[商管] [統計]假設檢定

看板Grad-ProbAsk作者 (你怎麼捨的我難過)時間14年前 (2011/12/13 23:21), 編輯推噓3(3推 0噓 13→)
留言16則, 3人參與, 最新討論串1/1
執法機構欲了解超速違規之比例，隨機抽取625張罰單， 若有286至339張罰單屬於超速，則稱50%之違規為超速， (1)假設50%之違規為超速，則犯型I錯誤之機率。 (2)假設60%之違規為超速，則犯型II錯誤之機率。 不知道該如何下手?請高手指點,謝謝 -- -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 218.174.200.229
推
12/13 23:41, , 1F
我猜你的則都要改成求
12/13 23:41, 1F
→
12/13 23:44, , 2F
(1) 題意蠻不清的 但應該是這樣算 X~Bin(n=625,p)~N(625p,
12/13 23:44, 2F
→
12/13 23:45, , 3F
625p(1-p) ) -> P(Type I error)=P(286≦X≦339|p=0.5)
12/13 23:45, 3F
→
12/13 23:47, , 4F
=P(286.5<X<339.5|p=0.5)≒P((286-312.5)/√156.25 ≦Z
12/13 23:47, 4F
→
12/13 23:48, , 5F
≦(339.5-312.5)/√156.25 )=...
12/13 23:48, 5F
→
12/13 23:49, , 6F
(2) P(Type II error)=P(X<286,X>339|p=0.6)=P(X≧285.5,
12/13 23:49, 6F
→
12/13 23:49, , 7F
上一行推錯= =
12/13 23:49, 7F
→
12/13 23:50, , 8F
(2) P(Type II error)=P(X<286,X>339|p=0.6)=P(X≦287,
12/13 23:50, 8F
推
12/13 23:52, , 9F
X≧338|p=0.6)=P(X<287.5,X>338.5|p=0.6)
12/13 23:52, 9F
→
12/13 23:53, , 10F
≒P(Z<(287.5-375)/√150 , Z>(338.5-375)/√150)=...
12/13 23:53, 10F
→
12/14 00:52, , 11F
H0：p＝0.5 H1：p≠0.5
12/14 00:52, 11F
→
12/14 00:52, , 12F
(1)P(拒絕Ho|Ho真)=P(X<286,X>339|p=0.5)
12/14 00:52, 12F
→
12/14 00:52, , 13F
(2)P(不拒絕Ho|Ho偽)=P(286≦X≦339|p=0.6)
12/14 00:52, 13F
→
12/14 00:52, , 14F
我覺得這樣才對
12/14 00:52, 14F
推
12/14 01:36, , 15F
是阿 推文太亂了 我下次還是回好=.=
12/14 01:36, 15F
→
09/11 14:40, , 16F
(1) 題意蠻不清的 https://daxiv.com
09/11 14:40, 16F
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