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Sampling Methods and Bias · Tutorial 34 of 996

Systematic Sampling With a Random Start

Learn to calculate a sampling interval, choose a random start, and construct a systematic sample while understanding the selection chances it gives each unit.

Beginner 9 min read

What You'll Learn

  • Identify the frame, target sample size, and conditions for the basic systematic-sampling procedure.
  • Calculate the interval when the frame size is an exact multiple of the desired sample size.
  • Select and record a random start from the valid starting positions.
  • Generate the sample by adding the interval repeatedly to the starting position.
  • Derive each unit’s inclusion probability and distinguish the design from a simple random sample.
  • Recognize common implementation errors, including choosing the start nonrandomly or replacing selected units.

Why Use a Random Start?

Suppose a clinic has an ordered list of 120 eligible patients and needs to contact 12 of them for a hypothetical survey. Selecting every tenth patient seems straightforward, but the choice of where to begin matters. If staff always start with the tenth name, some positions on the list can never be selected. A random start gives each eligible starting position a chance to determine the sample, while the fixed interval makes selection practical.

Systematic sampling is a probability sampling design: selection uses a specified random mechanism, rather than convenience or personal choice. To construct it correctly, first define the sampling frame and desired sample size, then calculate the sampling interval, randomly choose a start, and follow the interval through the ordered list.

Definition: In systematic sampling with a random start, the units in a sampling frame are placed in a specified order. A sampling interval \(k\) is calculated, a starting position \(r\) is selected at random from the permitted starting positions, and the sample consists of that unit and every \(k\)th unit after it.

Calculating the Interval and Constructing the Sample

Let \(N\) be the number of units in the frame and \(n\) the desired sample size. The basic procedure is easiest to define when \(N\) is an exact multiple of \(n\). In that case the interval is the number of frame positions per desired sample member:

$$ k = \frac{N}{n} $$

Here \(k\) is a positive integer. The random start \(r\) is chosen uniformly from the integers \(1,2,\ldots,k\). “Uniformly” means that each of these \(k\) positions has the same probability, \(1/k\), of being selected as the start. Once \(r\) is chosen, the selected positions are \(r, r+k, r+2k,\ldots\), stopping at position \(N\). Since \(N=nk\), this gives exactly \(n\) units.

1
Fix the frame and target size.
Count the eligible units, label their positions from 1 through \(N\), and decide the desired sample size \(n\). Each eligible unit should appear once in the frame.
2
Calculate the interval.
When \(N\) is divisible by \(n\), calculate \(k=N/n\). The interval must be an integer for this basic procedure.
3
Choose the random start.
Use a random-number method to select one integer from 1 through \(k\), giving each possible start the same chance.
4
Follow the interval.
Select the unit at the start position, then add \(k\) repeatedly to obtain the remaining positions. Keep the resulting selections as specified; do not substitute more convenient units.

Worked Example: Selecting 12 Patients From 120

A hypothetical clinic has a frame listing \(N=120\) eligible patients and wants a sample of \(n=12\). Assume the list is finalized before selection and contains each eligible patient once. Calculate the interval and select the sample if a random-number generator chooses start 7.

Calculate the interval. The frame size is an exact multiple of the target sample size:

$$ k = \frac{N}{n} = \frac{120}{12} = 10 $$

The allowable starts are therefore positions 1 through 10. The start 7 is valid. It is treated as one outcome of a uniform random selection among those ten positions.

Generate the selected positions. Begin at 7 and add 10 each time:

$$ 7,\ 7+10=17,\ 17+10=27,\ \ldots,\ 7+11(10)=117 $$

The full set of positions is 7, 17, 27, 37, 47, 57, 67, 77, 87, 97, 107, and 117. There are exactly 12 positions, as intended. The clinic would contact the patients recorded at those positions, not the patients nearest to them or those who are easiest to reach.

Why Each Unit Has an Equal Inclusion Probability

In the divisible case, the possible starts are \(1,\ldots,k\), so there are \(k\) equally likely samples. Each sample contains \(n\) positions separated by \(k\). Every position from 1 through \(N=nk\) belongs to exactly one of these samples. For example, position 1 belongs to the sample starting at 1, position 12 in a frame with \(k=10\) belongs to the sample starting at 2, and position 120 belongs to the sample starting at 10.

More generally, divide a unit’s position by \(k\) and consider its remainder. Its position can be written uniquely as \(r+jk\), where \(r\) is one of \(1,\ldots,k\) and \(j\) is a nonnegative integer. Thus that unit is included precisely when the random start is \(r\). Since each possible start has probability \(1/k\), the unit’s inclusion probability is \(1/k\). Because \(k=N/n\), this is also \(n/N\).

$$ P(\text{a given unit is selected}) = \frac{1}{k} = \frac{n}{N} $$

This equal inclusion probability is an important design property, but it does not make systematic sampling the same as simple random sampling without replacement. In a simple random sample, every possible set of \(n\) units has the same chance of selection. In this systematic design, only the \(k\) sets generated by the possible starts can be selected. The random start determines which one of those sets is chosen.

