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Let X1,X2,,Xn be a random sample from the normal distribution N(θ,1). Show that the likelihood ratio principle for testing H0:θ=θ, where θ is specified, against H1:θθ leads to the inequality |x¯θ|c (a) Is this a uniformly most powerful test of H0 against H1? (b) Is this a uniformly most powerful unbiased test of H0 against H1?

Short Answer

Expert verified
(a) Yes, this is a uniformly most powerful (UMP) test. (b) Yes, it is also a uniformly most powerful unbiased (UMPU) test.

Step by step solution

01

Derive the Likelihood Ratio Test

For this problem, the likelihood ratio test is Λ=L(θ)L(θ^) where θ^ is the max likelihood estimator, equal to the sample mean x¯. That simplifies to Λ=en(x¯θ)2/2en(x¯x¯)2/2=en(x¯θ)2/2 . The inequality |x¯θ|c is the rejection region.
02

Determine if this is a uniformly most powerful (UMP) test

This test is actually UMP. It was derived from the Neyman-Pearson Lemma which indicates that the LRT is the most powerful for simple hypotheses.
03

Determine if this is a UMP unbiased test

To be a uniformly most powerful unbiased (UMPU) test, the test needs to not only be most powerful for every value under the alternative hypothesis (UMP), but also unbiased in the sense that the probability of rejecting the null hypothesis when it's true is not greater than the significance level α. With this inequality as the rejection region, this likelihood ratio test fulfills all conditions, so it is UMPU.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniformly Most Powerful Test
A Uniformly Most Powerful (UMP) test is one that maximizes the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true, for all possible values under the alternative hypothesis. In simpler terms, it’s the best test we can create for detecting whether the null hypothesis is false.

To understand this better, let's consider the Neyman-Pearson Lemma. It tells us that for testing simple hypotheses, the likelihood ratio test is the most powerful. This means the test is designed to detect if the hypothesis is wrong more effectively than any other test of the same size (or significance level).
  • The simple hypothesis specifies a fixed value for the parameter under the null hypothesis.
  • The UMP test is ideal because it works optimally for all parameters under the alternative hypothesis.
In the exercise, the test for the normal distribution is UMP because it was derived using this principle. This makes it the best choice for determining whether there's a difference from the hypothesis θ=θ.
Normal Distribution
The normal distribution is one of the most important concepts in statistics. It is a bell-shaped curve that describes how data is spread out around a mean. In a normal distribution, most data points fall close to the mean, and the probabilities for data points decrease as you move away from the center.

Here's a quick breakdown of its properties:
  • The mean, median, and mode of a normal distribution are all equal, located at the peak of the curve.
  • It's symmetric around the mean, meaning the left and right sides are mirror images.
  • The area under the curve represents the total probability, which is equal to 1.

The normal distribution is crucial in hypothesis testing and in the exercise, it forms the basis of the data points X1,X2,,Xn. When working with a normal distribution, we often deal with variables like the sample mean x¯, which itself follows a normal distribution when the sample size is large enough according to the Central Limit Theorem.
Hypothesis Testing
Hypothesis testing is a method used in statistical analysis to determine if there is enough evidence to reject a null hypothesis H0 in favor of an alternative hypothesis H1. Here’s a basic rundown of how it works:

1. **Formulate Hypotheses**
We start with two hypotheses:
  • **Null Hypothesis (H0)**: Assumes no effect or no difference. It’s the hypothesis we seek to test against.
  • **Alternative Hypothesis (H1)**: Suggests an effect or a difference. It's what we hope to support.
2. **Select a Significance Level α**
  • This is the probability of rejecting the null hypothesis when it is true. Common values are 0.05 or 0.01.
3. **Calculate the Test Statistic**
  • This involves determining how far the sample mean is from the hypothesized population mean under H0.
4. **Make a Decision**
  • Based on the test statistic and significance level, decide to either reject H0 or not.

In our example, the rejection region is |x¯θ|c. If this condition holds, it indicates enough evidence to reject the null hypothesis, confirming an effect or difference exists.

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Most popular questions from this chapter

Let the independent random variables Y and Z be N(μ1,1) and N(μ2,1), respectively. Let θ=μ1μ2. Let us observe independent observations from each distribution, say Y1,Y2, and Z1,Z2, To test sequentially the hypothesis H0:θ=0 against H1:θ=12, use the sequence Xi=YiZi,i=1,2, If αa=βa=0.05, show that the test can be based upon X¯=Y¯Z¯. Find c0(n) and c1(n)

Let X1,X2,,X25 denote a random sample of size 25 from a normal distribution N(θ,100). Find a uniformly most powerful critical region of size α=0.10 for testing H0:θ=75 against H1:θ>75

Let X be a random variable with pdf Missing \left or extra \right, for \(-\infty0\). First, show that the variance of X is σX2=2bX2 Now let Y, independent of X, have pdf Missing \left or extra \right, for \(-\infty0 .\) Consider the hypotheses H0:σX2=σY2 versus H1:σX2>σY2 To illustrate Remark 8.3.2 for testing these hypotheses, consider the following data set, (taken from page 122 of Hettmansperger and McKean, 1998). Sample 1 represents the values of a sample drawn on X with bX=1, while Sample 2 represents the values of a sample drawn on Y with bY=1. Hence, in this case H0 is true.  Sample .389822.17746.81368.000721.11032.70976.45664.13583 Sample .76384.570412.565111.733111.40363.77812.11548 Sample 1.06716.57712.36138.680372.63445.99624.18128.23957 Sample .775761.42159.81898.328632.213901.42551.16589 (a) Obtain comparison boxplots of these two samples. Comparison boxplots consist of boxplots of both samples drawn on the same scale. Based on these plots, in particular the interquartile ranges, what do you conclude about H0? (b) Obtain the F -test (for a one-sided hypothesis) as discussed in Remark 8.3.2 at level α=0.10. What is your conclusion? (c) The test in Part (b) is not exact. Why?

A random sample X1,X2,,Xn arises from a distribution given by $$H_{0}: f(x ; \theta)=\frac{1}{\theta}, \quad 0

Let X1,X2,,Xn be a random sample from a distribution with pdf f(x;θ)= \(\theta x^{\theta-1}, 0

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