/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 A study reported in Newsweek (De... [FREE SOLUTION] | 91Ó°ÊÓ

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A study reported in Newsweek (December 23, 1991) involved a sample of 935 smokers. Each individual received a nicotine patch, which delivers nicotine to the bloodstream but at a much slower rate than cigarettes do. Dosage was decreased to 0 over a 12-week period. Suppose that 245 of the subjects were still not smoking 6 months after treatment (this figure is consistent with information given in the article). Estimate the percentage of all smokers who, when given this treatment, would refrain from smoking for at least 6 months.

Short Answer

Expert verified
The estimated percentage of all smokers who, when given this treatment, would refrain from smoking for at least 6 months is \( 26.2\% \).

Step by step solution

01

Identify the Relevant Sample and Outcome

The total number of smokers in the sample is 935, and the number of those who were successful in refraining from smoking for at least 6 months after the treatment is 245. These will be used to calculate the proportion.
02

Calculate the Proportion

The proportion of succeeding smokers can be calculated by dividing the number of successful outcomes by the total sample size, then multiplying by 100 to get the percentage. Thus, the formula is as follows: \( Proportion = \frac{Number\: of\: successes}{Total\: sample\: size} \times 100 \)
03

Substitute the Known Values

Now, substitute the values into the formula: \( Proportion = \frac{245}{935} \times 100 \)
04

Compute the Answer

Perform the calculations to get the final proportion. This represents an estimate of the percentage of all smokers who, when given this treatment, would refrain from smoking for at least 6 months.

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Key Concepts

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

Sampling Distribution
Sampling distribution is a fundamental concept in statistics that helps us understand how sample statistics behave when repeated samples are taken from a population. When we take samples, each sample can give us different results. The sampling distribution provides a way to quantify these variations.

Imagine repeatedly taking a sample of 935 smokers and noting how many quit after using the nicotine patch. Each sample might show a different number of successful quitters because of random variability. The sampling distribution of the proportion would represent the distribution of these proportions across many samples.

Key characteristics of sampling distributions are:
  • The mean of the sampling distribution will be equal to the true population proportion if the sampling process is unbiased.
  • The spread of the sampling distribution, measured by the standard error, decreases as sample size increases, leading to more precise estimates.
Understanding sampling distribution is crucial because it helps in making inferences about the population based on sample data. It allows us to calculate confidence intervals and conduct hypothesis tests.
Proportion Calculation
Proportion calculation is a method used to determine how a part relates to a total in percentage terms. It provides a clear understanding of the composition of the total by highlighting the size of that part relative to the whole.

In the context of our exercise, the proportion calculation is essential to estimate how many smokers in a broader population might successfully quit smoking using a nicotine patch. Here's how you calculate it:
  • Determine the number of successes (ex-smokers) and the total sample size. In our example, there are 245 successful quitters out of 935 participants.
  • Use the formula: \( \text{Proportion} = \frac{\text{Number of successes}}{\text{Total sample size}} \times 100 \) to convert this into a percentage.
The calculated proportion gives us a percentage of smokers who successfully ceased smoking after treatment, offering a predictive insight into the treatment's potential success rate.
Success Rate in Experiments
Success rate in experiments refers to the proportion of trials in which a predefined outcome occurs. Analyzing the success rate gives an indication of the effectiveness or efficiency of the treatment or intervention being tested.

In scientific studies, like the nicotine patch experiment in our exercise, measuring the success rate helps researchers:
  • Draw conclusions about the treatment's efficacy.
  • Compare the outcome against other competing treatments or control groups.
  • Make informed decisions on the potential for widescale application of the treatment.
If 245 smokers did not smoke 6 months post-treatment, this result provides a success rate based on the initial sample, giving an understanding of the treatment's impact. Success rates derived from experiments must be interpreted carefully, considering sample size and variability, to assure reliability and reproducibility in diverse populations.

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Most popular questions from this chapter

The formula used to compute a large-sample confidence interval for \(\pi\) is $$ p \pm(z \text { critical value }) \sqrt{\frac{p(1-p)}{n}} $$ What is the appropriate \(z\) critical value for each of the following confidence levels? a. \(95 \%\) d. \(80 \%\) b. \(90 \%\) e. \(85 \%\) c. \(99 \%\)

Recent high-profile legal cases have many people reevaluating the jury system. Many believe that juries in criminal trials should be able to convict on less than a unanimous vote. To assess support for this idea, investigators asked each individual in a random sample of Californians whether they favored allowing conviction by a 10-2 verdict in criminal cases not involving the death penalty. The Associated Press (San Luis Obispo TelegramTribune, September 13,1995 ) reported that \(71 \%\) supported the \(10-2\) verdict. Suppose that the sample size for this survey was \(n=900\). Compute and interpret a \(99 \%\) confidence interval for the proportion of Californians who favor the \(10-2\) verdict.

Given a variable that has a \(t\) distribution with the specified degrees of freedom, what percentage of the time will its value fall in the indicated region? a. \(10 \mathrm{df}\), between \(-1.81\) and \(1.81\) b. \(10 \mathrm{df}\), between \(-2.23\) and \(2.23\) c. \(24 \mathrm{df}\), between \(-2.06\) and \(2.06\) d. \(24 \mathrm{df}\), between \(-2.80\) and \(2.80\) e. 24 df, outside the interval from \(-2.80\) to \(2.80\) f. \(24 \mathrm{df}\), to the right of \(2.80\) g. \(10 \mathrm{df}\), to the left of \(-1.81\)

Why is an unbiased statistic generally preferred over a biased statistic for estimating a population characteristic? Does unbiasedness alone guarantee that the estimate will be close to the true value? Explain. Under what circumstances might you choose a biased statistic over an unbiased statistic if two statistics are available for estimating a population characteristic?

The Chronicle of Higher Education (January 13, 1993) reported that \(72.1 \%\) of those responding to a national survey of college freshmen were attending the college of their first choice. Suppose that \(n=500\) students responded to the survey (the actual sample size was much larger). a. Using the sample size \(n=500\), calculate a \(99 \%\) confidence interval for the proportion of college students who are attending their first choice of college. b. Compute and interpret a \(95 \%\) confidence interval for the proportion of students who are not attending their first choice of college. c. The actual sample size for this survey was much larger than 500 . Would a confidence interval based on the actual sample size have been narrower or wider than the one computed in Part (a)?

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