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Statistics · S.IC.A.1

Systematic sample

Boards come off a saw line one after another, and nobody has a list of them until the shift is over. How can chance choose a fair sample while they pass?

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Simple random samples and the sampling frameMath 7
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Systematic sample

When you reach for it

A simple random sample lets chance alone choose, so that every group of n individuals is equally likely to be the sample. To draw one you need the whole list in hand and n separate random numbers.

Often the individuals arrive in order instead: boards off a saw, cars through a gate. Nobody can number them in advance. A systematic sample needs only one random number: pick a random starting point, then take every k-th individual down the list or down the line. Every individual gets the same chance, and the sample behaves like a simple random sample whenever the order has nothing to do with what is measured.

The procedure

The interval. A population of N individuals stands in some order, and the sample is to hold n of them. The sampling interval is

k = \frac{N}{n}.

At the sawmill where Aitor checks boards for moisture, a shift produces N = 3{,}600 boards and he wants n = 120, so \displaystyle k = \frac{3600}{120} = 30: one board in every 30.

The random start. Choose a whole number s from 1 to k at random, each equally likely (a random digit table read in pairs, a calculator, or 30 slips in a hat). The sample is

s,\ s + k,\ s + 2k,\ \ldots,\ s + (n - 1)k.

The last one is at most k + (n-1)k = nk = N, so the sample never runs off the end of the list and always holds exactly n individuals.

The start must be random: if Aitor always began at board 1, no chance would be involved, and boards in the other 29 positions would never be checked.

Every individual has chance \displaystyle \frac{n}{N}. Each individual belongs to exactly one of the k possible samples: board 47 is in the sample with start 17 and in no other, since 47 = 17 + 30. The start is chosen with equal chances, so each board’s chance is

\frac{1}{k} = \frac{n}{N},

the same as in a simple random sample of the same size.

But it is not a simple random sample. There are only k possible samples. Every other group of n has chance 0: two boards next to each other, such as 16 and 17, can never be in the same sample. A simple random sample needs every group equally likely, not only every individual.

When \displaystyle \frac{N}{n} is not whole. Round k down to a whole number and take every k-th individual to the end of the list. The sample then holds n or a few more, and every individual still has chance \displaystyle \frac{1}{k} if the extras are kept; dropping them at random gives exactly n, with chances no longer quite equal. Rounding up gives a few fewer than n. A second remedy treats the list as a circle: choose the start at random from 1 to N, step forward by k, and when the count passes N carry on from the top of the list until n are chosen.

When it works as well as a simple random sample. If the order of the list has nothing to do with the variable measured, a column of the wrapped list in Figure 1 is as good as a scattered handful, as with boards in sawing order when moisture does not follow the clock.

start 17 boards 1 to 30 boards 31 to 60 boards 61 to 90 boards 91 to 120 boards 121 to 150 boards 151 to 180 boards 181 to 210 boards 211 to 240 and so on: 120 rows of 30 boards, one board taken from each row
Figure 1Wrap the line of boards into rows of k = 30. A systematic sample with start 17 is one column of this picture.

When it fails. If something in the list repeats with a period that matches k, or divides k, every individual in the sample sits at the same point of the cycle. The sample then sees one point of the cycle, never the rest. Example 5 shows how badly this can miss.

When the list is sorted by the response. Each sample then takes one individual from every block of k, so it spreads over the whole range and usually beats a simple random sample. The start leans every value the same way, but only within one block (Example 6).

Worked examples

Example 1

Every 30th board

Aitor’s shift will produce 3{,}600 boards and he wants 120 of them. His random start, from 1 to 30, is 17. List the first four boards and the last one in his sample, find each board’s chance, and say which start would include board 3{,}600.

Solution

The interval is \displaystyle k = \frac{3600}{120} = 30. Starting at 17 and adding 30 each time gives boards 17, 47, 77, 107, and so on. The 120th board is

17 + 119 \times 30 = 3587.

