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Chapter 22: Problem 15

The proportion of drivers who use seat belts depends on characteristics such as age, sex, and ethnicity. As part of a broader study, investigators observed a random sample of 117 female Hispanic drivers in Boston. Suppose that, in fact, \(60 \%\) of all female Hispanic drivers in the Boston area wear seat belts. In repeated samples, the sample proportion \(p^{\wedge \hat{p}}\) would follow approximately a Normal distribution with mean (a) \(70.2\). (b) \(0.6\) (C) \(0.002\).

Short Answer

Expert verified
(b) 0.6

Step by step solution

01

Understand the Problem

We need to find the expected mean of the sample proportion \( \hat{p} \) for a large sample size. We are given that the true proportion \( p = 0.6 \) for all female Hispanic drivers in a specific region who wear seat belts. We aim to find the mean of the sample proportion \( \hat{p} \) when \( p = 0.6 \).
02

Mean of Sample Proportion

The concept used here is that the sample proportion \( \hat{p} \) follows a Normal distribution as the sample size increases. For a sufficiently large sample, the mean of the sample proportion \( \hat{p} \) is equal to the population proportion \( p \). Therefore, the mean of \( \hat{p} \) when the sample is large is \( p = 0.6 \).
03

Compare with Given Choices

We have identified that the mean of the sample proportion is \( 0.6 \). Now, let's compare this value with the options provided in the exercise. Option (b) states the mean is \( 0.6 \).
04

Conclusion

The correct answer based on our calculation and understanding is option (b). The mean of the sample proportion \( \hat{p} \) for such a sample size, assuming that \( 60\% \) is the actual population proportion, is indeed \( 0.6 \).

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Key Concepts

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

sample proportion
A sample proportion is a statistical measure that represents the proportion of a particular characteristic within a sample. Suppose we have a group of 117 female Hispanic drivers and we want to find out how many wear seat belts. If 70 out of 117 drivers wear seat belts, the sample proportion \( \hat{p} \) is calculated as:\[ \hat{p} = \frac{70}{117} \]Sample proportions are useful because they serve as an estimate of the population proportion, which is the proportion of the entire population that has the characteristic. As the sample size increases, the sample proportion becomes a better estimate of the population proportion.
normal distribution
The normal distribution is a fundamental concept in statistics. It is often called the "bell curve" due to its shape. When discussing sample proportions, the normal distribution becomes particularly important when the sample size is sufficiently large. In our example, the sample size is 117, which is considered large enough.Why does this matter? Well, the beauty of the normal distribution is that it allows us to make predictions and assess probabilities about the data. For sample proportions, it means that as the size of our sample increases, the distribution of \( \hat{p} \) (the sample proportion) will approximate a normal distribution. This allows technicians to make inferences about the population proportion more confidently.
population proportion
A population proportion \( p \) indicates the ratio of individuals in the entire population that possess a certain characteristic. In this context, it tells us what percentage of the total group of female Hispanic drivers in Boston wears seat belts. We are given that \( p = 0.6 \), meaning 60% of all such drivers are seat belt users.The population proportion is often unknown and must be estimated by sampling, which is where the sample proportion comes in handy. Nevertheless, if the population proportion is known, like in this case, it can directly inform predictions about future samples.
random sampling
Random sampling is a technique used to ensure that every member of a population has an equal chance of being included in the sample. This process eliminates bias and contributes to the representativeness of the sample, which is essential for making accurate inferences about the population. In the exercise, a random sample of 117 female Hispanic drivers was observed. By ensuring that the sample is random, statisticians can be more confident that their findings about the sample proportion are valid and applicable to the entire population. Random sampling is a cornerstone of statistical practice as it ensures the reliability and validity of the insights drawn from sample data.

