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In an AP-AOL sports poll (Associated Press, December 18,2005\(), 394\) of 1000 randomly selected U.S. adults indicared that they considered themselves to be baseball fans. Of the 394 baseball fans, 272 stated that they thought the designated hitter rule should either be expanded to both baseball leagues or eliminated. a. Construct a \(95 \%\) confidence interval for the proportion of U.S. adults who consider themselves to be bascball fans. b. Construct a \(95 \%\) confidence interval for the proportion of those who consider themselves to be baseball fans who think the designated hitter rule should be expanded to both leagues or eliminated. c. Explain why the confidence intervals of Parts (a) and (b) are not the same width even though they both have a confidence level of \(95 \%\).

Short Answer

Expert verified
Part a: The 95% confidence interval for the proportion of U.S. adults who consider themselves to be baseball fans will be approximately \(0.394 \pm 1.96 \times \sqrt{ (0.394 \times (1-0.394)) / 1000 }\). Part b: The 95% confidence interval for the proportion of baseball fans who think the designated hitter rule should be expanded or eliminated would be \(0.690 \pm 1.96 \times \sqrt{ (0.690 \times (1-0.690)) / 394 }\). The difference in widths of the confidence intervals of parts (a) and (b) is because of the different sample sizes used in each scenario.

Step by step solution

01

Construct confidence interval for proportion of U.S. adults who are baseball fans

Assuming the sample is from a normal distribution, we can use the formula for the confidence interval, which is \( p \pm z \times \sqrt{ (p \times (1-p)) / n }\), where \(p\) is the estimated proportion, \(z\) is z-value (For \(95\%\) confidence interval, z is approximately \(1.96\)), and \(n\) is the sample size. The estimated proportion \(p\) for this case is \(394/1000 = 0.394\). Therefore, 95% confidence interval will be \( 0.394 \pm 1.96 \times \sqrt{ (0.394 \times (1-0.394)) / 1000 } \).
02

Construct confidence interval for proportion of baseball fans who think the designated hitter rule should be expanded or eliminated

Again, the formula for the confidence interval is used, where \(p\) is the estimated proportion, \(z\) is z-value, and \(n\) is the sample size. The estimated proportion \(p\) in this case is \(272/394 = 0.690\). So, the 95% confidence interval would be \( 0.690 \pm 1.96 \times \sqrt{ (0.690 \times (1-0.690)) / 394 } \).
03

Explain why the confidence intervals of parts (a) and (b) are not the same width

The difference in the width of the confidence intervals is due to the differing sample sizes used in determining the two proportions. While the first interval depends on a sample of 1000 adults, the second interval is based on a sample of only 394 baseball fans. The size of the sample directly influences the standard error, which forms part of the calculation for the interval. A larger sample size will attribute to a smaller standard error, hence a narrower confidence interval, supposing all other conditions remain the same.

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Key Concepts

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

Proportion
The concept of proportion is crucial in statistics, especially when determining how much of a population exhibits a certain characteristic. In this context, a proportion is the fraction of the sample that has a particular attribute. For the exercise mentioned, proportions help us understand two things:
  • What fraction of U.S. adults consider themselves baseball fans?
  • What fraction of those baseball fans think the designated hitter rule should be altered?
The calculated proportions offer insight into public opinion. Calculating a proportion is straightforward: divide the number of favorable outcomes by the total number of cases considered. It provides a snapshot of the population based on the sample and is foundational in determining confidence intervals.
Sample Size
Sample size, denoted by \( n \), is a critical component in statistical studies. It affects the accuracy and reliability of the confidence intervals we estimate. In the exercise provided:
  • A sample of 1000 was used to determine how many U.S. adults are baseball fans.
  • A smaller subset of 394 baseball fans was examined to see their opinions on the designated hitter rule.
The size of the sample is pivotal:
  • A larger sample size can lead to more accurate estimates, as it better represents the population.
  • Smaller samples might increase the margin of error and widen the confidence interval.
Thus, understanding the effect of sample size on the results is essential to accurately interpret confidence intervals and their widths.
Standard Error
Standard error is a measure that tells us how much sample mean estimates are likely to vary from the true population mean. It plays a vital part in confidence intervals, impacting their width. The formula for standard error is \( \sqrt{ (p \times (1-p)) / n } \), where:
  • \( p \) is the sample proportion.
  • \( n \) is the sample size.
In the exercise, calculating the standard error allowed us to frame how precisely we can estimate the true proportion for the population.
  • The larger the standard error, the wider the confidence interval.
  • Conversely, a smaller standard error results from larger sample sizes, which is why sample size is paramount in determining the reliability of confidence intervals.
This understanding helps in refining the confidence intervals to make them more reflective of reality.
Normal Distribution
In statistical analysis, the assumption of a normal distribution is a key concept because it allows us to use certain statistical methods, such as constructing confidence intervals. \(z\) values are derived from this assumption. When your data follows a normal distribution, it is symmetric, bell-shaped, and characterized by its mean and standard deviation. In this context:
  • The central limit theorem suggests that with a large enough sample size, the distribution of the sample proportion will approximate a normal distribution, regardless of the population's distribution.
  • This property is why the formula for a confidence interval includes a \( z \)-score, typically 1.96 for a 95% confidence level.
The use of normal distribution assumptions ensures that our statistical conclusions, like those about the proportion of baseball fans or their opinions on rule changes, are valid and reliable, given the sample data available.

