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A consumer agency wants to estimate the proportion of all drivers who wear seat belts while driving. Assume that a preliminary study has shown that \(76 \%\) of drivers wear seat belts while driving. How large should the sample size be so that the \(99 \%\) confidence interval for the population proportion has a margin of error of \(.03\) ?

Short Answer

Expert verified
The required sample size is 1067.

Step by step solution

01

Identify the Given Values

From the details provided, we have the population proportion (p) as 0.76, the desired confidence level (Z) which in this case is 2.57 (99% confidence level), and the margin of error (E) as 0.03.
02

Set up Margin of Error Formula and Input Given Values

The formula for margin of error (E) is \(E = Z\sqrt{{\frac{{p(1-p)}}{n}}}\). In this case, to find 'n', we need to rearrange the formula to be in terms of 'n': \(n = {\frac{{Z^2p(1-p)}}{E^2}}\). Now, plug in the given values and solve for 'n'.
03

Solve for 'n'

Substituting the appropriate values to the formula gives \(n = {\frac{{(2.57)^2 \times 0.76 \times (1-0.76)}}{(0.03)^2}}\). Solving this equation yields a value approximately equal to 1066.72.
04

Round up 'n' to the Nearest Whole Number

Since we cannot have a fractional sample size, we have to round up to the next whole number. Therefore, the sample size 'n' should be 1067 to achieve a 99% confidence interval with a margin of error of 0.03.

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Key Concepts

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

Confidence Interval
A confidence interval provides a range of values which is likely to include a population parameter with a certain level of confidence. In the context of estimating the proportion of drivers who wear seat belts, the confidence interval suggests a range where the actual population proportion is likely to fall. Confidence levels, like 99% in this case, indicate the degree of certainty we have about this range. The higher the confidence level, the wider the interval and the more certain we are that it contains the true population parameter. In practical terms, a 99% confidence interval means that if we were to take many random samples from the population, 99% of those intervals would contain the true population proportion.
Margin of Error
The margin of error determines how much the observed results are expected to differ from the true population value. It is an essential part of constructing confidence intervals. For example, a margin of error of 0.03 implies that the estimated proportion could vary by plus or minus 0.03 from the true population proportion. By controlling the margin of error with sample size calculations, researchers can ensure that their confidence intervals are precise enough for their purpose. A smaller margin of error requires a larger sample size, allowing more certainty in estimating the proportion of drivers who wear seat belts.
Population Proportion
The population proportion refers to the percentage of the entire group that displays a particular attribute or behavior. In this exercise, it is the proportion of all drivers who wear seat belts, which from preliminary information is about 76%, or 0.76 as a decimal. This value helps to estimate the overall proportion for the larger group from the sample. Knowing the population proportion is crucial in calculating the necessary sample size for a study, allowing researchers to make informed estimations about the population as a whole based on sample data.
Statistical Estimation
Statistical estimation involves making inferences about population parameters based on sample data. This process is vital in drawing meaningful conclusions in research fields. In this exercise, statistical estimation uses sample data to approximate the proportion of drivers wearing seat belts in the entire population. There are two main types of estimation: point estimation, which gives a single value estimate, and interval estimation, which provides a range of values — the confidence interval. Using statistical estimation techniques allows researchers to make accurate predictions about the population, aiding in policy-making and understanding trends.

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Most popular questions from this chapter

a. A sample of 300 observations taken from a population produced a sample proportion of .63. Make a \(95 \%\) confidence interval for \(p\). b. Another sample of 300 observations taken from the same population produced a sample proportion of .59. Make a \(95 \%\) confidence interval for \(p\). c. A third sample of 300 observations taken from the same population produced a sample proportion of .67. Make a \(95 \%\) confidence interval for \(p\). d. The true population proportion for this population is .65. Which of the confidence intervals constructed in parts a through c cover this population proportion and which do not?

What assumptions must hold true to use the \(t\) distribution to make a confidence interval for \(\mu ?\)

An economist wants to find a \(90 \%\) confidence interval for the mean sale price of houses in a state. How large a sample should she select so that the estimate is within \(\$ 3500\) of the population mean? Assume that the standard deviation for the sale prices of all houses in this state is \(\$ 31,500\)

A group of veterinarians wants to test a new canine vaccine for Lyme disease. (Lyme disease is transmitted by the bite of an infected deer tick.) In an area that has a high incidence of Lyme disease, 100 dogs are randomly selected (with their owners' permission) to receive the vaccine. Over a 12 -month period, these dogs are periodically examined by veterinarians for symptoms of Lyme disease. At the end of 12 months, 10 of these 100 dogs are diagnosed with the disease. During the same 12 -month period, \(18 \%\) of the unvaccinated dogs in the area have been found to have Lyme disease. Let \(p\) be the proportion of all potential vaccinated dogs who would contract Lyme disease in this area. a. Find a \(95 \%\) confidence interval for \(p\). b. Does \(18 \%\) lie within your confidence interval of part a? Does this suggest the vaccine might or might not be effective to some degree? c. Write a brief critique of this experiment, pointing out anything that may have distorted the results or conclusions.

Refer to Exercise \(8.116\). Assume that a preliminary sample has shown that \(63 \%\) of the adults in this city favor legalized casino gambling. How large should the sample size be so that the \(95 \%\) confidence interval for the population proportion has a margin of error of \(.05 ?\)

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