Answer to Question #256019 in Statistics and Probability for Dada

Question

Let x₁,x₂,..., Xn be a random sample from N(μ,sigma²), where u is known. Construct a (1-a) * 100% Confidence Interval for sigma².

Answer 1

For building confidence interval we will determine next random value:

$H = {\frac{(x{\scriptscriptstyle1}-u)^{2}+...+(x{\scriptscriptstyle n}-u)^{2}} {σ^{2}}}$ ~ $Xi^{2}(n)$, where u is μ(mean value)

Let ${(x{\scriptscriptstyle1}-u)^{2}+...+(x{\scriptscriptstyle n}-u)^{2}} = sum$

Since we have 1 - a confidence interval, then we have to find such m and n, that

$P(m<H<n)=1-a$

Technically, we can pick up any $m = Xi^{2}(k{\scriptscriptstyle 1},n), n = Xi^{2}(1-k{\scriptscriptstyle 2},n)$ , where $k{\scriptscriptstyle 1}+k{\scriptscriptstyle 2}=a$

For example, we can put $k{\scriptscriptstyle 1}={\frac a 2}, k{\scriptscriptstyle 2}=1-{\frac a 2}$ . Now we have:

$P(Xi^{2}({\frac a 2},n)<H<Xi^{2}(1-{\frac a 2},n))=1-a$

$P(Xi^{2}({\frac a 2},n)<{\frac{sum} {σ^{2}}} <Xi^{2}(1-{\frac a 2},n))=1-a$

$P({\frac {sum} {Xi^{2}({\frac a 2},n)}}<σ^{2} <{\frac {sum} {Xi^{2}(1-{\frac a 2},n)}})=1-a$

The 1-a confidence interval is for $σ^{2}$ is $({\frac {sum} {Xi^{2}({\frac a 2},n)}},{\frac {sum} {Xi^{2}(1-{\frac a 2},n)}})$