2qh11+qh3kh/kh−1+qh32kh/kh−1θ^2h−Th1+1−Th1ph11+ph3kh/kh−1+ph32kh/kh−1θ^3hDh1,where(39)Dh1=Th21−Th1ph21+ph3khkh−1−Th11−Th2qh21+qh3khkh−1+1−Th11−Th2ph2qh1−ph1qh21+ph3khkh−11+qh3khkh−1+1−Th11−Th2khkh−1qh32ph21+ph3khkh−1−ph32qh21+qh3khkh−1≠0.The variances of estimatorsπ^PIIh and π^Pxh are given as follows:(40)Vπ^PIIh=dh22ch32πh1−πh+ch31−ch3−2ch2πxhπh+ch2πxh1−ch2πxhmh1Ch12+ch22dh32πh1−πh+dh31−dh3−2dh2πxhπh+dh2πxh1−dh2πxhmh2Ch12,and(41)Vπ^Pxh=dh32ch32πh1−πh+ch31−ch3−2ch2πxhπh+ch2πxh1−ch2πxhmh1Dh12+ch32dh32πh1−πh+dh31−dh3−2dh2πxhπh+dh2πxh1−dh2πxhmh2Dh12,where(42)ch2=1−Th1ph21+ph3khkh−1,ch3=Th1+1−Th1ph11+ph3khkh−1+ph32khkh−1dh2=1−Th2qh21+qh3khkh−1,dh3=Th2+1−Th2qh11+qh3khkh−1+qh32khkh−1.The ratio of optimal sample sizesmh1 and mh2 for the two independent samples to minimize Vπ^PIIh in the hth stratum is given as follows:(43)mh1mh2π^PIIhopt=dh22ch32πh1−πh+ch31−ch3−2ch2πxhπh+ch2πxh1−ch2πxhch22dh32πh1−πh+dh31−dh3−2dh2πxhπh+dh2πxh1−dh2πxh.The minimum variance of the estimatorπ^PIIh in the hth stratum is given as follows:(44)Vπ^PIIhmin=dh22ch32πh1−πh+ch31−ch3−2ch2πxhπh+ch2πxh1−ch2πxh+ch22dh32πh1−πh+dh31−dh3−2dh2πxhπh+dh2πxh1−dh2πxh2mhCh12.The unbiased estimators forπ and πx are defined as follows:(45)π^PIIst=∑h=1LWhπ^PIIh.The variance of estimatorπ^PIIst is given as follows:(46)Vπ^PIIst=1n∑h=1LWh2VPII,where mh=mh1+mh2, ∑h=1Lmh=n, and(47)VPII=1Ch12dh22ch32πh1−πh+ch31−ch3−2ch2πxhπh+ch2πxh1−ch2πxh+ch22dh32πh1−πh+dh31−dh3−2dh2πxhπh+dh2πxh1−dh2πxh2.The variances of the Singh and Suman [1] RRT model when the proportion of unrelated innocuous attribute is unknown are decomposed under stratification as follows:(48)Vπ^SIIst=1n∑h=1LWh2VSIIh,where mh=mh1+mh2, ∑h=1Lmh=n, and(49)VSII=1Chdh22ch12πh1−πh+ch11−ch1−2ch2πxhπh+ch2πxh1−ch2πxh+ch22dh12πh1−πh+dh11−dh1−2dh2πxhπh+dh2πxh1−dh2πxh2,where(50)Ch=Th11−Th2qh21+qh3khkh−1−Th21−Th1ph21+ph3khkh−1+1−Th21−Th1ph1qh2−ph2qh11+ph3khkh−11+qh3khkh−1≠0,and(51)ch1=Th1+1−Th1ph11+ph3khkh−1,dh1=Th2+1−Th2qh11+qh3khkh−1.


