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L Lifestyle: Smoking In a survey of 1000 large corporations, 250 said that, given a choice between a job candidate who smokes and an equally qualified nonsmoker, the nonsmoker would get the job (USA Today). (a) Let \(p\) represent the proportion of all corporations preferring a nonsmoking candidate. Find a point estimate for \(p\). (b) Find a \(0.95\) confidence interval for \(p\). (c) Interpretation As a news writer, how would you report the survey results regarding the proportion of corporations that hire the equally qualified nonsmoker? What is the margin of error based on a \(95 \%\) confidence interval?

Short Answer

Expert verified
(a) Point estimate for \( p \) is 0.25. (b) The 95% CI for \( p \) is (0.223, 0.277). (c) Report: "25%, with a margin of error of 2.7%."

Step by step solution

01

Understand the Variables

We are given in the problem that out of 1000 large corporations surveyed, 250 prefer a nonsmoking candidate over a smoking candidate. We need to use this information to find the point estimate and construct a confidence interval.
02

Calculate Point Estimate for p

The point estimate for the proportion, \( p \), of all corporations preferring a nonsmoking candidate is calculated as the ratio of the number preferring nonsmokers to the total number of corporations surveyed. Therefore, \( \hat{p} = \frac{250}{1000} = 0.25 \).
03

Identify Confidence Interval Formula

To calculate the confidence interval for \( p \), we use the formula for the confidence interval for a proportion: \[ \hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]where \( Z \) is the Z-value corresponding to the desired confidence level (for 95% confidence, \( Z \approx 1.96 \)), \( \hat{p} \) is the sample proportion, and \( n \) is the sample size.
04

Compute the Confidence Interval

Using the calculated point estimate \( \hat{p} = 0.25 \) and with \( n = 1000 \), we plug into the formula:\[ 0.25 \pm 1.96 \sqrt{\frac{0.25 \times (1-0.25)}{1000}} \]Calculating the standard error:\[ \sqrt{\frac{0.25 \times 0.75}{1000}} = \sqrt{\frac{0.1875}{1000}} \approx 0.01369 \]Then calculate the margin of error:\[ 1.96 \times 0.01369 \approx 0.02684 \]Thus, the confidence interval is:\[ 0.25 \pm 0.02684 \]\[ (0.22316, 0.27684) \]
05

Interpret the Results

As a news writer, you would report: "In a survey, 25% of large corporations preferred a nonsmoker over an equally qualified smoker, with a 95% confidence interval from 22.3% to 27.7%." The margin of error for this confidence interval is approximately 2.7%.

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Key Concepts

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

Confidence Interval
A confidence interval is a statistical concept used to express the reliability of an estimate. In simple terms, it gives us a range within which we expect the true population parameter to lie with a certain level of confidence.
This range is constructed using the sample data, and it varies based on the size of the sample and the desired confidence level, usually 95% or 99%.
For instance, in survey analysis, if we conduct a survey and calculate that 25% of a population has a particular opinion, it's unlikely that this percentage is perfectly accurate.
To build a 95% confidence interval, we use the sample proportion (\(\hat{p}\)), the sample size \(n\), and the critical value \(Z\) based on the standard normal distribution.
  • Formula: \(\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\)
  • For a 95% confidence interval, \(Z \approx 1.96\)
In our example with corporations, the confidence interval was calculated as approximately \((0.22316, 0.27684)\). This gives us a cushion around our estimate, confirming that we're 95% confident that the true proportion of corporations preferring nonsmokers falls between 22.3% and 27.7%.
Proportion Estimation
Proportion estimation is a statistical process used to gauge the fraction of a whole that exhibits a specific characteristic.
This is particularly useful in survey analysis where we try to understand behaviors or preferences within a population.
In the given example, we want to estimate the proportion of corporations that prefer hiring a nonsmoker over a smoker.Proportion estimation involves using a sample to infer the proportion in the entire population. The point estimate, denoted as \(\hat{p}\), is calculated as the ratio of the desired outcome to the total number of observations.
  • In this case, \(\hat{p} = \frac{250}{1000} = 0.25\)
This means 25% of surveyed corporations prefer nonsmoking candidates. However, this value is an estimate of the true proportion, not an exact figure. Using this point estimate, we can then construct a confidence interval to better understand the range in which the true proportion likely falls, underscoring the estimate's potential variability.
Survey Analysis
Survey analysis is the powerful process of collecting and analyzing data from a sample group to make inferences about a larger population.
It's a key method in research and decision-making across many sectors, including business, healthcare, and social sciences.
These analyses are pivotal in understanding opinions, behaviors, and preferences. In the scenario of evaluating corporation preferences, a survey of 1000 corporations revealed that 250 preferred a nonsmoking candidate.
It's important to approach survey results critically as they provide a snapshot of the population that diverges slightly due to sampling variability. Survey analysis generally involves:
  • Designing a representative survey sample.
  • Collecting responses accurately and systematically.
  • Analyzing the data with statistical methods to draw meaningful insights.
Survey analysis often involves point estimation and constructing confidence intervals to appropriately measure sentiment and behavior, thus mitigating the impact of sampling variability on the results.
This ensures a more reliable reflection of the larger population's preferences.
In this case, the proper use of statistical concepts like proportion estimation and confidence intervals can lend significant weight to the understanding and reporting of the survey outcomes.

