I have been trying to solve the following exercise, but there is something unclear about it. Namely, it goes like this:
In the solutions it says: $X_i ∈ N(µ_i + µ_δ, \sqrt{σ^2 + σ^2_δ})$, and $Yi ∈ N(µ_i , σ)$. Furthermore, the paired difference is $Z_i = X_i − Y_i $ and corresponds to $N(µ_δ, \sqrt{2σ^2 + σ^2_δ})$. Now, let $\tilde{σ}= \sqrt{2σ^2 + σ^2_δ}$. We estimate $µ_δ$ with $\overline{z} = −0.774$, and $\tilde{σ}$ with $s_z=0.9546$.
My question: How did they calculate that $s_z=0.9546$? I tried calculating first $$Var\left(\frac{X}{n_1}-\frac{Y}{n_2}\right)=\frac{1}{n_1^2}\cdot Var(X)+\frac{1}{n_2^2}\cdot Var(Y)= \frac{1}{n_1^2} \cdot \sum_{i=0}^n (x_i-\overline{x})^2 + \frac{1}{n_2^2} \cdot \sum_{i=0}^n (y_i-\overline{y})^2$$ but this led me to a wrong solution. Why cannot I calculate the variance in this way?
The unbiased estimate for the sample variance is $s_z^2=\frac{n}{n-1}(\bar{z^2}-\bar{z}^2)$. Substituting the values in the table yields $s_z=0.954557$.