Each start from 1 to 30 is equally likely and each board belongs to exactly one of these 30 samples, so every board’s chance is \displaystyle \frac{1}{30} = \frac{120}{3600}. Board 3{,}600 is 30 + 119 \times 30, so it is in the sample only when the start is 30.

Example 2

Counting the possible samples

Unai tries the method on the first 12 boards of a shift, taking a sample of 3. Find k, list every possible systematic sample, and compare with a simple random sample of 3 of the 12.

Solution

Here \displaystyle k = \frac{12}{3} = 4, so the start is 1, 2, 3 or 4, giving

\begin{aligned}&\{1, 5, 9\},\quad \{2, 6, 10\},\\ &\{3, 7, 11\},\quad \{4, 8, 12\}.\end{aligned}

Each board is in exactly one of these, so its chance is \displaystyle \frac{1}{4} = \frac{3}{12}, the same as in a simple random sample. But 12 boards make \binom{12}{3} = 220 different groups of 3, and a simple random sample gives each the chance \displaystyle \frac{1}{220}. The systematic method can produce only 4 of them, so 220 - 4 = 216 groups, such as \{1, 2, 3\}, have chance 0. For Aitor’s full shift the contrast is far larger: 30 possible samples, against a number of groups of 120 from 3{,}600 that has 228 digits.

Example 3

A shift that does not divide evenly

The next shift produces 3{,}610 boards, and Aitor still wants 120. What interval should he use, and how many boards does he get?

Solution

Here \displaystyle \frac{3610}{120} \approx 30.08, not a whole number. Rounding down to k = 30, the start runs from 1 to 30, and boards s, s + 30, \ldots continue while they are at most 3{,}610. For a start from 1 to 10 the last board is s + 120 \times 30 = s + 3600 \le 3610, so the sample holds 121 boards; for a start from 11 to 30 it holds 120. Keeping the extra board leaves every chance exactly \displaystyle \frac{1}{30}; dropping one at random leaves boards in the first 10 places of each block of 30 a little less likely, \displaystyle \frac{1}{30} \times \frac{120}{121}.

Rounding up to k = 31 instead gives 117 boards for a start from 1 to 14 and 116 for a start from 15 to 31, short of the 120 he wanted.

The circular remedy also works: a start from 1 to 3{,}610 at random, then every 30th board, carrying on from board 1 after board 3{,}610. The 120 boards are always different, because 119 \times 30 = 3570 is less than 3{,}610. Each board is reached from exactly 120 of the 3{,}610 starts, so its chance is \displaystyle \frac{120}{3610} = \frac{12}{361}, exactly \displaystyle \frac{n}{N}.

Example 4

Sampling a line nobody has listed

Ainhoa samples on a day when the boards are bundled and shipped as they come off the line, so there is no stack to draw from at the end. The shift is expected to run to 3{,}600 boards. Explain why a simple random sample is impractical, describe a systematic sample of about 120, and find its size when her random start is 9 and the shift actually runs to 3{,}690 boards.

Solution

A simple random sample needs N. Ainhoa could draw 120 random numbers from 1 to 3{,}600 before the shift, sort them, and pull those boards as they pass, but 3{,}600 is only a guess. If the shift runs short, the picks past its last board are wasted and the sample is too small; if it runs long, the boards after 3{,}600 have no chance at all.

A systematic sample needs only \displaystyle k = \frac{3600}{120} = 30 and one random start, 9: she pulls boards 9, 39, 69, and so on as they pass. Every board made, however long the shift, has chance \displaystyle \frac{1}{30}. With 3{,}690 boards the last one she pulls is the largest 9 + 30j that is at most 3{,}690, which is j = 122, so the sample holds 123 boards: a few more than planned, with no change to anyone’s chance.