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Most popular questions from this chapter

The interval from -2.33 to 1.75 captures an area of .95 under the \(z\) curve. This implies that another large-sample \(95 \%\) confidence interval for \(\mu\) has lower limit \(\bar{x}-2.33 \frac{\sigma}{\sqrt{n}}\) and upper limit \(\bar{x}+1.75 \frac{\sigma}{\sqrt{n}},\) Would you recommend using this \(95 \%\) interval over the \(95 \%\) interval \(\bar{x} \pm 1.96 \frac{\sigma}{\sqrt{n}}\) discussed in the text? Explain. (Hint: Look at the width of each interval.)

The report "2005 Electronic Monitoring \& Surveillance Survey Many Companies Monitoring. Recording. Videotaping-and Firing-Employees" (American Management Association, 2005\()\) summarized the results of a survey of 526 U.S. businesses. The report stated that 137 of the 526 businesses had fired workers for misuse of the Internet and 131 had fired workers for e-mail misuse. For purposes of this exercise, assume that it is reasonable to regard this sample as representative of businesses in the Unired Srates. a. Construct and interpret a \(95 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of the Internet. b. What are two reasons why a \(90 \%\) confidence interval for the proportion of U.S. businesses that have fired workers for misuse of e-mail would be narrower than the \(95 \%\) confidence interval computed in Part (a)?

Discuss how each of the following factors affects the width of the confidence interval for \(P\) : a. The confidence level b. The sample size c. The value of \(\hat{p}\)

The authors of the paper "Short-Term Health and Economic Benefits of Smoking Cessation: Low Birth Weight" (Pediatrics [1999]: \(1312-1320\) ) investigated the medical cost associated with babies born to mothers who smoke. The paper induded estimates of mean medical cost for low-birth-weight babies for different cthnic groups. For a sample of 654 Hispanic lowbirth-weight babies, the mean medical cost was \(\$ 55,007\) and the standard error \((s / \sqrt{n})\) was \(\$ 3011\). For a sample of 13 Native American low-birth-weight babies, the mean and standard error were \(\$ 73,418\) and \(\$ 29,577\), respectively. Explain why the two standard errors are so different.

The authors of the paper "Deception and Design . The Impact of Communication Technology on Lying Behavior" (Proceedings of Computer Human Interaction [2004]\()\) asked 30 students in an upper division communications course at a large university to keep a journal for 7 days, recording each social interaction and whether or not they told any lies during that interaction. A lie was defined as "any time you intentionally try to mislead someone." The paper reported that the mean number of lies per day for the 30 students was 1.58 and the standard deviation of number of lies per day was 1.02 a. What assumption must be made in order for the \(t\) confidence interval of this section to be an appropriate method for estimating \(\mu,\) the mean number of lies per day for all students at this university? b. Would you recommend using the \(t\) confidence interval to construct an estimate of \(\mu\) as defined in Part (a)? Explain why or why not.

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