/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Let \(Y_{4}\) be the nth order s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 40

Let \(Y_{4}\) be the nth order statistic of a random sample, \(n=4\), from a continuous-type uniform distribution on the interval \((0, \theta)\). Let \(0

Short Answer

Expert verified
The values \(c_1 = \sqrt[4]{0.05}\) and \(c_2 = 1\) indeed satisfy the given condition. The 95% confidence for \(\theta\) is \(\left(\frac{Y_4}{1}, \frac{Y_4}{\sqrt[4]{0.05}}\right)\). These values denote that with 95% probability, the real value of \(\theta\) lies between \(Y_4\) and \(\frac{Y_4}{\sqrt[4]{0.05}}\).

Step by step solution

01

Confirmation of given Probability

The probability of \(Y_4\), the 4th order statistic, to be between \(c_1\theta\) and \(c_2\theta\) is given as 0.95 according to the exercise. The cumulative distribution function (CDF) of the 4th order statistic in a sample of size 4 from a uniform distribution (0, \(\theta\)) is given by \(F_{Y_4}(y) = \left(\frac{y}{\theta}\right)^4\). The probability \(Pr(c_1\theta < Y_4 < c_2\theta)\) can be expressed in terms of the CDF as \(F_{Y_4}(c_2\theta) - F_{Y_4}(c_1\theta) = 0.95. By plugging \(c_1 = \sqrt[4]{0.05}\) and \(c_2 = 1\) into the equation, we get \((c_2)^4 - (c_1)^4 = 0.95\). Now, substitute the values into this expression. We get \(1 - 0.05 = 0.95\) which verifies that \(c_1\) and \(c_2\) satisfy the given condition.
02

Derive 95% confidence interval for theta

A 95% confidence interval for \(\theta\) can be derived from the given probability. Confidence interval is defined by two boundaries, lower and upper, between which a parameter lies with a certain probability. Here, the boundaries are defined by \(c_1\theta\) and \(c_2\theta\). Rearranging these equations we get \(\theta > \frac{Y_4}{c_2}\) and \(\theta < \frac{Y_4}{c_1}\). Therefore, the 95% confidence interval for \(\theta\) can be written as \(\left(\frac{Y_4}{1}, \frac{Y_4}{\sqrt[4]{0.05}}\right)\) as we are given \(c_2=1\) and \(c_1=\sqrt[4]{0.05}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Uniform Distribution
Uniform distribution is a type of probability distribution where all outcomes are equally likely within a specific range. For a continuous uniform distribution on an interval (0, \(\theta\)), every point within this interval has the same likelihood of occurring. This means that if you pick a number at random from this interval, each number is equally probable. This distribution is useful in statistics when you know the minimum and maximum values, but nothing else is known.
  • Range: Defined by the interval (0, \(\theta\)).
  • Probability Density: Constant for all values in this interval.
  • Area under the probability density function within the interval equals 1.
These features make the uniform distribution simple but effective for certain statistical methods.
Confidence Interval
A confidence interval provides an estimated range of values which is likely to include an unknown parameter. This range is derived from a given level of confidence, often expressed as a percentage like 95%. The idea is to generate a range that would contain the true parameter value a certain proportion of times if you were to take many random samples. For instance, in the exercise, the confidence interval for \(\theta\) is set between \(c_{1}\theta\) and \(c_{2}\theta\) such that there's a 95% probability the true \(\theta\) falls within this range.
  • Calculated using statistical techniques that account for sample data variability.
  • A broader range implies more uncertainty about the parameter's actual value.
  • Not a guarantee, but rather an expression of likelihood or probability.
Understanding confidence intervals is crucial in making inferences from sample data.
Cumulative Distribution Function
A cumulative distribution function (CDF) represents the probability that a random variable takes on a value less than or equal to a certain point. For a continuous random variable, the CDF describes the probability that the variable falls within a specific range. In the given exercise, the CDF is used to calculate the probability that the 4th order statistic, \(Y_{4}\), falls between two points.
  • For a uniform distribution on \((0,\theta)\), the CDF is expressed as\(F(y) = \left( \frac{y}{\theta} \right)^4\) for the 4th order statistic.
  • Helps to convert ranges of values into probabilities.
  • Essential for deriving probability-based measures like confidence intervals.
CDFs are vital in understanding how probabilities accumulate across ranges of outcomes.
Random Sample
A random sample is a subset of individuals chosen randomly from a population, in such a way that each individual has an equal chance of being selected. This ensures that the sample represents the population well, minimizing bias and allowing for generalization of results to the broader population. In statistical exercises, like the current one, working with a random sample helps ensure that the conclusions drawn are as close to reality as possible.
  • Key for obtaining a representative sample.
  • Enables unbiased estimation of population parameters.
  • Essential for the validity of statistical inference methods such as the calculation of order statistics and confidence intervals.
Random samples provide a foundation for many statistical methods, allowing researchers to make valid inferences about populations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(Y_{1}

Let a random sample of size 17 from the normal distribution \(n\left(\mu, \sigma^{2}\right)\) yield \(x=4.7\) and \(s^{2}=5.76 .\) Determine a 90 per cent confidence interval for \(\mu\).

Let \(Y_{n}\) be the nth order statistic of a random sample of size \(n\) from a distribution with p.d.f. \(f(x \mid \theta)-1 / \theta, 00, \beta>0\). Find the Bayes' solution \(w\left(y_{n}\right)\) for a point estimate of \(\theta\).

Let \(X\) have a Poisson distribution with parameter \(\theta\). Assume that the unknown \(\theta\) is a value of a random variable \(\Theta\) that has a gamma distribution with parameters \(\alpha=r\) and \(\beta=(1-p) / p_{r}\) where \(r\) is a positive integer and \(0

Let \(Y_{1}\) and \(Y_{2}\) be two stochastically independent unbiased estimators of \(\theta\). Say the variance of \(Y_{1}\) is twice the variance of \(Y_{2}\). Find the constants \(k_{1}\) and \(k_{2}\) so that \(k_{1} Y_{1}+k_{2} Y_{2}\) is an unbiased estimator with smallest possible variance for such a linear combination.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.