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Chapter 9: Problem 76

The National Highway Traffic Safety Administration found that, among the crashes with recorded times, EMS notification times exceeded 10 minutes in \(19.4 \%\) of rural fatal crashes. A random sample of 500 reported fatal crashes in Kentucky showed \(21.8 \%\) of the notification times exceeded 10 minutes. Construct the \(95 \%\) confidence interval for the true proportion of fatal crashes in Kentucky whose elapsed notification time exceeded 10 minutes.

Short Answer

Expert verified
The \(95\%\) confidence interval for the true proportion of fatal crashes in Kentucky whose elapsed notification time exceeded 10 minutes is between \(17.8\%\) and \(25.8\%\).

Step by step solution

01

Identify the Sample Size and Sample Proportion

Given a sample of 500 reported fatal crashes in Kentucky, and among them \(21.8 \% = 0.218\) exceeded the notification time of 10 minutes. So, the sample size \(n = 500\) and the sample proportion \(\hat{p} = 0.218\).
02

Calculate Standard Error

To calculate standard error of the proportion, use the formula \(\sqrt{\hat{p}(1 - \hat{p}) / n} = \sqrt{0.218 * (1 - 0.218) / 500} = 0.0202\).
03

Determine Critical Value

For a \(95\%\) confidence interval, the critical value \(z^*\) corresponds to \(1.96\) (this value comes from the standard normal distribution table).
04

Calculate the Confidence Interval

Use the formula for the confidence interval: \(\hat{p} \pm z^* \times \text{Standard Error}\). Substituting the known values, we get \(0.218 \pm 1.96 * 0.0202\) which results in an interval from \(0.178\) to \(0.258\) or \(17.8\%\) to \(25.8\%\).

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Key Concepts

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

Sample Proportion
When working with statistics, the sample proportion is a measure that helps us understand the attribute of a certain subset within a larger population. In this exercise about fatal crashes, the sample proportion is derived from a specific scenario where 21.8% of the sample has a particular characteristic鈥攕pecifically, the notification time exceeded 10 minutes.

This sample proportion is represented as \( \hat{p} \). It is calculated by dividing the number of favorable outcomes by the total number of observations. Here, it is \( \hat{p} = 0.218 \), meaning that out of 500 fatal crashes in the sample, approximately 109 had delayed notifications.

  • Computation: \( \hat{p} = \text{Number of favorable outcomes} / \text{Total sample size} \)
  • It reflects the estimated proportion of the population showing a certain feature.
  • Insights from the sample proportion allow us to infer potential patterns in the wider population.
Standard Error
The standard error is an important calculation in statistics that measures the accuracy of a sample mean by examining the level of variation or "spread" of the data points around that mean. Think of it as a way to gauge how much your sample proportion might differ from the actual population proportion.

In our analysis, the standard error of the sample proportion was calculated using the formula \( \sqrt{\hat{p}(1 - \hat{p}) / n} \), where \( n \) is the sample size, and \( \hat{p} \) is the sample proportion. Plugging in the numbers, we find the standard error to be 0.0202.

  • Shows variability among sample means
  • Helps in constructing confidence intervals
  • The lower the standard error, the more reliable the sample proportion represents the population
Understanding standard error helps in constructing accurate confidence intervals and drawing meaningful conclusions about the population from a sample.
Critical Value
Critical value is a key concept in statistics that helps determine the "line" or threshold we use to estimate the boundaries of our confidence interval. Imagine it as a multiplier that helps adjust our intervals to reflect the desired confidence level. For a \( 95\% \) confidence interval, this critical value is often found by referring to a standard normal distribution table and, in most cases, is 1.96.

When constructing the confidence interval for our sample, the critical value of 1.96 is used to multiply the standard error and provide the margin of error. This means that our interval will capture the true population proportion 95 times out of 100 if we repeated the process.

  • Essential to calculate the range of a confidence interval
  • Chosen based on the desired confidence level
  • Larger critical values are used for higher confidence levels
'Determining the correct critical value ensures the confidence interval is appropriately adjusted to the desired level of certainty.
Proportion of Fatal Crashes
In this statistical exercise, understanding the proportion of fatal crashes with delayed EMS notifications provides crucial insights into emergency response efficiency. Here, "proportion" refers to the fraction of the sample being studied鈥攊n this case, the percentage of incidents where the EMS notification time exceeded 10 minutes.

