<第一題> A certain change in a manufacturing procedure for componet parts is being considered. Samples are taken using both the4 existing and the new procedure in order to determin if the newprocedure results in an improvement. If 75 of 1500 items from the existing procedure were found to be defective and 80 of 2000 items from the new procedure were found to be defective, find a 90% confidence interval for the true difference in the fraction of defectives between the existin and the new process. (九三台科工工) 我知道C.I沒問題是[(0.05-0.04)+-Z √(0.05*0.95/1500+0.04*0.96/2000)] 0.05 =[-0.00163,0.0216] 結果是non-reject 我的問題是為什麼我想用下面的假設檢定去算 結果卻是reject呢? 這是我的算式 ------------- H0:P2-P1>0 H1:P2-P1<0 C:{Z>Z =-1.28} 0.1 z=(0.04-0.05)/√0.04429*0.95571*(1/1500+1/2000) =-1.42 ------------- 為什麼兩種檢定法的結果得出來的是不一樣的呢 這困擾了我很久 不知道有沒有人能幫我看一下呢 謝謝 <第二題> Yearly demand on a supplier for a given item has been 10 in the past, and demands seem to be distributed as a negative binomial random variable with parameter r=2 and p=0.2 .An inventory manager stocks 8 items for the coming year. What is the probability that he will not be able to supply the demand this year? (九一清大工工) 我知道答案是P(X>8)=1-P(X≦8) 8 2 x-2 =1-∫ (x-1)0.2 0.8 2 2-1 =0.5033 我的問題是我看不懂題目的英文是什麼@@ 為什麼需求可以用負二項來表示呢 我的想法是他跟supplier要了x個item 而有可能有的失敗 直到第二個item成功的才可以 所以第一行的10跟本也不用考慮 是這樣嗎 <第三題> 抱歉 問題有點多 http://0rz.tw/4e25g 第三題的第三小題 解答寫可以 但是兩者的1-α不是不同嗎 為什麼假設檢定能直接用上面的C.I法呢 若要可以的話 第二小題是不是要改用5% level of significance呢 謝謝大家~ --

※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 218.170.97.95 ※ 編輯: qmorewhy 來自: 218.170.97.95 (11/14 22:49)