How does jsPerf determine which of the code snippets is fastest?

I wrote Benchmark.js, which jsPerf uses.

1. "ops/sec" stands for operations per second. That is how many times a test is projected to execute in a second.

2. A test is repeatedly executed until it reaches the minimum time needed to get a percentage uncertainty for the measurement of less than or equal to 1%. The number of iterations will vary depending on the resolution of the environment’s timer and how many times a test can execute in the minimum run time. We collect completed test runs for 5 seconds (configurable), or at least 5 runs (also configurable), and then perform statistical analysis on the sample. So, a test may be repeated 100,000 times in 50 ms (the minimum run time for most environments), and then repeated 100 times more (5 seconds). A larger sample size (in this example, 100), leads to a smaller margin of error.

3. We base the decision of which test is faster on more than just ops/sec by also accounting for margin of error. For example, a test with a lower ops/sec but higher margin of error may be statistically indistinguishable from a test with higher ops/sec and lower margin of error.

   We used a welch t-test, similar to what SunSpider uses, but switched to an unpaired 2-sample t-test for equal variance (the variance is extremely small) because the welch t-test had problems comparing lower ops/sec and higher ops/sec with small variances which caused the degrees of freedom to be computed as less than 1. We also add a 5.5% allowance on tests with similar ops/sec because real world testing showed that identical tests can swing ~5% from test to re-test. T-tests are used to check that differences between tests are statistically significant.

You can read Bulletproof JavaScript benchmarks article from the authors. It uses Benchmark.js btw, which is Open Source.