The analysis seeks to create a confidence interval around the sample proportion of 21.8% to estimate this characteristic in the entire population of fatal crashes in Kentucky. Using the calculated interval of 17.8% to 25.8%, we can be 95% confident that this range includes the true proportion of fatal crashes with delayed notifications.

  • Important for evaluating public safety measures
  • Provides target for improvement efforts in EMS response times
  • Helps in policy formulation and resource allocation
By accurately estimating this proportion, policymakers and stakeholders can better understand and address areas needing improvement in emergency response mechanisms.

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Most popular questions from this chapter

A recent survey conducted by Lieberman Research Worldwide and Charles Schwab reported that the "High cost of Living" was the top concern that most surprised young adults as they began life on their own. Twenty-six percent reported "High cost of Living" as their top concern. A disbeliever of this information took his own random sample of 500 young adults starting out on their own in an attempt to show that the true percentage for this top concern is actually higher. a. Find the \(p\) -value if 148 of the young adults surveyed put down "High cost of Living" as their top concern. b. Explain why it is important for the level of significance to be established before the sample results are known.

Find the test statistic for the hypothesis test: a. \(H_{o}: \sigma^{2}=532\) versus \(H_{a}: \sigma^{2}>532\) using sample information \(n=18\) and \(s^{2}=785\) b. \(H_{o}: \sigma^{2}=52\) versus \(H_{a}: \sigma^{2} \neq 52\) using sample information \(n=41\) and \(s^{2}=78.2\)

Determine the critical region and critical value(s) that would be used to test the following using the classical approach: a. \(H_{o}: \sigma=0.5\) and \(H_{a}: \sigma>0.5,\) with \(n=18\) and \(\alpha=0.05\) b. \(H_{o}: \sigma^{2}=8.5\) and \(H_{a}: \sigma^{2}<8.5,\) with \(n=15\) and \(\alpha=0.01\) c. \(H_{o}: \sigma=20.3\) and \(H_{a}: \sigma \neq 20.3,\) with \(n=10\) and \(\alpha=0.10\) d. \(H_{o}: \sigma^{2}=0.05\) and \(H_{a}: \sigma^{2} \neq 0.05,\) with \(n=8\) and \(\alpha=0.02\) e. \(H_{o}: \sigma=0.5\) and \(H_{a}: \sigma<0.5,\) with \(n=12\) and \(\alpha=0.10\)

Julia Jackson operates a franchised restaurant that specializes in soft ice cream cones and sundaes. Recently she received a letter from corporate headquarters warning her that her shop is in danger of losing its franchise because the average sales per customer have dropped "substantially below the average for the rest of the corporation." The statement may be true, but Julia is convinced that such a statement is completely invalid to justify threatening a closing. The variation in sales at her restaurant is bound to be larger than most, primarily because she serves more children, elderly, and single adults rather than large families who run up big bills at the other restaurants. Therefore, her average ticket is likely to be smaller and exhibit greater variability. To prove her point, Julia obtained the sales records from the whole company and found that the standard deviation was \(2.45\)dollar per sales ticket. She then conducted a study of the last 71 sales tickets at her store and found a standard deviation of \(2.95\)dollar per ticket. Is the variability in sales at Julia's franchise, at the 0.05 level of significance, greater than the variability for the company?

Although most people are aware of minor dehydration symptoms such as dry skin and headaches, many are less knowledgeable about the causes of dehydration. According to a poll done for the Nutrition Information Center, the results of a random sample of 3003 American adults showed that \(20 \%\) did not know that caffeine dehydrates. The survey listed a margin of error of plus or minus \(1.8 \%\). a. Describe how this survey of 3003 American adults fits the properties of a binomial experiment. Specifically identify \(n,\) a trial, success, \(p,\) and \(x .\) b. What is the point estimate for the proportion of all Americans who did not know that caffeine dehydrates? Is it a parameter or a statistic? c. Calculate the \(95 \%\) confidence maximum error of estimate for a binomial experiment of 3003 trials that result in an observed proportion of \(0.20 .\) d. How is the maximum error, found in part c, related to the \(1.8 \%\) margin of error quoted in the survey report? e. Find the \(95 \%\) confidence interval for the true proportion \(p\) based on a binomial experiment of 3003 trials that results in an observed proportion of 0.20.
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