Take-off ground roll: six runs and one number to defend
Your company is certifying a 9-seat twin turboprop. The agreed means of compliance requires the demonstrated take-off ground roll at maximum take-off mass, sea level ISA, zero wind, to be shown below 1,250 m with 95% confidence in the true mean. Six runs were flown: same aircraft, mass within 10 kg, flaps 15°, wind never above 3 kt, all timed with GPS and a nose-wheel accelerometer. The measured ground rolls, in metres, are 1,180 / 1,225 / 1,195 / 1,210 / 1,310 / 1,205. The 1,310 m run was the only one flown by the second test pilot and carries no recorded anomaly. The chief test pilot wants to report the arithmetic mean of the six runs and declare compliance; the flight test engineer is not comfortable. Twenty flight hours remain in the certification budget and the authority's technical review is in five weeks.
1. Compute the sample mean, the sample standard deviation and the one-sided 95% upper confidence bound on the true mean. State, with the numbers, whether compliance is demonstrated.
Read the worked answer
Mean. Sum = 1180 + 1225 + 1195 + 1210 + 1310 + 1205 = 7325 m; mean = 7325/6 = 1220.8 m.
Standard deviation. Deviations: −40.8, +4.2, −25.8, −10.8, +89.2, −15.8 m. Sum of squares = 10,672 m². s² = 10,672/5 = 2134 m²; s = 46.2 m.
One-sided 95% upper bound. Standard error = s/√n = 46.2/2.449 = 18.9 m. With 5 degrees of freedom, t(0.95) = 2.015, so the bound is 1220.8 + 2.015 × 18.9 = 1258.9 m.
Conclusion. The point estimate is 29 m under the limit, but the 95% bound overshoots it by 9 m, so compliance is not demonstrated. The test is under-powered for the question being asked: a 2% effect is being resolved with six samples whose scatter is 3.8% of the mean. Check that this is not an artefact of the distributional choice: even the normal quantile (1.645) gives 1220.8 + 31.1 = 1251.9 m, still a fail. Reporting the bare mean as the demonstrated value would misrepresent the performance the AFM is about to promise.
2. The 1,310 m run is the obvious suspect. Should it be excluded, and on what basis? Give the statistical test you would apply and the decision rule you would actually use.
Read the worked answer
Test. Grubbs statistic G = |1310 − 1220.8| / 46.2 = 1.93. The two-sided 5% critical value for n = 6 is 1.82, so the run is flagged as an outlier.
But a flag is not a justification. On a compliance test, deleting a point changes the demonstrated value, so exclusion has to rest on a documented assignable cause identified independently of the answer it produces. Here the only distinguishing feature is that a second pilot flew it - plausible (rotation rate, technique, acceleration onset) but not established.
Rule. Debrief the pilot, pull the run time history (rotation speed, pitch rate), and look for a procedural cause. If one is found and recorded, exclude with justification: the five remaining runs give mean 1203 m, s = 16.8 m, bound = 1203 + 2.132 × (16.8/√5) = 1219 m, which passes. If no cause is found, retain the point and add runs.
Warning. Excluding the worst point because it fails the requirement, with no cause, is data manipulation. It would also hide genuine pilot-to-pilot scatter that belongs in the AFM data.
3. Assuming the mean and standard deviation stay as they are, how many additional runs are needed for the 95% one-sided bound to fall under 1,250 m? State the caveats that belong in the test plan.
Read the worked answer
Requirement: 1220.8 + t × 46.2/√n ≤ 1250, i.e. t × 46.2/√n ≤ 29.2 m.
- n = 7: t = 1.943 → 1.943 × 17.46 = 33.9 → 1254.7 m, fails.
- n = 8: t = 1.895 → 1.895 × 16.34 = 30.9 → 1251.7 m, fails.
- n = 9: t = 1.860 → 1.860 × 15.40 = 28.6 → 1249.4 m, passes.
So three more runs, nine in total. (A normal-approximation shortcut would wrongly suggest only one more - the widening t quantile for small samples matters at this margin.)
Caveats for the plan: the new runs change both the mean and s, so nine is a projection, not a guarantee - plan nine and stop when the bound is met; if the extra runs confirm the 1,310 m level of scatter, s rises and more than nine will be needed. The cheaper lever is variance reduction: a single dedicated test pilot, tighter mass and CG control, and early-morning still-air slots. Halving s would let the existing six runs meet the bound.