01 · Interval and start
A frame has 360 units and a sample of 30 is required. Find the interval and the range for a random start.
Hint
Use N/n.
Worked solution
The interval is 12. Choose the start uniformly from integers 1–12.
Understand · explore · practise
Systematic sampling
Calculate a sampling interval, choose a random start and select a fixed-size sample. Explore how repeating patterns in a list affect systematic samples.
Before you startSampling frames, simple random sampling and arithmetic sequences.
01 / Regular spacing, random start
Select at fixed intervals through an ordered frame.
For N = nk, choose a random start r from 1 to k.
Select labels r, r + k, r + 2k, …, r + (n − 1)k.
For 24 records and a sample of 6, the interval is 24 ÷ 6 = 4. Choose the first label uniformly from 1, 2, 3 and 4, then take every fourth record. There are four possible systematic samples under this rule.
Try all four starts. Compare the repeating values with the rearranged list.
Every fourth recordExplore
Selected labels: 1, 5, 9, 13, 17, 21.
Observed values: 4, 4, 4, 4, 4, 4.
Sample mean = 4; population mean = 10.
Full list: 4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16, 4, 8, 12, 16.
The labels are positions in the list. This is a manual comparison of starts, not a random draw. For actual sampling, choose the start randomly. Both lists contain the same invented values in different orders.
02 / Calculate the interval
Divide the population size by the sample size.
Select 15 records systematically from a frame of 120.Worked example
k = 120 ÷ 15 = 8
The interval is eight positions.
Choose r uniformly at random from 1–8
The start must be chosen from one full interval.
If r = 6, take 6, 14, 22, …, 118
There are 15 labels, since the final label is 6 + 14 × 8.
02 · Last label
With interval 12, start 5 and sample size 30, find the last label.
Hint
There are 29 jumps after the first selected unit.
Worked solution
5 + 29 × 12 = 353.
03 / Write a complete procedure
State the frame, order, start, interval and stopping point.
“Take every tenth person” leaves the first selection unspecified. A reproducible method identifies the eligible list, numbers it in its existing or stated order, chooses a random start from the first interval and stops after the required number of units.
03 · A full description
Describe a systematic sample of 20 from 200 numbered customer records.
Hint
There should be ten equally possible starts.
Worked solution
Check the frame contains the 200 eligible records once. Choose a uniform random integer r from 1–10. Select r, r + 10, …, r + 190, giving 20 distinct records.
04 · A fixed first item
A researcher always starts at record 1 and takes every fifth record. What part of a random-start systematic design is missing?
Hint
The start has been predetermined.
Worked solution
The start should be chosen randomly from 1–5. Always starting at 1 gives only one fixed sample under the stated list order.
04 / Not a simple random sample
Most subsets cannot arise from fixed spacing.
In the 24-record model, only four six-record subsets are possible. A simple random sample would allow every six-record subset. A random-start systematic design can give every individual the same inclusion probability while still assigning probability zero to many subsets.
05 · Individual inclusion
For N = 24, n = 6 and k = 4 with a fair start, what is the probability that record 10 is selected?
Hint
Which start produces 2, 6, 10, …?
Worked solution
Only start 2, so the probability is 1/4. Every record belongs to exactly one of the four possible samples.
06 · An impossible subset
Could a sample containing both records 1 and 2 arise under the interval-four rule?
Hint
Selected labels all have the same remainder modulo four.
Worked solution
No. Record 1 requires start 1 and record 2 requires start 2. One sample uses only one start.
05 / Watch for repeating patterns
The interval can repeatedly land on the same kind of unit.
The values 4, 8, 12, 16 repeat six times in the model.Worked example
The population mean is (4 + 8 + 12 + 16)/4 = 10
Each value occurs six times.
Start 1 selects six 4s; start 4 selects six 16s
The sample means are 4 and 16.
All four possible means are 4, 8, 12 and 16
The realised estimate may be far from 10.
Their average over equally likely starts is 10
In this complete divisible example, the issue is high sampling variability, not a systematic shift of the average estimator.
