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nadya68 [22]
2 years ago
7

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
Mathematics
1 answer:
Alex777 [14]2 years ago
4 0

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

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Keywords: Coordinate geometry, scaling

Learn more about scaling at:

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