Worked Example: A Random Start and Inclusion Chances

A hypothetical training program has a frame of \(N=84\) participants and wants a systematic sample of \(n=7\). The list contains each eligible participant exactly once. Suppose a valid random procedure selects start 5.

Calculate the interval.

$$ k = \frac{N}{n} = \frac{84}{7} = 12 $$

A start must be selected uniformly from positions 1 through 12. Start 5 is allowed. Adding 12 repeatedly gives positions 5, 17, 29, 41, 53, 65, and 77. The final position is within the frame, and the list contains the required seven participants.

Find each participant’s inclusion probability. There are 12 possible starts, each equally likely. Every participant appears in exactly one of the 12 samples, so for any particular participant:

$$ P(\text{participant is selected}) = \frac{1}{12} = \frac{7}{84} \approx 0.08333 $$

Thus each participant has about an 8.333% chance of being selected. This statement is about the selection procedure, not about the participant’s chance of responding or completing the survey.

Conditions and a Practical Boundary

The calculation and inclusion-probability argument above rely on specific conditions. The frame must have a known, fixed number \(N\) of positions; the desired sample size \(n\) must divide \(N\) exactly for this version of the procedure; and the random start must be selected uniformly from 1 through \(k=N/n\). The ordering must be set before the random selection, and the selection must follow the stated interval.

In practice, \(N\) is not always divisible by the desired \(n\). Then \(N/n\) is not an integer, and the exact procedure described here cannot be applied as written while also guaranteeing exactly \(n\) selections. Simply rounding the interval and still claiming that this procedure gives a fixed sample size with inclusion probability \(n/N\) is not justified. A nondivisible case requires a separately specified systematic-sampling method and its own account of sample size and selection probabilities. The key is to state the method rather than silently treat a rounded interval as the exact design.

Worked Example: Checking Whether the Basic Interval Works

A hypothetical library has an ordered frame of \(N=250\) members and wants \(n=20\) in a sample. Determine whether the basic integer-interval procedure applies directly.

Calculate the ratio:

$$ \frac{N}{n} = \frac{250}{20} = 12.5 $$

The ratio is not an integer. Therefore, there is no integer \(k\) with \(N=nk\), and the exact fixed-interval procedure in this tutorial does not directly produce exactly 20 selections from a start chosen uniformly among \(1,\ldots,k\). Rounding 12.5 to 12 or 13 and asserting an exact sample of 20 would not follow from the stated construction. The library must specify a method designed for this nondivisible case before selection, rather than altering the interval after seeing which members were selected.

Common Mistakes and Why the Design Matters

Systematic sampling can be convenient when a frame is already in a usable order: after the random start, the rule is simple to apply and audit. It also makes the random mechanism visible. Someone checking the procedure can verify the frame count, interval, random start, and sequence of selected positions.

  • Starting at position 1 automatically. This is not a random start. In the divisible procedure, the start must be selected randomly from all positions 1 through \(k\), not fixed by habit.
  • Choosing a start outside the allowed range. If \(k=10\), the start must be an integer from 1 to 10. Choosing position 17 as the start and then stepping by 10 is a different procedure.
  • Using a noninteger interval as if it were exact. The formula \(k=N/n\) gives an integer interval only when \(N\) is a multiple of \(n\). A fractional result requires a separately defined method.
  • Replacing a selected unit who cannot be reached. Replacing that unit with the next convenient person changes the selection procedure. Record the issue and follow a prespecified protocol instead of making an ad hoc substitution.
  • Calling the result a simple random sample. Although every unit has the same individual inclusion probability in the divisible design, not every group of \(n\) units can be selected. The design is systematic sampling, not simple random sampling.

A well-documented systematic sample records the frame and its ordering, \(N\), \(n\), \(k\), the method used to generate the start, the selected positions, and any departures from the protocol. These details make it possible to distinguish the intended random design from the way it was actually carried out.

Key takeaway: When \(N\) is divisible by \(n\), set \(k=N/n\), choose a uniform random start from 1 through \(k\), and select that position and every \(k\)th position afterward. Each unit then has inclusion probability \(n/N\), but the design is not a simple random sample.

Check Your Understanding

Use the basic systematic-sampling procedure and its assumptions to answer these questions.

  1. A frame has 180 units and the target sample size is 15. Calculate \(k\), state the permitted random starts, and list the selected positions if the start is 8.
  2. For a frame with \(N=96\) and desired sample size \(n=8\), what is the inclusion probability of any specified unit under the basic design? Explain why.
  3. A researcher has \(N=140\) and wants \(n=12\). Does the integer-interval procedure described here apply directly? Show the calculation and explain your conclusion.
  4. Why is selecting the first position and then every \(k\)th position not equivalent to choosing a random start?
  5. How can every unit have the same inclusion probability even though systematic sampling is not simple random sampling without replacement?