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91Ó°ÊÓ

USA Today (October 14,2002 ) reported that \(36 \%\) of adult drivers admit that they often or sometimes talk on a cell phone when driving. This estimate was based on data from a sample of 1004 adult drivers, and a bound on the error of estimation of \(3.1 \%\) was reported. Explain how the given bound on the error can be justified.

Short Answer

Expert verified
The error bound of 3.1% can be justified by using the Margin of Error formula with a 95% confidence level. The calculated result closely aligns with the given error bound.

Step by step solution

01

Understanding Margin of Error

The margin of error is an indicator of the estimation error. Assuming a confidence level of 95%, the z-score is approximately 1.96. The formula for margin of error (E) is \( E = Z * \sqrt{ \frac{P*(1-P)}{n} } \) where P is the sample proportion and n the sample size.
02

Apply values to Margin of Error formula

Let's substitute P = 0.36 (the proportion given in the exercise) and n = 1004 (the sample size) into the formula. The Z value is 1.96 for a 95% confidence interval.
03

Calculate the error bound

Perform the calculation: \( E = 1.96 * \sqrt{ \frac{0.36*(1-0.36)}{1004} } \). The calculated result will be the error estimation in decimal, you must convert it to percentage by multiplying the result by 100.
04

Validate the given error bound

After deriving the margin of error from the given data, compare it with the bound error given in the question (3.1%). If they match, it justifies why the error bound is valid.

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Key Concepts

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

Sampling Error
When conducting a study, researchers often use a sample—a subset of the population—to draw conclusions about the population as a whole. A sampling error occurs due to the natural variation between the sample and the actual population. It's the difference between the sample statistic (like a sample mean or proportion) and the population parameter it's meant to estimate. The size of the sampling error can fluctuate depending on the size of the sample and the variance within the population.

In the case of the USA Today survey, the sample consisted of 1004 adult drivers. The sampling error in this context reflects the uncertainty that arises because only a portion of all adult drivers were surveyed. Although every adult driver has an equal chance of being included in the sample, each sample can yield a slightly different result. This is why the bound on the error of estimation is crucial—it gives us a way to measure and report this uncertainty.
Confidence Interval
A confidence interval is a range of values that is likely to contain a population parameter with a certain degree of confidence. It’s constructed around a sample statistic to capture the true population parameter. For example, if researchers report a 95% confidence interval for a proportion, it means that if they were to take numerous random samples and calculate a confidence interval for each sample, they expect about 95% of those intervals to contain the true population proportion.

The width of the confidence interval is determined by several factors, including the size of the sample, the level of confidence one desires, and the variability of the data. In the example question provided, the margin of error reported (3.1%) is part of the confidence interval. When we say that 36% of adult drivers admit to using their cell while driving with a margin of error of 3.1%, we are expressing a 95% confidence interval that the true proportion in the entire population would fall between 32.9% (36% - 3.1%) and 39.1% (36% + 3.1%).
Sample Proportion Calculation
Calculating the sample proportion involves taking the number of occurrences of a certain event in a sample and dividing it by the total sample size. It is represented by 'P' in statistical formulas. The formula for the sample proportion is \( P = \frac{x}{n} \) where 'x' is the number of successes and 'n' is the total number of observations in the sample.

The sample proportion calculation in the context of the USA Today survey indicated that 36% of adult drivers often or sometimes talk on a phone while driving, which translates to 0.36 as the sample proportion (P). To verify this with error bounds, statisticians use the margin of error formula: \( E = Z * \sqrt{ \frac{P*(1-P)}{n} } \). By inserting the values of the sample proportion (P=0.36) and sample size (n=1004) into the formula and multiplying by the z-score that corresponds to the desired confidence level, we can calculate the error bound, which helps gauge the precision of the estimated proportion. This calculated bound should ideally be close to the reported bound on the error of estimation, which is a measure of reliability for the reported percentage.

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Most popular questions from this chapter

The Chronicle of Higher Education (January 13, 1993) reported that \(72.1 \%\) of those responding to a national survey of college freshmen were attending the college of their first choice. Suppose that \(n=500\) students responded to the survey (the actual sample size was much larger). a. Using the sample size \(n=500\), calculate a \(99 \%\) confidence interval for the proportion of college students who are attending their first choice of college. b. Compute and interpret a \(95 \%\) confidence interval for the proportion of students who are not attending their first choice of college. c. The actual sample size for this survey was much larger than 500 . Would a confidence interval based on the actual sample size have been narrower or wider than the one computed in Part (a)?

For each of the following choices, explain which would result in a wider large-sample confidence interval for \(\pi\) : a. \(90 \%\) confidence level or \(95 \%\) confidence level b. \(n=100\) or \(n=400\)

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Seventy-seven students at the University of Virginia were asked to keep a diary of a conversation with their mothers, recording any lies they told during these conversations (San Luis Obispo Telegram-Tribune, August 16 , 1995 ). It was reported that the mean number of lies per conversation was \(0.5 .\) Suppose that the standard deviation (which was not reported) was \(0.4\). a. Suppose that this group of 77 is a random sample from the population of students at this university. Construct a \(95 \%\) confidence interval for the mean number of lies per conversation for this population. b. The interval in Part (a) does not include 0 . Does this imply that all students lie to their mothers? Explain.

Recent high-profile legal cases have many people reevaluating the jury system. Many believe that juries in criminal trials should be able to convict on less than a unanimous vote. To assess support for this idea, investigators asked each individual in a random sample of Californians whether they favored allowing conviction by a 10-2 verdict in criminal cases not involving the death penalty. The Associated Press (San Luis Obispo TelegramTribune, September 13,1995 ) reported that \(71 \%\) supported the \(10-2\) verdict. Suppose that the sample size for this survey was \(n=900\). Compute and interpret a \(99 \%\) confidence interval for the proportion of Californians who favor the \(10-2\) verdict.

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