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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 sample size needed for the 99% confidence interval with a margin of error of 0.03 is 1193.

Step by step solution

01

Identify proportion, confidence level, margin of error

From the problem, we know that the proportion (\(p\)) from the preliminary study is 0.76, the confidence level is 99% which corresponds to a Z-score (\(z\)) of 2.576 (for a two-tailed test), and the margin of error (E) is 0.03.
02

Apply the formula for sample size

The sample size (\(n\)) can be calculated using the formula \( n = (z^2*p*(1-p))/E^2 \). Plugging in the values we got from Step 1, this gives us \( n = (2.576^2*0.76*(1-0.76))/(0.03^2) \).
03

Calculate the sample size

Upon computation, we get \( n \approx 1192 \). Since we can't have a fraction of a person wearing a seatbelt, we need to round this number up to the next whole number. So, the needed sample size should be 1193.

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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 is an essential concept in statistics. It provides a range of values, derived from a sample, that is likely to contain the true parameter of the population. In other words, it estimates the uncertainty of an estimate. Confidence intervals are expressed with a confidence level, usually in percentage terms, such as 95% or 99%, indicating the probability that the interval contains the population parameter.
  • The wider the confidence interval, the more uncertain the estimate is.
  • The higher the confidence level, the wider the interval becomes, because we are saying "we are more sure that the parameter lies within this range."
In our context, the 99% confidence interval means that we are 99% certain that the true proportion of drivers wearing seat belts falls within our calculated range. This high level of confidence typically requires more data, which explains why the sample size can be substantial.
Population Proportion
The population proportion is a measure of how common a particular trait or behavior is in a population. It is expressed as a fraction or percentage. For instance, if we know that the proportion of drivers wearing seat belts is 0.76, it means that 76% of the entire driving population regularly wears seat belts.
When dealing with population proportions, samples are often taken to infer about the broader population. The sample proportion serves as an estimate of the true population proportion. Our exercise hinges on understanding this idea because it is the starting point for all sample and interval calculations.
  • The larger the sample size and the closer the sample proportion is to 0.5, the more accurately it is expected to reflect the true population proportion.
  • This accuracy is directly linked to the concept of confidence intervals, which is why careful measurement and correct calculation of sample sizes are critical.
Sample Size Calculation
Sample size calculation is crucial for ensuring that study results are valid and reliable. It refers to determining how large a sample should be to draw dependable conclusions. To calculate sample size, several factors must be taken into account, including given proportions, confidence levels, and margins of error.
  • The formula \( n = \frac{z^2 \times p \times (1-p)}{E^2} \) is employed for this calculation, where \( z \) is the Z-score corresponding to the desired confidence level, \( p \) is the preliminary estimate of population proportion, and \( E \) is the margin of error.
  • For example, with a confidence level of 99% and a margin of error of 0.03, the necessary computation reveals a sample size of around 1193.
Accurate sample size ensures the study has enough power to discern the true population proportion within our margins of error, maintaining the integrity of the conclusion.

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Most popular questions from this chapter

The mean time taken to design a house plan by 40 architects was found to be 23 hours with a standard deviation of \(3.75\) hours. a. Construct a \(98 \%\) confidence interval for the population mean \(\mu .\) b. Suppose the confidence interval obtained in part a is too wide. How can the width of this interval be reduced? Describe all possible alternatives. Which alternative is the best and why?

What is the point estimator of the population proportion, \(p ?\)

a. A sample of 1100 observations taken from a population produced a sample proportion of .32. Make a \(90 \%\) confidence interval for \(p\). b. Another sample of 1100 observations taken from the same population produced a sample proportion of .36. Make a \(90 \%\) confidence interval for \(p\). c. A third sample of 1100 observations taken from the same population produced a sample proportion of .30. Make a \(90 \%\) confidence interval for \(p\). d. The true population proportion for this population is \(.34 .\) Which of the confidence intervals constructed in parts a through c cover this population proportion and which do not?

A drug that provides relief from headaches was tried on 18 randomly selected patients. The experiment showed that the mean time to get relief from headaches for these patients after taking this drug was 24 minutes with a standard deviation of \(4.5\) minutes. Assuming that the time taken to get relief from a headache after taking this drug is (approximately) normally distributed, determine a \(95 \%\) confidence interval for the mean relief time for this drug for all patients.

For each of the following, find the area in the appropriate tail of the \(t\) distribution. a. \(t=-1.302\) and \(d f=42\) b. \(t=2.797\) and \(n=25\) c. \(t=1.397\) and \(n=9\) d. \(t=-2.383\) and \(d f=67\)

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