Question

Use the given information to find the number of degrees of​ freedom, the critical values chi Subscript Upper L Superscript 2χ2L and chi Subscript Upper R Superscript 2χ2R​, and the confidence interval estimate of sigmaσ. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Nicotine in menthol cigarettes 95​% ​confidence; nequals=30​, sequals=0.27 mg.

Answer 1

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 29 df are 45.7223 , 16.047
s.d( s )=0.27
sample size(n)=30
confidence interval for σ^2= [ 29 * 0.0729/45.7223 < σ^2 < 29 * 0.0729/16.047 ]
= [ 2.1141/45.7223 < σ^2 < 2.1141/16.0471 ]
[ 0.0462 < σ^2 < 0.1317 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (0.0462) < σ < sqrt(0.1317), ]
= [ 0.215 < σ < 0.363 ]