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Chapter 22: Problem 39

College-Educated Parents. The National Assessment of Educational Progress (NAEP) includes a "longterm trend" study that tracks reading and mathematics skills over time and obtains demographic information. In the 2012 study (the most recent available as of 2020), a random sample of 9000 17-year-old students was selected. \(\underline{27}\) The NAEP sample used a multistage design, but the overall effect is quite similar to an SRS of 17-year-olds who are still in school. a. In the sample, \(51 \%\) of students had at least one parent who was a college graduate. Estimate, with \(99 \%\) confidence, the proportion of all 17-year-old students in 2012 who had at least one parent graduate from college. b. The sample does not include 17-year-olds who dropped out of school, so your estimate is valid only for students. Do you think the proportion of all 17-year- olds with at least one parent who was a college graduate would be higher or lower than \(51 \%\) ? Explain.

Short Answer

Expert verified
Estimate: (0.4966, 0.5234); Total proportion likely lower than 51%.

Step by step solution

01

Identify the Given Information

We are given that 51% of the sample of 17-year-old students had at least one parent who was a college graduate. The sample size is 9000 students. We need to estimate the proportion of all 17-year-old students with at least one college-educated parent with 99% confidence.
02

Calculate the Sample Proportion

The sample proportion \( \hat{p} \) is calculated as:\[\hat{p} = \frac{51}{100} = 0.51\]
03

Determine the Standard Error

The standard error (SE) for the sample proportion is calculated using the formula:\[SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.51 \times 0.49}{9000}}\approx 0.0052\]
04

Find the Critical Value for 99% Confidence

For a 99% confidence interval, the critical value (\( z^* \)) is approximately 2.576. This value is determined from the standard normal distribution.
05

Calculate the Confidence Interval

The 99% confidence interval for the proportion is calculated using the formula:\[\hat{p} \pm z^* \times SE = 0.51 \pm 2.576 \times 0.0052\]This results in:\[0.51 \pm 0.0134\]Therefore, the confidence interval is (0.4966, 0.5234).
06

Interpret the Confidence Interval in Context

The 99% confidence interval (0.4966, 0.5234) suggests that we are 99% confident that the true proportion of all 17-year-old students in 2012 who had at least one parent graduate from college lies between 49.66% and 52.34%.
07

Consider the Effect on the Entire Population

Since the sample only includes students who haven't dropped out, it might underestimate the overall proportion of 17-year-olds with a college-educated parent because students from less privileged backgrounds are more likely to drop out.

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Key Concepts

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

Sample Proportion
The sample proportion is a way to estimate a ratio or percentage out of a specific population from a sample group. In this case, the sample consists of 9,000 17-year-old students surveyed by the National Assessment of Educational Progress (NAEP). Out of these, 51% were found to have at least one parent who is a college graduate. This is our sample proportion. Mathematically, you calculate the sample proportion (denoted as \( \hat{p} \)) by dividing the number of successful outcomes by the total number of observations in the sample.
  • Successful outcomes refer to those who met the criteria (here, having a college-educated parent).
  • The formula is \( \hat{p} = \frac{\text{number of successes}}{\text{total sample size}} \).
In our case, \( \hat{p} = \frac{51}{100} = 0.51 \). This proportion is critical in estimating population parameters like confidence intervals.
Standard Error
Standard error (SE) helps us understand the precision of our sample statistic as an estimate of the population parameter. It tells us how much the sample proportion might differ from the true population proportion. The formula for standard error when dealing with sample proportions is:
  • \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \)
  • Where \( \hat{p} \) is the sample proportion.
  • \( n \) represents the total sample size.
In the context of our exercise, \( SE = \sqrt{\frac{0.51 \times 0.49}{9000}} \approx 0.0052 \).

This means that if we took many samples of 9,000 students, the sample proportion would typically deviate by around 0.52% from the true population proportion. The smaller the SE, the more precise is our estimate.
Critical Value
When constructing a confidence interval, the critical value indicates how many standard errors to go from the sample proportion to cover the desired confidence level鈥攊n this case, 99%. This critical value is based on the standard normal distribution, often found in z-tables or standard normal distribution tables.
  • For a 99% confidence level, the critical value (\( z^* \)) is commonly about 2.576.
  • It controls the width of the confidence interval; a higher critical value leads to a wider interval.
This means our estimate will be within approximately 2.576 standard errors of the true proportion 99% of the time. We use this critical value along with the standard error to calculate the confidence interval and better understand the estimate's reliability.
Multistage Sampling
Multistage sampling is a complex yet systematic sampling method involving several stages of sampling. It helps when dealing with large populations, where a simple random sample might be impractical or costly. The NAEP study in question used this method to survey 17-year-olds in 2012.
  • In the first stage, larger groups or primary sampling units are sampled, such as schools or regions.
  • Subsequently, within each primary unit, secondary units鈥攁 sample like students鈥攁re selected.
This layered approach leads to increased efficiency, allowing researchers to manage logistics and costs better in comprehensive studies. Although the sample was collected using multistage sampling, its results are similar to what might be expected from a simple random sample of the same population. This method is especially useful in considerable population studies due to its practicality and cost-efficiency.

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Most popular questions from this chapter

Black Raspberries and Cancer. Sample surveys usually contact large samples, so we can use the large-sample confidence interval if the sample design is close to an SRS. Scientific studies often use smaller samples that require the plus four method. For example, Familial Adenomatous Polyposis (FAP) is a rare inherited disease characterized by the development of an extreme number of polyps early in life and colon cancer in virtually \(100 \%\) of patients before the age of 40 . A group of 14 people suffering from FAP being treated at the Cleveland Clinic drank black raspberry powder in a slurry of water every day for nine months. The numbers of polyps were reduced in 11 out of 14 of these patients. 19 a. Why can't we use the large-sample confidence interval for the proportion \(p\) of patients suffering from FAP that will have the number of polyps reduced after nine months of treatment? b. The plus four method adds four observations: two successes and two failures. What are the sample size and the number of successes after you do this? What is the plus four estimate \(\tilde{p}\) of \(p\) ? c. Give the plus four \(90 \%\) confidence interval for the proportion of patients suffering from FAP who will have the number of polyps reduced after nine months of treatment.

Cannabidiol- (CBD-) based products are widely touted for their therapeutic benefits without any psychoactive effects. A 2019 Gallup Poll asked a random sample of 2543 U.S. adults if they personally use CBD products or not. Suppose that, in fact, \(15 \%\) of all U.S. adults have used CBD products. In repeated samples, the sample proportion \(\widehat{p}\) would follow approximately a Normal distribution with mean a. \(381.45\). b. \(0.15\). c. \(0.007\).

Book Reading. Although an increasing share of Americans are reading e-books on tablets and smartphones rather than dedicated e-readers, print books continue to be much more popular than books in digital format (digital format includes both e-books and audio books). A Pew Research Center survey of 1502 adults nationwide conducted January 8-February 7, 2019, found that 1081 of those surveyed had read a book in either print or digital format in the preceding 12 months. 28 (You may regard the 1081 adults in the survey who had read a book in the preceding 12 months as a random sample of readers.) a. What can you say with \(95 \%\) confidence about the percentage of all adults who had read a book in either print or digital format in the preceding 12 months? b. Of the 1081 surveyed who had read a book in the preceding 12 months, 105 had read only digital books. Among those adults who had read a book in the preceding 12 months, find a \(95 \%\) confidence interval for the proportion that had read digital books exclusively.

Running Red Lights. 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. 25 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?

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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