統計學, 高手請進!! Thanks!!

2012-05-11 6:58 pm

統計學功課, 唔係甘識做, 請各位幫幫忙, Thanks!!!!

Metal castings are manufactured on two production lines, A and B. Their masses are distributed normally with μA andμB and standard deviations 0.18kg and 0.21 kg respectively. A random sample of 8 casting from production line A has mean mass 5.43kg and a random sample of 12 casting from production line B has means 5.58kg . 問題 : (a) Calculate a 98% confidence interval for μA andμB.(b) State, giving a reason, whether there is a difference between the means at 2% significance level.(c) 呢題點解用z-test而唔係用t-test 去計呢..? 答(a) & (b) 時, 應該注意咩地方? 以咩方法去計呢條數呢..? Thanks!!!! ^ u ^

更新1:

唔該晒你丫, 解釋得好詳盡, thanks!~~ ^u^

回答 (1)

用戶25975

2012-05-11 8:02 pm

✔ 最佳答案

a) μA follows N(5.43, 0.18/8) = N(5.43, 0.0225)

Hence standard dev. of sample from A = √0.0225 = 0.15, and the 98% confidence interval of μA is 5.43 +/- 2.33 x 0.15 = 5.43 +/- 0.3495

μB follows N(5.58, 0.21/12) = N(5.58, 0.0175)

Hence standard dev. of sample from A = √0.0175 = 0.132, and the 98% confidence interval of μA is 5.58 +/- 2.33 x 0.132 = 5.58 +/- 0.308

b) H0: μA = μB

H1: μA =/= μB

Now μA - μB follows N(5.43 - 5.58, 0.0225 + 0.0175) = N(-0.15, 0.04)

Standard dev. of μA - μB = √0.04 = 0.2

Hence z-test result is -0.15/0.2 = -0.75 > -2.33

Thus H0 is accepted under 2% sig. level.

c) Pls. refer to:

http://hk.knowledge.yahoo.com/question/question?qid=7007040301627

參考: 原創答案

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