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\(\hat{p}\). Greenville County, South Carolina, has 461,299 adult residents, of which 59,969 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) 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) \( p \approx 0.1299 \). (b) The range is (0.1033, 0.1565). (c) Possible sampling bias; may not be an SRS.

Step by step solution

01

Calculate the Proportion p

To find the proportion of Greenville county adult residents who are 65 years or older, divide the number of adults 65 and older by the total number of adult residents.\[ p = \frac{\text{Number of adults 65 or older}}{\text{Total number of adult residents}} = \frac{59,969}{461,299} \approx 0.1299 \]
02

Find the Standard Deviation of the Sample Proportion

The standard deviation for the sample proportion is given by:\[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \]Substituting the values:\[ \sigma_{\hat{p}} = \sqrt{\frac{0.1299 \times (1-0.1299)}{689}} \approx 0.0133 \]
03

Apply the 68-95-99.7 Rule for the Sample Proportion

The 95% range according to the 68-95-99.7 rule, which is approximately the empirical rule, spans from:\[ \hat{p} \pm 2 \times \sigma_{\hat{p}} \]Calculating the interval:\[ 0.1299 \pm 2 \times 0.0133 = (0.1033, 0.1565) \]
04

Assess the Simple Random Sampling Assumption

The sample of 689 respondents corresponds to roughly 30% response rate according to the problem statement. Because the method used involves a database of exchanges without cell phone numbers, this may exclude a significant segment of the population, potentially introducing selection bias, particularly if older residents are more likely to have landlines.

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Key Concepts

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

Proportion
A proportion is a way to express a part of a total or whole. It is often used in statistics to represent the fraction of a population having a specific characteristic. In this exercise, the proportion we are interested in is the number of adults aged 65 and older in Greenville County relative to the total number of adult residents. Calculating this is straightforward:

\[ p = \frac{\text{Number of adults 65 or older}}{\text{Total number of adult residents}} = \frac{59,969}{461,299} \approx 0.1299 \]

Here, 0.1299 is interpreted as about 13% of the adult residents being aged 65 and older. This is a crucial statistic as it provides insight into the demographic makeup of the county.

Understanding proportions is vital because it allows us to compare different subsets within a population or across different populations. They are also used in determining probabilities, making them fundamental in various statistical methods.
Simple Random Sample
A simple random sample is a method used in statistics where every individual has an equal chance of being selected from the population. This helps minimize biases or influences that might affect the outcome of a study. It ensures that the sample accurately represents the whole population.

In the Greenville County survey, residents were contacted using random digit dialing based on landline numbers. However, this method raises concerns about the sample's randomness. Excluding cell phone numbers might lead to a biased sample since it is possible that different age groups have distinct preferences for landlines over cell phones. For instance, older adults might rely more on landlines, thus not truly representing other age groups adequately.

When conducting a survey, ensuring the sample is as random as possible helps validate the results. This concept underscores the importance of precise sample selection processes in statistical studies.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. In the context of sample proportions, the standard deviation helps us understand how much these proportions would vary if we took multiple samples from the same population.

For the sample proportion in Greenville County, the standard deviation is computed as:

\[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \approx \sqrt{\frac{0.1299 \times (1-0.1299)}{689}} \approx 0.0133 \]

This value tells us that the sample proportion will typically vary by about 1.33% from the true population proportion if we sample repeatedly.

Having a smaller standard deviation indicates that the sample proportions are more tightly clustered around the average, leading to more reliable results.
Empirical Rule
The empirical rule is a statistical guideline suggesting how data should be distributed in a normal distribution. It states that:

  • 68% of the data falls within one standard deviation of the mean.
  • 95% falls within two standard deviations.
  • 99.7% falls within three standard deviations.


For the Greenville County survey, the 95% confidence interval for the proportion of residents aged 65 and older can be calculated using this rule. The interval is given by:

\[ \hat{p} \pm 2 \times \sigma_{\hat{p}} \]

Substituting the values gives us:

\[ 0.1299 \pm 2 \times 0.0133 = (0.1033, 0.1565) \]

This interval means we are 95% confident that the true proportion of adults 65 and older falls between 10.33% and 15.65%.

The empirical rule helps in understanding the likely range of variation in statistical estimates, making it an essential tool for interpreting data in research.

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Most popular questions from this chapter

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 \(\mathrm{p}^{\wedge \hat{p}}\) ?

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 \(\mathrm{p}^{\wedge \hat{p}}=0.5\). What would the margin of error (in percent) be if the sample had resulted in \(\mathrm{p}^{\wedge \hat{p}}=0.5\) ? (c) Why do you think that Harris announces a \(\pm 3\) margin of error for all samples of about this size?

A random digit dialing telephone survey of 880 drivers asked, "Recalling the last 10 traffic lights you drove through, how many of them were red when you entered the intersections?" Of the 880 respondents, 171 admitted that at least one light had been red. \({ }^{24}\) (a) Give a \(95 \%\) confidence interval for the proportion of all drivers who ran one or more of the last 10 red lights they met. (b) Nonresponse is a practical problem for this survey-only \(21.6 \%\) of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: do you think more or fewer than 171 of the 880 respondents really ran a red light? Why?

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

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) \(0.002\).

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