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Chapter 10: Problem 14

A random variable is assumed to have the given probability density function. Find the observed significance level if the random variable is equal to the given value in an experiment. $$ f(x)=2 / x^{3} \text { over }[1, \infty) ; \text { observed value of } 20 $$

Short Answer

Expert verified
The observed significance level is 0.0025.

Step by step solution

01

Understanding the Given Information

The problem provides a probability density function (pdf) for a random variable: \( f(x) = \frac{2}{x^3} \) for \( x \in [1, \, \infty) \). We are required to find the observed significance level for the observed value \( x = 20 \).
02

Cumulative Distribution Function (CDF) Integration

To find the observed significance level, first, find the cumulative distribution function (CDF) F(x). Integrate the pdf from the lower bound to x: \( F(x) = \int_1^x \frac{2}{t^3} dt \).
03

Solving the Integral

Compute the integral: \( F(x) = \left[ -\frac{1}{t^2} \right]_1^x = 1 - \frac{1}{x^2} \). So, \( F(x) = 1 - \frac{1}{x^2} \).
04

Evaluate the CDF at the Observed Value

Find the cumulative probability up to the observed value \( x = 20 \): \( F(20) = 1 - \frac{1}{20^2} = 1 - \frac{1}{400} = 1 - 0.0025 = 0.9975 \).
05

Calculate the Significance Level

The observed significance level is the complement of the cumulative distribution value at 20: \( 1 - F(20) = 1 - 0.9975 = 0.0025 \).

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Key Concepts

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

probability density function
A probability density function (pdf) is a function that describes the likelihood of a random variable to take on a particular value. In the given exercise, the probability density function is defined as: \[ f(x) = \frac{2}{x^3} \text{ for } x \text{ in } [1, \, \infty) \] The pdf must be non-negative for all values within its domain and integrate to 1 over its entire range. This ensures that the total probability is 1. This specific form of pdf helps us understand how probable different values of the variable are.
cumulative distribution function
The cumulative distribution function (CDF) is a function that gives the probability that a random variable is less than or equal to a certain value. For the given probability density function, the CDF helps us find the total probability up to a particular value. To find the CDF, we integrate the pdf from the lower bound of the domain to the point of interest. Here, we need to integrate: \[ F(x) = \int_1^x \frac{2}{t^3} dt \] Solving this integral, we get: \[ F(x) = 1 - \frac{1}{x^2} \] This expression now gives us the cumulative probability up to any value x in the domain.
significance level
The significance level in this context represents the probability that the observed value is at least as extreme as our interest value. In the exercise, we compute the significance level for the observed value x = 20. First, find the cumulative probability up to x = 20: \[ F(20) = 1 - \frac{1}{20^2} = 1 - \frac{1}{400} = 0.9975 \] The observed significance level is obtained by subtracting this cumulative probability from 1: \[ 1 - F(20) = 1 - 0.9975 = 0.0025 \] This means there is a 0.25% probability that the observed value is as extreme or more extreme than 20.
integral calculation
Integral calculation plays a critical role in finding the cumulative distribution function (CDF). The exercise requires us to compute an integral to transition from the pdf to the CDF. The integral of our given pdf function is: \[ F(x) = \int_1^x \frac{2}{t^3} dt \] To solve this, we use the fundamental theorem of calculus. The integrated form is: \[ F(x) = \left[ -\frac{1}{t^2} \right]_1^x = -\frac{1}{x^2} - ( -\frac{1}{1^2}) = 1 - \frac{1}{x^2} \] This calculation enables us to move from the pdf to the CDF, and is essential in finding probabilities and significance levels in the context of continuous random variables.

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Most popular questions from this chapter

Suppose that an individual is randomly selected from a population that satisfies the Hardy-Weinberg Law. Let \(p\) and \(q\) denote the proportions of \(A\) and \(a\) alleles in the population. a) The allele inherited by the offspring may be visualized with the following multiplication tree. Fill in the probabilities of the unmarked arrows. b) In the multiplication tree, there are three ways corresponding to the transmission of the \(A\) allele. Show that the probability that the \(A\) allele is transmitted is \(p\). c) Show that the probability that the \(a\) allele is Iransmitted is \(q .\) d) Use the result of Example 9 to show that, if one generation satisfies the Hardy-Weinberg Law and survives to mate, then the offspring generation should also satisfy it.

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A significance level \(\alpha\) and a tail of the standard normal distribution are given. Use the normal table to approximately determine the critical value. \(\alpha=0.005\), right tail
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