Here is an example of when the sampling frame does not match up perfectly with the target population.

What would be another example where an important part of the target population would be missed? A serious non-sampling error might have occurred for non-response.

In a well-designed research study, handling non-responses is important. How do we handle this common phenomenon?

An important use of double sampling...

Double sampling can be used to adjust for non-response in the form of callbacks.

Non-response is an important problem to consider in any survey. We can consider the two groups: response and non-response in two strata.

The two steps of double sampling for non-response:

1. Step 1:

   n' initial simple random samples are selected from a population of N units. These units are classified into two strata: response and non-response.This is the thing you need to do.

   \(n'_1\) of these respond - stratum 1
   \(n'_2\) of these do not respond - stratum 2

2. Step 2:

   Call back \(n_2\) samples by simple random sampling from the \(n'_2\) non-respondents (by giving more incentives, etc.)

   Thus, we are in a double sampling setting where \(n_1=n'_1,n_2\) is the number of callbacks.

• \(c_0\): the initial cost of sampling each respondent (the set-up cost for each respondent)
• \(c_1\): the cost of a standard response (cost of producing the response)
• \(c_2\): the cost of a call-back response

Total cost =\(\left(n^{\prime} \times c_{0}\right)+\left(n_{1}^{\prime} \times c_{1}\right)+\left(n_{2} \times c_{2}\right)\)

We want to determine the value k(k > 1) where: \(n_2=\dfrac{n'_2}{k}\).

As \(\bar{y}_d= \sum\limits_{h=1}^2 w_h \bar{y}_h\), its variance can be derived and one can find the value of k and n' that minimizes the expected cost of sampling for a desired fixed value of \(\hat{V}ar(\bar{y}_d)\), which we denote as V₀.

When N is large, the optimal value of k and n' are:

\(k=\sqrt{\dfrac{c_2(\sigma^2-w_2\sigma^2_2)}{\sigma^2_2(c_0+c_1w_1)}}\)

\(n'=\dfrac{N(\sigma^2+(k-1)w_2\sigma^2_2)}{NV_0+\sigma^2}\)

where \(\sigma^2\) is the variance of the entire population and \(\sigma_2^2\)is the variance of the non-response group. [Sometimes, we are only given the variance of the response group and non-response group. In such a case, we can use the formula for the variance of a mixture distribution provided after the Example to compute the variance of the entire population.] 

\(X \sim\left(\mu_{X}, (\sigma_{X})^{2}\right)\)
\(Y \sim\left(\mu_{Y}, (\sigma_{Y})^{2}\right)\)

W is a mixture of pX and (1 - p)Y

\(W=pX\oplus (1-p)Y\)

where \(0\leq p \leq 1\)

Then:

\(\mu_{W}=p \mu_{X}+(1-p) \mu_{Y}\)

What is the variance of the overall population?

\((\sigma_{W})^{2}=p (\sigma_{X})^{2}+(1-p) (\sigma_{Y})^{2}+p(1-p)\left(\mu_{X}-\mu_{Y}\right)^{2}\)

You can think of these p's as weights of the two variables. When one wants to apply this formula to find the mean and variance of a population that consists of the response and the nonresponse group, then p is the proportion of the response group and (1-p) the proportion of the nonresponse group.

Good survey practice is to discover why the non-response occurs and resolve as many of the problems as possible before commencing the survey.

There are k interviewers and they are each different in their manner of interviewing and hence may obtain slightly different responses. To make the notation simple, we assume that each interviewer conducts the same number of interviews. Let n denote the total sample size and n = k * m. There are k subsamples and each interviewer will be assigned m subjects.

Objective: to use simple random sampling to estimate \(\mu\)

• Interviewer \(1-y_{11}, y_{12}, y_{13},...,y_{1m}\)
• Interviewer \(2-y_{21}, y_{22}, y_{23},...,y_{2m}\)
• Interviewer \(3-y_{31}, y_{32}, y_{33},...,y_{3m}\)

• Interviewer \(k-y_{k1}, y_{k2}, y_{k3},...,y_{km}\)

The average for the ith interviewer is denoted as:

\(\bar{y}_i=\dfrac{1}{m}\sum\limits_{j=1}^m y_{ij}\)

The grand average is denoted as:

\(\bar{y}=\dfrac{1}{k}\sum\limits_{i=1}^k \bar{y}_i\)

The grand average \(\bar{y}\) is unbiased for μ and the estimated variance of \(\bar{y}\) is:

\(\hat{V}ar(\bar{y})=\dfrac{N-n}{N}\cdot \dfrac{s^2_k}{k}\)

\(\text{where } s^2_k=\dfrac{\sum\limits_{i=1}^k (\bar{y}_i-\bar{y})^2}{k-1}\)

The technique of interpenetrating the subsample gives an estimate of the variance of ybar that accounts for interviewer biases. In practice, the estimated variance given in the above formula is usually larger than the estimate of the variance by using simple random sampling.

Quite often, obtaining a frame that lists only those elements of the population that one is interested in is impossible. For example, perhaps you want to sample households with children, however, the best frame available is a list of all households. Therefore, we wish to estimate the parameters of a subpopulation of the population represented in the frame.

Main Issue: You do not know the size of the subpopulation.

Notation:

• N - the number of elements in the population
• \(N_1\)- the number of elements in the subpopulation
• n - sample size from the population
• \(n_1\) - the number of sampled elements from the subpopulation
• \(y_{1j}\) - the jth sampled observation that falls in the subpopulation

Usually, we do not know \(N_1\), so we will estimate the finite population correction factor as :

\(\dfrac{N_1-n_1}{N_1} \text{ by } \dfrac{N-n}{N}\)