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Chapter 9: Problem 24

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.)

Short Answer

Expert verified
The 95% confidence interval for the mean travel time is [19.076, 29.324] minutes.

Step by step solution

01

Identify the Data Given

Identify the pieces of information provided in the problem:Sample size (\(n\)) = 50Sample mean (\(\bar{x}\)) = 24.2 minutesPopulation standard deviation (\(\sigma\)) = 18.5 minutes
02

Determine the Z-Score for the 95% Confidence Level

For a 95% confidence level, the Z-score is approximately 1.96 (you can find this value using a Z-table or standard normal distribution table).
03

Calculate the Standard Error of the Mean

The standard error of the mean (\(SE\)) is calculated using the formula:\[ SE = \frac{\sigma}{\sqrt{n}} \]Substitute the values:\[ SE = \frac{18.5}{\sqrt{50}} \approx 2.615 \] minutes.
04

Calculate the Margin of Error

The margin of error (\(E\)) is calculated using the formula:\[ E = Z \times SE \]Substitute the values:\[ E = 1.96 \times 2.615 \approx 5.124 \] minutes.
05

Construct the Confidence Interval

The confidence interval is calculated using the formula:\[ CI = \bar{x} \pm E \]Substitute the values:\[ 24.2 \pm 5.124 \]This gives the interval:\[ 19.076 \text{ minutes to } 29.324 \text{ minutes} \]
06

Interpret the Confidence Interval

We are 95% confident that the true mean travel time to work for the population is between 19.076 minutes and 29.324 minutes.

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Key Concepts

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

Sample Mean
In statistics, the sample mean, also known as \(\bar{x}\), represents the average value of a sample. It is computed by adding up all the values in the sample and then dividing by the number of values (sample size, n).
For our exercise, the urban economist obtained a sample mean of 24.2 minutes. This means that, on average, people in this sample spend 24.2 minutes traveling to work.
The sample mean serves as a point estimate for the population mean, which is what we ultimately want to estimate.
Standard Deviation
Standard deviation, denoted as \(\sigma\) for population or \(s\) for a sample, measures the amount of variation or dispersion in a set of values.
In this exercise, the population standard deviation is given as 18.5 minutes. This tells us how much individual travel times vary from the average travel time.
A high standard deviation means that travel times are quite spread out from the mean, which is true in our scenario due to the presence of both zero-minute commutes and commutes exceeding one hour.
Margin of Error
The margin of error (\(E\)) provides a range that is likely to contain the population mean. It accounts for sampling error and is calculated by multiplying the standard error by the Z-score corresponding to the desired confidence level.
In our example, the margin of error is determined using the formula: \[ E = Z \times SE \]
After substituting the values: \[ E = 1.96 \times 2.615 \approx 5.124 \text{ minutes} \]
This tells us that the estimate of the mean travel time could vary by roughly 5.124 minutes on either side of the sample mean.
Z-Score
The Z-score is a statistic that represents the number of standard errors a data point is from the mean. It is used to determine the critical value for confidence intervals.
For a 95% confidence level, the Z-score is approximately 1.96. This value can be obtained from a Z-table or standard normal distribution table.
Using the Z-score, we ensure that our confidence interval provides a range where we can be 95% confident that the true population mean lies.
Standard Error
The standard error (SE) measures the variability of the sample mean from the population mean and is essential for constructing confidence intervals. It decreases with larger sample sizes, reflecting the law of large numbers.
The standard error is calculated using the formula: \[ SE = \frac{\rho}{u\text{n}} \]
In our case: \[ SE = \frac{18.5}{u\text{50}} \approx 2.615 \text{ minutes} \]
This value will be used to calculate the margin of error and the confidence interval, offering an estimate of how much the sample mean tends to vary around the true population mean.

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Most popular questions from this chapter

In a survey of 1010 adult Americans, the Gallup Organization asked, "Are you worried or not worried about having enough money for retirement?" Of the 1010 surveyed, 606 stated that they were worried about having enough money for retirement. Construct a \(90 \%\) confidence interval for the proportion of adult Americans who are worried about having enough money for retirement.

Construct the appropriate confidence interval. In a February 2005 Harris Poll, 769 of 1010 randomly selected adults said that they always wear their seat belt. Construct and interpret a \(98 \%\) confidence interval for the proportion of adults who always wear their seat belt.

A simple random sample of size \(n\) is drawn. The sample mean, \(\bar{x},\) is found to be \(18.4,\) and the sample standard deviation, \(s\), is found to be \(4.5 .\) (a) Construct a \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is 35 (b) Construct a \(95 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(50 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(99 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(35 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the margin of error, \(E ?\) (d) If the sample size is \(n=15,\) what conditions must be satisfied to compute the confidence interval?

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?

A simple random sample of size \(n\) is drawn. The sample mean, \(\bar{x},\) is found to be 35.1 , and the sample standard deviation, \(s,\) is found to be 8.7 (a) Construct a \(90 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is 40 (b) Construct a \(90 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(100 .\) How does increasing the sample size affect the margin of error, \(E ?\) (c) Construct a \(98 \%\) confidence interval about \(\mu\) if the sample size, \(n,\) is \(40 .\) Compare the results to those obtained in part (a). How does increasing the level of confidence affect the margin of error, \(E ?\) (d) If the sample size is \(n=18,\) what conditions must be satisfied to compute the confidence interval?
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