curlxp
请高人指点一下,小弟在统计学的回归分析部分有点问题求教。
如果线性回归方程y=a+bx中,y(i)服从分布:N(a+b*x(i),σ^2).
y.hat(i)=a.hat+b.hat*x(i),
a.hat=y.bar-b.hat*x.bar,
b.hat=lxy/lxx,
回归平方和SR=∑(i=1...n)(y.hat-y.bar)^2
则如果假设H0:b=0成立情况下,SR/σ^2服从自由度为1的卡方分布。
为什么自由度是1?请高手指点下这是怎么算出来的?万分感谢~
curlxp
这问题是不是太弱智了以至于没人回答??
statax
你要看看卡方分布的定义。
定义!定义!定义!概念!概念!概念!
rtist
Well, it might need a bit more derivations beyond the definitions, although not too much. Although classical, the question is stated in a non-traditional way (at least, non-traditional to me).
I suggest to try it first by yourself, if you know what Chi-squared distribution is and what relationship it has with normal distribution. If you don't know, then, as statax suggested, get to know the definitions first.
lwzh0629
∑e=0, ∑xe=0,因受这两个条件的限制,所以残差平方和的自由度为(n-2)
longoR
[quote]引用第4楼lwzh0629于2007-10-12 05:50发表的“”:
∑e=0, ∑xe=0,因受这两个条件的限制,所以残差平方和的自由度为(n-2)[/quote]
恩,但是H0下两个限制其实是同一个限制。
而楼主的量是n-1与n-2之间的差值。