Example 5

Kiln carts with thirty layers

The boards reach the saw line from kiln carts, each stacked in 30 layers and unloaded top layer first, so in each group of 30 boards the j-th board came from layer j. The lower layers dry less: in these made-up data a board from layer j holds

8.0 + 0.2(j - 1)

percent moisture. Unai takes every 30th board. Find the shift’s mean moisture, his sample’s mean for a start of 4 and of 27, and the average over all 30 starts.

Solution

Every layer is equally common, so the shift mean is the mean over j = 1 to 30. The average of j - 1 over 0, 1, \ldots, 29 is 14.5, so the shift mean is

8.0 + 0.2 \times 14.5 = 10.9\%.

With k = 30 the cart’s cycle and the interval match, so a start of 4 takes the 4th board of every cart: all 120 boards come from layer 4, and the sample mean is 8.0 + 0.2 \times 3 = 8.6\%. A start of 27 takes only layer 27, for 8.0 + 0.2 \times 26 = 13.2\%.

8 9 10 11 12 13 14 shift mean 10.9% mean moisture of the sample, percent
Figure 2The 30 possible systematic samples when kiln carts hold 30 layers. Each dot is one sample’s mean moisture (made-up data). They average 10.9\%, but every single one sits at one layer’s value.

The 30 possible sample means run from 8.0\% to 13.8\% and average 10.9\%: right on average, since every board has the same chance, but each sample is one layer and can miss by almost 3 percentage points. A simple random sample would mix the layers. Another remedy is one board chosen at random from each cart, a stratified random sample with the carts as strata.

Example 6

A list sorted by moisture

Twelve boards were read with a quick meter and listed from driest to wettest. Their readings are 8, 9, 10, \ldots, 19 percent. Ainhoa takes a systematic sample of 3. Find every possible sample mean, and compare with the extremes a simple random sample of 3 could give.

Solution

The population mean is \displaystyle \frac{8 + 19}{2} = 13.5\%, the middle of twelve equally spaced values. With \displaystyle k = \frac{12}{3} = 4, the four possible samples are

\begin{aligned}&\{8, 12, 16\},\quad \{9, 13, 17\},\\ &\{10, 14, 18\},\quad \{11, 15, 19\},\end{aligned}

with means 12\%, 13\%, 14\% and 15\%. They average 13.5\%. A simple random sample of 3 could be the three driest, mean 9\%, or the three wettest, mean 18\%. Sorting by the response makes every systematic sample take one board from each third of the list, so its mean stays within 1.5 points of 13.5\%. The start still leans the whole sample: a start of 1 takes the driest board of every block of four.

Boundaries

The start must be random, and from 1 to k. A chosen start, or a start from a larger range that is not wrapped around, gives some individuals more chance than others.

Equal chances for individuals, not for groups. Only k samples are possible, so methods built for simple random samples apply only approximately, soundly when the order is unrelated to the response.

A cycle that matches k or divides it is fatal. Then every sample sees one point of the cycle. Knowing how the list was built, or a quick look at the data in order, is how to catch it.

Where this goes next

A stratified random sample samples inside groups chosen to be alike, the fix for the kiln carts. A cluster sample takes whole groups and weighs the designs against each other. Any draw speaks only for the list or stream it came from, the subject of undercoverage and nonresponse.

Exercises

The interval and the start

  1. Find k, the range of the random start, and each individual’s chance. (a) 2{,}400 boards, a sample of 60. (b) 900 boards, a sample of 45. (c) 5{,}000 boards, a sample of 125.
  2. A short shift makes 1{,}800 boards and Ainhoa wants 75. Her random start is 11. List the first four boards in her sample and the last one.
  3. A shift makes 2{,}050 boards and Unai wants 80. (a) Find \displaystyle \frac{N}{n}. (b) With k = 25, how many boards does a start of 1 give, and a start of 25? (c) With k = 26, how many does a start of 22 give, and a start of 23? (d) Which interval guarantees at least 80, and what is each board’s chance with it?
  4. A crate holds 50 boards, numbered 1 to 50, and Aitor wants 8 by the circular method with k = 6. His random start, from 1 to 50, is 47. List the 8 boards.
  5. Ainhoa plans every 40th board, expecting 2{,}000 boards, and her random start is 23. How many boards does she pull if the shift runs to 2{,}030? To 1{,}950?

