<第一题> 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没问题是[(0.05-0.04)+-Z √(0.05*0.95/1500+0.04*0.96/2000)] 0.05 =[-0.00163,0.0216] 结果是non-reject 我的问题是为什麽我想用下面的假设检定去算 结果却是reject呢? 这是我的算式 ------------- 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 ------------- 为什麽两种检定法的结果得出来的是不一样的呢 这困扰了我很久 不知道有没有人能帮我看一下呢 谢谢 <第二题> 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 我的问题是我看不懂题目的英文是什麽@@ 为什麽需求可以用负二项来表示呢 我的想法是他跟supplier要了x个item 而有可能有的失败 直到第二个item成功的才可以 所以第一行的10跟本也不用考虑 是这样吗 <第三题> 抱歉 问题有点多 http://0rz.tw/4e25g 第三题的第三小题 解答写可以 但是两者的1-α不是不同吗 为什麽假设检定能直接用上面的C.I法呢 若要可以的话 第二小题是不是要改用5% level of significance呢 谢谢大家~ --

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