### 3.3. Relative Efficiency

The relative efficiency (RE) of proposed two-stage unrelated question RRT model with respect to the Singh and Suman [1] RRT model under stratification is defined as follows:(52)RE1=Vπ^SIstVπ^PIstandRE2=Vπ^SIIstVπ^PIIst.To calculate the relative efficiency, we assume that the population has two strata. We have chosenn=1000 with the weights of stratum W1=0.3,W2=0.7 and W1=0.7,W2=0.3, respectively, with different values of π1, π2 and πx=πx1=πx2. The design parameters were fixed in each stratum when the proportion of unrelated innocuous attribute is known as p1=p11=p12, p2=p21=p22, p3=p31=p32, k=k1=k2, and T=T1=T2, respectively. Similarly, when the proportion of unrelated innocuous attribute is unknown, the design parameters were fixed in each stratum as p1=p11=p12, p2=p21=p22, p3=p31=p32, q1=q11=q12, q2=q21=q22, q3=q31=q32, k=k1=k2, T1=T11=T12, and T2=T21=T22, respectively.From Table5, we observed that RE1 at different values of T is always greater than 1, which indicates that the proposed estimator π^PIst is more efficient than the Singh and Suman [1] estimator π^SIst in stratification. The results in Table 6 at assumed values of design parameters indicate that the proposed estimator, π^PIIst, is more efficient as compared to π^SIIst.Table 5
Relative efficiencyRE1 of proposed estimator π^PIst with respect to the Singh and Suman [1] estimator π^SIst, when k=45, n=1000, and at various choices of T, πx, and p1, p2=1−p1/3, and p3=1−p1−p2, respectively.
p1π1π2W1W2πx=0.2πx=0.4πx=0.6πx=0.8TTTT0.30.50.70.30.50.70.30.50.70.30.50.70.20.080.130.30.71.671.331.151.771.381.161.821.411.181.861.431.190.70.31.691.341.151.781.391.171.841.421.181.871.431.190.280.330.30.71.641.341.161.731.381.171.811.411.181.871.431.190.70.31.641.341.161.731.371.171.801.401.181.861.431.190.480.530.30.71.741.421.211.841.461.221.931.491.232.031.531.250.70.31.721.411.211.821.441.221.911.481.232.001.521.240.680.730.30.71.941.571.322.091.641.342.251.711.362.461.791.390.70.31.921.551.302.051.611.322.201.671.342.401.751.360.880.930.30.72.391.971.642.702.151.743.192.431.874.162.942.080.70.32.321.901.582.602.061.663.032.291.763.812.681.910.40.080.130.30.71.241.141.071.271.161.081.291.171.081.311.181.090.70.31.251.141.071.281.161.081.301.171.081.311.181.090.280.330.30.71.251.161.081.281.171.091.301.181.091.311.191.090.70.31.251.151.081.271.171.091.291.181.091.311.181.090.480.530.30.71.311.201.111.341.211.121.361.221.121.391.241.120.70.31.301.191.111.331.211.111.351.221.111.381.231.120.680.730.30.71.431.291.171.471.311.181.521.341.191.581.371.200.70.31.411.281.161.451.301.171.501.321.181.551.351.190.880.930.30.71.741.571.391.871.661.442.081.791.502.452.001.590.70.31.691.521.351.801.591.391.971.701.432.261.851.490.60.080.130.30.71.081.051.031.091.061.031.101.061.031.101.061.030.70.31.081.051.031.091.061.031.101.061.031.101.061.030.280.330.30.71.091.061.031.101.061.041.101.071.041.111.071.040.70.31.091.061.031.091.061.031.101.061.041.101.071.040.480.530