Question

The following data represent the battery life, in hours, for a random sample of 5 full charges on a fifth
generation iPod music player. The data are normally distributed.

7.3 10.2 12.9 10.8 12.1
(a) Construct a 90% confidence interval for the mean number of hours that the battery will last on
this player.
(b) Construct a 90% confidence interval for the population standard deviation of the battery life.

Answer 1

10.6600	mean
2.1571	std. dev.

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 5- 1 ) = 2.132
10.66 ± t(0.1/2, 5 -1) * 2.1571/√(5)
Lower Limit = 10.66 - t(0.1/2, 5 -1) 2.1571/√(5)
Lower Limit = 8.6035
Upper Limit = 10.66 + t(0.1/2, 5 -1) 2.1571/√(5)
Upper Limit = 12.7165
90% Confidence interval is ( 8.6035 , 12.7165 )

Part b)

(0.1/2, 5 - 1 ) = 9.4877
(1 - 0.1/2, 5 - 1) ) = 0.7107
Lower Limit = 1.4006
Upper Limit = 5.1175
90% Confidence interval is ( 1.4006 , 5.1175 )
( 1.4006 < σ < 5.1175 )