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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
First, randomly select a sample of 30 or more cars and record their ages. Compute the mean age of this sample. Then, use the formula for the estimate of the population mean with known standard deviation to calculate an estimate of the mean age of cars owned by people in the United States: \( \bar{x} \pm Z\frac{\sigma}{\sqrt{n}} \).

Step by step solution

01

Understanding the Procedure

To estimate the mean age of cars, a study needs to be conducted. The study would need to get a sample of car owners, with a sample size of at least 30 (as mentioned in the problem). The 30 car ownerships would be chosen randomly to represent a valid cross-section. Each member of the population must have an equal chance of being selected in the sample.
02

Data Collection

From the chosen sample of 30 or more cars, data regarding the age of each car needs to be collected and recorded. Once the data is collected, calculate the mean (average) age of the sample. Let's denote the sample mean as \(\bar{x}\). In general, \(\bar{x} = \sum_{i=1}^{n}x_i/n\), where \(x_i\) is the age of each car in the sample, and \(n\) is the sample size.
03

Estimation of Population Mean

Once the sample mean is calculated, the population mean can be estimated using it in the interval estimate for a known standard deviation, using the formula: \( \bar{x} \pm Z\frac{\sigma}{\sqrt{n}} \), where \( Z \) is the Z-score for the desired confidence level (1.96 for 95%), \( \sigma \) is the population standard deviation, and \( n \) is the sample size. The calculated value will be the estimated mean age of cars owned by people in the United States.

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Most popular questions from this chapter

A sample of 18 observations taken from a normally distributed population produced the following data: \(\begin{array}{lllllllll}28.4 & 27.3 & 25.5 & 25.5 & 31.1 & 23.0 & 26.3 & 24.6 & 28.4\end{array}\) \(\begin{array}{llllllllll}37.2 & 23.9 & 28.7 & 27.9 & 25.1 & 27.2 & 25.3 & 22.6 & 22.7\end{array}\) a. What is the point estimate of \(\mu\) ? b. Make a \(99 \%\) confidence interval for \(\mu .\) c. What is the margin of error of estimate for \(\mu\) in part b?

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