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 80​% ​confidence; n = 25​, s = 0.27 mg.

df = (Type a whole​ number.)

χ2L = ​(Round to three decimal places as​ needed.)

χ2R = ​(Round to three decimal places as​ needed.)

The confidence interval estimate of σ is __ mg < σ < __ mg. ​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

s = 0.27

s² = 0.0729

Degrees of freedom = df = n - 1 = 24

$\chi$²_L = $\chi$ ²_{$\alpha$/2,df} = 33.196

$\chi$²_R = $\chi$ ²_{1 - $\alpha$ /2,df} = 15.659

The 80% confidence interval for $\sigma$ is,

$\sqrt$(n - 1)s² / $\chi$ ²_$\alpha$/2 < $\sigma$ < $\sqrt$ (n - 1)s² / $\chi$ ²_{1 - $\alpha$ /2}

  $\sqrt$24 * 0.0729 / 33.196 < $\sigma$ < $\sqrt$ 24 * 0.0729 / 15.659

0.23 mg < $\sigma$ < 0.33 mg

(0.23 , 0.33)