The following data represents the battery life in hours for

The following data represents the battery life, in hours, for a random sample of 10 fully charged batteries on a 5^th generation iPod. Assume battery life is normally distributed. Battery Life (hrs): 7.3, 6.6, 10.2, 10.2, 12.9, 9.0, 10.8, 8.5, 12.1, 7.1

a.      Calculate a 90% CI for

Solution

(a) sample mean=9.47

sample standard deviation =2.140639

The degree of freedom =n-1=10-1=9

Given a=1-0.9=0.1, t(0.05, df=9) =1.83 (from student t table)

So the lower bound is

xbar -t*s/vn=9.47-1.83*2.140639/sqrt(10) =8.231219

So the upper bound is

xbar + t*s/vn=9.47+1.83*2.140639/sqrt(10)=10.70878

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(b) We have 90% confident that the population mean will be within this interval.

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(c) No, it is not reasonable to conclude that