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The article "Nine Out of Ten Drivers Admit in Survey to Having Done Something Dangerous" (Knight Ridder Newspapers, July 8,2005 ) reported the results of a survey of 1100 drivers. Of those surveyed, 990 admitted to careless or aggressive driving during the previous 6 months. Assuming that it is reasonable to regard this sample of 1100 as representative of the population of drivers, use this information to construct a \(99 \%\) confidence interval to estimate \(p\), the proportion of all drivers who have engaged in careless or aggressive driving in the previous 6 months.

Short Answer

Expert verified
The 99% confidence interval for the proportion of drivers who have engaged in careless or aggressive driving in the past 6 months is [0.876, 0.923].

Step by step solution

01

Calculate the Sample Proportion

The sample proportion \(p\) is calculated by dividing the number of successes by the total number of observations, which results in \(p = x/n = 990/1100 = 0.9\). This proportion represents the observed probability of a driver admitting to dangerous driving.
02

Identify the Z-Score

By consulting a standard normal distribution table, we find that a 99% confidence interval corresponds to a Z score of 2.576. This Z score reflects how many standard deviations an element is from the mean in a normal distribution.
03

Compute the Standard Error

The standard error (SE) is determined by the formula \(\sqrt{p(1 - p)/n} = \sqrt{0.9 * 0.1 / 1100} = 0.0091287\). The standard error measures the statistical accuracy of an estimate or a prediction.
04

Calculate the Confidence Interval

The calculation is completed by subtracting and adding the product of the Z score and the standard error (Z*SE) from the observed proportion \(p\). The resulting interval, \([p - Z*SE, p + Z*SE] = [0.9 - 2.576*0.0091287, 0.9 + 2.576*0.0091287] = [0.876,0.923]\), is the 99% confidence interval.

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Key Concepts

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

Sample Proportion
When conducting surveys or experiments, we often need to summarize the results with a single value that reflects the entire sample. This is where the concept of **Sample Proportion** becomes crucial. The sample proportion, denoted by \( \hat{p} \), is a way to express the ratio of occurrences of a certain event to the total number of observations in a sample.

For example, in a survey of 1100 drivers, if 990 admitted to dangerous driving, the sample proportion \( \hat{p} \) would be \( \frac{990}{1100} = 0.9 \). This 0.9, or 90%, tells us that 90% of the sampled drivers confessed to having engaged in careless or aggressive driving.

Sample proportion is not only a measure of what happened in the sample but also serves as a crucial step in making inferences about the whole population — this population being all drivers in this context. Accurate calculation of the sample proportion is the first step in finding a confidence interval, which estimates the range within which the true population proportion is likely to be.
Z-Score
Z-Score is a statistical term that might sound intimidating at first, but it’s surprisingly straightforward and incredibly powerful. A Z-Score is essentially a measure of how many standard deviations an element is from the mean.

In our confidence interval exercise, the Z-Score helps us understand the spread of our data points around the mean in a standard normal distribution. For instance, when constructing a 99% confidence interval, we utilize a Z-Score of 2.576. This Z-Score indicates that the stochastic fluctuations of our dataset are usually within 2.576 standard deviations of the mean.

In practical terms, by using Z-Scores, we can accommodate variations and build a range (confidence interval) to predict the true population parameter with high certainty. Specifically, a 99% confidence interval guarantees that if we were to take 100 different samples, we’d expect the true population proportion to lie within our calculated interval in 99 of those samples.
Standard Error
The **Standard Error** is a statistical metric that plays an essential role in the estimation process of how well our sample statistic represents the population parameter. It quantifies the variability or dispersion of the sample proportion \( \hat{p} \).

In our example, we used the formula \( \text{SE} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \), where \( \hat{p} = 0.9 \) and \( n = 1100 \). By calculating \( \text{SE} = \sqrt{\frac{0.9 \times 0.1}{1100}} = 0.0091287 \), we find the standard error to be approximately 0.0091.

This small Standard Error indicates that there is not much variation expected between our sample proportion and the true population proportion. The standard error is crucial as it directly affects the width of the confidence interval: a smaller SE results in a narrower and more precise interval, while a larger SE would lead to a wider interval, indicating more uncertainty in our estimate. Through understanding the standard error, one gains insights into the confidence level and reliability of the survey results.

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Most popular questions from this chapter

A random sample of 10 houses heated with natural gas in a particular area, is selected, and the amount of gas (in therms) used during the month of January is determined for each house. The resulting observations are as follows: \(\begin{array}{llllllllll}103 & 156 & 118 & 89 & 125 & 147 & 122 & 109 & 138 & 99\end{array}\) a. Let \(\mu_{J}\) denote the average gas usage during January by all houses in this area. Compute a point estimate of \(\mu_{J}\) b. Suppose that 10,000 houses in this area use natural gas for heating. Let \(\tau\) denote the total amount of gas used by all of these houses during January. Estimate \(\tau\) using the given data. What statistic did you use in computing your estimate? c. Use the data in Part (a) to estimate \(p\), the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median us- age based on the sample of Part (a). Which statistic did you use?

Suppose that each of 935 smokers received a nicotine patch, which delivers nicotine to the bloodstream but at a much slower rate than cigarettes do. Dosage was decreased to 0 over a 12 -week period. Suppose that 245 of the subjects were still not smoking 6 months after treatment. Assuming it is reasonable to regard this sample as representative of all smokers, estimate the percentage of all smokers who, when given this treatment, would refrain from smoking for at least 6 months.

Samples of two different models of cars were selected, and the actual speed for each car was determined when the speedometer registered \(50 \mathrm{mph}\). The resulting \(95 \%\) confidence intervals for mean actual speed were \((51.3,52.7)\) and \((49.4,50.6)\). Assuming that the two sample standard deviations are equal, which confidence interval is based on the larger sample size? Explain your reasoning.

Discuss how each of the following factors affects the width of the confidence interval for \(p\) : a. The confidence level b. The sample size c. The value of \(\hat{p}\)

Data consistent with summary quantities in the article referenced in Exercise \(9.3\) on total calorie consumption on a particular day are given for a sample of children who did not eat fast food on that day and for a sample of children who did eat fast food on that day. Assume that it is reasonable to regard these samples as representative of the population of children in the United States. $$\begin{aligned} &\text { No Fast Food }\\\ &\begin{array}{llllllll} 2331 & 1918 & 1009 & 1730 & 1469 & 2053 & 2143 & 1981 \\ 1852 & 1777 & 1765 & 1827 & 1648 & 1506 & 2669 & \end{array} \end{aligned}$$ $$\begin{aligned} &\text { Fast Food } \\ &\begin{array}{rrrrrrrr} 2523 & 1758 & 934 & 2328 & 2434 & 2267 & 2526 & 1195 \\ 890 & 1511 & 875 & 2207 & 1811 & 1250 & 2117 & \end{array} \end{aligned}$$ a. Use the given information to estimate the mean calorie intake for children in the United States on a day when no fast food is consumed. b. Use the given information to estimate the mean calorie intake for children in the United States on a day when fast food is consumed. c. Use the given information to produce estimates of the standard deviations of calorie intake for days when no fast food is consumed and for days when fast food is consumed.

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