/*! 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 1 For a normal population with kno... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 1

For a normal population with known variance \(\sigma^{2}\), answer the following questions: (a) What is the confidence level for the interval \(\bar{x}-2.14 \sigma / \sqrt{n}\) \(\leq \mu \leq \bar{x}+2.14 \sigma / \sqrt{n} ?\) (b) What is the confidence level for the interval \(\bar{x}-2.49 \sigma / \sqrt{n}\). \(\leq \mu \leq \bar{x}+2.49 \sigma / \sqrt{n} ?\) (c) What is the confidence level for the interval \(\bar{x}-1.85 \sigma / \sqrt{n}\). \(\leq \mu \leq \bar{x}+1.85 \sigma / \sqrt{n} ?\)

Short Answer

Expert verified
(a) 96.38%, (b) 93.60%, (c) 86.68%

Step by step solution

01

Understanding Confidence Intervals

The given intervals are confidence intervals for the population mean \( \mu \) given that the variance \( \sigma^2 \) is known. The formula \( \bar{x} \pm z \left( \sigma / \sqrt{n} \right) \) is used, where \( z \) is the z-score corresponding to the desired confidence level.
02

Using Z-Scores for Confidence Intervals

For each part of the question, identify the \( z \)-score used in the interval:- (a) uses \( z = 2.14 \)- (b) uses \( z = 2.49 \)- (c) uses \( z = 1.85 \)
03

Calculating Confidence Level for Part (a)

To find the confidence level for \( z = 2.14 \), look up this \( z \)-score in the standard normal distribution table, which shows the probability that a standard normal random variable is less than this value. The two-tailed confidence level is \( P(-2.14 < Z < 2.14) \). This is approximately 96.38%.
04

Calculating Confidence Level for Part (b)

For \( z = 2.49 \), use the standard normal distribution table to find \( P(-2.49 < Z < 2.49) \). This yields a two-tailed confidence level of approximately 93.60%.
05

Calculating Confidence Level for Part (c)

For \( z = 1.85 \), use the standard normal distribution table to find \( P(-1.85 < Z < 1.85) \). This yields a two-tailed confidence level of approximately 86.68%.

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.

Normal Distribution
A normal distribution, often depicted as a bell-shaped curve, is fundamental in statistics because it describes how data is distributed around a central value.
It is symmetrical and has key properties that make it useful for various statistical analyses.
  • The peak of the bell curve represents the mean (average) of the data.
  • Most of the data points are clustered around the mean.
  • The further you move away from the mean, the fewer data points you encounter.
  • It is defined by two parameters: the mean ( ) and the standard deviation ( ).
Normal distribution is crucial when working with confidence intervals, as it helps to assess probabilities for any given range of data.
Z-Scores
Z-scores are a way of standardizing individual data points relative to the mean of the dataset.
They help to assess how far, in standard deviations, a particular value is from the mean.
To calculate a z-score, use the formula: \( z = \frac{(X - \mu)}{\sigma} \), where
  • \( X \) is the value of the data point,
  • \( \mu \) is the mean of the data set,
  • \( \sigma \) is the standard deviation.
Z-scores are essential in determining probabilities using the standard normal distribution.
In the context of confidence intervals, different z-scores correspond to different levels of confidence.
Population Mean
The population mean () represents the average value of a particular dataset if the entire population data is available.
Calculating the population mean is straightforward:\[ \mu = \frac{\sum X}{N} \]where
  • \( \sum X \) is the sum of all the data points,
  • \( N \) is the number of data points.
The population mean is often unknown in practice, especially for large populations.
In such cases, sample means () are used as an estimator to infer about the population mean with a certain degree of confidence.
Standard Normal Distribution
The standard normal distribution is a specific type of normal distribution that has a mean of 0 and a standard deviation of 1.
This makes it a useful tool for comparing different sets of data on a standardized scale.
  • Any normal distribution can be transformed into a standard normal distribution using z-scores.
  • The standard normal distribution table or Z-table is used to understand the probability of a specific z-score or range of scores.
In practical application, the standard normal distribution is invaluable when determining confidence intervals, as it allows statisticians to use z-scores to calculate the probability that a given range of values will contain the true population mean.

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

8-34. The solar energy consumed (in trillion BTU) in the U.S. by year from 1989 to 2004 (source: U.S. Department of Energy Web site, http://www.eia.doe.gov/emeu) is shown in the table below. Read down, then right for year. $$ \begin{array}{llll} \hline 55.291 & 66.458 & 70.237 & 65.454 \\ 59.718 & 68.548 & 69.787 & 64.391 \\ 62.688 & 69.857 & 68.793 & 63.62 \\ 63.886 & 70.833 & 66.388 & 63.287 \\ \hline \end{array} $$ Check the assumption of normality in the population. Construct a \(95 \%\) confidence interval for the mean solar energy consumed.

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 circuits is tested, revealing 13 defectives. (a) Calculate a \(95 \%\) two-sided CI on the fraction of defective circuits produced by this particular tool. (b) Calculate a \(95 \%\) upper confidence bound on the fraction of defective circuits.

Consider the one-sided confidence interval expressions for a mean of a normal population. (a) What value of \(z_{\alpha}\) would result in a \(90 \%\) CI? (b) What value of \(z_{\alpha}\) would result in a \(95 \%\) CI? (c) What value of \(z_{\alpha}\) would result in a \(99 \% \mathrm{CI} ?\)

An article in the Journal of Composite Materials (December 1989 , Vol. \(23,\) p. 1200 ) describes the effect of delamination on the natural frequency of beams made from composite laminates. Five such delaminated beams were subjected to loads, and the resulting frequencies were as follows (in hertz): $$ 230.66,233.05,232.58,229.48,232.58 $$ Check the assumption of normality in the population. Calculate a \(90 \%\) two- sided confidence interval on mean natural frequency.

An article in the Journal of the American Statistical Association (1990, Vol. 85, pp. 972-985) measured the weight of 30 rats under experiment controls. Suppose that there are 12 underweight rats. (a) Calculate a \(95 \%\) two-sided confidence interval on the true proportion of rats that would show underweight from the experiment. (b) Using the point estimate of \(p\) obtained from the preliminary sample, what sample size is needed to be \(95 \%\) confident that the error in estimating the true value of \(p\) is less than \(0.02 ?\) (c) How large must the sample be if we wish to be at least \(95 \%\) confident that the error in estimating \(p\) is less than 0.02, regardless of the true value of \(p ?\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

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.