probability 2

👨🏻‍🏫 學術台數學

[匿名]

2011-10-04 6:15 am

The mean life of a log of eco-friendly batteries is 375 hours.
The process standard deviation is 100 hours.
A random sample of 64 batteries has a sample mean life of 350 hours.

(i) tell the null & alternative hypotheses.

(ii) at 0.05 level of significance, is there evidence showing the mean life is different from 375 hours?

(iii) set up a 95% confidence interval estimate of the population mean life of the batteries.

(iv) compare the outcomes of (ii) & (iii) & draw the conclusions.

p.s. Pls kindly explain. Tks for your time!

更新1:

To Herman Tam: (iv) pls compare. your answer doesn't make sense. tks!

更新2:

To the advisor Freda: 答題不求分數,情操高! 感謝意見唷!

回答 (2)

Herman Tam

2011-10-04 6:34 am

✔ 最佳答案

Part i
null hypothesis: mean life of the battery =375
alternative hypothesis: mean life of the battery not equal to 375

Part ii
Test statistic = (350-375)/100 = -0.25
Since the test statistic should belong to N(0, 1)
Z (0.025) = 1.96
Since -1.96< -0.25 <1.96
null hypothesis is not rejected.
No evidence shows that the mean life is different from 375hrs

Part (iii)
CI = {350-1.96*100 , 350+1.96*100}
= {104, 546}

Part (iv)
The random sample of battery is eco-friendly battery.

2011-10-05 00:37:06 補充：
Part (iv)
Since the Confidence Interval contains 375 hours and the null hypothesis (mean life=375 hours) is not rejected, we can draw the conclusion that it is believed the mean life of the battery is 375 hours.

參考: ME, ME

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用戶46949

2011-10-05 11:46 pm

test statistic = 350-375/ (100/ √64 ) =-25/12.5 =-2
z(0.025)=1.96
since -2 <-1.96
this showing the mean life is different from 375 hours at 0.05 level of signifcance

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