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Most popular questions from this chapter

(a) Suppose a \(95 \%\) confidence interval for the difference of means contains both positive and negative numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain both positive and negative numbers? Explain. What about a \(90 \%\) confidence interval? Explain. (b) Suppose a \(95 \%\) confidence interval for the difference of proportions contains all positive numbers. Will a \(99 \%\) confidence interval based on the same data necessarily contain all positive numbers as well? Explain. What about a \(90 \%\) confidence interval? Explain.

Myers-Briggs: Marriage Counseling Isabel Myers was a pioneer in the study of personality types. She identified four basic personality preferences, which are described at length in the book A Guide to the Development and Use of the Myers-Briggs Type Indicator by Myers and McCaulley (Consulting Psychologists Press). Marriage counselors know that couples who have none of the four preferences in common may have a stormy marriage. Myers took a random sample of 375 married couples and found that 289 had two or more personality preferences in common. In another random sample of 571 married couples, it was found that only 23 had no preferences in common. Let \(p_{1}\) be the population proportion of all married couples who have two or more personality preferences in common. Let \(p_{2}\) be the population proportion of all married couples who have no personality perferences in common. (a) Check Requirements Can a normal distribution be used to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(99 \%\) confidence interval for \(p_{1}-p_{2}\). (c) Interpretation Explain the meaning of the confidence interval in part (a) in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you (at the \(99 \%\) confidence level) about the proportion of married couples with two or more personality preferences in common compared with the proportion of married couples sharing no personality preferences in common?

Pro Football and Basketball: Weights of Players Independent random samples of professional football and basketball players gave the following information (References: Sports Encyclopedia of Pro Football and Official NBA Basketball Encyclopedia). Note: These data are also available for download at the Online Study Center. Assume that the weight distributions are moundshaped and symmetric. $$ \begin{aligned} &\text { Weights (in lb) of pro football players: } x_{1} ; n_{1}=21\\\ &\begin{array}{lllllllllll} 245 & 262 & 255 & 251 & 244 & 276 & 240 & 265 & 257 & 252 & 282 \\ 256 & 250 & 264 & 270 & 275 & 245 & 275 & 253 & 265 & 270 & \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { Weights (in lb) of pro basketball players: } x_{2} ; n_{2}=19 \\ &\begin{array}{llllllllll} 205 & 200 & 220 & 210 & 191 & 215 & 221 & 216 & 228 & 207 \\ 225 & 208 & 195 & 191 & 207 & 196 & 181 & 193 & 201 & \end{array} \end{aligned} $$ (a) Use a calculator with mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 259.6, s_{1} \approx 12.1, \bar{x}_{2} \approx 205.8\), and \(s_{2} \approx 12.9 .\) (b) Let \(\mu_{1}\) be the population mean for \(x_{1}\) and let \(\mu_{2}\) be the population mean for \(x_{2}\). Find a \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\). (c) Interpretation Examine the confidence interval and explain what it means in the context of this problem. Does the interval consist of numbers that are all positive? all negative? of different signs? At the \(99 \%\) level of confidence, do professional football players tend to have a higher population mean weight than professional basketball players? (d) Which distribution (standard normal or Student's \(t\) ) did you use? Why?

Consider a \(90 \%\) confidence interval for \(\mu\). Assume \(\sigma\) is not known. For which sample size, \(n=10\) or \(n=20\), is the critical value \(t_{c}\) larger?

Confidence Intervals: Sample Size A random sample is drawn from a population with \(\sigma=12\). The sample mean is 30 . (a) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size 49 . What is the value of the margin of error? (b) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size 100 . What is the value of the margin of error? (c) Compute a \(95 \%\) confidence interval for \(\mu\) based on a sample of size 225 . What is the value of the margin of error? (d) Compare the margins of error for parts (a) through (c). As the sample size increases, does the margin of error decrease? (e) Critical Thinking Compare the lengths of the confidence intervals for parts (a) through (c). As the sample size increases, does the length of a \(90 \%\) confidence interval decrease?

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