Statistical Inference II J Lee Assignment 4 Problem 1 249309
Statistical Inference Ii J Lee Assignment 4problem 1 Exercise 12 I
Derive a likelihood ratio test of H0 : θ = θ0 versus HA : θ 6= θ0, and show that the rejection region is of the form {Xe−θ0X ≤ c}.
Suppose under H0, X has the uniform distribution over (0,1) and under HA, X has the pdf of f(x) = 2x, 0 ≤ x ≤ 1. (a) Find the most powerful level α = 0.10 test of H0 vs. HA. (b) Calculate the type II error probability of the test in (a).
Formulate the hypothesis testing problem appropriate for analyzing whether the manufacturer is in compliance and explain how one would test at a significance level of .05. (b) Suppose that X = 24.8 and the sample standard deviation s = 1. What would you conclude about compliance if you were testing at a significance level of .10?
Paper For Above instruction
Statistical inference plays a pivotal role in determining the validity of hypotheses based on sample data. Among the various methods employed, the likelihood ratio test (LRT) is particularly powerful and widely used due to its optimality properties under certain conditions. This paper explores the derivation of the likelihood ratio test for an exponential distribution, examines hypotheses involving uniform and exponential distributions, and discusses practical applications such as environmental standards compliance.
Likelihood Ratio Test for Exponential Distribution
Consider a random sample X₁, X₂, ..., Xₙ drawn from an exponential distribution with the density function f(x|θ) = θe^(-θx), where θ > 0. The primary goal is to test the null hypothesis H₀: θ = θ₀ against the alternative HA: θ ≠ θ₀. The likelihood function based on the sample is:
L(θ) = ∏_{i=1}^n θ e^(-θx_i) = θ^n e^{-θ ∑_{i=1}^n x_i}
Computing the likelihood ratio involves the ratio of maximized likelihoods under H₀ and the entire parameter space:
Λ(x) = \frac{L(θ₀)}{L(\hat{θ})}
where the maximum likelihood estimator (MLE) of θ is:
\hat{θ} = \frac{n}{\sum_{i=1}^n x_i}
Substituting into the likelihoods yields:
L(θ₀) = θ₀^n e^{-θ₀ \sum x_i}
L(\hat{θ}) = \hat{θ}^n e^{-\hat{θ} \sum x_i} = \left(\frac{n}{\sum x_i}\right)^n e^{-n}
Hence, the likelihood ratio becomes:
Λ(x) = \frac{θ₀^n e^{-θ₀ \sum x_i}}{\left(\frac{n}{\sum x_i}\right)^n e^{-n}} = \left( \frac{θ₀ \sum x_i}{n} \right)^n e^{n - θ₀ \sum x_i}
Rearranged, the rejection region for a level α test is based on the statistic:
X̄ = \frac{1}{n} \sum_{i=1}^n x_i
The test rejects H₀ when Λ(x) ≤ c, which simplifies to the form {X̄ e^{-θ₀X̄} ≤ c'} for some constant c'. Equivalently, the test can be expressed as:
Reject H₀ if X̄ ≤ c''
indicating a threshold on the sample mean, aligned with the properties of exponential distributions.
Hypotheses Involving Uniform and Exponential Distributions
Distinguishing whether hypotheses are simple or composite is vital for choosing the appropriate testing procedure. A simple hypothesis specifies a single distribution completely, whereas a composite hypothesis encompasses multiple possible distributions.
- (a) X follows a uniform distribution on [0,1]. — Simple, as the distribution is fully specified with parameters fixed at [0,1].
- (b) A die is unbiased. — Simple, assuming the probability of each outcome is known as 1/6.
- (c) X follows a normal distribution with mean 0 and variance σ² > 10. — Composite, as the variance σ² varies over a range greater than 10.
- (d) X follows a normal distribution with mean µ = 0. — Simple, the mean is fixed at zero, but variance may vary, so depending on context, it might be considered composite.
Likelihood Ratio Test for Double Exponential Distribution
For i.i.d. samples X₁, ..., Xₙ from a double exponential (Laplace) distribution with density f(x) = (1/2λ) e^{−λ|x|}, the likelihood function is:
L(λ) = (1/2λ)^n e^{−λ ∑_{i=1}^n |x_i|}
The MLE of λ is obtained by maximizing L(λ). Setting the derivative to zero yields:
\hat{λ} = \frac{n}{\sum_{i=1}^n |x_i|}
Testing H₀: λ = λ₀ versus HA: λ = λ₁ > λ₀ involves forming the likelihood ratio:
Λ = \frac{L(λ₀)}{L(\hat{λ})} = \left( \frac{λ₀}{\hat{λ}} \right)^n e^{−(λ₀ - \hat{λ}) \sum |x_i|}
Because the density is symmetric and depends solely on the absolute value, the likelihood ratio simplifies to comparing the sum of absolute deviations. The test's power depends on whether λ > λ₀, with higher λ indicating more peakedness around zero, and the likelihood ratio serves as the basis for the rejection region.
Testing for Compliance in Environmental Standards
In environmental regulatory contexts, hypotheses often concern whether a parameter such as mean miles per gallon meets specified standards. Given data X from 16 cars, where the mileages are normally distributed, the hypotheses can be formulated as:
- H₀: μ ≥ 25 miles per gallon (compliance)
- H₁: μ
Assuming the known standard deviation s, the test statistic follows:
t = \frac{\bar{X} - μ_0}{s/√n}
At a significance level α = 0.05, critical values are obtained from the t-distribution with n−1 degrees of freedom. If the observed t-value falls below the critical value, we reject H₀, concluding the manufacturer is not in compliance. Conversely, if the sample mean is sufficiently high, we fail to reject H₀, indicating compliance.
Similarly, when the sample mean is 24.8 with s=1, testing at α=0.10 involves calculating the corresponding t-statistic and comparing it to the critical t-value. If the t-value exceeds the critical value, the data does not provide sufficient evidence to conclude non-compliance.
Conclusion
Likelihood ratio testing provides a rigorous framework for hypothesis testing across various distributions. Its application to exponential and double exponential distributions demonstrates its versatility. In practical scenarios such as environmental regulation, accurately framing hypotheses and understanding the distributional properties of test statistics are fundamental for correct inference. Proper application of these techniques ensures sound scientific and policy decisions, safeguarding environmental and public health.
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