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The Wall Street Journal reported that automobile crashes cost the United States \(\$ 162\) billion annually (The Wall Street Journal, March 5,2008 ). The average cost per person for crashes in the Tampa, Florida, area was reported to be \(\$ 1599 .\) Suppose this average cost was based on a sample of 50 persons who had been involved in car crashes and that the population standard deviation is \(\sigma=\$ 600 .\) What is the margin of error for a \(95 \%\) confidence interval? What would you recommend if the study required a margin of error of \(\$ 150\) or less?

Short Answer

Expert verified
The margin of error for a 95% confidence interval for the average cost per person for car crashes in Tampa, Florida is approximately \$166.30. To achieve a margin of error of \$150 or less, an increase in the sample size is recommended.

Step by step solution

01

Find the standard error

The standard error is calculated using the following formula: \[SE = \frac{\sigma}{\sqrt{n}}\] where: - \(SE\) is the standard error - \(\sigma\) is the population standard deviation (\(\$600\)) - \(n\) is the sample size (50) Using the given values, we can calculate the standard error: \[SE = \frac{600}{\sqrt{50}} = 600 \times \frac{1}{\sqrt{50}} \approx 84.85\]
02

Calculate the margin of error for a 95% confidence interval

To calculate the margin of error for a 95% confidence interval, we need to find the critical value (also known as the z-score) corresponding to 95% confidence. For a 95% confidence interval, the critical value is approximately 1.96. Now, we can calculate the margin of error using the following formula: \[Margin\_of\_error = z \times SE\] where: - \(Margin\_of\_error\) is the desired margin of error - \(z\) is the critical value (1.96 for a 95% confidence interval) - \(SE\) is the standard error (\(84.85\)) Plugging in the values, we get the margin of error: \[Margin\_of\_error = 1.96 \times 84.85 \approx 166.30\]
03

Make a recommendation based on the margin of error requirement

Since the calculated margin of error for the 95% confidence interval (\(\$166.30\)) is larger than the required margin of error of \$150 or less, we can recommend increasing the sample size. Increasing the sample size will decrease the standard error, which in turn will reduce the margin of error. In conclusion, the margin of error for a 95% confidence interval for the average cost per person for car crashes in the Tampa, Florida area is approximately \$166.30. To achieve a margin of error of \$150 or less, an increase in the sample size is recommended.

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Key Concepts

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

Confidence Interval
Understanding the confidence interval (CI) is essential when analyzing data and making inferences about a population. A CI is a range of values, derived from the sample data, that is likely to contain the value of an unknown population parameter.

In our example, we're focusing on estimating the average cost of car crashes per person in Tampa, Florida. The CI gives us a range that we are 95% confident contains the actual average cost for the entire population. It is important to note that '95%' refers to the confidence level, which indicates how sure we are that the interval includes the true value. If we were to take many samples and create a CI from each one, we expect about 95% of those intervals to contain the true population parameter. The actual average cost is not a range, but the CI provides a way to express the uncertainty inherent in our estimate.
Standard Error
The standard error (SE) measures the amount of variability or dispersion of a sample mean in relation to the true population mean.

In this context, the SE tells us how far the sample mean (The standard error is calculated from the population standard deviation and the sample size, as seen in the exercise, where we use the formula
\[SE = \frac{\sigma}{\sqrt{n}} \]. By knowing the SE, we can gauge the accuracy of our sample mean in representing the population mean—the smaller the SE, the more reliable the sample mean. However, it's worth noting that the SE alone doesn't provide a direct answer to our final question; it's an intermediate step towards calculating the confidence interval's margin of error.
Population Standard Deviation
The population standard deviation (\(\sigma\)), a key component of our formula, is a measure of the dispersion or variation in a set of data values.

A small standard deviation indicates that the values tend to be close to the mean of the set, while a larger one shows that the values are spread out over a wider range. In the given exercise, the population standard deviation is \(\sigma=\$600\), signifying the variability in individual costs of car crashes in the population under study. This figure is crucial for calculating the standard error, which in turn is vital for determining the margin of error for the confidence interval.
Sample Size
Sample size (\(n\)) plays a pivotal role in statistical analysis. It refers to the number of observations or data points included in a sample drawn from a population.

As demonstrated in the step-by-step solution, increased sample size generally results in a lower standard error and thereby a more narrow margin of error, leading to a more precise confidence interval. If the survey of Tampa's car crash costs aimed for a margin of error less than \(\$150\), then we would recommend a larger sample size than the initial 50 participants. This change would provide more reliability and precision in the results.
Z-score
The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean.

In the context of our confidence interval, the z-score corresponds to the desired level of confidence and is used to capture the tail areas of the normal distribution needed for our estimation. For a 95% confidence interval, the appropriate z-score is approximately 1.96. This means that our margin of error extends 1.96 standard deviations from the sample mean on both sides, establishing the range where we are 95% confident the true mean lies.

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Most popular questions from this chapter

Which would be hardest for you to give up: Your computer or your television? In a survey of 1677 U.S. Internet users, \(74 \%\) of the young tech elite (average age of 22 ) say their computer would be very hard to give up (PC Magazine, February 3, 2004). Only 48\% say their television would be very hard to give up. a. Develop a \(95 \%\) confidence interval for the proportion of the young tech elite that would find it very hard to give up their computer. b. Develop a \(99 \%\) confidence interval for the proportion of the young tech elite that would find it very hard to give up their television. c. In which case, part (a) or part (b), is the margin of error larger? Explain why.

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