Find a 95 confidence interval for \(\displaystyle\theta\)

Jazmyn Holden

Jazmyn Holden

Answered question

2022-04-04

Find a 95 confidence interval for θ based on inverting the test statistic statistic θ^.
For our data we have YiN(θxi,1) for i=1,,n.
Therefore it can be proven that the MLE for θ is given by
θ^=xiYixi2
To find the confidence interval I should invert the test statistic θ^.
The most powerful unbiased size α=0,05 test for testing
H0:μ=μ0 vs. H1:μμ0
where X1,,Xn  iid  n(μ,σ2) has acceptance region
A(μ0)=x:|x¯-μ0|1,96σ/n.
Substituting my problem (I think) we get that the most powerful unbiased size α=0,05 test for testing
H0:θ=θ^ vs. H1:θθ^
has acceptance region {A(θ^)={y:|yθ^|1,96n}
or equivalently, A(θ^)=y:ny¯-1,96nxiθ^ny¯+1,96nxi
Substituting θ^=xiYixi2 we obtain
A(θ^)=y:ny¯-1,96nxiΣxiYiΣxi2ny¯+1,96nxi
This means that my 10,05=0,95(95%) confindence interval is defined to be
C(y)={θ^:yA(θ^)}
But I can't seem to find anything concrete and I feel that I've made mistakes somewhere. What to do?

Answer & Explanation

Malia Booth

Malia Booth

Beginner2022-04-05Added 16 answers

You have θ^=ixiYiixi2.
Although non-linear in (x1,,xn), this is linear in (Y1,,Yn), and that's why it's a "linear" model (it's not because one is fitting a straight line; if that were true then least-squares fitting of polynomials would be considered non-linear regression, but it's linear regression). So you have a linear combination of independent normally distributed random variables, where the coefficients are constant (i.e. not random). Being a linear combination of independent normally distributed random variables with constant coefficients makes its distribution easy to find:
θ^~N(i(xiE(Yi))ixi2,i(xi2var(Yi))(ixi2)2)=N(θ,1ixi2) since E(Yi)=xiθ and var(Yi)=1.
So Pr(θA1ixi2<θ^<θ+A1ixi2)=0.95
where A is such that Pr(A<Z<A)=0.95. Consequently
Pr(θ^A1ixi2<θ<θ^+A1ixi2)=0.95
and that is your confidence interval.

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