Basically, I have that $\ Y_i = \alpha +\beta(x_i-x_{bar}) + \epsilon_i $ where $\epsilon_i$ are i.i.d normally distributed with mean 0 variance $\sigma^2$ $\ Y_i ~~has ~a~normal~distribution~as ~follows~; N(\alpha + \beta(x_i-x_{bar}),\sigma^2 )$ Where the $\ x_i $ are just given numbers $\ \alpha^{hat}$ has distribution $\ N(\alpha,\sigma^2/n)$ AND $\beta^{hat}$ has a distribution $\ N(\beta,\sigma^2/\Sigma x_i^2)$
Let Z=AY where A an orthogonal matrix.
Show what $\Sigma (Y_i -\alpha_{hat} -\beta_{hat}(x_i))^2 ~~~{i=1...n} ~~~~= ~~\Sigma Z_i^2 ~~~~{i=3...n} $ and show then that $\Sigma Z_i^2 ~~~~{i=3...n} $ has a chi sqared distribution with n-2 degrees of freedom.
The thing is, I can do this up until the last part, what I dont understand is the degrees of freedom, I can see by showing this equation that $\Sigma Z_i^2 ~~~~{i=3...n} $ is a sum of squared standard normals, namely each of the terms $\ (Y_i -\alpha_{hat} -\beta_{hat}(x_i))^2$ are standard normals, but if the left = the right,. then isnt it the sum of n squared standard normals not n-2?!?!
Really confused by this.
Thanks