作者loveliver (糗了糗了)
站内Statistics
标题Re: [问题] F分配求信赖区间的上界
时间Sun Feb 15 00:10:48 2009
※ 引述《ltisld ()》之铭言：
: 是研究所考古题
: 和室友两人研究许久後仍不得其法
: 希望有先进能指点一二^^
: 93台大商研
: Let X and Y equal the concentration in parts per billion of chromium in the
: blood for healthy persons and for persons with a suspected disease, repectively.
: Assume that the diftribution of X and Y are N(μ_x, σ^2_x) and
: N(μ_y, σ^2_y). Using n=8 observationsof X: 15, 23, 12, 18, 9, 28, 11, 10
: and m=10 observations of Y:25, 20, 35, 15, 40, 16, 10, 22, 18, 32.
Find a
one sided 95% condifence interval that is an
upper bound for
σ^2_x/σ^2_y
由於是要算两母体变异数比之CI,使用的是F分配
所以CI的下界一定是0
故只要算上界就好
由题知 P( F(0.95,7,9)<F<无穷大 )=0.95
则 Upper bound = (Sx^2/ Sy^2)*F(0.05,9,7)
故 σ^2_x/σ^2_y之单尾95%的CI为 (0,1.835) #
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也许我们都已经活得太自由 忘了该预习自惭形秽的本能......
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1F：推 ltisld:谢谢你喔^^ 02/15 20:26