Question

A random sample of n = 15 heat pumps of a certain type yielded the following observations on lifetime (in years):

2.0	1.4	6.0	1.6	5.1	0.4	1.0	5.3
15.7	0.7	4.8	0.9	12.3	5.3	0.6

Assume that the lifetime distribution is exponential and use an argument parallel to that of this example to obtain a 95% CI for expected (true average) lifetime. (Round your answers to two decimal places.)

( ,    ) years

What is a 95% CI for the standard deviation of the lifetime distribution? [Hint: What is the standard deviation of an exponential random variable?] (Round your answers to two decimal places.)

( , ) years

Answer 1

Given
X bar	4.00625(AVERAGE())
S	4.431323(STDEV())
n	16

a)

95% CI (Average)

α=	0.05
df	15	n-1

CI
tc	2.13145	T.INV.2T(alpha,df)
Upper	6.367535	X bar + tc*(S/SQRT(n))
Lower	1.644965	X bar - tc*(S/SQRT(n))

95% CI = (1.65, 6.37)

b)

95% CI (Standard deviation)

CI (Variance) = ((n-1)S^2/X^2 Upper, (n-1)S^2/X^2 Lower)

Mean	4.00625	Critical values:
Sd	4.431323166	X^2 Upper	27.48839	CHISQ.INV(1-(0.05/2),15)
Var	19.636625	X^2 Lower	6.262138	CHISQ.INV(0.05/2,15)
n	16
CI	Variance
Upper	47.03655279	(n-1)S^2/X^2 Lower
Lower	10.7154091	(n-1)S^2/X^2 Upper
CI	Standard deviation
Upper	6.858319969	SQRT(Var upper limit)
Lower	3.273439949	SQRT(Var Lower limit)

95% CI = (3.27, 6.86)