.30.71.121.081.051.121.081.051.131.091.051.141.091.050.70.31.111.081.051.121.081.051.131.081.051.131.091.050.680.730.30.71.181.131.081.191.131.081.201.141.081.211.141.080.70.31.171.121.071.181.121.071.191.131.081.201.141.080.880.930.30.71.361.291.201.421.321.211.491.371.241.601.431.260.70.31.331.251.171.371.281.181.431.321.201.511.361.210.60.080.130.30.71.021.011.011.021.011.011.021.011.011.021.011.010.70.31.021.011.011.021.011.011.021.011.011.021.011.010.280.330.30.71.021.011.011.021.011.011.021.011.011.021.011.010.70.31.021.011.011.021.011.011.021.011.011.021.011.010.480.530.30.71.031.021.011.031.021.011.031.021.011.031.021.010.70.31.031.021.011.031.021.011.031.021.011.031.021.010.680.730.30.71.041.031.021.041.031.021.051.031.021.051.031.020.70.31.041.031.021.041.031.021.041.031.021.041.031.020.880.930.30.71.111.091.061.121.091.061.141.101.061.151.111.070.70.31.101.071.051.111.081.051.111.081.051.121.091.05Table 6
Relative efficiencyRE2 of proposed estimator π^PIIst with respect to the Singh and Suman [1] estimator π^SIIst when k=45, p1=0.35, p2=0.2, p3=1−p1−p2 and q1=0.25, q2=0.45, q3=1−q1−q2, and various choices of T1 and T2, respectively.
T1π1π2W1W2π=0.2π=0.4π=0.6π=0.8T2T2T2T20.30.50.30.50.30.50.30.50.20.080.130.30.71.826.101.896.281.926.381.956.460.70.31.846.151.916.331.946.431.966.490.280.330.30.71.826.121.886.271.926.391.966.490.70.31.826.091.876.251.926.361.956.460.480.530.30.71.906.361.966.522.026.682.086.840.70.31.886.311.946.462.006.602.056.750.680.730.30.72.076.872.167.112.267.392.407.760.70.32.036.752.116.972.207.222.327.520.880.930.30.72.438.002.648.632.989.623.6511.510.70.32.347.722.518.222.778.973.2410.240.40.080.130.30.71.331.601.361.631.381.651.401.670.70.31.341.601.371.641.391.661.401.670.280.330.30.71.341.611.361.641.391.661.411.680.70.31.331.601.361.631.381.651.401.670.480.530.30.71.391.671.421.701.451.731.481.760.70.31.381.661.411.681.441.711.461.740.680.730.30.71.511.801.561.851.611.901.681.970.70.31.481.771.521.811.571.861.631.910.880.930.30.71.802.131.932.272.142.502.532.900.70.31.722.041.832.151.982.312.232.570.60.080.130.30.71.131.181.151.201.161.211.171.220.70.31.141.191.161.201.171.211.181.220.280.330.30.71.151.201.161.211.171.221.181.220.70.31.141.191.151.201.171.211.171.220.480.530.30.71.191.241.201.251.211.261.231.270.70.31.181.231.191.241.201.251.221.260.680.730.30.71.271.321.301.341.321.371.361.390.70.31.251.301.271.321.301.341.321.360.880.930.30.71.531.571.611.661.741.781.961.990.70.31.451.501.521.561.601.641.741.770.80.080.130.30.71.041.051.051.051.051.061.061.060.70.31.041.051.051.061.051.061.061.060.280.330.30.71.051.061.061.061.061.071.061.070.70.31.051.061.051.061.061.061.061.070.480.530.30.71.071.081.081.091.081.091.091.090.70.31.071.081.071.081.081.081.081.090.680.730.30.71.131.131.131.141.141.151.161.160.70.31.111.121.121.131.131.131.141.140.880.930.30.71.331.331.371.371.431.431.511.500.70.31.261.261.291.291.321.321.371.37In the next section, we consider a real data set application.


