Let \(\displaystyle{X}_{{{1}}},\ \cdots,{X}_{{{n}}}\) be a random sample

Deegan Chase

Deegan Chase

Answered question

2022-03-23

Let X1, ,Xn be a random sample from a normal distribution with known mean μ and unknown variance σ2. Three possible confidence intervals for σ2 are
a) (i=1n(XiX)2a1, i=1n(XiX)2a2)
b) (i=1n(Xiμ)2b1, i=1n(Xiμ)2b2)
c) (n(Xμ)2c1, n(Xμ)2c2)
where a1, a2, b1, b2, c1, c2 are constants.
Find values of these six constants which give confidence level 0.90 for each of the three intervals when n=10 and compare the expected widths of the tree intervels in this case
With σ2=1, what value of n is required to achieve a 90% confidence interval of expected width less than 1 in cases (b) and (c) above?

Answer & Explanation

Leonardo Mcpherson

Leonardo Mcpherson

Beginner2022-03-24Added 13 answers

Step 1
X1,,Xn are i.i.d. N(μ,σ2) random variables and
X=n1i=1nXi The distributions and confidence intervals are as follow.
a) 1σ2i=1n(XiX)2χn12
and
Prl(i=1n(XiX)2χn1,α22σ2i=1n(XiX)2χn1,1α22r)=1α.
b) 1σ2i=1n(Xiμ)2χn2
and
Prl(i=1n(Xiμ)2χn,α22σ2i=1n(Xiμ)2χn,1α22r)=1α.
c) n(Xμ)2σ2χ12
and
Prl(n(Xμ)2χ1,α22σ2n(Xμ)2χ1,1α22r)=1α.
I hope this helps.

kachnaemra

kachnaemra

Beginner2022-03-25Added 16 answers

Step 1
I believe you have the denominators switched. Also, say c1=χ{1,1α2}2 and c2=χ{1,α2}2 then for (c), the expected width of the interval is
E[n(Xμ)2(1c21c1)]=
n(1c21c1)×E(Xμ)2=
n(1c21c1);;×σ2n=(1c21c1)σ2
since nσ2(Xμ)χ12
and thus E(Xμ)2=σ2n
strange the expected width does not depend on n.

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