Describe the actual pattern risk rather than claiming any ordered list must be biased. Random start is important, but it does not ensure a balanced realised sample.
Watch: the interval follows a repeating pattern
Pause, replay or seek freely. The notes explain the same idea and stay in view.
07 · Match a production cycle
Four machines contribute one product each in a repeating order. A sampler takes every fourth product. What could happen?
Hint
The interval matches the cycle length.
Worked solution
The sample can contain products from only one machine. If machines produce different results, the sample may poorly represent overall production.
06 / Use context to assess the order
An alphabetical list is not automatically biased.
Systematic sampling spreads selections through a list and is easy to carry out. Its quality depends on the relationship between list order, interval and measured variable. A pattern aligned with the interval can matter; the mere fact that a list is alphabetical does not establish bias.
A simple random design avoids locking every selection to one fixed interval. Randomly shuffling a complete frame first can also remove an existing order pattern, although it adds practical work.
08 · A precise criticism
Improve the claim “The sample is biased because the list is ordered.”
Hint
Name a possible relationship between the ordering and interval.
Worked solution
For example: “If the list repeats groups every k positions and the sampling interval is k, one start may repeatedly select the same group, producing an unbalanced sample.” Without such context, ordering alone is not enough to establish bias.
07 / When N/n is not an integer
Do not silently round the interval and assume equal coverage.
If N = 50 and n = 8, N/n = 6.25. Simply using interval 6 and starts 1–6 with eight observations covers only labels 1–48; labels 49 and 50 cannot be selected. State how the design handles the remainder.
One fixed-size alternative is a fractional interval: choose u uniformly in (0, 6.25], then take the ceilings of u, u + 6.25, …, u + 7 × 6.25. This gives eight distinct valid labels and treats all 50 labels symmetrically in inclusion probability. Use a method appropriate to the question and explain it clearly.
Illustrate the fractional rule with u = 2.5.Worked example
Positions: 2.5, 8.75, 15, 21.25, 27.5, 33.75, 40, 46.25
Add 6.25 each time.
Ceilings: 3, 9, 15, 22, 28, 34, 40, 47
The ceiling is the smallest integer at least as large as the position.
09 · Do not claim complete coverage
What is wrong with claiming the rounded-interval-six method above gives every unit a chance?
Hint
Find the largest possible final label.
Worked solution
The largest final label is 6 + 7 × 6 = 48. Labels 49 and 50 have zero chance.
08 / Selected records may lack measurements
Sample size and usable data count can differ.
A weather table may contain a missing gust reading on a selected date. Selecting 20 distinct dates does not guarantee 20 usable gust values. Report missing records and follow a defensible preplanned rule; do not quietly substitute convenient dates or treat missing values as zero.
10 · Missing readings
A systematic sample selects 18 dates; 3 have no reading for the variable studied. How many usable values remain?
Hint
Count only records with a reading.
Worked solution
15 usable values. The requested sample contained 18 dates; its usable measurement count is smaller.
11 · An invalid zero
Why should “not available” not automatically become a zero?
Hint
Missingness does not state the measured quantity.
Worked solution
Zero is an actual value. Replacing missing readings by zero invents observations and can distort a mean or other statistic.
09 / Interval, start and ordering
Describe the design and its limitations.
- Use a suitable complete frame.
- Calculate the interval from population and sample sizes.
- Choose a random start in the first interval.
- List the rule and stop at the intended size.
- Check for patterns matching the interval.
- Handle non-integer intervals and missing data explicitly.
12 · Final method
A frame contains 480 invoices and you need 40. Describe the selection, including the last label if the random start is 9.
Hint
There are 39 jumps after the start.
Worked solution
Use interval 480/40 = 12. Randomly choose a start from 1–12, then take every twelfth invoice until 40 are selected. For start 9, the last label is 9 + 39 × 12 = 477. Check that invoice order has no problematic cycle aligned with 12.
Section 1 of 9 · Regular spacing, random start