Question

Construct the confidence interval for the population standard deviation for the given values. Round your answers to one decimal place. n=21 , s=3.3, and c=0.9

Answer 1

Answer:

The correct answer is "$2.633< \sigma < 4.480$".

Step-by-step explanation:

Given:

n = 21

s = 3.3

c = 0.9

now,

$df = n-1$

    $=20$

⇒ $x^2_{\frac{\alpha}{2}, n-1 }$ = $x^2_{\frac{0.9}{2}, 21-1 }$

                  = $31.410$

⇒ $x^2_{1-\frac{\alpha}{2}, n-1 }$ = $10.851$

hence,

The 90% Confidence interval will be:

= $\sqrt{\frac{(n-1)s^2}{x^2_{\frac{\alpha}{2}, n-1 }} } < \sigma < \sqrt{\frac{(n-1)s^2}{x^2_{1-\frac{\alpha}{2}, n-1 }}$

= $\sqrt{\frac{(21-1)3.3^2}{31.410} } < \sigma < \sqrt{\frac{(21.1)3.3^2}{10.851} }$

= $\sqrt{\frac{20\times 3.3^2}{31.410} } < \sigma < \sqrt{\frac{20\times 3.3^2}{10.851} }$

= $2.633< \sigma < 4.480$