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Given the following values of x, s, and n, form a 90% confidence interval forσ2

a. x=21,s=2.5,n=50

b. x=1.3,s=0.02,n=15

c. x=167,s=31,n=22

d.x=9.4,s=1.5,n=5


Short Answer

Expert verified

a. 4.536≤σ2≤8.809

b. 0.00023644≤σ2≤0.00085228

c. 617.713228≤σ2≤1741.04716

d.0.9485936≤σ2≤12.663197

Step by step solution

01

Forming a 90% confidence interval for σ2 in part a.

We know that, for a 90% confidence interval that is for a 0.90 confidence coefficient,

α=0.10andα2=0.05.So,(1-α2)=0.95.

The formula for forming a confidence interval is given by,

(n-1)s2χα22≤σ2≤(n-1)s2χ(1-α2)2............(i)

We will find the values of χα22and χ(1-α2)2from table IV, Appendix D of the book as per the degrees of freedom which is calculated as(n-1).

a. Substituting the given values x=21,s=2.5,n=50in equation (i), we get the required interval as,

(50-1)(2.5)267.5048≤σ2≤(50-1)(2.5)234.7642306.2567.5048≤σ2≤306.2534.76424.536≤σ2≤8.809

02

Forming a 90% confidence interval for σ2 in part b.

b. Substituting the given values x=1.3,s=0.02,n=15 in equation (i), we get the required interval as,


role="math" localid="1661407112611" (15-1)(0.02)223.6848≤σ2≤(15-1)(0.02)26.570630.005623.6848≤σ2≤0.00566.570630.00023644≤σ2≤0.00085228

03

Forming a 90% confidence interval for σ2 in part c.

c. Substituting the given values x=167,s=31,n=22in equation (i), we get the required interval as,

role="math" localid="1661407085643" (22-1)(31)232.6705≤σ2≤(22-1)(31)211.59132018132.6705≤σ2≤2018111.5913617.713228≤σ2≤1741.04716

04

Forming a 90% confidence interval for σ2 in part d.

d. Substituting the given values x=9.4,s=1.5,n=5in equation (i), we get the required interval as,

(5-1)(1.5)29.48773≤σ2≤(5-1)(1.5)20.71072199.48773≤σ2≤90.7107210.9485936≤σ2≤12.663197

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