People may lie about sensitive questions such as: "Have you used cocaine before?"

For these types of questions, a question form that encourages truthful answers and makes people comfortable is useful.

Horvitz (1967) based on the idea from Warner (1965), suggests using two questions - a sensitive question and an unrelated question - and uses a randomization device to determine which is the question the respondent should answer.

## Example

Q1: Have you used cocaine before?

Q2: Is the second hand of your watch between 0 and 30?

The respondent will flip a coin and decide which question to answer whereas the interviewer does not know the outcome of the coin.

The randomization device can be anything but it must have:

- known probability
*t*that the person is asked the sensitive question and probability 1 -*t*that the person is asked other questions. - the probability that the person responds yes to the other question is known.

##
Example 12-2: Tax return question
Section* *

Q1: Have you ever falsified your tax return? Yes or no.

Q2: Flip a book and answer: is the page number odd? Yes or no.

The interviewer merely records the answer and does not know whether the respondent is answering Q1 or Q2.

We will conduct this survey on *n* subjects, *n*_{1} denotes the number of respondents who respond yes. How are we going to estimate the population proportion *p*?

*t* = 1/2

Here we write out what this tree diagram is expressing in terms of the probability of yes:

\(P(\text{yes})=\dfrac{1}{2}\times p+\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{p}{2}+\dfrac{1}{4}\)

Let \(n_1\) denote the number of yes in *n* subjects

\(\dfrac{n_1}{n}=\dfrac{\hat{p}}{2}+\dfrac{1}{4}\)

\(\hat{p}=2\left(\dfrac{n_1}{n}-\dfrac{1}{4}\right)\)

If the sample size is small compared to the population size, the finite correction factor can be omitted and the variance formula is:

\(\hat{V}ar(\hat{p})=\dfrac{4}{n}\times \dfrac{n_1}{n}\times (1-\dfrac{n_1}{n})\)

### Try it!

*n* = 400

\(n_1=128\)

\(\hat{p}=2\left(\dfrac{128}{400}-\dfrac{1}{4}\right)=0.14\)

\(\hat{V}ar(\hat{p})=\dfrac{4}{400}\times \dfrac{128}{400}\times \left(1-\dfrac{128}{400}\right)=0.0022\)

\(\hat{s}.d.(\hat{p})=0.047\)