CT3 IFoA october 2010 Q10

[SURESH SHARMA]

In the collection of questionnaire data, randomised response sampling is a method which is used to obtain answers to sensitive questions. For example a company is interested in estimating the proportion, p, of its employees who falsely take days off sick. Employees are unlikely to answer a direct question truthfully

and so the company uses the following approach.

Each employee selected in the survey is given a fair six-sided die and asked to throw it. If it comes up as a 5 or 6, then the employee answers yes or no to the question “have you falsely taken any days off sick during the last year?”. If it comes up as a 1, 2, 3 or 4, then the employee is instructed to toss a coin and answer yes or no to the question “did you obtain heads?”. So an individual’s answer is either yes or no, but it is not known which question the individual has answered.

For the purpose of the following analysis you should assume that each employee
answers the question truthfully.

(i) Show that the probability that an individual answers yes is 1/3(p+1)

in this question probability of getting 5 or 6 in a fair dice is 2/6 , i am not getting what would be the probability of answering yes to the main question.
as per me the probability of saying yes to the main question would be (1/2) .

as total employee is p so the probability would be (2/6)*p*(1/2)


please clarify

rgards
Suresh sharma

 

[KaustavSen]

In the collection of questionnaire data, randomised response sampling is a method which is used to obtain answers to sensitive questions. For example a company is interested in estimating the proportion, p, of its employees who falsely take days off sick. Employees are unlikely to answer a direct question truthfully

and so the company uses the following approach.

Each employee selected in the survey is given a fair six-sided die and asked to throw it. If it comes up as a 5 or 6, then the employee answers yes or no to the question “have you falsely taken any days off sick during the last year?”. If it comes up as a 1, 2, 3 or 4, then the employee is instructed to toss a coin and answer yes or no to the question “did you obtain heads?”. So an individual’s answer is either yes or no, but it is not known which question the individual has answered.

For the purpose of the following analysis you should assume that each employee
answers the question truthfully.

(i) Show that the probability that an individual answers yes is 1/3(p+1)

in this question probability of getting 5 or 6 in a fair dice is 2/6 , i am not getting what would be the probability of answering yes to the main question.
as per me the probability of saying yes to the main question would be (1/2) .

as total employee is p so the probability would be (2/6)*p*(1/2)


please clarify

rgards
Suresh sharma

Hello Suresh,

The probability that an individual answers yes can be broken down into 2 mutually exclusive events:

1. Getting 5 or 6 in the dice throw and then the event that the individual falsely takes sick leaves.
2. Getting 1, 2, 3 or 4 in the dice throw and then the event of getting a Heads in the coin toss.

Combining the above two events to get the overall probability, we have:
\[
\begin{align}
\mathbb{P}[individual\ answers\ yes] &= \Big(\frac{2}{6}\times p\Big) + \Big(\frac{4}{6}\times\frac{1}{2}\Big)\\
&=\frac{p+1}{3}
\end{align}
\]

Hope this helps.

Regards,
Kaustav.

 

[SURESH SHARMA]

Hello Suresh,

The probability that an individual answers yes can be broken down into 2 mutually exclusive events:

1. Getting 5 or 6 in the dice throw and then the event that the individual falsely takes sick leaves.
2. Getting 1, 2, 3 or 4 in the dice throw and then the event of getting a Heads in the coin toss.

Combining the above two events to get the overall probability, we have:
\[
\begin{align}
\mathbb{P}[individual\ answers\ yes] &= \Big(\frac{2}{6}\times p\Big) + \Big(\frac{4}{6}\times\frac{1}{2}\Big)\\
&=\frac{p+1}{3}
\end{align}
\]

Hope this helps.

Regards,
Kaustav.

thanks for clarification