Not a simple random sample

  1. Twenty boards, a systematic sample of 4. (a) Find k and list every possible sample. (b) How many groups of 4 can be made from 20 boards? (c) What fraction of them can the systematic method produce? (d) Find each board’s chance under both methods.
  2. For the method in exercise 6, which of these groups can be the sample? (a) \{3, 8, 13, 18\} (b) \{1, 2, 3, 4\} (c) \{5, 10, 15, 20\} (d) \{4, 9, 14, 20\}
  3. A shift of 3{,}600 boards is sampled with k = 30. Can boards 100 and 130 both be in the sample? Boards 100 and 145? Give the start in each case where it is possible.

When it works and when it fails

  1. The saw is reset every 30 boards, and after a reset each board is cut a little thicker: in made-up data the j-th board after a reset is 0.02(j - 1) millimeters over the target. (a) Find the mean amount over target for the whole shift. (b) Unai takes every 30th board. What does his sample show for a start of 1, and for a start of 30? (c) He doubles the sample to 240 with k = 15. Find the lowest and highest possible sample means. (d) Would k = 60 help?
  2. Fifteen boards are listed from driest to wettest, with readings 10, 11, 12, \ldots, 24 percent. (a) Find the mean of the fifteen. (b) List the possible systematic samples of 3 and their means. (c) Find the lowest and highest means a simple random sample of 3 could give.
  3. In the kiln carts of Example 5, suppose Unai takes every 31st board instead. (a) If the first board taken comes from layer 5, which layers do the next three come from? (b) Among the first 90 boards he takes, how many come from each layer?
  4. For each plan, say whether a systematic sample is likely to behave like a simple random sample, and why. (a) The mill’s 600 workers, listed alphabetically, asked how far they drive to work, every 20th taken. (b) A year of daily delivery records, one row per day starting on a Monday, every 7th row taken to estimate the average number of trucks a day. (c) Boards in the order they leave the saw, when logs are fed in no planned order, every 30th taken for moisture.