## 3.1. When the Proportion of Unrelated Innocuous Attribute Is Known

The response from each respondent in thehth stratum is taken through two randomized devices Rh1,Rh2. The layout of two-stage randomized response model in the hth stratum is given as follows: first-stage randomized device Rh1 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Go to the randomized deviceRh2The probabilities areTh and 1−Th, respectively. Second-stage randomized device Rh2 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Do you possess nonsensitive attributeXh?(iii)
Draw one more cardThe probabilities areph1, ph2, and ph3, respectively, such that ∑i=13phi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute Ah. The probability of “yes” answer in stratum h is as follows:(26)θ1h=Thπh+1−Thph1πh+ph2πxh1+ph3khkh−1+ph32khkh−1πh,where kh be the total number of cards in the proposed deck, Rh2, in stratum h. Solving equation (26) for πh, we have an estimator for πh as follows:(27)π^PIh=θ^1h−1−Thph21+ph3kh/kh−1πxhTh+1−Thph1+ph3kh/kh−1ph1+ph3.An unbiased estimator ofπ is given as follows:(28)π^PIst=∑h=1LWhπ^PIh,where Wh=Nh/N for h=1,2,..,L; Nh is the total number of individuals in the hth stratum; and N is total number of individuals in the population.The variance ofπ^P1st is given as follows:(29)Vπ^PIst=∑h=1LWh2nhVPIh,where(30)VPIh=πh1−πh+1−Th+1−Thph1+ph3kh/kh−1ph1+ph3−21−Thph21+ph3kh/kh−1πxhTh+1−Thph1+ph3kh/kh−1ph1+ph3+1−Thph21+ph3kh/kh−1πxh1−1−Thph21+ph3kh/kh−1πxhTh+1−Thph1+ph3kh/kh−1ph1+ph32.The variances of the Singh and Suman [1] RRT model when the proportion of unrelated innocuous attribute is known are given as follows:(31)Vπ^SIst=∑h=1LWh2nhVSIh,where(32)VSI=πh1−πh+1−Th+1−Thph11+ph3kh/kh−1−21−Thph21+ph3kh/kh−1πxhTh+1−Thph11+ph3kh/kh−1+1−Thph21+ph3kh/kh−1πxh1−1−Thph21+ph3kh/kh−1πxhTh+1−Thph11+ph3kh/kh−12.Using Theorem 5.7 (see Cochran [9]), the variance, i.e., Vπ^ist, is minimized under the Neyman allocation after ignoring FPC as follows:(33)Vπ^istmin=1n∑h=1LWhVih2,i=SI,PI.


