A repeat tests whether you can get the same reading again with the same method. An independent method tests whether a different route gives the same answer. The first checks precision, the second can reveal a systematic fault.
This lesson builds on identifying confounding variables. Controlling variables stops the comparison being unfair, but it does not tell you whether the measurement itself is trustworthy.
Why can good repeats still be wrong?
Random errors push readings above and below the true value, so repeats scatter and a mean reduces the effect. A systematic error pushes every reading the same way, such as a ruler with a worn end or a balance that needs zeroing.
Repeat that method ten times and you get ten tidy, wrong answers. Only a different route can expose the offset.
Step by step
- Find the scatter within the repeats: the range, or half the range.
- Calculate the mean and the result you need.
- Choose a second method that does not share the same instrument or assumption.
- Compare the gap between the two results with the scatter.
- Decide: if the gap is within the scatter, the methods agree. If the gap is larger, look for a systematic cause.
Worked example
All readings are invented.
Task: Find the density of a small metal block. Mass from a balance: 54.0 g.
Method A, ruler dimensions. The block volume is calculated from length × width × height. Three attempts give 18.0 cm³, 18.1 cm³ and 17.9 cm³.
- Mean volume = (18.0 + 18.1 + 17.9) ÷ 3 = 54.0 ÷ 3 = 18.0 cm³.
- Range = 18.1 − 17.9 = 0.2 cm³, about 1% of 18.0.
- Density = 54.0 g ÷ 18.0 cm³ = 3.00 g/cm³.
Method B, water displacement. Three readings give 18.3 cm³, 18.5 cm³ and 18.4 cm³.
- Mean volume = (18.3 + 18.5 + 18.4) ÷ 3 = 55.2 ÷ 3 = 18.4 cm³.
- Range = 0.2 cm³.
- Density = 54.0 g ÷ 18.4 cm³ = 2.93 g/cm³ (3 significant figures).
Compare: The two mean volumes differ by 18.4 − 18.0 = 0.4 cm³. That is twice the range inside either method. The scatter is small in both, so the gap is real.
A reasonable evaluation says that the repeats are precise, but the methods disagree by more than the scatter. One method has a systematic error, for example rounded block edges that make the ruler volume too large, or a small air bubble clinging to the block that changes the displaced volume. The next step is to investigate, not to average 3.00 and 2.93.
The mistake to watch for
A frequent claim after getting matching repeats is that the result is accurate.
Mistaken answer: “The three readings were close together, so the density value is accurate.”
Close repeats show precision only. The correction is “The repeats are precise, but an independent method would be needed to check for a systematic error.”
A second slip is repeating the same ruler measurement and calling it an independent check.
Check yourself
1. A student times 20 swings of a pendulum three times and gets 28.2 s, 28.4 s and 28.6 s. Calculate the mean and the range.
Show answer
Mean = (28.2 + 28.4 + 28.6) ÷ 3 = 85.2 ÷ 3 = 28.4 s. Range = 28.6 − 28.2 = 0.4 s.
2. A balance is not set to zero and always reads 2.0 g too high. Would repeating the mass measurement five times reveal this?
Show answer
No. Every reading is shifted by the same 2.0 g, so the repeats still agree with each other. Checking against a second balance or a known mass would reveal it.
3. Method A gives 3.00 g/cm³ and Method B gives 2.93 g/cm³, each with small scatter. Name one thing to do next.
Show answer
Investigate a possible systematic error in each method (for example edge measurement, trapped air or zero error) rather than averaging the two values. Then repeat the method with the fault removed.
Where this leads next
With a trustworthy measurement in hand, explaining what a dataset can and cannot establish asks how far a conclusion can go. The scientific investigation critic reminds you that repeating a biased reading does not fix calibration.
If these distinctions feel clear in class but blur in a written answer, online one-to-one Co-ordinated Sciences tuition gives you a place to rehearse the wording with a teacher.