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Fans A survey of 81 randomly selected people standing in line to enter a football game found that 73 of them were home team fans. a) Explain why we cannot use this information to construct a confidence interval for the proportion of all people at the game who are fans of the home team. b) Construct a "plus-four" confidence interval and interpret it in this context.

Short Answer

Expert verified
Can't construct a CI because the sample isn't random; Plus-four CI: [81.48%, 95%].

Step by step solution

01

Understand the Scenario

We surveyed 81 people in a line for a football game and found that 73 were home team fans. The issue arises when trying to generalize this proportion to all attendees at the game.
02

Identify the Reason for Inapplicability

The sample may not be randomly representative of all people at the game because it only includes those standing in line. Therefore, the sample might not reflect the diversity of all attendees, introducing bias.
03

Introduce the Plus-Four Method

The plus-four method is used to estimate confidence intervals and is more robust when sample sizes are small or proportions are near 0 or 1. It involves adding 'four imaginary observations': two successes and two failures.
04

Modify Data for Plus-Four

Add 2 home team fans and 2 non-home team fans to the survey data, making it 75 home team fans out of 85 total people surveyed.
05

Calculate the Adjusted Proportion

Compute the new proportion of home team fans: \[ \hat{p} = \frac{75}{85} \approx 0.8824 \]
06

Calculate the Standard Error

The standard error is calculated as:\[ SE = \sqrt{\frac{0.8824 \times (1 - 0.8824)}{85}} \approx 0.0345 \]
07

Find the Critical Value

For a 95% confidence interval, the critical value (z*) is approximately 1.96.
08

Compute the Confidence Interval

The confidence interval is:\[ \hat{p} \pm (z^* \times SE) = 0.8824 \pm (1.96 \times 0.0345) \] \[ = 0.8824 \pm 0.0676 \]This results in the interval [0.8148, 0.9500].
09

Interpret the Confidence Interval

We are 95% confident that the true proportion of home team fans among those surveyed, with the plus-four adjustment, is between 81.48% and 95%.

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Key Concepts

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

Plus-Four Method
The Plus-Four Method is a simple yet powerful technique to ensure more reliable confidence intervals, especially for small sample sizes or sample proportions closely approaching 0 or 1. This method involves adding four extra imaginary observations – two successes and two failures – to your data.
For example, if you observed 73 home team fans out of 81 people, the original calculation might look skewed due to its proximity to the higher end of possible proportion values (near 1 or 100%). By adding these four extra observations, your sample becomes 85 in total, with 75 successes (home team fans) providing a more balanced perspective.
This adjustment ensures that the calculations are less susceptible to skewed results that might be misleading. It brings increased stability and more truthful representation in cases where the sample isn't entirely reflective, or when the boundaries of normal distribution assumptions might weaken.
Sample Bias
Sample bias is a critical problem when trying to estimate a population parameter like the proportion of home team fans. When the sample isn't representative of the entire population, it introduces bias.
In the provided exercise, selecting fans only standing in line at a football game may inadvertently capture more fervent supporters, skewing the proportion data. These people might want to grab seats early, suggesting higher enthusiasm levels than the average attendee.
To avoid sample bias, it's essential to ensure the sample is randomly selected from the entire population. This randomness helps in capturing the diverse behaviors and inclinations otherwise overlooked, leading to a more reliable data set for proportion estimation.
Proportion Estimation
Proportion estimation is a technique used to infer the percentage of a specific trait within an entire population based on a sample. It's like having a snapshot that tries to capture the essence of a larger scenario.
In the scenario presented, we aim to estimate the proportion of total game attendees supporting the home team using the observed data (fans in line). The initial calculation was the ratio of 73 fans out of 81, resulting in 0.9 (or 90%). However, such high proportions can be tricky when using standard methods, due to potential deviation from normality in small samples.
Thus the use of the Plus-Four method becomes beneficial as it stabilizes the proportion estimation, allowing us to derive more robust confidence intervals, and ensuring that our snapshot is as clear and accurate as possible.
Standard Error
Standard error measures the variability or precision of an estimator from the true population parameter; essentially, it tells us how much an estimated proportion might fluctuate if different samples were taken.
Calculating standard error involves the adjusted proportion: for instance, using the Plus-Four Method, the calculation modifies both the proportion and sample size, making the standard error \[ SE = \sqrt{\frac{0.8824 \times (1 - 0.8824)}{85}} \approx 0.0345 \] This calculation helps us understand how much our sample proportion (88.24% in this scenario) could vary if repeated samples were observed from the entire population.
In constructing confidence intervals, this variability is crucial as it essentially informs the range within which the true parameter is believed to lie, ensuring the estimated intervals are not overly optimistic or misleading.

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Most popular questions from this chapter

More \(P\) -values Which of the following are true? If false, explain briefly. a) A very low \(P\) -value provides evidence against the null hypothesis. b) A high P-value is strong evidence in favor of the null hypothesis. c) A P-value above 0.10 shows that the null hypothesis is true. d) If the null hypothesis is true, you can't get a P-value below 0.01.

Hypotheses and parameters As in Exercise \(1,\) for each of the following situations, define the parameter and write the null and alternative hypotheses in terms of parameter values. a) Seat-belt compliance in Massachusetts was \(65 \%\) in \(2008 .\) The state wants to know if it has changed. b) Last year, a survey found that \(45 \%\) of the employees were willing to pay for on-site day care. The company wants to know if that has changed. c) Regular card customers have a default rate of \(6.7 \%\) A credit card bank wants to know if that rate is different for their Gold card customers.

Which alternative? In each of the following situations, is the alternative hypothesis one-sided or two-sided? What are the hypotheses? a) A college dining service conducts a survey to see if students prefer plastic or metal cutlery. b) In recent years, \(10 \%\) of college juniors have applied for study abroad. The dean's office conducts a survey to see if that's changed this year. c) A pharmaceutical company conducts a clinical trial to see if more patients who take a new drug experience headache relief than the \(22 \%\) who claimed relief after taking the placebo. d) At a small computer peripherals company, only \(60 \%\) of the hard drives produced passed all their performance tests the first time. Management recently invested a lot of resources into the production system and now conducts a test to see if it helped.

Hard times In June \(2010,\) a random poll of 800 working men found that \(9 \%\) had taken on a second job to help pay the bills. (www.careerbuilder.com) a) Estimate the true percentage of men that are taking on second jobs by constructing a \(95 \%\) confidence interval. b) A pundit on a TV news show claimed that only \(6 \%\) of working men had a second job. Use your confidence interval to test whether his claim is plausible given the poll data.

Ads A company is willing to renew its advertising contract with a local radio station only if the station can prove that more than \(20 \%\) of the residents of the city have heard the ad and recognize the company's product. The radio station conducts a random phone survey of 400 people. a) What are the hypotheses? b) The station plans to conduct this test using a \(10 \%\) leve of significance, but the company wants the significance level lowered to \(5 \%\). Why? c) What is meant by the power of this test? d) For which level of significance will the power of this test be higher? Why? e) They finally agree to use \(\alpha=0.05,\) but the company proposes that the station call 600 people instead of the 400 initially proposed. Will that make the risk of Type II error higher or lower? Explain.

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