Question

Show Work a Construct a 95% confidence interval for the population standard deviations of a random sample of 15 crates which have a mean weight of 165 2 pounds and a standard deviation of 11.6 pounds. Round to the nearest thousandth. Assume the population is normally distributed O.A. (8.493,18.294) O B. (8.918,16.932) OC (2.494,5.371) OD. (72.125,334.667) Jick to select your answer. Sample tests and quizzes can be taken for practice or to build your study plan. Sample tests and quizzes do not 31 A MacBook 10

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 14 df are 26.119 , 5.629
s.d( s )=11.6
sample size(n)=15
confidence interval for σ^2= [ 14 * 134.56/26.119 < σ^2 < 14 * 134.56/5.629 ]
= [ 1883.84/26.119 < σ^2 < 1883.84/5.629 ]
[ 72.125 < σ^2 < 334.667 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (72.125) < σ < sqrt(334.667), ]
= [ 8.493 < σ < 18.294 ]
option:A