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Let's say I have 100 schools and each has a different number of students. I want to estimate which % of students are in schools with electricity. Simulation and theory indicate it is more efficient to give a higher probability to schools with a larger number of students.

If I do a proportion-to-size with replacement sample, I am able to get a self weighted sample, where my estimator is a plain average of the sample (no weighted needed). Below is my logic (if you already know this is correct, you can skip to the questions):

Let's say we have:

  • A sample selected proportional to the student size of the school (with replacement)
  • A sample size 2

Based on this, let's say we got the following sample: 1, 0 (with 1 being the school has electricity); a good estimate of the proportion of students with electricity I think would be:

$s_i =$ number of students school i

$s_t =$ total number of students

$\bar{y} =$ estimate of proportion of students that have electricity $$ \bar{y} =\frac{\frac{s_1 \frac{1}{\frac{s_1}{s_t}} + s_2 \frac{0}{\frac{s_2}{s_t}}}{2}}{s_t} $$

which is the same as $$ \bar{y} =\frac{\frac{s_t\cdot1 + s_t\cdot0}{2}}{s_t} = \frac{1 + 0}{2} = 0.5 $$ Which is the sample average.

Question 1: Does this look right? Basically, the selection probability compensates for the school size, and the average now represents the proportion of students rather than the proportion of schools.

Question 2: Is there a way that I can design my sample so the calculation is a plain average of the sample with sampling without replacement? I have read a few papers, and I haven't found an answer to this question.

Thanks in advance!

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1 Answer 1

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As I understand it, you are describing one-stage sampling with probability proportionate to size (PPS) in which case neither sampling with replacement nor sampling without replacement result in a self-weighting survey. Instead, in both cases, you need to weight the selected units (schools) by the inverse of the probability of selection.

Taking a design-based approach where $t_i$ represents the total number of students in each school and is considered fixed, $p_i$ is the probability of selecting school $i$, and $W_i$ is a (random) variable that indicates whether a school was selected we can easily show that the weighted sample total is an unbiased estimator for the total.

$$ \begin{aligned} \mathbb{E}(\hat{t}) &= \sum^N_{i = 1} W_i \cdot \frac{t_i}{p_i} \\ &= \sum^N_{i = 1} \mathbb{E}(W_i) \cdot \frac{t_i}{p_i} \\ &= \sum^N_{i = 1} p_i \cdot \frac{t_i}{p_i} \\ & = t \end{aligned} $$

This holds whether sampling is with replacement or without replacement as long as $p_i$ is fixed. Of course, the variance for the two cases is not the same.

It's important to note, though, that you need to use a special algorithm for sampling without replacement. A function like 'sample()' in R will not sample without replacement with a fixed probability when the 'prob' argument is specified.

(Two-stage sampling where primary sampling units (PSU) are selected with PPS and a fixed number of secondary units are selected from each selected PSU does result in a self-weighting sample. So perhaps you saw something on this that confused you.)

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