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The Associated Press (December 16,1991 ) reported that in a random sample of 507 people, only 142 correctly described the Bill of Rights as the first 10 amendments to the U.S. Constitution. Calculate a \(95 \%\) confidence interval for the proportion of the entire population that could give a correct description.

Short Answer

Expert verified
Therefore, with 95% confidence, the proportion of the entire population that could give a correct description of the Bill of Rights is between 23.67% and 32.33%.

Step by step solution

01

Calculate the Sample Proportion (p̂)

The formula to calculate the population proportion (p̂) using a sample subgroup is: \(p̂ = x / n\), where x is the number of successes in the sample and n is the total size of the sample. In this situation, x is the number of people who correctly described the Bill of Rights (142) and n is the total size of the sample (507). Therefore, the sample proportion is \(p̂ = 142 / 507 = 0.28\) or 28%.
02

Calculate the Standard Error

The standard error of the sample proportion (SEp̂) is calculated using the formula: \(SEp̂ = sqrt{[p̂(1 - p̂) / n]}\). Substituting the calculated sample proportion from Step 1 and the total sample size into the formula, we get \(SEp̂ = sqrt{[0.28(1 - 0.28) / 507]} = 0.022\) or 2.2%.
03

Calculate the Confidence Interval

A 95% confidence interval for the population proportion is calculated by the formula: \(CI = p̂ ± Z * SEp̂\), where Z is the z-score matching the desired confidence level. The z-score for a 95% confidence interval is 1.96. To establish the confidence interval, compute \(CI = 0.28 ± 1.96 * 0.022\). The lower bound of the confidence interval is \(0.28 - 1.96*0.022 = 0.2367\) and the upper bound is \(0.28 + 1.96*0.022 = 0.3233\). Thus, the confidence interval is (0.2367, 0.3233).

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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, often denoted as \( \hat{p} \), is a statistic representing the fraction of successes in a sample in relation to the total number of observations. It serves as an estimate for the true population proportion. For instance, in a survey where 142 out of 507 participants identified the Bill of Rights correctly, the sample proportion is the number of successes (142) divided by the total number of participants (507), which equals approximately 0.28 or 28%. This tells us that in our sample, 28% of respondents could correctly describe the Bill of Rights.

Understanding and calculating the sample proportion is fundamental as it forms the basis for confidence intervals, hypothesis testing, and various other statistical conclusions. It's important to remember that the sample proportion is only an estimate, and its precision relies heavily on the size and randomness of the sample.
Standard Error of the Proportion
Standard error, often abbreviated as SE, measures the variability or precision of a sampling statistic such as the sample proportion. In the context of a proportion, the standard error can be understood as the standard deviation of the distribution of the sample proportion. It provides an indication of how much the sample proportion could differ from the true population proportion if different samples were taken.

The formula for the standard error of the sample proportion is \( SE_{\hat{p}} = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \), where \( \hat{p} \) is the sample proportion and \( n \) is the sample size. Using our exercise example, with a sample proportion of 0.28 and a sample size of 507, the standard error is calculated to be 0.022 or 2.2%. This small value suggests our estimate is relatively precise, meaning that our sample proportion is likely to be a good representation of the population proportion.
Z-score and Confidence Intervals
A z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. When it comes to confidence intervals, the z-score helps determine how much the sample proportion could vary from the true population proportion.

The z-score is used to capture the area under the normal distribution curve that corresponds to a certain confidence level. For example, a z-score of 1.96 is associated with a 95% confidence interval, implying that if you were to repeat the experiment multiple times, 95% of the constructed confidence intervals would contain the true population proportion.

In the textbook exercise, a 95% confidence interval for the proportion of people who know the Bill of Rights is determined by multiplying the standard error (0.022) with the z-score (1.96). The resulting values are then added to and subtracted from the sample proportion (0.28) to find the lower and upper bounds of the confidence interval, yielding the interval (0.2367, 0.3233). These bounds tell us that we can be 95% confident that the true population proportion lies within this range.

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Most popular questions from this chapter

The article "National Geographic, the Doomsday Machine," which appeared in the March 1976 issue of the Journal of Irreproducible Results (yes, there really is a journal by that name-it's a spoof of technical journals!) predicted dire consequences resulting from a nationwide buildup of National Geographic magazines. The author's predictions are based on the observation that the number of subscriptions for National Geographic is on the rise and that no one ever throws away a copy of National Geographic. A key to the analysis presented in the article is the weight of an issue of the magazine. Suppose that you were assigned the task of estimating the average weight of an issue of National Geographic. How many issues should you sample to estimate the average weight to within 0.1 oz with \(95 \%\) confidence? Assume that \(\sigma\) is known to be 1 oz.

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The Center for Urban Transportation Research released a report stating that the average commuting distance in the United States is \(10.9 \mathrm{mi}\) (USA Today, August \(13 .\) 1991). Suppose that this average is actually the mean of a random sample of 300 commuters and that the sample standard deviation is \(6.2 \mathrm{mi}\). Estimate the true mean commuting distance using a \(99 \%\) confidence interval.

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