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Most popular questions from this chapter

The value of the \(z\) statistic for the Exercise \(22.22\) is \(2.53\). This test is (a) not significant at either \(\alpha=0.05\) or \(\alpha=0.01\). (b) significant at \(\alpha=0.05\) but not at \(\alpha=0.01\). (c) significant at both \(\alpha=0.05\) and \(\alpha=0.01\).

Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. Only half of all mice complete one particular maze in less than 18 seconds. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures the proportion of 40 mice that completed the maze in less than 18 seconds with noise as a stimulus. The proportion of mice that completed the maze in less than 18 seconds is \(\mathrm{p}^{\wedge \hat{p}}=0.7\). The hypotheses for a test to answer the researcher's question are (a) \(H_{0}: p=0.5, H_{a}: p>0.5\). (b) \(H_{0}: p=0.5, H_{a}: p<0.5\). (c) \(H_{0}: p=0.5, H_{a}: p \neq 0.5\).

Although more than \(50 \%\) of American adults believe the maxim that breakfast is the most important meal of the day, only about \(30 \%\) eat breakfast daily. \({ }^{6}\) A cereal manufacturer contacts an SRS of 1500 American adults and calculates the proportion \(\mathrm{p}^{\wedge \hat{p}}\) in this sample who eat breakfast daily. (a) What is the approximate distribution of \(\mathrm{p}^{\wedge \hat{p}}\) ? (b) If the sample size were 6000 rather than 1500 , what would be the approximate distribution of \(\mathrm{p}^{\wedge \hat{p}}\) ?

Sample surveys usually contact large samples, so we can use the large-sample confidence interval if the sample design is close to an SRS. Scientific studies often use smaller samples that require the plus four method. For example, Familial Adenomatous Polyposis (FAP) is a rare inherited disease characterized by the development of an extreme number of polyps early in life and colon cancer in virtually \(100 \%\) of patients before the age of 40 . A group of 14 people suffering from FAP being treated at the Cleveland Clinic drank black raspberry powder in a slurry of water every day for nine months. The numbers of polyps were reduced in 11 out of 14 of these patients. \({ }^{18}\) (a) Why can't we use the large-sample confidence interval for the proportion \(p\) of patients suffering from FAP that will have the number of polyps reduced after nine months of treatment? (b) The plus four method adds four observations-two successes and two failures. What are the sample size and the number of successes after you do this? What is the plus four estimate \(\mathrm{p}^{-\bar{p}}\) of \(p\) ? (c) Give the plus four \(90 \%\) confidence interval for the proportion of patients suffering from FAP who will have the number of polyps reduced after nine months of treatment.

Throughout Europe, more than 8000 pedestrians are killed each year in road accidents, with approximately \(25 \%\) of these dying when using a pedestrian crossing. Although failure to stop for pedestrians at a pedestrian crossing is a serious traffic offense in France, more than half of drivers do not stop when a pedestrian is waiting at a crosswalk. In this experiment, a female research assistant was instructed to stand at a pedestrian crosswalk and stare at the driver's face as a car approached the crosswalk. In 400 trials, the research assistant maintained a neutral expression, and in a second set of 400 trials, the research assistant was instructed to smile. The order of smiling or not smiling was randomized, and several pedestrian crossings were used in a town on the coast in the west of France. Research assistants were dressed in normal attire for their age (jeans/T-shirt/trainers/sneakers). \({ }^{23}\) (a) In the 400 trials in which the assistant maintained a neutral expression, the driver stopped in 229 out of the 400 trials. Find a \(95 \%\) confidence interval for the proportion of drivers who would stop when a neutral expression is maintained. (b) In the 400 trials in which the assistant smiled at the driver, the driver stopped in 277 out of the 400 trials. Find a 95\% confidence interval for the proportion of drivers who would stop when the assistant is smiling. (c) What do your results in parts (a) and (b) suggest about the effect of a smile on a driver stopping at a pedestrian crosswalk? Explain briefly. (In Chapter 23, we will consider formal methods for comparing two proportions.)
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