Pathology Turnaround Time: Skewed Data, a Normality Assumption, and a Unit Trap
The pathology service of a 700-bed hospital audited turnaround time (TAT, hours from specimen receipt to verified report) for 240 consecutive specimens in March. The analyst reports: mean 7.4 h, SD 9.8 h, median 4.1 h, IQR 2.6–6.0 h. Inspection shows 228 specimens between 1.5 and 9.0 h and 12 specimens between 30 and 70 h; all 12 were sent to an external reference laboratory for confirmation. The draft states: "Average TAT 7.4 h; assuming approximate normality, 95% of specimens fall within mean ± 2 SD, i.e. −12.2 to 27.0 h, so performance is acceptable." The accompanying histogram uses 1-hour bins over the full range 0–70 h. A second section compares TAT across two laboratories and concludes that Laboratory B is "twice as consistent" because its SD is 0.05 against 1.2 for Laboratory A. Laboratory A reports in hours (mean 4.1, SD 1.2); Laboratory B reports in days (mean 0.17, SD 0.05). The quality committee must decide what to publish.
1. Critique the "mean ± 2 SD covers 95%" statement and the choice of summary measures for this TAT distribution. State what you would report instead, and why.
Read the worked answer
The ±2 SD rule is a normal-theory statement: about 95% of a normal distribution lies within mean ± 2 SD. It cannot simply be applied to any data.
Why it fails here:
- Mean 7.4 h is far above median 4.1 h, and the IQR is only 3.4 h wide. That gap is the signature of strong right skew, so symmetry - the premise of the rule - is absent.
- The resulting interval, −12.2 to 27.0 h, contains impossible negative times and excludes all 12 specimens between 30 and 70 h. It describes neither the body nor the tail the committee actually cares about.
- The SD (9.8) exceeds the mean (7.4); CV = 9.8/7.4 = 1.32. For a non-negative, right-skewed variable this is expected and is not a defect of the data, but it makes the SD a poor dispersion summary.
What to report instead:
- Median 4.1 h with IQR 2.6–6.0 h for the typical specimen.
- The 90th and 95th percentiles, and the proportion exceeding a clinical threshold (e.g. 24 h): 12/240 = 5.0%.
- The 12 send-out cases listed separately, since they arise from a different process (external referral), not from ordinary variation.
- Keep the mean, but for a different purpose: total workload is a sum of times, so the mean is the right building block for staffing/capacity, while the median and percentiles describe service to a patient. Give both, each labelled with its purpose.
A box plot or empirical CDF displays body and tail together; a mean ± 2 SD interval on skewed data does not.
2. Evaluate the two-laboratory comparison. Is Laboratory B twice as consistent as Laboratory A? What is the correct conclusion and what caveats remain?
Read the worked answer
First convert to a common unit. Laboratory B: mean 0.17 days × 24 = 4.08 h; SD 0.05 days × 24 = 1.20 h. So B is (4.08 h, 1.20 h) and A is (4.1 h, 1.2 h) - essentially the same numbers.
The SD carries the units of the measurement scale, so a smaller numerical SD in days is simply a coarser unit, not better performance. Comparing SDs across scales is meaningless.
Compare relative dispersion instead:
- CV_A = 1.2/4.1 = 0.293 (29.3%)
- CV_B = 1.2/4.08 = 0.294 (29.4%)
Identical to three decimal places. The "twice as consistent" claim is a unit-conversion error, not a finding.
Correct conclusion: the two laboratories have the same relative variability and almost the same mean. Choose between them on the median and 95th percentile, and on case mix (specimen types, priority split, proportion of send-outs), not on the SD.
Remaining caveats:
- CV is only a sound relative-dispersion measure if both distributions have similar shape. Both TAT distributions are likely right-skewed; CVs of skewed non-negative data are inflated by the mean–variance relationship. Compare IQR/median, or compare medians and upper percentiles directly, before drawing any conclusion about consistency.
- Both means should be accompanied by n and a measure of precision; a mean without n cannot support a ranking.
3. The histogram uses 1-hour bins across 0–70 h. Explain why this display misleads and how you would redesign it.
Read the worked answer
With x spanning 0–70 h and a bin width of 1 h, the 12 long-TAT specimens each produce a bar of height 1 out of 240 (0.4%). Meanwhile the 3–7 h region puts roughly 60 specimens into a single bin (25%). On one linear vertical scale the isolated bars are invisible, and the empty stretch from 10 to 70 h reads as "nothing happens there" - the opposite of the intended message.
Second problem: above 30 h no bin can hold more than one observation, so bar height reflects bin width and the accident of where bin boundaries fall, not local density. Any rule for the number of bins (Sturges, Freedman–Diaconis) computed from the SD of such a heavy-tailed sample will be dominated by the outliers and give a poor display.
Redesign options:
- Split the figure: a main panel zoomed to 0–10 h (where 95% of the data are) plus a second panel or table for the send-out group.
- Log-scale the x axis so both clusters are visible at once; label the axis with the original units.
- Plot density rather than counts, with wider bins in the tail, or use a kernel density on the log scale.
- Use a box plot with individual outliers, or an empirical CDF with 90th/95th percentiles marked.
- Always state n, the bin width, and annotate the 12 external referrals and their route, since they belong to a different generating process.
The aim is a display that shows both the concentrated body and the long tail on a scale a reader can interpret.