Answer:
The correct answer is "[tex]2.633< \sigma < 4.480[/tex]".
Step-by-step explanation:
Given:
n = 21
s = 3.3
c = 0.9
now,
[tex]df = n-1[/tex]
[tex]=20[/tex]
⇒ [tex]x^2_{\frac{\alpha}{2}, n-1 }[/tex] = [tex]x^2_{\frac{0.9}{2}, 21-1 }[/tex]
= [tex]31.410[/tex]
⇒ [tex]x^2_{1-\frac{\alpha}{2}, n-1 }[/tex] = [tex]10.851[/tex]
hence,
The 90% Confidence interval will be:
= [tex]\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 }}[/tex]
= [tex]\sqrt{\frac{(21-1)3.3^2}{31.410} } < \sigma < \sqrt{\frac{(21.1)3.3^2}{10.851} }[/tex]
= [tex]\sqrt{\frac{20\times 3.3^2}{31.410} } < \sigma < \sqrt{\frac{20\times 3.3^2}{10.851} }[/tex]
= [tex]2.633< \sigma < 4.480[/tex]
