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Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a sample from an exponential distribution with parameter \(\lambda\). It can be shown that \(2 \lambda \sum_{i=1}^{n} X_{i}\) has a chi-square distribution with \(2 n\) degrees of freedom. Use this fact to devise a test statistic and critical region for \(H_{0}: \lambda=\lambda_{0}\) versus the three usual alternatives.

Short Answer

Expert verified
Use the test statistic \(T = 2 \lambda_{0} \sum X_{i}\), with chi-square critical values for decision-making.

Step by step solution

01

Understand the Problem

We have a sample from an exponential distribution with parameter \(\lambda\). The problem states that \(2 \lambda \sum_{i=1}^{n} X_{i}\) follows a chi-square distribution with \(2n\) degrees of freedom. We need to create a test statistic and find the critical region to test the null hypothesis \(H_{0}: \lambda=\lambda_{0}\) against the alternatives.
02

Identify the Test Statistic

Given the problem, the statistic \(2 \lambda \sum_{i=1}^{n} X_{i}\) is distributed as \(\chi^2(2n)\). Under \(H_0: \lambda=\lambda_{0}\), the test statistic becomes \(T = 2 \lambda_{0} \sum_{i=1}^{n} X_{i}\), which under \(H_0\) is \(\chi^2(2n)\)-distributed.
03

Design the Critical Regions

To test \(H_0: \lambda=\lambda_{0}\) against the three types of alternatives, use the chi-square distribution as follows:- For \(H_a: \lambda > \lambda_{0}\), reject \(H_0\) if \(T\) is greater than the chi-square critical value corresponding to the desired significance level.- For \(H_a: \lambda < \lambda_{0}\), reject \(H_0\) if \(T\) is less than the chi-square critical value corresponding to the lower tail.- For \(H_a: \lambda eq \lambda_{0}\), reject \(H_0\) if \(T\) is outside the chi-square critical values corresponding to the upper and lower tails.
04

Execute the Test

Calculate the test statistic \(T = 2 \lambda_{0} \sum_{i=1}^{n} X_{i}\) from the sample data. Compare \(T\) against the critical values for the specific alternative hypothesis you are testing. Based on this comparison, decide whether to reject \(H_0\) or not.

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Key Concepts

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

Exponential Distribution
In statistics, the exponential distribution is a continuous probability distribution often used as a model for the time until an event, such as the failure of a device or the arrival of a new call. It's particularly popular in reliability engineering and queuing theory.

The exponential distribution is defined by a single parameter \( \lambda \), known as the rate parameter. In essence, \( \lambda \) dictates the rate at which the events, say failures, occur. A high \( \lambda \) means events happen rapidly, while a low \( \lambda \) suggests they're spread out.

Some useful properties of the exponential distribution include:
  • Memoryless Property: The probability of an event happening in the next interval is independent of previous intervals.
  • Constant Failure Rate: Particularly useful in reliability contexts where devices don't age.
  • Relation to Poisson Process: It describes the time between events in a Poisson process.
  • Probability Density Function (PDF): Given by \( f(x;\lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \).
Understanding these helps one determine its applicability, e.g., whether the assumption of constant failure rate fits the context.
Chi-Square Distribution
The chi-square distribution is a key concept in hypothesis testing and statistical analysis. It arises from the summation of square values, notably in the context of normal distributions. When examining the exponential distribution, the transformation \( 2 \lambda \sum_{i=1}^{n} X_{i} \) forms a chi-square distribution with \( 2n \) degrees of freedom. This relationship is foundational in statistical tests involving variance.

Key characteristics of the chi-square distribution include:
  • Non-negative values: Only takes values from zero to positive infinity.
  • Degrees of Freedom (df): Determines the shape of the distribution, with more df leading towards a normal distribution shape.
Applications include:
  • Goodness-of-fit tests: To determine if sample data matches a population with a specific distribution.
  • Test of independence: Often seen in contingency tables.
In hypothesis testing, knowing the chi-square distribution aids in understanding the critical values and test boundaries, essential for making informed decisions based on sample data.
Null and Alternative Hypotheses
Understanding null and alternative hypotheses is crucial for conducting hypothesis tests. These hypotheses form the backbone of statistical testing. They provide a clear framework for making decisions based on data.

The null hypothesis \( (H_0) \) generally states that there is no effect or difference, or that a parameter equals a certain value, e.g., \( \lambda = \lambda_0 \). It serves as the default assumption that a test seeks to challenge.

The alternative hypothesis \( (H_a) \), on the other hand, is what you might believe to be true or is aiming to prove. Commonly considered alternatives include:
  • One-tailed hypotheses: Asserts that a parameter is either less than or greater than a certain value.
  • Two-tailed hypotheses: Proposes that a parameter is not equal to a specific value.
In our context, to test against these hypotheses, a test statistic such as \( T = 2 \lambda_{0} \sum_{i=1}^{n} X_{i} \) is used and compared to the chi-square distribution's critical values. If the test statistic lies beyond a critical value, it provides grounds to reject the null hypothesis, concluding the sample data sufficiently supports the alternative.

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Most popular questions from this chapter

Consider the hypothesis test of \(H_{0}: \sigma^{2}=10\) against \(H_{1}: \sigma^{2}>10 .\) Approximate the \(P\) -value for each of the following test statistics. (a) \(x_{0}^{2}=25.2\) and \(n=20\) (b) \(x_{0}^{2}=15.2\) and \(n=12\) (c) \(x_{0}^{2}=4.2\) and \(n=15\)

For the hypothesis test \(H_{0}: \mu=10\) against \(H_{1}: \mu>10\) with variance unknown and \(n=15\), approximate the \(P\) -value for each of the following test statistics. (a) \(\mathrm{t}_{0}=2.05\) (b) \(\mathrm{t}_{0}=-1.84\) (c) \(\mathrm{t}_{0}=0.4\)

Suppose that eight sets of hypotheses of the form $$H_{0}: \mu=\mu_{0} \quad H_{1}: \mu \neq \mu_{0}$$ have been tested and that the \(P\) -values for these tests are \(0.15,\) \(0.06 .0 .67,0.01,0.04,0.08,0.78,\) and \(0.13 .\) Use Fisher's procedure to combine all of the \(P\) -values. What conclusions can you draw about these hypotheses?

For the hypothesis test \(H_{0}: \mu=7\) against \(H_{1}: \mu \neq 7\) with variance unknown and \(n=20\), approximate the \(P\) -value for each of the following test statistics. (a) \(t_{0}=2.05\) (b) \(t_{0}=-1.84\) (c) \(t_{0}=0.4\)

A textile fiber manufacturer is investigating a new drapery yarn, which the company claims has a mean thread elongation of 12 kilograms with a standard deviation of 0.5 kilograms. The company wishes to test the hypothesis \(H_{0}: \mu=12\) against \(H_{1}: \mu<12,\) using a random sample of four specimens. (a) What is the type I error probability if the critical region is defined as \(\bar{x}<11.5\) kilograms? (b) Find \(\beta\) for the case in which the true mean elongation is 11.25 kilograms. (c) Find \(\beta\) for the case in which the true mean is 11.5 kilograms.

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