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Do Smokers Know That Smoking is Bad for Them? The Harris Poll asked a sample of smokers, "Do you believe that smoking will probably shorten your life, or not?" Of the 1010 people in the sample, 848 said "yes." a. Harris called residential telephone numbers at random in an attempt to contact an SRS of smokers. Based on what you know about national sample surveys, what is likely to be the biggest weakness in the survey? b. We will nonetheless act as if the people interviewed are an SRS of smokers. Give a \(95 \%\) confidence interval for the percent of smokers who agree that smoking will probably shorten their lives.

Short Answer

Expert verified
a) Non-response bias is likely a significant issue. b) The 95% confidence interval is (81.68%, 86.24%).

Step by step solution

01

Understanding the Sampling Method

The Harris Poll used residential telephone numbers selected at random to contact smokers. This method assumes that each smoker has an equal probability of being selected, aiming for a Simple Random Sample (SRS). However, the biggest weakness is potential non-response bias, where individuals who do not answer phone calls or do not have landlines (which is more common today) might differ systematically from those who do answer, possibly affecting the results.
02

Identifying Given Data

From the sample of 1010 smokers, 848 responded 'yes' to the belief that smoking would shorten their life. We need to calculate the proportion of smokers who believe smoking is harmful to their longevity.
03

Calculate the Sample Proportion

Calculate the sample proportion \( \hat{p} \) using the formula: \( \hat{p} = \frac{x}{n} \), where \( x = 848 \) and \( n = 1010 \). Substitute the values into the formula to find \( \hat{p} = \frac{848}{1010} \approx 0.8396 \).
04

Determine the Confidence Interval Formula

We will use the formula for the confidence interval of a population proportion: \( \hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \). For a 95% confidence level, the critical value \( z \) is approximately 1.96.
05

Calculate the Margin of Error

Calculate the standard error (SE) using \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.8396 \times 0.1604}{1010}} \approx 0.0116 \). The margin of error (ME) is then \( ME = 1.96 \times 0.0116 \approx 0.0228 \).
06

Construct the Confidence Interval

Add and subtract the margin of error from the sample proportion: Lower bound = \( 0.8396 - 0.0228 = 0.8168 \). Upper bound = \( 0.8396 + 0.0228 = 0.8624 \). Therefore, the 95% confidence interval is \((0.8168, 0.8624)\).

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Key Concepts

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

Simple Random Sample (SRS)
A Simple Random Sample, or SRS, is a foundational concept in statistics. It involves selecting a sample in such a way that every individual of the population has an equal chance of being chosen. This method aims to ensure that the sample is representative of the broader population.

The Harris Poll attempted to contact an SRS of smokers by randomly dialing residential telephone numbers. The idea here was that by reaching out to random numbers, each potential respondent—a smoker in this context—would have an equal likelihood of inclusion in the sample.

This method is often used because it helps eliminate selection bias, which might occur if the selection process inadvertently favors certain groups over others. However, it's critical to note that in real-world applications, achieving a truly simple random sample can be challenging due to various factors, such as non-response bias, which we will explore next.
Non-response bias
Non-response bias is a type of bias that occurs when certain individuals from a sample do not respond or cannot be reached, and these non-respondents differ significantly from those who do respond.

In the context of the Harris Poll, which used residential phone calls to reach smokers, non-response bias could emerge if certain smokers do not have landlines or are unavailable to answer the phone. Perhaps these individuals have characteristics or habits that differ from those who can be reached, such as a different awareness or concern about the health effects of smoking.

When non-response bias is present, it can skew the results. The opinions of non-respondents might vary significantly from those who participated, potentially leading to inaccurate conclusions about the entire population. As such, acknowledging and attempting to mitigate non-response bias is crucial when designing and interpreting surveys.
Population proportion
The population proportion refers to a percentage or fraction of individuals in a population that share a certain characteristic. In this exercise, the characteristic under study is whether smokers believe that smoking will likely shorten their lives.

From the original sample data, 848 out of 1010 smokers answered "yes" to this question. To find the sample proportion, we use the formula \( \hat{p} = \frac{x}{n} \), where \( x \) is the number of successful outcomes and \( n \) is the total number of observations. Substituting the given values, we find \( \hat{p} = \frac{848}{1010} \approx 0.8396 \).

This sample proportion serves as an estimate of the true population proportion. The accuracy of this estimate can be further refined by calculating a confidence interval, which will be discussed in conjunction with the margin of error.
Margin of Error (ME)
The Margin of Error (ME) is a statistic expressing the amount of random sampling error in a survey's results. It indicates the range within which the true population parameter is expected to lie with a certain level of confidence.

For the Harris Poll, we calculate the margin of error to determine the reliability of our estimate of the population proportion. The formula for ME in a confidence interval for a proportion is determined using the critical value from the standard normal distribution, which is approximately 1.96 for a 95% confidence interval.

To calculate ME, first, we determine the standard error (SE) using \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( \hat{p} \) is the sample proportion, and \( n \) is the sample size. For our data, \( SE \approx 0.0116 \).
  • Multiply the SE by the critical value, \( ME = 1.96 \times 0.0116 \approx 0.0228 \).

The resulting ME tells us how much the sample results can be expected to vary from the true population proportion, reinforcing the reliability of the study's conclusions.

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Most popular questions from this chapter

Reporting cheating. Students are reluctant to report cheating by other students. A student project put this question to an SRS of 172 undergraduates at a large university: "You witness two students cheating on a quiz. Do you go to the professor?" Only 19 answered "yes." 21 Give a \(95 \%\) confidence interval for the proportion of all undergraduates at this university who would report cheating.

Do You Eat Red Meat? About \(60 \%\) of American adults include beef and other red meat in their diets. \(-6\) A plant-based meat company contacts an SRS of 1500 American adults and calculates the proportion \(\widehat{p}\) in this sample who eat red meat. a. What is the approximate distribution of \(\widehat{p}\) ? b. If the sample size were 6000 rather than 1500 , what would be the approximate distribution of \(\widehat{p}\) ?

Coin Tossing. The French naturalist Count Buffon (1707-1788) tossed a coin 4040 times. The result was 2048 heads, with \(\frac{2048}{4040}=0.5069\) the proportion of heads. Is this evidence that the coin was not fair? State the appropriate hypotheses and give the P-value.

Stopping Traffic with a Smile! 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 assist ant 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. The research assistant was dressed in normal attire for her age (jeans, t-shirt, and sneakers). \({ }^{24}\) 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.)

Speeding. It often appears that most drivers on the road are driving faster than the posted speed limit. Situations differ, of course, but here is one set of data. Researchers studied the behavior of drivers on a rural interstate highway in Maryland where the speed limit was 55 miles per hour. They measured speed with an electronic device hidden in the pavement and, to eliminate large trucks, considered only vehicles less than 20 feet long. They found that 5690 out of 12,931 vehicles exceeded the speed limit. Is this good evidence, at a significance level of \(0.05\), that (at least in this location) fewer than half of all drivers are speeding?

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