Statistics: mean, SD, CV, accuracy and precision

11 DOH Abu Dhabi (HAAD) lab licensing practice questions on Statistics: mean, SD, CV, accuracy and precision, part of General Laboratory, Quality and Safety. The first 11 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Hard

   A laboratory reviews its monthly peer-group report for a TSH control. Laboratory: 30 results, mean 2.38 mIU/L, SD 0.048 mIU/L. Peer group (same analyzer, reagent and control lot): mean 2.20 mIU/L, SD 0.07 mIU/L. The SDI is (laboratory mean - peer mean) / peer SD, and the CV ratio is laboratory CV / peer CV. The laboratory investigates bias when the SDI is beyond ±2.0, and imprecision when the CV ratio is above 1.5. Internal QC showed no rule violations this month. Which conclusion is correct?

   1. AInvestigate bias and precision (SDI +2.6, CV ratio 1.58)
   2. BNo investigation needed (SDI +1.3, CV ratio 0.63)
   3. CInvestigate bias only (SDI +2.6, CV ratio 0.63)
   4. DNo investigation needed (SDI +0.08, CV ratio 0.63)

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   Answer: C · Investigate bias only (SDI +2.6, CV ratio 0.63)

   The SDI is (2.38 - 2.20) / 0.07 = +2.6, beyond the ±2.0 limit, so the laboratory reads high against its peers and the bias must be investigated, starting with calibration. The laboratory CV is 0.048 / 2.38 × 100 = 2.0% and the peer CV is 0.07 / 2.20 × 100 = 3.2%, so the CV ratio is 2.0 / 3.2 = 0.63: the laboratory is more precise than its peers. Internal QC can stay in control while a bias like this persists, because the control limits are built around the laboratory's own mean. Peer comparison is how such a bias is found between EQA events.

   • A: 1.58 is the inverted ratio; the laboratory CV is lower than the peer CV.
   • B: Dividing the difference by twice the peer SD halves the SDI and hides the bias.
   • D: 0.08 is the difference divided by the peer mean, a relative bias, not an SDI.

2. Question 2 · Medium

   Replicate glucose results from four methods on one certified reference material are drawn as frequency curves. Which method has a small random error but a large systematic error?

   

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   1. AMethod R
   2. BMethod P
   3. CMethod S
   4. DMethod Q

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   Answer: D · Method Q

   Random error shows as the spread of a curve (its SD), and systematic error as the distance of the curve's centre from the certified value (bias). Method Q's curve is as narrow as P's but centred near 5.60 mmol/L, a bias of about +12% from 5.00 mmol/L. Such a method is precise but inaccurate, and recalibration can often remove the bias. Method R is the reverse: centred on the target but widely spread.

   • A: Method R is centred on the target but wide: large random error, little bias.
   • B: Method P is narrow and centred on the certified value: both errors are small.
   • C: Method S is wide and off target: both random and systematic error are large.

3. Question 3 · Hard

   A calcium precision study gives a repeatability (within-run) SD of 0.06 mmol/L and a between-day component SD of 0.08 mmol/L. What is the within-laboratory (total) SD?

   1. A0.07 mmol/L
   2. B0.01 mmol/L
   3. C0.10 mmol/L
   4. D0.14 mmol/L

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   Answer: C · 0.10 mmol/L

   Independent sources of random error add as variances, not as SDs: 0.06² + 0.08² = 0.0036 + 0.0064 = 0.0100. The total SD is the square root of this, 0.10 mmol/L. Adding the SDs directly would overstate the imprecision. CLSI precision protocols such as EP05 and EP15 combine the components in this way to report within-laboratory precision.

   • A: 0.07 mmol/L is the average of the two SDs, which has no statistical meaning here.
   • B: 0.01 is the total variance; its square root must be taken to get the SD.
   • D: 0.14 mmol/L adds the two SDs instead of their variances.

