I have a few (e.g. 10) observations of travel times for a route.

A commuter needs to arrive to his destination before t0 (given).

I want to say: The commuter should depart before `t0 - x`

to arrive to his destination at `t0`

with probability 95%.

Can I do this just by calculating the sample mean (m) and standard deviation (s) and say that

```
x = m + 2*s
```

(the 2 is an approximation of the 2 sigma)

? Thanks.

## Best Answer

Assuming that the travel time is close enough to normally-distributed, you are on the right track. However, there are two issues that you have not got quite right.

First, your description makes it sound like a one-tailed interval is what you need (95% probability of arriving before the cutoff time). Your calculation is an approximation of a 95% two-tailed interval.

Second, your calculation x = m * 2s for a confidence interval of the mean assumes that the standard deviation is known (i.e. s = sigma) or is accurate (i.e. the number of observations is large enough that s can be assumed to be very close to sigma). In reality there is almost never a good reason to use that approach instead of the one that correctly assumes that s is an observed estimate of sigma. Therefore, use the calculation of t * s instead of 2 * s, where t is the value from Student's t-distribution relevant to the experiment. For your experiment use a one-tailed t value (0.05 area in the tail) for 9 degrees of freedom, 1.83.

Thus you need to start traveling at time t0 - 1.83 * s to have a 95% probability of arriving before time t0.