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Chapter 22: Problem 2

The 68-95-99.7 Rule and \(\widehat{\boldsymbol{p}}\). Greenville County, South Carolina, has 396,183 adult residents, of which 80,987 are 65 years or older. A survey wants to contact \(n=689\) residents. \({ }^{5}\) a. Find \(p\), the proportion of Greenville County adult residents who are 65 years or older. b. If repeated simple random samples of 689 residents are taken, what would be the range of the sample proportion of adults over 65 in the sample according to the 95 part of the 68-95-99.7 rule? c. Suppose the actual survey contacted 689 adults using random digit dialing of residential numbers using a database of exchanges, with no cell phone numbers contacted. The 689 respondents represent a response rate of approximately \(30 \%\). In the sample obtained, 253 of the 689 adults contacted were over 65 . Do you have any concerns treating this as a simple random sample from the population of adult residents of Greenville County? Explain briefly.

Short Answer

Expert verified
a. 0.2045; b. [0.1721, 0.2369]; c. Yes, due to exclusion and low response rate.

Step by step solution

01

Calculate Proportion p

To find the proportion \( p \) of Greenville County adult residents who are 65 years or older, divide the number of adults aged 65 and older by the total number of adult residents. So, \( p = \frac{80,987}{396,183} \approx 0.2045 \). This tells us that approximately 20.45% of the adults in Greenville County are 65 years or older.
02

Standard Deviation of Sample Proportion

The standard deviation of the sample proportion \( \widehat{p} \) can be calculated using the formula \( \sigma_{\widehat{p}} = \sqrt{\frac{p(1-p)}{n}} \). Substituting the values, this becomes \( \sigma_{\widehat{p}} = \sqrt{\frac{0.2045(1-0.2045)}{689}} \approx 0.0162 \).
03

Apply the 68-95-99.7 Rule

According to the 95% part of the 68-95-99.7 rule, approximately 95% of the sample proportions will fall within \( \pm 2 \times \sigma_{\widehat{p}} \) of the population proportion \( p \). Therefore, the range is \( 0.2045 \pm 2 \times 0.0162 \). Calculating, this gives us \( 0.2045 \pm 0.0324 \), or a range of approximately \([0.1721, 0.2369]\).
04

Evaluate Concerns about the Sample

The sample obtained may not perfectly represent the population of adult residents due to the exclusion of cell phone numbers and the low response rate of 30%. Therefore, there's a concern that the sample may not be entirely random since households solely with landlines may have different demographics than those with cell phones, potentially skewing the sample.

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Key Concepts

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

68-95-99.7 Rule
The 68-95-99.7 Rule, also known as the Empirical Rule, is a statistical guideline that helps us understand data distribution. It is particularly useful for describing normal distributions in statistics. This rule states that:
  • About 68% of the data falls within one standard deviation (\( \sigma \)) of the mean.
  • Approximately 95% of the data falls within two standard deviations of the mean.
  • Nearly 99.7% of the data falls within three standard deviations of the mean.
When dealing with sample proportions, this rule helps predict where the majority of sample proportions are likely to fall. In the context of the exercise, we used the 68-95-99.7 Rule to determine the range within which 95% of the sample proportions should lie. This allowed us to establish a confidence interval for the sample proportion, guiding us in estimating the range for practical applications.
Sample Proportion
The sample proportion, denoted by \( \widehat{p} \), represents the ratio of a specific subgroup within a sample to the total number of subjects in the sample. It is calculated by dividing the number of subgroup members by the total sample size. Applying this to the exercise scenario, if a survey of 689 adults identified 253 aged 65 and older, then the sample proportion \( \widehat{p} \) of adults over 65 is \( \frac{253}{689} \approx 0.367 \).This metric is crucial because it serves as an estimate for the actual proportion of the subgroup in the population. By comparing the sample proportion to the population proportion \( p \), we can assess the representativeness of the sample.
Simple Random Sample
A Simple Random Sample (SRS) is a method of selecting subjects from a population such that each individual has an equal probability of being chosen. It is essential in ensuring the data is representative of the whole population, reducing the likelihood of bias. In the exercise, the survey aimed to contact residents through random digit dialing to achieve a simple random sample. However, there are concerns when using this method:
  • Excluding cell phone numbers might mean some population segments are underrepresented.
  • A low response rate of 30% can also skew the randomness, as non-respondents may have different characteristics from those who participated.
These factors could affect how well the sample resembles the population, questioning the assumption that the sample is indeed a simple random sample.
Standard Deviation of Sample Proportion
The standard deviation of the sample proportion, \( \sigma_{\widehat{p}} \), measures the variability of a sample proportion estimate. It indicates how much the sample proportion \( \widehat{p} \) is expected to vary from the actual population proportion \( p \) across different samples. The formula for this standard deviation is:\[\sigma_{\widehat{p}} = \sqrt{\frac{p(1-p)}{n}}\]This formula implies that the standard deviation decreases as sample size increases, providing a more stable estimate. In the exercise, given the known population proportion \( p \) and sample size \( n \), we calculated the standard deviation to be a small value, emphasizing that most sample proportions should closely match the population proportion. Understanding this concept is key to applying the 68-95-99.7 Rule effectively, as it determines the range of likely sample outcomes.

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Most popular questions from this chapter

Harris Announces a Margin of Error. Exercise \(22.25\) describes a Harris Poll survey of smokers in which 848 of a sample of 1010 smokers agreed that smoking would probably shorten their lives. Harris announces a margin of error of \(\pm 3\) percentage points for all samples of about this size. Opinion polls announce the margin of error for \(95 \%\) confidence. a. What is the actual margin of error (in percent) for the large-sample confidence interval from this sample? b. The margin of error is largest when \(\hat{p}=0.5\). What would the margin of error (in percent) be if the sample had resulted in \(\widehat{p}=0.5\) ? c. Why do you think that Harris announces a \(\pm 3 \%\) margin of error for all samples of about this size?

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.)

An opinion poll asks an SRS of 100 college seniors how they view their job prospects. In all, 53 say "Good." The largesample \(95 \%\) confidence interval for estimating the proportion of all college seniors who think their job prospects are good is a. \(0.530 \pm 0.082\). b. \(0.530 \pm 0.098\). c. \(0.530 \pm 0.049\).

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 \(\widehat{p}=0.7\). 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. \(H_{0}: p=0.5, H_{a}: p \neq 0.5\).

A Gallup Poll in November 2019 found that \(55 \%\) of the people in the sample said they wanted to lose weight. The poll's margin of error for \(95 \%\) confidence was \(4 \%\). This means that a. the poll used a method that gets an answer within \(4 \%\) of the truth about the population \(95 \%\) of the time. b. we can be sure that the percentage of all adults who want to lose weight is between \(50 \%\) and \(58 \%\). c. if Gallup takes another poll using the same method, the results of the second poll will lie between \(51 \%\) and \(59 \%\).
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