Consider an i.i.d. sample X1,…,Xn∼Poiss(λ) for λ>0 .
Starting from the Central Limit Theorem, find a confidence interval I=[A,B] with asymptotic level 1−α that is centered about Xn using the plug-in method.
now consider the following hypothesis with a fixed number λ0>0 :
H0:λ=λ0vsH1:λ≠λ0.
Define a test for the above hypotheses with asymptotic level α , and rewrite it in the form
ψ=1{λ0∉J},
for some interval J=[C,D] .
( Write barX_n for Xn . If applicable, type abs(x) for |x| , Phi(x) for Φ(x)=P(Z≤x) where Z∼N(0,1) , and q(alpha) for qα , the 1−α quantile of a standard normal variable. )
C=?
D=?
Calm down, I'll try to help you with that. A clown's thoughts on statistical inference can be quite amusing! Let's start with the confidence interval using the plug-in method.
According to the Central Limit Theorem, as n (the sample size) goes to infinity, the distribution of the sample mean follows a normal distribution. For a Poisson distribution with a mean λ, the sample mean can be approximated by a normal distribution with mean λ and variance λ/n.
To find a confidence interval, we need to estimate λ. The plug-in method suggests substituting the estimated value of λ, denoted by 𝑎̅𝑋_𝑛, into the formula. So, our estimated value for λ is just the sample mean.
Now, for a confidence level of 1-α, we need to find the appropriate z value such that the area between -z and z under the standard normal curve is 1-α. The z value is obtained from q(α/2).
Finally, the confidence interval is defined as [𝑋̅_𝑛−𝑧*(√(𝑋̅_𝑛/𝑛)), 𝑋̅_𝑛+𝑧*(√(𝑋̅_𝑛/𝑛))], where 𝑧 is the z value.
Now let's move on to the hypothesis test. The null hypothesis 𝐻₀ states that λ equals 𝜆₀, while the alternative hypothesis 𝐻₁ states that λ is not equal to 𝜆₀.
To set up the test, we need to construct a test statistic. In this case, the test statistic is (𝑋̅_𝑛−𝜆₀)/√(𝑋̅_𝑛/𝑛). This follows a standard normal distribution under the null hypothesis.
To find the rejection region for the test, we should reject the null hypothesis if the test statistic is too unlikely to occur by chance alone. We reject 𝐻₀ if the test statistic falls outside the critical region, which is determined by the interval [C, D].
Now, let's discuss C and D.
C represents the critical value on the left side of the interval. It can be found by calculating 𝜆₀−𝑧(𝛼/2)√𝜆₀/n, where 𝑧(𝛼/2) is the quantile corresponding to the significance level 𝛼/2.
D represents the critical value on the right side of the interval. It can be found by calculating 𝜆₀+𝑧(𝛼/2)√𝜆₀/n.
In summary, C is 𝜆₀−𝑧(𝛼/2)√𝜆₀/n, and D is 𝜆₀+𝑧(𝛼/2)√𝜆₀/n.
Remember, laughter is the best way to approach statistics!
To find the confidence interval using the plug-in method, we can start by applying the Central Limit Theorem to the sample mean. Since we have an i.i.d. sample X1, ..., Xn ∼ Poiss(λ) for λ > 0, the sample mean X̄n is approximately normally distributed with mean λ and variance λ/n.
Since we don't know the true value of λ, we can estimate it with the sample mean: λ̂ = X̄n. Thus, we can construct the confidence interval by plugging in the estimated value λ̂ for λ in the asymptotic formulas.
Using the central limit theorem, we have:
X̄n ~ N(λ, λ/n)
By standardizing the sample mean, we get:
(Z = (X̄n - λ) / sqrt(λ/n)) ~ N(0, 1)
To construct a confidence interval with asymptotic level 1-α, we need to find the critical values z_(α/2) and z_(1-α/2), which correspond to the (α/2)th and (1-α/2)th quantiles of the standard normal distribution.
The confidence interval can be written as:
I = [X̄n - z_(1-α/2) * sqrt(λ̂/n), X̄n + z_(1-α/2) * sqrt(λ̂/n)]
Now let's move to the hypothesis test. We are given the following hypotheses:
H0: λ = λ0
H1: λ ≠ λ0
To construct a test at level α, we can reject the null hypothesis if the true value λ0 falls outside a confidence interval for λ constructed using the plug-in method.
By substituting λ0 into the confidence interval formula, we have:
J = [X̄n - z_(1-α/2) * sqrt(λ0/n), X̄n + z_(1-α/2) * sqrt(λ0/n)]
The test can be written as:
ψ = 1{λ0 ∉ J}
Therefore,
C = X̄n - z_(1-α/2) * sqrt(λ0/n)
D = X̄n + z_(1-α/2) * sqrt(λ0/n)
To obtain a confidence interval using the plug-in method, we can start with the Central Limit Theorem (CLT) for the Poisson distribution.
According to the CLT, for a large sample size n, the sample mean Xn follows a normal distribution with mean λ and variance σ^2/n, where σ^2 is the variance of the Poisson distribution (which is also equal to λ).
Using this, we can approximate Xn as a normal distribution:
Z = (Xn - λ) / sqrt(λ/n) ~ N(0,1)
Now, for a confidence level of 1-α, we can find the corresponding quantiles of the standard normal distribution. The confidence interval is then given by:
I = [ Xn - z(α/2) * sqrt(Xn/n), Xn + z(α/2) * sqrt(Xn/n) ]
where z(α/2) is the (1-α/2) quantile of the standard normal distribution.
Therefore, for the given problem:
A = Xn - z(α/2) * sqrt(Xn/n)
B = Xn + z(α/2) * sqrt(Xn/n)
Now, moving on to the hypothesis test:
To test the hypothesis H0: λ = λ0 vs H1: λ ≠ λ0, we can use the likelihood ratio test (LRT) based on the Poisson distribution.
The test statistic for the LRT is given by:
-2 * log (L(λ0) / L(barX_n))
Here, L(λ0) represents the likelihood under the null hypothesis, and L(barX_n) represents the likelihood using the observed sample mean.
Under the null hypothesis H0, we have λ = λ0, so the likelihood can be written as:
L(λ0) = exp(-λ0) * (λ0)^(sum(Xi)) / (Product(Xi!))
Similarly, under the observed sample mean barX_n, the likelihood is:
L(barX_n) = exp(-barX_n) * (barX_n)^(sum(Xi)) / (Product(Xi!))
Taking the logarithm and simplifying, we get:
-2 * log (L(λ0) / L(barX_n)) = 2 * (λ0 - barX_n + sum(log(Xi!)))
To define the test, we need to compare this test statistic with the critical value from a chi-square distribution with 1 degree of freedom at level α.
Therefore, we can write the test as follows:
ψ = 1{2 * (λ0 - barX_n + sum(log(Xi!))) > q(α)}
where q(α) is the α quantile of the chi-square distribution with 1 degree of freedom.