4. Question 4 · Hard

   A magnesium control has a mean of 0.80 mmol/L, an SD of 0.040 mmol/L and a CV of 5.0%. A calibration error then makes every result exactly 10% higher. The new control statistics will be:

   1. AMean 0.88, SD 0.044, CV 5.0%
   2. BMean 0.88, SD 0.040, CV 4.5%
   3. CMean 0.80, SD 0.044, CV 5.5%
   4. DMean 0.88, SD 0.044, CV 5.5%

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   Answer: A · Mean 0.88, SD 0.044, CV 5.0%

   Multiplying every result by 1.10 multiplies both the mean (0.80 to 0.88 mmol/L) and the SD (0.040 to 0.044 mmol/L) by the same factor. Their ratio, the CV, therefore stays at 5.0%. A proportional error shifts the control mean without changing relative imprecision, so it shows on a Levey-Jennings chart as a shift rather than as wider scatter. A constant error of +0.08 mmol/L would also move the mean to 0.88 but leave the SD at 0.040, giving a CV of 4.5%.

   • B: An unchanged SD with a CV of 4.5% is what a constant shift of +0.08 mmol/L would produce.
   • C: The mean cannot stay at 0.80 when every result rises by 10%.
   • D: The CV is a ratio; when mean and SD both rise by 10%, it does not change.

5. Question 5 · Medium

   An HbA1c control has a mean of 48 mmol/mol and an SD of 1.2 mmol/mol over 20 runs. Its coefficient of variation is:

   1. A2.5%
   2. B40%
   3. C5.0%
   4. D0.025%

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   Answer: A · 2.5%

   CV = SD / mean × 100 = 1.2 / 48 × 100 = 2.5%. Because the CV expresses the SD as a percentage of the mean, it allows precision to be compared between control levels, methods and units. At least 20 runs are normally used to set a control's mean and SD. The laboratory then compares the CV with its goal for allowable imprecision.

   • B: 40 is the mean divided by the SD, the ratio inverted.
   • C: 5.0% uses 2 SD instead of 1 SD in the numerator.
   • D: 0.025 is the ratio before multiplying by 100; it is not a percentage.

6. Question 6 · Medium

   A laboratory measures a certified calcium material, assigned value 2.50 mmol/L (10.0 mg/dL), ten times and obtains a mean of 2.40 mmol/L. Its percent bias is:

   1. A-4.2%
   2. B+4.0%
   3. C-0.10%
   4. D-4.0%

   Answer and explanation in LearnFree account

7. Question 7 · Hard

   A glucose method has a bias of +3.0% and a CV of 2.5%. The laboratory estimates total error as |bias| + 2 CV, and its allowable total error is 10%. What is the total error, and is the method acceptable?

   1. A8.0%, acceptable
   2. B11.0%, not acceptable
   3. C5.5%, acceptable
   4. D13.0%, not acceptable

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8. Question 8 · Medium

   A creatinine control has a target mean of 180 µmol/L (2.04 mg/dL), and the laboratory's CV for it is 4.0%. What are its ±2 SD limits?

   1. A172.0 to 188.0 µmol/L
   2. B158.4 to 201.6 µmol/L
   3. C165.6 to 194.4 µmol/L
   4. D172.8 to 187.2 µmol/L

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9. Question 9 · Medium

   A certified glucose reference material has an assigned value of 5.00 mmol/L (90 mg/dL). Twenty replicates give: method A, mean 5.02 mmol/L, SD 0.20 mmol/L; method B, mean 5.40 mmol/L, SD 0.05 mmol/L. How do the two methods compare?

   1. AB has less bias and better precision
   2. BA has less bias and better precision
   3. CA has less bias; B has better precision
   4. DA has better precision; B has less bias

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10. Question 10 · Hard

    Five replicate potassium results on one serum are 4.0, 4.2, 4.1, 3.9 and 4.3 mmol/L. The CV, using the sample SD (n - 1), is:

    1. A0.6%
    2. B2.9%
    3. C3.9%
    4. D3.4%

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11. Question 11 · Easy

    Repeated results are very close to each other but far from the true value. How are they described?

    1. AAccurate but not precise
    2. BNeither precise nor accurate
    3. CPrecise but not accurate
    4. DBoth precise and accurate

    Answer and explanation in LearnFree account