/*! 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 50 By measuring the amount of time ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 50

By measuring the amount of time it takes a component of a product to move from one workstation to the next, an engineer has estimated that the standard deviation is 5 seconds. a. How many measurements should be made to be \(95 \%\) certain that the maximum error of estimation will not exceed 1 second? b. What sample size is required for a maximum error of 2 seconds?

Short Answer

Expert verified
For a maximum error of estimation of 1 second, 97 measurements should be made. For a maximum error of estimation of 2 seconds, 25 measurements should be made.

Step by step solution

01

Identify known variables

The standard deviation \( \sigma \) is 5 seconds. The desired error for the first part of the question (part a) is 1 second, and the desired error for the second part of the question (part b) is 2 seconds. The desired confidence level is 95%, so the Z score is about 1.96.
02

Apply the sample size formula for part a

Substitute known variables into the sample size formula for part a to get: \( n = (1.96 *\frac{5}{1})^2 \). Computing the above gives n approximately equal to 96.04. But since we can't have a fraction of a measurement, this needs to be rounded up to the nearest whole number, therefore n = 97.
03

Apply the sample size formula for part b

Substitute known variables into the sample size formula for part b to get: \( n = (1.96 * \frac{5}{2})^2 \). Computing the above gives n approximately equal to 24.01. But since we can't have a fraction of a measurement, this needs to be rounded up to the nearest whole number, therefore n = 25.

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.

Standard Deviation
Standard deviation is a key concept in statistics that measures the amount of variation or dispersion in a set of values. It tells you how much the individual data points differ from the mean (average) of the data set.
In this exercise, we're dealing with a standard deviation of 5 seconds. This figure indicates that the time it takes for the component to move from one workstation to the next varies by about 5 seconds from the average time.
  • The formula for standard deviation is: \[\sigma = \sqrt{\frac{\sum (x_i - \mu)^2}{N}}\]where \( \sigma \) is the standard deviation, \( x_i \) are the data points, \( \mu \) is the mean, and \( N \) is the number of data points.
  • A smaller standard deviation means the data points are close to the mean, while a larger standard deviation indicates more spread out data points.
Confidence Interval
Confidence interval is a range of values, derived from the sample data, that is likely to contain the value of an unknown population parameter. When we say we are 95% confident, we mean that if we were to take 100 different samples and compute a confidence interval for each sample, approximately 95 of the 100 confidence intervals will contain the true mean.
  • The confidence interval provides a way to assess the reliability of an estimate.
  • In the context of this exercise, we aim for 95% certainty, which corresponds to a confidence level, commonly used in statistical analysis to ensure our estimates are reliable.
Computationally, confidence intervals are calculated as follows:\[\text{Confidence Interval} = \bar{x} \pm Z * \frac{\sigma}{\sqrt{n}}\]where \( \bar{x} \) is the sample mean, \( Z \) is the z-score, \( \sigma \) is the standard deviation, and \( n \) is the sample size.
Maximum Error of Estimation
The maximum error of estimation, also known as the margin of error, is the maximum expected difference between the true population parameter and a sample estimate of that parameter. It reflects the degree of accuracy in our measurements and estimates.
  • It is crucial in determining sample size, especially when precision is needed.
  • Maximal allowable error impacts the sample size needed: the smaller the error, the larger the sample size required.
In our exercise, we examined two cases: For part a, the maximum error of estimation desired is 1 second, requiring a much larger sample to maintain accuracy than a maximum error of 2 seconds for part b.This relationship is represented by the formula:\[E = Z * \frac{\sigma}{\sqrt{n}}\]with \( E \) representing the maximum error of estimation.
Z-score
The z-score represents the number of standard deviations a data point is from the mean. In statistical terms, it measures the position of a sample mean relative to the population mean in terms of standard deviations. For confidence intervals, the z-score corresponds to a confidence level.
  • A common z-score for a 95% confidence level is 1.96, as utilized in the original exercise. This means that 95% of sample means will lie within 1.96 standard deviations of the true population mean.
  • Z-scores are derived from the standard normal distribution, which is a Gaussian distribution with a mean of 0 and a standard deviation of 1.
Mathematically, the z-score is calculated as:\[Z = \frac{X - \mu}{\sigma}\]where \( X \) is the value of the element, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. Knowing how to use z-scores helps in quantitatively comparing data points from different normal distributions.

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

Jack Williams is vice president of marketing for one of the largest natural gas companies in the nation. During the past 4 years, he has watched two major factors erode the profits and sales of the company. First, the average price of crude oil has been virtually flat, and many of his industrial customers are burning heavy oil rather than natural gas to fire their furnaces, regardless of added smokestack emissions. Second, both residential and commercial customers are still pursuing energy-conservation techniques (e.g., adding extra insulation, installing clockdrive thermostats, and sealing cracks around doors and windows to eliminate cold air infiltration). In previous years, residential customers bought an average of 129.2 mcf of natural gas from Jack's company \((\sigma=18 \mathrm{mcf})\) based on internal company billing records, but environmentalists have claimed that conservation is cutting fuel consumption up to \(3 \%\) per year. Jack has commissioned you to conduct a spot check to see if any change in annual usage has transpired before his next meeting with the officers of the corporation. A sample of 300 customers selected randomly from the billing records reveals an average of \(127.1 \mathrm{mcf}\) during the past 12 months. Is there a significant decline in consumption? a. Complete the appropriate hypothesis test at the 0.01 level of significance using the \(p\) -value approach so that you can properly advise Jack before his meeting. b. Because you are Jack's assistant, why is it best for you to use the \(p\) -value approach?

Calculate the \(p\) -value, given \(H_{a}: \mu \neq 245\) and \(z \star=1.1\)

A manufacturer of automobile tires believes it has developed a new rubber compound that has superior antiwearing qualities. It produced a test run of tires made with this new compound and had them road tested. The data values recorded were the amount of tread wear per 10,000 miles. In the past, the mean amount of tread wear per 10,000 miles, for tires of this quality, has been 0.0625 inch. The null hypothesis to be tested here is "The mean amount of wear on the tires made with the new com- pound is the same mean amount of wear with the old compound, 0.0625 inch per 10,000 miles," \(H_{o}: \mu=0.0625\) Three possible alternative hypotheses could be used: \(H_{a}: \mu<0.0625,(2) H_{a}: \mu \neq 0.0625,(3) H_{a}: \mu > 0.0625\) a. Explain the meaning of each of these three alternatives. b. Which one of the possible alternative hypotheses should the manufacturer use if it hopes to conclude that "use of the new compound does yield superior wear"?

Suppose a hypothesis test is conducted using the \(p^{-}\) value approach and assigned a level of significance of \(\alpha=0.01\) a. How is the 0.01 used in completing the hypothesis test? b. If \(\alpha\) is changed to \(0.05,\) what effect would this have on the test procedure?

A normally distributed population is known to have a standard deviation of \(5,\) but its mean is in question. It has been argued to be either \(\mu=80\) or \(\mu=90,\) and the following hypothesis test has been devised to settle the argument. The null hypothesis, \(H_{o}: \mu=80,\) will be tested by using one randomly selected data value and comparing it with the critical value of \(86 .\) If the data value is greater than or equal to \(86,\) the null hypothesis will be rejected. a. Find \(\alpha,\) the probability of the type I error. b. Find \(\beta,\) the probability of the type II error.
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

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.