For example compute a variance for the sample of 10 coin flips: 4 Tails (0) and 6 Heads (1) (estimated mean p=0.6):
octave-3.2.4.exe>(4*(0-0.6)**2 + 6*(1-0.6)**2)/(10-1)This calculator really works for me. Sometimes I have a Python window and it works just as well:
ans = 0.15360
>>> (4*(0-0.6)**2 + 6*(1-0.6)**2)/(10-1)Having said that Octave beats Python in easiness when calculating with vectors (or time series or sequences or anything that can be represented as a vector). Let's suppose we test if certain coin is not loaded (is fair) by flipping it 14 times. We would like to be 95% certain that coin is fair (i.e. p=0.5 which is equivalent to two-tailed test). Suppose that 14 flips resulted in only 3 heads. First, we build a critical interval - the number of tails that would result in rejecting coin fairness given number of heads:
critical_interval = binopdf(0:14,14,0.5) > 0.025 | binocdf(0:14,14,0.5)- binopdf(0:14,14,0.5) > 0.975critical_interval is a Boolean vector where i-th element corresponds to (i-1) number of tails: if it's true then with 95% certainty it is a fair coin. This expression is a logical OR of 2 expressions: first for left tail and second for right tail. Octave seamlessly handles any vectors just as if it were a number: I can change this to 1000 flips with minimum keystrokes.
Thus, given number of tails 3 we get
octave-3.2.4.exe> critical_interval(3+1)We use (3+1) as 1st element corresponds to 0 tails. Hence, we accept the fact that our coin is fair with 95% certainty because 3 heads do not belong to the critical interval. Similarly we have to reject this coin as loaded if number of tails happens to be 12:
ans = 0
ans = 1
I leave Python example as an exercise to a reader but I am certain that result won't be close to Octave in neither transparency nor conciseness. The Octave solution does leave me with the right to keep claiming this use similar to a calculator - not a programming.