/*! 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 9 Construct the appropriate confid... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 9

Construct the appropriate confidence interval. A simple random sample of size \(n=40\) is drawn from a population. The sample mean is found to be \(\bar{x}=120.5\) and the sample standard deviation is found to be \(s=12.9 .\) Construct a \(99 \%\) confidence interval about the population mean.

Short Answer

Expert verified
The \(99\%\) confidence interval is \(115.25 \leq \text{population mean} \leq 125.75\).

Step by step solution

01

Determine the Sample Mean and Standard Deviation

Identify the sample mean \(\bar{x}\) and the sample standard deviation \(s\). Here, \(\bar{x}=120.5\) and \(s=12.9\).
02

Identify the Sample Size

Determine the sample size \(n\). Given \(n=40\).
03

Determine the Confidence Level and Find the Corresponding z-value

For a \(99\%\) confidence interval, the z-value is approximately \(2.576\).
04

Calculate the Standard Error

The standard error (SE) is calculated using the formula \(\text{SE} = \frac{s}{\root n}\). Plug in the values: \(\text{SE} = \frac{12.9}{\root 40} = \frac{12.9}{6.3246} \approx 2.04\).
05

Compute the Margin of Error

Calculate the margin of error (ME) using the formula \(\text{ME} = z \times \text{SE}\). Substituting the values: \( \text{ME} = 2.576 \times 2.04 \approx 5.25 \).
06

Compute the Confidence Interval

The confidence interval is given by \(\bar{x} \pm \text{ME} \). Therefore: \( 120.5 - 5.25 \leq \text{population mean} \leq 120.5 + 5.25 \), which results in the interval \( 115.25 \leq \text{population mean} \leq 125.75 \).

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.

sample mean
The sample mean, often represented by \(\bar{x}\), is a key concept in statistics. It represents the average value of a sample, which is a subset of a population. For example, in the given problem, the sample mean is 120.5. This means that if we were to sum up all the values in our sample of 40 observations and then divide this sum by 40, we would get 120.5. The sample mean is crucial as it is used to estimate the population mean (the average of the entire population). Calculating the sample mean is the initial step in determining the confidence interval.
sample standard deviation
The sample standard deviation, denoted as \(s\), measures the amount of variation or dispersion in a sample data set. In our problem, the sample standard deviation is given as 12.9. This value indicates how much the individual data points in the sample deviate, on average, from the sample mean. A higher standard deviation signifies more spread out data, whereas a lower standard deviation indicates that the data points are closer to the mean. The sample standard deviation helps to calculate the standard error, which is necessary for constructing the confidence interval.
confidence level
The confidence level represents the degree of certainty that the population parameter lies within the confidence interval. In this problem, we use a 99% confidence level. This means we can be 99% confident that the true population mean falls within the calculated interval. The confidence level influences the z-value, which is derived from the standard normal distribution. Higher confidence levels lead to wider intervals because they require more margin for error to ensure the population parameter is within the range.
standard error
The standard error (SE) gauges the accuracy with which a sample mean estimates the population mean. It is calculated using the formula: \(\text{SE} = \frac{s}{\sqrt{n}} \), where \(s\) is the sample standard deviation and \(n\) is the sample size. For our problem, SE is calculated as \(\frac{12.9}{\root 40} = \frac{12.9}{6.3246} \approx 2.04\). The standard error decreases with larger sample sizes, providing a more accurate estimate of the population mean.
margin of error
The margin of error (ME) quantifies the range within which we can expect the true population mean to fall, considering the sample data. It is computed with the formula: \(\text{ME} = z \times \text{SE}\), where \(z\) is the z-value corresponding to the desired confidence level, and \(SE\) is the standard error. In our example, the ME is \(2.576 \times 2.04 \approx 5.25\). The margin of error is added to and subtracted from the sample mean to create the confidence interval, ensuring the interval accounts for possible sampling variability.
z-value
The z-value is a statistical measurement corresponding to the desired confidence level, derived from the standard normal distribution. It indicates how many standard deviations an element is from the mean. For a 99% confidence level, the z-value is approximately 2.576. This value is critical for calculating the margin of error and, subsequently, the confidence interval. A higher z-value leads to a wider confidence interval, reflecting greater certainty that the interval contains the 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

An urban economist wishes to estimate the mean amount of time people spend traveling to work. He obtains a random sample of 50 individuals who are in the labor force and finds that the mean travel time is 24.2 minutes. Assuming the population standard deviation of travel time is 18.5 minutes, construct and interpret a \(95 \%\) confidence interval for the mean travel time to work. Note: The standard deviation is large because many people work at home (travel time \(=0\) minutes) and many have commutes in excess of 1 hour. (Source: Based on data obtained from the American Community Survey.)

Suppose a professor in a class of 20 students wants to estimate the mean pulse rate of students prior to the final exam. Would it make sense for the professor to construct a \(90 \%\) confidence interval about the population mean? Explain.

(a) Find the \(t\) -value such that the area in the right tail is 0.10 with 25 degrees of freedom. (b) Find the \(t\) -value such that the area in the right tail is 0.05 with 30 degrees of freedom. (c) Find the \(t\) -value such that the area left of the \(t\) -value is 0.01 with 18 degrees of freedom. [Hint: Use symmetry. (d) Find the critical \(t\) -value that corresponds to \(90 \%\) confidence. Assume 20 degrees of freedom.

Packer Fans In a Harris Poll conducted October \(20-25\) 2004,381 of 2114 randomly selected adults who follow professional football said the Green Bay Packers were their favorite team. (a) Verify that the requirements for constructing a confidence interval about \(\hat{p}\) are satisfied. (b) Construct a \(90 \%\) confidence interval for the proportion of adults who follow professional football who say the Green Bay Packers is their favorite team. Interpret this interval. (c) Construct a \(99 \%\) confidence interval for the proportion of adults who follow professional football who say the Green Bay Packers is their favorite team. Interpret this interval. (d) What is the effect of increasing the level of confidence on the width of the interval?

Fifty rounds of a new type of ammunition were fired from a test weapon, and the muzzle velocity of the projectile was measured. The sample had a mean muzzle velocity of 863 meters per second, with a standard deviation of 2.7 meters per second. Construct and interpret a \(99 \%\) confidence interval for the mean muzzle velocity.
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Calculus

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.