Reasoning

  1. Unai says: “Every board has chance \displaystyle \frac{1}{30}, so my sample is a simple random sample.” Explain what is right and what is wrong.
  2. Explain why the start must be chosen at random, and why from 1 to k rather than from 1 to N when the list is not treated as a circle.
  3. Ainhoa says: “Sorting the list by the response ruins a systematic sample.” Using exercise 10, explain what is right and what is wrong in this.
  4. Explain why, when the boards must be sampled as they come off the line, a systematic sample is the practical choice, and what Aitor should find out about the line before trusting it.
Answers
  1. (a) \displaystyle k = \frac{2400}{60} = 40; start 1 to 40; chance \displaystyle \frac{1}{40}. (b) \displaystyle k = \frac{900}{45} = 20; start 1 to 20; chance \displaystyle \frac{1}{20}. (c) \displaystyle k = \frac{5000}{125} = 40; start 1 to 40; chance \displaystyle \frac{1}{40}.
  2. \displaystyle k = \frac{1800}{75} = 24. Boards 11, 35, 59, 83, and last 11 + 74 \times 24 = 1787.
  3. (a) \displaystyle \frac{2050}{80} = 25.625. (b) 82 for both: the last board is s + 81 \times 25 = s + 2025, at most 2{,}050 for every start from 1 to 25. (c) 79 for a start of 22, since 22 + 78 \times 26 = 2050; 78 for a start of 23, since 23 + 78 \times 26 = 2051 is past the end. (d) k = 25, which gives 82; each board’s chance is \displaystyle \frac{1}{25} = \frac{82}{2050}.
  4. Boards 47, then 53 - 50 = 3, then 9, 15, 21, 27, 33, 39.
  5. 51 boards: the last is 23 + 50 \times 40 = 2023 \le 2030. 49 boards: the last is 23 + 48 \times 40 = 1943 \le 1950, and 1983 is past the end.
  6. (a) k = 5: \{1, 6, 11, 16\}, \{2, 7, 12, 17\}, \{3, 8, 13, 18\}, \{4, 9, 14, 19\}, \{5, 10, 15, 20\}. (b) \binom{20}{4} = 4845. (c) \displaystyle \frac{5}{4845} = \frac{1}{969}. (d) \displaystyle \frac{1}{5} under both, since \displaystyle \frac{4}{20} = \frac{1}{5}.
  7. (a) and (c) only. (b) holds neighbors, and (d) has 20 - 14 = 6, not 5, between its last two.
  8. 100 and 130: yes, 130 - 100 = 30; the start is 10, since 100 = 10 + 3 \times 30. 100 and 145: no, 145 - 100 = 45 is not a multiple of 30.
  9. (a) 0.02 \times 14.5 = 0.29 millimeters. (b) Start 1: every board is the first after a reset, so the sample shows 0 millimeters over. Start 30: every board shows 0.02 \times 29 = 0.58 millimeters over. (c) Start s takes positions s and s + 15, mean 0.02(s + 6.5): lowest 0.15 millimeters (s = 1), highest 0.43 millimeters (s = 15), so the result still depends heavily on the start. (d) No: 60 is a multiple of 30, so every board is again at the same point of the cycle.
  10. (a) \displaystyle \frac{10 + 24}{2} = 17\%. (b) k = 5: \{10, 15, 20\} mean 15\%, \{11, 16, 21\} mean 16\%, \{12, 17, 22\} mean 17\%, \{13, 18, 23\} mean 18\%, \{14, 19, 24\} mean 19\%. (c) \{10, 11, 12\}, mean 11\%, and \{22, 23, 24\}, mean 23\%.
  11. (a) Layers 6, 7 and 8: 31 boards is one full cart plus one, so each pick moves down one layer. (b) 3 from each: the picks run through all 30 layers in turn, three times.
  12. (a) Likely fine: the alphabet has nothing to do with driving distance. (b) Fails: 7 matches the week, so every row taken is the same weekday, and deliveries differ by weekday. (c) Likely fine: the order carries no pattern in moisture, so this is the good case.
  13. Right: each board’s chance is \displaystyle \frac{1}{30} = \frac{120}{3600}, as in a simple random sample. Wrong: only 30 samples are possible, and every other group of 120, including any group with two neighboring boards, has chance 0, while a simple random sample gives every group the same chance.
  14. A chosen start lets a person decide who is in the sample, and a fixed one makes the sample certain, with no chance at all. A start drawn from 1 to 3{,}600 without wrapping could be 3{,}000, giving only the 21 boards 3000, 3030, \ldots, 3600. Board 10 would be reached only from a start of 10, chance \displaystyle \frac{1}{3600}, while board 3{,}600 is reached from the 120 starts 30, 60, \ldots, 3600, chance \displaystyle \frac{120}{3600}: early boards would be badly underweighted.
  15. Wrong: the samples in exercise 10 have means from 15\% to 19\%, while a simple random sample ranges from 11\% to 23\%, so sorting makes the method more reliable, not less. Right: the start leans every board the same way, so a start of 1 gives a mean below 17\%, by at most 2 points.
  16. A simple random sample needs the whole list first, and on a running line the boards are gone before the list exists. A systematic sample needs one random start and a count, with every board’s chance still \displaystyle \frac{1}{k}. Aitor should check that nothing on the line repeats every k boards, or a divisor of k, such as kiln-cart layers or saw resets, since that is when the method fails.
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