2 ms·
Let's say there are 6 trials -- name them x1, x2, x3, y1, y2, y3. In the case of checking if x is faster than y 3 times, you're doing x1 < y1 AND x2 < y2 AND x
by JoshuaDavid 8y ago
Let's say there are 6 trials -- name them x1, x2, x3, y1, y2, y3.
In the case of checking if x is faster than y 3 times, you're doing x1 < y1 AND x2 < y2 AND x3 < y3.
In the case of checking if all three measurements of x are smaller than all three measurements of y, you're checking x1 < y1 AND x1 < y2 AND x1 < y3 AND x2 < y1 AND x2 < y2 AND x2 < y3 AND x3 < y1 AND x3 < y2 AND x3 < y3.
In other words, the latter case is checking whether the slowest x of three trials is faster than the fastest y of three trials.
- comboy 8y agoYou're right. I actually edited before you replied. Should have deleted it or preferably think more before typing. But now it's forever.
- Waterluvian 8y agoI'm glad you kept it. The explanation was enlightening to me.