/*! 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 44 Let \(Y\) have probability densi... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(Y\) have probability density function $$f_{Y}(y)=\left\\{\begin{array}{ll} \frac{2(\theta-y)}{\theta^{2}}, & 0

Short Answer

Expert verified
a. The given CDF matches the integration result. b. \(Y/\theta\) is a pivotal quantity. c. 90% lower confidence limit for \(\theta\) is \(\frac{y}{1-\sqrt{0.1}}\).

Step by step solution

01

Understand the Density Function

The given probability density function for the random variable \(Y\) is defined as follows: \(f_Y(y) = \frac{2(\theta-y)}{\theta^2}\) for \(0 < y < \theta\) and \(f_Y(y) = 0\) elsewhere. This density function is defined only within this range, meaning that \(Y\) is a continuous random variable that can take values between 0 and \(\theta\).
02

Derive the Distribution Function (CDF)

To find the cumulative distribution function (CDF) \(F_Y(y)\), we integrate the density function up to \(y\): $$F_Y(y) = \int_{0}^{y} \frac{2(\theta-t)}{\theta^2} \, dt.$$ Evaluating this integral, we get $$F_Y(y) = \left[\frac{2\theta t}{\theta^2} - \frac{t^2}{\theta^2}\right]_{0}^{y} = \frac{2\theta y}{\theta^2} - \frac{y^2}{\theta^2} = \frac{2y}{\theta} - \frac{y^2}{\theta^2}.$$ So, the CDF is as follows: for \(0 < y < \theta\), \(F_Y(y) = \frac{2y}{\theta} - \frac{y^2}{\theta^2}\), and for other ranges as given in the problem statement.
03

Define Pivotal Quantity for Part (b)

A pivotal quantity is a function of the sample data and parameter \(\theta\) whose distribution does not depend on \(\theta\). Here, we aim to show that \(\frac{Y}{\theta}\) is a pivotal quantity. Given the transformation \(U = \frac{Y}{\theta}\), calculate the distribution of \(U\) by finding its PDF. The joint transformation, \(Y = U\theta\), replaces the integration bounds in the PDF of \(Y\). Differentiating, \(U\) has a distribution independent of \(\theta\).
04

Calculate the PDF of U

Since \(U = \frac{Y}{\theta}\), its range is from 0 to 1. If \(Y\) is in the interval \(0 < Y < \theta\), then \(0 < U < 1\). The PDF of \(U\) can be found by adjusting \(f_Y(y)\) using the density transformation technique \(f_U(u) = \theta \cdot f_Y(u \theta)\). Substituting gives \(f_U(u) = \frac{2(1-u)}{\theta} \cdot \theta = 2(1-u)\), which is free of \(\theta\). Therefore, \(U\) is a pivotal quantity.
05

Find Lower Confidence Limit for \(\theta\)

With a confidence level of 90%, the goal is to find a value \(L\) such that \(P(\theta > L) = 0.9\) given observed \(y\). If \( \theta = \frac{Y}{U} \), interpret \(U\) that leads to an inverse of the pivotal transformation. Using the cumulative distribution of \(U\), \(P(U > u) = 1 - (1-u)^2\), for 90% confidence solve: \(1-(1-u)^2 = 0.1\). Solving before gives \(u = 1-\sqrt{0.1}\), thus \(L = \frac{y}{1-\sqrt{0.1}}\).

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Key Concepts

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

Cumulative Distribution Function
When we talk about the Cumulative Distribution Function (CDF), we are referring to a function that provides the probability of a random variable being less than or equal to a particular value. In simpler terms, it maps every possible outcome to a probability indicating how likely the random variable is to be below or at that outcome.
For a continuous random variable like in the given exercise, the CDF is essential because it accumulates the probabilities up to a specific point by integrating the Probability Density Function (PDF).
The CDF can be given by:
  • For values less than the minimum of the range: 0.
  • For values within the range: the result of the integral of the PDF.
  • For values greater than the maximum of the range: 1.
To achieve this, you calculate the integral of the PDF from the start of the range to your current value. This gives you the CDF, helping you understand the distribution of the random variable fully.
Pivotal Quantity
A pivotal quantity is a unique concept in statistics. It is a function that combines your sample data and some unknown parameters, but crucially, it has a known distribution that is independent of those unknown parameters. This characteristic makes pivotal quantities incredibly useful in constructing confidence intervals.
In the context of the exercise, \( \frac{Y}{\theta} \) is shown to be a pivotal quantity. By transforming \( Y \) into \( U = \frac{Y}{\theta} \), we see that \( U \) has a distribution that does not depend on \( \theta \). Since pivotal quantities remove dependence on unknown variables, they are fundamental tools for statistical estimation and hypothesis testing. Here, they allow us to derive estimates for other parameters without being skewed by unknown variables.
Confidence Interval
Confidence intervals provide us with a range of values in which we believe an unknown parameter, like \( \theta \), lies with a certain degree of confidence. They combine our sample data analysis and pivotal quantities to find this plausible range.
Typically, confidence intervals are expressed in terms of a confidence level, such as 90%, which indicates that 90% of similarly constructed intervals will contain the parameter. To construct a confidence interval using a pivotal quantity, you solve the distribution of the pivotal quantity to find its critical values that correspond to your desired confidence level.
For instance, in our example, the 90% lower confidence limit gives you a sense of how low \( \theta \) might be, backed up by statistical theory. This approach offers a statistically sound way to make inferences about population parameters, even when they cannot be directly observed.
Continuous Random Variable
A continuous random variable is a variable that can take on an infinite number of possible values within a given range. Unlike discrete random variables, which have distinct and separate values, continuous random variables are associated with probability density functions (PDFs) rather than probability mass functions (PMFs).
In many real-world scenarios, such as measuring weight, time, or in our particular exercise, continuous random variables allow us to model these measurements smoothly across their possible outcomes. The key characteristic is that, while individual outcomes have zero probability (since their exact probability cannot be pinpointed), the probability is distributed across a range of values. This means that probability is defined over intervals.
Continuous random variables provide the foundational knowledge for understanding more complex statistical concepts, such as probability density functions, cumulative distribution functions, and how these are used to analyze statistical data effectively.

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Most popular questions from this chapter

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The Mars twin rovers, Spirit and Opportunity, which roamed the surface of Mars in the winter of 2004, found evidence that there was once water on Mars, raising the possibility that there was once life on the plant. Do you think that the United States should pursue a program to send humans to Mars? An opinion poll \(^{\star}\) indicated that \(49 \%\) of the 1093 adults surveyed think that we should pursue such a program. a. Estimate the proportion of all Americans who think that the United States should pursue a program to send humans to Mars. Find a bound on the error of estimation. b. The poll actually asked several questions. If we wanted to report an error of estimation that would be valid for all of the questions on the poll, what value should we use? [Hint: What is the maximum possible value for \(p \times q ?]\)

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Organic chemists often purify organic compounds by a method known as fractional crystallization. An experimenter wanted to prepare and purify \(4.85 \mathrm{g}\) of aniline. Ten 4.85 -gram specimens of aniline were prepared and purified to produce acetanilide. The following dry yields were obtained: $$\begin{array}{llllllll} 3.85, & 3.88, & 3.90, & 3.62, & 3.72, & 3.80, & 3.85, & 3.36, & 4.01, & 3.82 \end{array}$$ Construct a \(95 \%\) confidence interval for the mean number of grams of acetanilide that can be recovered from 4.85 grams of aniline.

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