/*! 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 15 Let \(\bar{x}\) be the observed ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 15

Let \(\bar{x}\) be the observed mean of a random sample of size \(n\) from a distribution having mean \(\mu\) and known variance \(\sigma^{2}\). Find \(n\) so that \(\bar{x}-\sigma / 4\) to \(\bar{x}+\sigma / 4\) is an approximate \(95 \%\) confidence interval for \(\mu\).

Short Answer

Expert verified
The sample size needed for \(\bar{x}-\sigma / 4\) to \(\bar{x}+\sigma / 4\) to be an approximate \(95 \%\) confidence interval for \(\mu\) is \(n = 62\).

Step by step solution

01

Understanding Confidence Intervals

A common form of the confidence interval for a mean is \(\bar{x} \pm z_{\alpha / 2} \times \sigma / \sqrt{n}\), where \(z_{\alpha / 2}\) is the critical value from the standard normal distribution required for a \(95 \%\) confidence interval, \(\sigma\) is the standard deviation and \(\sqrt{n}\) is the standard error of the mean.
02

Setup the Comparison

The task gives an interval for the confidence interval in form of \(\bar{x} \pm \sigma / 4\). This has to be compared with the common form of the confidence interval which includes the standard error of the mean and the critical z-value from the standard normal distribution. This comparison allows to find n.
03

Solve for \(n\)

Set up the comparison: \(\sigma / 4 = z_{\alpha / 2} \times \sigma / \sqrt{n}\). One can solve this equation for \(n\), which will give the desired sample size.
04

Substitute \(z_{\alpha / 2}\)

In a \(95 \%\) confidence interval, \(z_{\alpha / 2} = 1.96\). Substitute this value for \(z_{\alpha / 2}\) in the equation.
05

Complete Calculation

Solve the equation for \(n\). This will result in \(n = (4*z_{\alpha / 2})^{2}\), which simplifies to \(n = (4*1.96)^{2} = 61.4656\). As you cannot have a fraction of an observation, round up to the nearest whole number: \(n = 62\).

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!

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 \(x_{1}, x_{2}, \ldots, x_{n}\) be the values of a random sample. A bootstrap sample, \(\mathbf{x}^{* \prime}=\left(x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}\right)\), is a random sample of \(x_{1}, x_{2}, \ldots, x_{n}\) drawn with replacement. (a) Show that \(x_{1}^{*}, x_{2}^{*}, \ldots, x_{n}^{*}\) are iid with common cdf \(\widehat{F}_{n}\), the empirical \(\mathrm{cdf}\) of \(x_{1}, x_{2}, \ldots, x_{n}\) (b) Show that \(E\left(x_{i}^{*}\right)=\bar{x}\). (c) If \(n\) is odd, show that median \(\left\\{x_{i}^{*}\right\\}=x_{((n+1) / 2)}\). (d) Show that \(V\left(x_{i}^{*}\right)=n^{-1} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}\).

Determine a method to generate random observations for the extreme-valued pdf which is given by $$ f(x)=\exp \left\\{x-e^{x}\right\\}, \quad-\infty

Let \(Y_{1}

In the baseball data set discussed in the last exercise, it was found that out of the 59 baseball players, 15 were left-handed. Is this odd, since the proportion of left-handed males in America is about \(11 \% ?\) Answer by using \((4.2 .7)\) to construct a \(95 \%\) approximate confidence interval for \(p\), the proportion of left-handed baseball players.

In Exercise 4.1.2, the weights of 26 professional baseball pitchers were given. From the same data set, the weights of 33 professional baseball hitters (not pitchers) are given below. Assume that the data sets are independent of one another. \(\begin{array}{lllllllllllll}155 & 155 & 160 & 160 & 160 & 166 & 170 & 175 & 175 & 175 & 180 & 185 & 185 \\ 185 & 185 & 185 & 185 & 185 & 190 & 190 & 190 & 190 & 190 & 195 & 195 & 195 \\ 195 & 200 & 205 & 207 & 210 & 211 & 230 & & & & & & \end{array}\) Use expression (4.2.13) to find a \(95 \%\) confidence interval for the difference in mean weights between the pitches and the hitters. Which group (on the average) appears to be heavier? Why would this be so? (The sample means and variances for the weights of the pitchers and hitters are, respectively, Pitchers \(201,305.68\) and Hitters \(185.4,298.13 .)\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

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.