## 3.2. When the Proportion of Unrelated Innocuous Attribute Is Unknown

To estimate the proportion of sensitive attribute and unrelated innocuous attribute simultaneously, the population in each stratumh is split into two sub-sample sizes such that mh=mh1+mh2, and in each subsample, two randomized response devices are provided to respondent. The layout of proposed two-stage RRT model in the hth stratum for two subsamples is described as follows.First-stage randomized deviceRh11 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Go to the randomized deviceRh12The probabilities areTh1 and 1−Th1, respectively. Second-stage randomized device Rh12 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Do you possess nonsensitive attributeXh?(iii)
Draw one more cardThe probabilities areph1, ph2, and ph3, respectively, such that ∑i=13phi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute Ah.The description of two-stage randomized response model for the second sample of sizemh2 in stratum h is given as follows.First-stage randomized deviceRh21 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Go to the randomized deviceRh22The probabilities areTh2 and 1−Th2, respectively. Second-stage randomized device Rh22 consists of the following statements:(i)
Do you possess sensitive attributeAh?(ii)
Do you possess nonsensitive attributeXh?(iii)
Draw one more cardThe probabilities areqh1, qh2, and qh3, respectively, such that ∑i=13qhi=1. If statement (iii) is appeared on card of respondent, then it is required to repeat the process without replacing the card. In the second draw, if statement (iii) is reappeared, then the respondent is suggested to report his/her actual status about sensitive attribute Ah.The probability of getting answer “yes” from the respondent in thehth stratum is as follows:(34)θ2h=Th1πh+1−Th1ph1πh+ph2πxh1+ph3khkh−1+ph32khkh−1πh,and(35)θ3h=Th2πh+1−Th2qh1πh+qh2πxh1+qh3khkh−1+qh32khkh−1πh.Solving equations (35) and (36) for πh and πxh, we have estimators for πh and πxh, respectively, as follows:(36)π^PIIh=1−Th2qh21+qh3kh/kh−1θ^2h−1−Th1ph21+ph3kh/kh−1θ^3hCh1,where(37)Ch1=Th11−Th2qh21+qh3khkh−1−Th21−Th1ph21+ph3khkh−1+1−Th11−Th2ph1qh2−ph2qh11+ph3khkh−11+qh3khkh−1+1−Th11−Th2khkh−1ph32qh21+qh3khkh−1−qh32ph21+ph3khkh−1≠0,and(38)π^Pxh=Th2+1−Th2qh11+qh3kh/kh−1+qh32kh/kh−1θ^2h−Th1+1−Th1ph11+ph3kh/kh−1+ph32kh/kh−1θ^3hDh1,where(39)Dh1=Th21−Th1ph21+ph3khkh−1−Th11−Th2qh21+qh3khkh−1+1−Th11−Th2ph2qh1−ph1qh21+ph3khkh−11+qh3khkh−1+1−Th11−Th2khkh−1qh32ph21+ph3khkh−1−ph32qh21+qh3khkh−1≠0.The variances of estimatorsπ^PIIh and π^Pxh are given as follows:(40)Vπ^PIIh=dh22ch32πh1−πh+ch31−ch3−2ch2πxhπh+ch2πxh1−ch2πxhmh1Ch12+ch22dh32πh1−πh+dh31−dh3−2dh2πxhπh+dh2πxh1−dh2πxhmh2Ch12,and(41)Vπ^Pxh=dh32ch32πh1−πh+ch31−ch3−2ch2πxhπh+ch2πxh1−ch2πxhmh1Dh12+ch32dh32πh1−πh+dh31−dh3−2dh2πxhπh+dh2πxh1−dh2πxhmh2Dh12,where(42)ch2=1−Th1ph21+ph3khkh−1,ch3=Th1+1−Th1ph11+ph3khkh−1+ph32khkh−1dh2=1−Th2qh21+qh3khkh−1,dh3=Th2+1−Th2qh11+qh3khkh−1+qh32khkh−1.The ratio of optimal sample sizesmh1 and mh2 for the two independent samples to minimize Vπ^PIIh in the hth stratum is given as follows:(43)mh1mh2π^PIIhopt=dh22ch32πh1−πh+ch31−ch3−2ch2πxhπh+ch2πxh1−ch2πxhch22dh32πh1−πh+dh31−dh3−2dh2πxhπh+dh2πxh1−dh2πxh.The minimum variance of the estimatorπ^PIIh in the hth stratum is given as follows:(44)Vπ^PIIhmin=dh22ch32πh1−πh+ch31−ch3−2ch2πxhπh+ch2πxh1−ch2πxh+ch22dh32πh1−πh+dh31−dh3−2dh2πxhπh+dh2πxh1−dh2πxh2mhCh12.The unbiased estimators forπ and πx are defined as follows:(45)π^PIIst=∑h=1LWhπ^PIIh.The variance of estimatorπ^PIIst is given as follows:(46)Vπ^PIIst=1n∑h=1LWh2VPII,where mh=mh1+mh2, ∑h=1Lmh=n, and(47)VPII=1Ch12dh22ch32πh1−πh+ch31−ch3−2ch2πxhπh+ch2πxh1−ch2πxh+ch22dh32πh1−πh+dh31−dh3−2dh2πxhπh+dh2πxh1−dh2πxh2.The variances of the Singh and Suman [1] RRT model when the proportion of unrelated innocuous attribute is unknown are decomposed under stratification as follows:(48)Vπ^SIIst=1n∑h=1LWh2VSIIh,where mh=mh1+mh2, ∑h=1Lmh=n, and(49)VSII=1Chdh22ch12πh1−πh+ch11−ch1−2ch2πxhπh+ch2πxh1−ch2πxh+ch22dh12πh1−πh+dh11−dh1−2dh2πxhπh+dh2πxh1−dh2πxh2,where(50)Ch=Th11−Th2qh21+qh3khkh−1−Th21−Th1ph21+ph3khkh−1+1−Th21−Th1ph1qh2−ph2qh11+ph3khkh−11+qh3khkh−1≠0,and(51)ch1=Th1+1−Th1ph11+ph3khkh−1,dh1=Th2+1−Th2qh11+qh3khkh−1.


