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Chapter 8: Problem 35

Abstain from drinking In a Harvard School of Public Health survey, 2105 of 10,904 randomly selected U.S. college students were classified as abstainers (nondrinkers). (a) Construct and interpret a 99% confidence interval for p. Follow the four- step process. (b) A newspaper article claims that 25% of U.S. college students are nondrinkers. Use your result from (a) to comment on this claim.

Short Answer

Expert verified
(a) 99% CI: 18.34% to 20.28%. (b) The claim of 25% nondrinkers is not supported as it is outside the CI.

Step by step solution

01

Calculate the Sample Proportion

To find the sample proportion \( \hat{p} \), divide the number of abstainers by the total number of students surveyed. This is \( \hat{p} = \frac{2105}{10904} \approx 0.1931 \).
02

Calculate the Standard Error (SE)

The standard error (SE) for the sample proportion is found using the formula \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \). Substitute \( \hat{p} = 0.1931 \) and \( n = 10904 \) into the formula: \( SE \approx \sqrt{\frac{0.1931 \times 0.8069}{10904}} \approx 0.0038 \).
03

Determine the Critical Value for 99% Confidence

For a 99% confidence interval, use the standard normal (Z) distribution to find the critical value. This critical value is approximately 2.576.
04

Construct the Confidence Interval

The confidence interval is calculated using the formula \( \hat{p} \pm Z \times SE \). Using \( \hat{p} = 0.1931 \), \( Z = 2.576 \), and \( SE = 0.0038 \), the interval is \( 0.1931 \pm 2.576 \times 0.0038 \). Calculating this gives \( 0.1834 \) to \( 0.2028 \).
05

Interpret the Confidence Interval

We are 99% confident that the true proportion of U.S. college students who are nondrinkers is between 18.34% and 20.28%.
06

Compare Newspaper Claim with Confidence Interval

The newspaper claims that 25% of U.S. college students are nondrinkers. Since 25% is outside the 99% confidence interval, the survey results do not support this claim.

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Key Concepts

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

Sample Proportion
The sample proportion is an essential component when dealing with surveys and confidence intervals. In this scenario, it represents the ratio of college students who abstain from drinking, based on the surveyed sample.
To calculate it, divide the number of students identified as abstainers (2105) by the total number of students surveyed (10904). This gives a sample proportion, denoted as \( \hat{p} \), calculated as:
  • \( \hat{p} = \frac{2105}{10904} \approx 0.1931 \)
This value, approximately 0.1931, indicates that about 19.31% of the students in the sample do not drink. Understanding the sample proportion helps in drawing inferences about the whole population based on this smaller group.
Standard Error
The standard error (SE) is a measure that indicates the variability or precision of the sample proportion estimate. It tells us how much the sample proportion can be expected to vary from the true population proportion. The formula to compute the standard error for a proportion is:
  • \( SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \)
In this exercise, substituting \( \hat{p} = 0.1931 \) and \( n = 10904 \) into the formula gives:
  • \( SE \approx \sqrt{\frac{0.1931 \times 0.8069}{10904}} \approx 0.0038 \)
A smaller standard error, like 0.0038 in this case, implies a more precise estimate of the population proportion.
Critical Value
The critical value is a point on the distribution chart which aids in computing confidence intervals. For confidence intervals, we often use the z-distribution, which relates to the standard normal distribution. The critical value determines the probability threshold at which we can say how sure we are that the true population parameter will fall within the confidence interval. For a 99% confidence level, the critical value is:
  • Approximately 2.576.
This value comes from the z-table, and its magnitude implies a high level of confidence (99%) that the interval calculated will truly contain the population proportion. It's vital because it directly influences the width of the confidence interval.
Confidence Level
A confidence level reflects the degree of certainty with which the interval estimate contains the population parameter. In simple terms, it's the probability that the interval calculated from repeated sampling will cover the actual population proportion. In this exercise, a 99% confidence level was chosen, indicating very high certainty.
  • This means if we were to survey students multiple times, 99% of the calculated confidence intervals from these samples would contain the true proportion of nondrinkers.
  • Higher confidence levels yield wider intervals, as they require covering more of the distribution curve.
The interpretation of the confidence interval (18.34% to 20.28%) means we are confident that the true proportion of nondrinking students in the entire population lies within this range with 99% certainty. Nonetheless, choosing a 99% confidence level ensures high reliability of the estimates, but also means dealing with larger intervals.

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Most popular questions from this chapter

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