confidence intervalstatistical significance
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.
Assuming your confidence interval is for the mean, you can work backwards from the formula for the confidence interval margin of error: $$MOE=\frac{SD}{\sqrt{n}}*t_{crit}(C,n-1)$$ And knowing from this example that $MOE=115-100$, $SD=45$, and $n=36$, we can fill in the following to solve for $C$: $$15=\frac{45}{\sqrt{36}}*t_{crit}(35,C)$$ $$t_{crit}(35,C)=2$$ Then we can use a critical $t$ table or calculator to see what $C$ level corresponds to 2.00 for 35 degrees of freedom.
Here, $C=95$% or $\alpha=.05$ for two tailed tests
qnorm is the quantile function for the normal distribution. More details are available by typing ?qnorm. You pick 0.975 to get a two-sided confidence interval. This gives 2.5% of the probability in the upper tail and 2.5% in the lower tail, as in the picture.
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.