## 3.3. Relative Efficiency

The relative efficiency (RE) of proposed two-stage unrelated question RRT model with respect to the Singh and Suman [1] RRT model under stratification is defined as follows:(52)RE1=Vπ^SIstVπ^PIstandRE2=Vπ^SIIstVπ^PIIst.To calculate the relative efficiency, we assume that the population has two strata. We have chosenn=1000 with the weights of stratum W1=0.3,W2=0.7 and W1=0.7,W2=0.3, respectively, with different values of π1, π2 and πx=πx1=πx2. The design parameters were fixed in each stratum when the proportion of unrelated innocuous attribute is known as p1=p11=p12, p2=p21=p22, p3=p31=p32, k=k1=k2, and T=T1=T2, respectively. Similarly, when the proportion of unrelated innocuous attribute is unknown, the design parameters were fixed in each stratum as p1=p11=p12, p2=p21=p22, p3=p31=p32, q1=q11=q12, q2=q21=q22, q3=q31=q32, k=k1=k2, T1=T11=T12, and T2=T21=T22, respectively.From Table5, we observed that RE1 at different values of T is always greater than 1, which indicates that the proposed estimator π^PIst is more efficient than the Singh and Suman [1] estimator π^SIst in stratification. The results in Table 6 at assumed values of design parameters indicate that the proposed estimator, π^PIIst, is more efficient as compared to π^SIIst.Table 5
Relative efficiencyRE1 of proposed estimator π^PIst with respect to the Singh and Suman [1] estimator π^SIst, when k=45, n=1000, and at various choices of T, πx, and p1, p2=1−p1/3, and p3=1−p1−p2, respectively.
p1π1π2W1W2πx=0.2πx=0.4πx=0.6πx=0.8TTTT0.30.50.70.30.50.70.30.50.70.30.50.70.20.080.130.30.71.671.331.151.771.381.161.821.411.181.861.431.190.70.31.691.341.151.781.391.171.841.421.181.871.431.190.280.330.30.71.641.341.161.731.381.171.811.411.181.871.431.190.70.31.641.341.161.731.371.171.801.401.181.861.431.190.480.530.30.71.741.421.211.841.461.221.931.491.232.031.531.250.70.31.721.411.211.821.441.221.911.481.232.001.521.240.680.730.30.71.941.571.322.091.641.342.251.711.362.461.791.390.70.31.921.551.302.051.611.322.201.671.342.401.751.360.880.930.30.72.391.971.642.702.151.743.192.431.874.162.942.080.70.32.321.901.582.602.061.663.032.291.763.812.681.910.40.080.130.30.71.241.141.071.271.161.081.291.171.081.311.181.090.70.31.251.141.071.281.161.081.301.171.081.311.181.090.280.330.30.71.251.161.081.281.171.091.301.181.091.311.191.090.70.31.251.151.081.271.171.091.291.181.091.311.181.090.480.530.30.71.311.201.111.341.211.121.361.221.121.391.241.120.70.31.301.191.111.331.211.111.351.221.111.381.231.120.680.730.30.71.431.291.171.471.311.181.521.341.191.581.371.200.70.31.411.281.161.451.301.171.501.321.181.551.351.190.880.930.30.71.741.571.391.871.661.442.081.791.502.452.001.590.70.31.691.521.351.801.591.391.971.701.432.261.851.490.60.080.130.30.71.081.051.031.091.061.031.101.061.031.101.061.030.70.31.081.051.031.091.061.031.101.061.031.101.061.030.280.330.30.71.091.061.031.101.061.041.101.071.041.111.071.040.70.31.091.061.031.091.061.031.101.061.041.101.071.040.480.530