# Dan's Segment 24

## To Calculate

1. We know that the sum of N normal deviates is normally distributed itself, so we can easily calculate the mean and variance of X.

$<X> = \sum_{k=1}^{20} \alpha_k <T_k> = 0$

And

$Var(x) = \sum_{k=1}^{20} \alpha_k^2 Var(T_k) = \sum_{k=1}^{20} \alpha_k^2$

So we can easily see that X is $N(0,\sum_{k=1}^{20} \alpha_k^2)$distributed.

This means the t-value is $t = \frac{X}{\sqrt{\sum_{k=1}^{20} \alpha_k^2}}$ which follows the N(0,1) distributions t-value. Thus, we can see that $t^2$ is chisquare(1) distributed.