Question

The mean replacement time for a random sample of 15 washing machine is 10.7 years and the standard deviation is 2.5 years. Construct 98% confidence interval for the standard deviation of the replacement of all washing machines of this type. What would be the value if the standard deviation is 2.7 years.

Answer 1

a.
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 =0.02
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.98)/2 = 0.02/2 = 0.01
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.01 = 0.99
the two critical values ᴪ^2 left, ᴪ^2 right at 14 df are 29.1412 , 4.66
s.d( s )=2.5
sample size(n)=15
confidence interval for σ^2= [ 14 * 6.25/29.1412 < σ^2 < 14 * 6.25/4.66 ]
= [ 87.5/29.1412 < σ^2 < 87.5/4.6604 ]
[ 3.0026 < σ^2 < 18.7752 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (3.0026) < σ < sqrt(18.7752), ]
= [ 1.7328 < σ < 4.333 ]

b.
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 =0.02
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.98)/2 = 0.02/2 = 0.01
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.01 = 0.99
the two critical values ᴪ^2 left, ᴪ^2 right at 14 df are 29.1412 , 4.66
s.d( s )=2.7
sample size(n)=15
confidence interval for σ^2= [ 14 * 7.29/29.1412 < σ^2 < 14 * 7.29/4.66 ]
= [ 102.06/29.1412 < σ^2 < 102.06/4.6604 ]
[ 3.5023 < σ^2 < 21.8994 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (3.5023) < σ < sqrt(21.8994), ]
= [ 1.8714 < σ < 4.6797 ]