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You are interested in estimating the mean age of cars owned by all people in the United States. Briefly explain the procedure you will follow to conduct this study. Collect the required data on a sample of 30 or more cars and then estimate the population mean at a \(95 \%\) confidence level. Assume that the population standard deviation is \(2.4\) years,

Short Answer

Expert verified
After obtaining the sample data, calculate the sample mean and use it along with the given population standard deviation to compute the \(95\%\) confidence interval for the population mean. The population mean age of cars in the US is expected to lie within this interval.

Step by step solution

01

Understanding the Given Parameter

Before we begin, let's understand the parameters we have. The exercise states that the population standard deviation is \(2.4\) years. A sample of 30 or more cars will be used to estimate the population mean. It's also given that we need to estimate the mean at a \(95\%\) confidence level.
02

Collect the Sample Data

To estimate the population mean, first collect the sample data. In this case, it would require obtaining the ages of a sample of 30 cars or more from the population.
03

Calculate the Sample Mean

Once the sample data is collected, calculate the sample mean (denoted by \(\bar{X}\)). This can be calculated by summing all the values (ages of cars in the sample) and dividing by the number of data points (number of cars in the sample).
04

Determine the Confidence Interval

With a confidence level of \(95\%\), we calculate a confidence interval for the population mean. The formula for the confidence interval is given by \(\bar{X} ± Z_{α/2} × \(\frac{σ}{\sqrt{n}}\)\) where \(σ\) is the population standard deviation, \(n\) is the sample size, \(\bar{X}\) is the sample mean and \(Z_{α/2}\) is the z-score corresponding to the desired degree of confidence. For a \(95\%\) confidence level, \( Z_{α/2}\) is \(1.96\) (from z-table). Substitute the values into the equation to get the confidence interval.
05

Interpret the Confidence Interval

The interval obtained in the last step is the confidence interval. This means that we're \(95\%\) confident that the population mean age of cars in the US falls within this interval. Remember that this is an estimate, and there will always be uncertainty associated with it.

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Key Concepts

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

Population Mean
The population mean is a statistical measure that represents the average value of a dataset consisting of all members of a particular group or category. In the context of our exercise, it refers to the average age of all cars owned by people in the United States. This value provides a general idea of the overall age of cars. Calculating the population mean would require data from every single car in the US, which is often impractical due to time and resource constraints. Instead, researchers collect data from a sample of cars to estimate the population mean. This estimate helps in making generalizations or predictions about the entire population based on the sample data.
Z-score
The z-score is a statistical metric that expresses the number of standard deviations a data point is from the mean of a dataset. In the context of confidence intervals, the z-score helps us determine the range within which we expect the true population mean to fall. When estimating the population mean with a 95% confidence level, a specific z-score known as the critical value is used. For a 95% confidence level, this z-score is 1.96. This value is obtained from a standard normal distribution table, often referred to as the z-table. The z-score forms a part of the formula that calculates the confidence interval, ensuring that there is a 95% chance the interval correctly approximates the population mean.
Sample Mean
The sample mean, denoted as \( \bar{X} \), is the average of the data points collected in a sample. In our study of car ages, it involves calculating the average age of the cars in our sample of 30 or more.To find the sample mean, you sum up the ages of all the cars in your sample and divide by the total number of cars. This value serves as an estimate for the population mean. It is crucial because it provides the central point of the confidence interval and is used to assess how close the sample mean is to the population mean.
Population Standard Deviation
Population standard deviation, denoted as \( \sigma \), is a measure that indicates the spread or dispersion of all data points in a population. In our exercise, it is given as 2.4 years, representing how much the ages of cars typically vary around the mean age.This measurement is critical when calculating the confidence interval for the population mean. The formula for the confidence interval incorporates the standard deviation, which helps determine the margin of error. It affects the width of the confidence interval: larger standard deviations result in wider intervals, reflecting greater uncertainty in our estimate. Understanding this concept allows researchers to interpret how variable the dataset is, relative to the mean.

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Most popular questions from this chapter

A local gasoline dealership in a small town wants to estimate the average amount of gasoline that people in that town use in a 1-week period. The dealer asked 44 randomly selected customers to keep a diary of their gasoline usage, and this information produced the following data on gas used (in gallons) by these people during a 1-week period. \(\begin{array}{rrrrrrrrrr}23.1 & 13.6 & 25.8 & 10.0 & 7.6 & 18.9 & 26.6 & 23.8 & 12.3 & 15.8 & 21.0 \\ 26.9 & 22.9 & 18.3 & 23.5 & 21.6 & 15.5 & 23.5 & 11.8 & 15.3 & 11.9 & 19.2 \\ 14.5 & 9.6 & 12.1 & 18.0 & 20.6 & 14.2 & 7.1 & 13.2 & 5.3 & 13.1 & 10.9 \\ 10.5 & 5.1 & 5.2 & 6.5 & 8.3 & 10.5 & 7.4 & 7.4 & 5.3 & 10.6 & 13.0\end{array}\) Construct a \(95 \%\) confidence interval for the average weekly gas usage by people in this town. Use the \(t\) distribution.

a. Find the value of \(t\) for the \(t\) distribution with a sample size of 21 and area in the left tail equal to \(.10\). b. Find the value of \(t\) for the \(t\) distribution with a sample size of 14 and area in the right tail equal to \(.025\). c. Find the value of \(t\) for the \(t\) distribution with 45 degrees of freedom and \(.001\) area in the right tail. d. Find the value of \(t\) for the \(t\) distribution with 37 degrees of freedom and \(.005\) area in the left tail.

a. How large a sample should be selected so that the margin of error of estimate for a \(98 \%\) confidence interval for \(p\) is \(.045\) when the value of the sample proportion obtained from a preliminary sample is \(.53\) ? b. Find the most conservative sample size that will produce the margin of error for a \(98 \%\) confidence interval for \(p\) equal to \(.045\).

A jumbo mortgage is a mortgage with a loan amount above the industry-standard definition of conyentional conforming loan limits. As of January 2009, approximately \(2.57 \%\) of people who took out a jumbo mortgage during the previous 12 months were at least 60 days late on their payments. Suppose that this percentage is based on a random sample of 1430 people who took out a jumbo mortgage during the previous 12 months. a. Construct a \(95 \%\) confidence interval for the proportion of all people who took out a jumbo mortgage during the previous 12 months and were at least 60 days late on their payments. b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Discuss all possible alternatives. Which alternative is the best?

An insurance company selected a sample of 50 auto claims filed with it and investigated those claims carefully. The company found that \(12 \%\) of those claims were fraudulent. a. What is the point estimate of the percentage of all auto claims filed with this company that are fraudulent? b. Make a \(99 \%\) confidence interval for the percentage of all auto claims filed with this company that are fraudulent.

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