Question

A random sample of n = 15 heat pumps of a certain type yielded the following observations on lifetime (in years):
2.0 1.2 6.0 1.9 5.1 0.4 1.0 5.3 15.6 0.9 4.8 0.9 12.2 5.3 0.6

a.) Assume that the lifetime distribution is exponential and use an argument parallel to that of this example to obtain a 96% CI for expected (true average) lifetime (round answer to two decimal places)

(________, _________) years

c.) What is a 95% CI for the standard deviation of the lifetime distribution?
(_______, _______) years

*You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

a.
TRADITIONAL METHOD
given that,
sample mean, x =4.2133
standard deviation, s =4.4668
sample size, n =15
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4.4668/ sqrt ( 15) )
= 1.15
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.04
from standard normal table, two tailed value of |t α/2| with n-1 = 14 d.f is 2.264
margin of error = 2.264 * 1.15
= 2.61
III.
CI = x ± margin of error
confidence interval = [ 4.2133 ± 2.61 ]
= [ 1.6 , 6.82 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =4.2133
standard deviation, s =4.4668
sample size, n =15
level of significance, α = 0.04
from standard normal table, two tailed value of |t α/2| with n-1 = 14 d.f is 2.264
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 4.2133 ± t a/2 ( 4.4668/ Sqrt ( 15) ]
= [ 4.2133-(2.264 * 1.15) , 4.2133+(2.264 * 1.15) ]
= [ 1.6 , 6.82 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 96% sure that the interval [ 1.6 , 6.82 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 96% of these intervals will contains the true population mean
c.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s = standard deviation
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since alpha =0.05
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.025 = 0.975
the two critical values ᴪ^2 left, ᴪ^2 right at 14 df are 26.1189 , 5.629
s.d( s )=4.4668
sample size(n)=15
confidence interval for σ^2= [ 14 * 19.9523/26.1189 < σ^2 < 14 * 19.9523/5.629 ]
= [ 279.3322/26.1189 < σ^2 < 279.3322/5.6287 ]
[ 10.6946 < σ^2 < 49.6264 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (10.6946) < σ < sqrt(49.6264), ]
= [ 3.2703 < σ < 7.0446 ]