I'm not sure this approach is correct. You cannot say that Y = X^86,381 because toss successions are correlated.
If we consider Enuma example and take the particular result HTH then knowing that HT is not 2 heads entails that TH is not two heads. We should introduce some kind of correlation...
Hello,
you want to predict X_{n+1} from X_{1}...X_{n} without knowing \mu and \sigma.
So, as you said,
1. the estimated variance of \bar{X}_{n} is S_{n}^{2}/n
2. the estimated variance of X_{i} is S_{n}
3. as a result the estimated variance of X_{n + 1} - \bar{X}_{n} is S_{n}^2 (1 + 1/n)...
If you know exactly the pdf (probability density function) f(x), the formula for the mean is
\mu = E[x] = \int x f(x) dx
and for the variance
\sigma^{2} = E[(x - E[x])^{2}] = \int (x - E[x])^{2} f(x) dx
If you only have experimental data, you can estimate the mean and variance of the...
But you do not know the exact frequency of the old technique. So I would suggest to tests :
H0 : there is no difference between the two techniques
As you observe a large number of events, you can assume gaussian distribution and use a Pearson chi2 test like :
chi2 = (250 - 175)^2/(250 +...