.30.71.121.081.051.121.081.051.131.091.051.141.091.050.70.31.111.081.051.121.081.051.131.081.051.131.091.050.680.730.30.71.181.131.081.191.131.081.201.141.081.211.141.080.70.31.171.121.071.181.121.071.191.131.081.201.141.080.880.930.30.71.361.291.201.421.321.211.491.371.241.601.431.260.70.31.331.251.171.371.281.181.431.321.201.511.361.210.60.080.130.30.71.021.011.011.021.011.011.021.011.011.021.011.010.70.31.021.011.011.021.011.011.021.011.011.021.011.010.280.330.30.71.021.011.011.021.011.011.021.011.011.021.011.010.70.31.021.011.011.021.011.011.021.011.011.021.011.010.480.530.30.71.031.021.011.031.021.011.031.021.011.031.021.010.70.31.031.021.011.031.021.011.031.021.011.031.021.010.680.730.30.71.041.031.021.041.031.021.051.031.021.051.031.020.70.31.041.031.021.041.031.021.041.031.021.041.031.020.880.930.30.71.111.091.061.121.091.061.141.101.061.151.111.070.70.31.101.071.051.111.081.051.111.081.051.121.091.05Table 6
Relative efficiencyRE2 of proposed estimator π^PIIst with respect to the Singh and Suman [1] estimator π^SIIst when k=45, p1=0.35, p2=0.2, p3=1−p1−p2 and q1=0.25, q2=0.45, q3=1−q1−q2, and various choices of T1 and T2, respectively.
T1π1π2W1W2π=0.2π=0.4π=0.6π=0.8T2T2T2T20.30.50.30.50.30.50.30.50.20.080.130.30.71.826.101.896.281.926.381.956.460.70.31.846.151.916.331.946.431.966.490.280.330.30.71.826.121.886.271.926.391.966.490.70.31.826.091.876.251.926.361.956.460.480.530.30.71.906.361.966.522.026.682.086.840.70.31.886.311.946.462.006.602.056.750.680.730.30.72.076.872.167.112.267.392.407.760.70.32.036.752.116.972.207.222.327.520.880.930.30.72.438.002.648.632.989.623.6511.510.70.32.347.722.518.222.778.973.2410.240.40.080.130.30.71.331.601.361.631.381.651.401.670.70.31.341.601.371.641.391.661.401.670.280.330.30.71.341.611.361.641.391.661.411.680.70.31.331.601.361.631.381.651.401.670.480.530.30.71.391.671.421.701.451.731.481.760.70.31.381.661.411.681.441.711.461.740.680.730.30.71.511.801.561.851.611.901.681.970.70.31.481.771.521.811.571.861.631.910.880.930.30.71.802.131.932.272.142.502.532.900.70.31.722.041.832.151.982.312.232.570.60.080.130.30.71.131.181.151.201.161.211.171.220.70.31.141.191.161.201.171.211.181.220.280.330.30.71.151.201.161.211.171.221.181.220.70.31.141.191.151.201.171.211.171.220.480.530.30.71.191.241.201.251.211.261.231.270.70.31.181.231.191.241.201.251.221.260.680.730.30.71.271.321.301.341.321.371.361.390.70.31.251.301.271.321.301.341.321.360.880.930.30.71.531.571.611.661.741.781.961.990.70.31.451.501.521.561.601.641.741.770.80.080.130.30.71.041.051.051.051.051.061.061.060.70.31.041.051.051.061.051.061.061.060.280.330.30.71.051.061.061.061.061.071.061.070.70.31.051.061.051.061.061.061.061.070.480.530.30.71.071.081.081.091.081.091.091.090.70.31.071.081.071.081.081.081.081.090.680.730.30.71.131.131.131.141.141.151.161.160.70.31.111.121.121.131.131.131.141.140.880.930.30.71.331.331.371.371.431.431.511.500.70.31.261.261.291.291.321.321.371.37In the next section, we consider a real data set application.


