OpenSpecEngineering

Tells you how many cycles to time in a stopwatch study so that the mean cycle time is known to a set precision, at a set confidence. Time a few cycles first, enter their mean and standard deviation, and the page gives the total count you need. If you have already timed some cycles, it also gives how many more to take and how precise your mean is so far. Time study analysts use it before setting a standard time [1], [2].

Equations

n = (z · s / (k · x̄))²

x̄ is the mean of the preliminary readings and s their sample standard deviation, both in the same time unit (any unit works, since only their ratio s/x̄, the coefficient of variation, matters). k is the desired relative precision as a fraction: ±5 % is k = 0.05. z is the two-sided standard normal value for the confidence level: 1.96 for 95 %, 1.645 for 90 %. The page rounds n up to the next whole cycle. With N cycles already taken, the precision they give is

k_achieved = z · s / (√N · x̄)

Limits and common mistakes

• The page uses z, the normal value. With a small preliminary sample (under about 30 readings), Student’s t is larger than z, so the true count needed is somewhat higher [1]. Recompute once more readings are in.
• Precision is relative to the mean, not an absolute time. ±5 % on a 0.5 min cycle is ±0.025 min.
• Use the sample standard deviation (divide by N − 1) of single cycle times, not of averages.
• Work out n for each element of the job. The element with the most spread sets the count for the study.
• Throw out readings with a known cause, such as a dropped part, before you compute s. Don’t throw out readings just because they are long.

Worked example

An analyst has timed 10 cycles of a packing operation. The readings have a mean of 0.62 min and a standard deviation of 0.08 min. The plant wants the mean within ±5 % at 95 % confidence. How many more cycles must she time?

1. Enter Mean of Preliminary Readings 0.62 and Sample Standard Deviation 0.08.
2. Enter Confidence Level 95 and Desired Relative Precision 5.
3. Enter Observations Already Taken 10.
4. Press Solve. The Solution shows: z value for confidence level: 1.95996
   Coefficient of variation (s / x̄): 0.129032
   Required number of observations n: 26
   Additional observations still needed: 16
   Relative precision achieved with observations to date: 7.99736 ± %

Check by hand. 1.96 × 0.129 / 0.05 = 5.06, and 5.06² = 25.6, so 26 cycles. With 10 cycles, the precision is 1.96 × 0.129 / √10 = 8.0 %, short of the 5 % target. The count grows with the square of the precision: halving the target to ±2.5 % would need four times as many cycles.