## 4. Real Data Application

A small-scale survey is conducted to assess the effectiveness of proposed two-stage RRT model in field and also examined proposed model superiority over the Singh and Suman [1] RRT model and direct questioning (DQ) survey method. In this study, we collected data from truck drivers that transport raw material to industries, and a total of 75 drivers have participated in the survey. The participants were requested verbally to volunteer themselves for this survey and were assured that their identity will not be disclosed in anyway. Before, starting the data collection process, we highlighted the importance of randomized response methods and explained the proposed two-stage RRT model and competitor method to the individuals who volunteer themselves for this study.To collect data from the proposed two-stage RRT model when the proportion of unrelated innocuous attribute is known asπx=0.35, two decks of cards were used: Deck I, a deck of yellow cards, and Deck II, a deck of blue cards. A deck of yellow cards bear two different types of statements by setting T=0.3 and a deck of blue cards bear three different types of statements by setting p1=0.6, p2=0.2, and p3=0.2, respectively, to get the randomized responses. Each participant was asked to choose a card at random from the well-shuffled deck and report “yes” (or “no”) if his/her actual status matches (unmatches) with the statement on the card.In Deck I or deck of yellow cards having 20 cards, 6 cards had type I statement and the remaining 14 cards had type II statements as described in Figure2. In a deck of blue cards or Deck II consisting of 54 cards, 32 cards had type I statement, 11 cards had type II, and the remaining 11 cards had type III statement as described in Figure 3. The respondent was requested to draw the first card from Deck I and then follow the statements given on the drawn card, and each respondent was given a blank white paper and advised to write down their response on it, and after completing the randomization process, it is folded and put in the black box lying on the table.Figure 2
Deck I, a deck of yellow cards having two types of statements.Figure 3
Deck II, a deck of blue cards having three types of statements.Similarly, in the Singh and Suman [1] RRT model the respondent was asked to use a deck of yellow and deck of orange cards. A deck of orange cards is similar to blue deck except in the second draw, and if type III statement reappeared, then the respondent is suggested to report “no” irrespective of his/her actual status. The respondent was provided a green paper for reporting their response and put it in black box.Lastly, each respondent was also requested to provide response through the direct question (DQ) method. In the DQ method, each participant was requested to mark their actual status on given questionnaire as follows.Question: did you smoke during driving?(a)
Yes(b)
NoNo further information was collected from any participant. The responses obtained from the survey are present in Table7 and their estimates in Table 8.Table 7
Responses from the survey.
MethodsYesNoProposed method3441Singh and Suman [1] method2847Direct question method2154Table 8
Estimates along with the variance and standard errors.
Methodsπ^V^π^S.E^π^Proposed method0.47280.00480.0694Singh and Suman [1] method0.39020.00490.0698Direct question method0.28000.00270.0522


## 5. Discussion of the Results

From the estimates given in Table8, one can conclude that the estimates of proportion π^PI=0.4728 and π^SI=0.3902, respectively, are observed higher than the direct question (DQ) method estimate π^DQ=0.2800. If these estimates reflect the true state of affairs, they indicate that the respondents are more reluctant to admit smoking during driving when using the DQ technique than when using the randomized response methods. The proportion estimate of sensitive attribute of proposed two-stage RRT model is higher than the estimates obtained from the Singh and Suman [1] RRT model, which indicate the superiority of proposed method over the competitor randomized response method. Further, we observed that the variance estimate of proposed method V^π^PI=0.0048 is smaller than V^π^SI=0.0049, indicating the efficiency of proposed method. We noted that the standard error values are large due to small sample sizes, and thus, we suggest that a large-scale sample survey should be conducted to reach at more realistic outcomes in future.


## 6. Concluding Remarks

It is clear from numerical comparisons that the proposed two-stage unrelated question RRT model is more efficient than the Singh and Suman [1] RRT model under simple and stratified random sampling when the proportion of unrelated innocuous attribute is known and unknown. An extensive simulation study shows that the empirical mean and variance of proposed estimators, i.e., π^PIst and π^PIIst, are in good agreement with the corresponding theoretical values. Moreover, we also observed that design parameters play an important role in increasing or decreasing the efficiency of suggested two-stage unrelated question RRT model. Thus, the proposed two-stage unrelated question RRT model is therefore recommended for its use in practice as an alternative of Singh and Suman [1] RRT model.



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*Source: 4559663-2022-05-17.xml*