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Chapter 22: Problem 35

The National Assessment of Educational Progress (NAEP) includes a "long-term trend" study that tracks reading and mathematics skills over time and obtains demographic information. In the 2012 study, a random sample of 9000 17-year- old students was selected. \({ }^{26}\) 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 only valid 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
The 99% confidence interval for the proportion is approximately (49.6%, 52.4%). The overall proportion could be lower than 51%.

Step by step solution

01

Identify the Sample Proportion

The sample proportion \( \hat{p} \) is given as 51%, which corresponds to \( \hat{p} = 0.51 \). This represents the proportion of students in the sample whose at least one parent graduated from college.
02

Determine the Sample Size

The sample size is \( n = 9000 \), as stated in the problem. This will be used in the calculation of the standard error.
03

Calculate the Standard Error

The standard error (SE) of the sample proportion is calculated using the formula:\[SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} = \sqrt{\frac{0.51 \times (1-0.51)}{9000}} \approx 0.00527\]
04

Find the Z-Score for 99% Confidence

For a 99% confidence interval, the Z-score is 2.576. This value is derived from the standard normal distribution corresponding to a 99% confidence.
05

Compute Confidence Interval

The confidence interval is calculated as follows:\[CI = \hat{p} \pm Z \times SE = 0.51 \pm 2.576 \times 0.00527\]This results in:\[CI = (0.51 - 0.01357, 0.51 + 0.01357)\]\[CI = (0.49643, 0.52357)\]
06

Interpretation of Confidence Interval

The 99% confidence interval for the proportion of 17-year-old students, who have at least one parent as a college graduate, in the population is approximately (49.6%, 52.4%).
07

Consider Exclusion of Dropouts

The sample excludes 17-year-olds who dropped out of school. It is plausible that students whose parents have not graduated college might have a higher dropout rate, potentially lowering the proportion with at least one college-graduate parent in the overall 17-year-old population compared to the 51% figure for students.

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Key Concepts

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

Sample Proportion
In statistics, the sample proportion is a way to express how often a certain characteristic appears within a sample. Here, it's represented by the symbol \( \hat{p} \).
In the context of this exercise, the sample proportion \( \hat{p} \) is 0.51, or 51%. This indicates that 51% of students in the surveyed sample had at least one parent who graduated from college.
Calculating the sample proportion involves dividing the number of favorable outcomes by the total number of observations in the sample: \( \hat{p} = \frac{x}{n} \), where \(x\) represents the count of favorable results and \(n\) is the sample size.
The sample proportion is crucial as it serves as an estimate of the actual proportion of the population that exhibits the attribute in question.
Standard Error
The standard error (SE) is a vital measure that tells us how much the sample proportion \( \hat{p} \) deviates from the true population proportion. It provides insight into the accuracy and reliability of the sample proportion as an estimate of the population proportion. Specifically, it reflects the sample's variability.
The formula for the standard error of a sample proportion is \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \]
In this exercise, with a sample proportion \( \hat{p} = 0.51 \) and sample size \( n = 9000 \), the standard error calculation results in approximately 0.00527.
The smaller the standard error, the more precise the estimate of the population proportion is considered. This precision is crucial when constructing confidence intervals.
Z-Score
The Z-score is a statistical measure that describes a data point's relationship to the mean of a group of data points. It's expressed in terms of standard deviations from the mean.
For confidence intervals, the Z-score helps determine the multiplier for the standard error.
Different confidence levels require different Z-scores:
  • For a 95% confidence level, the Z-score would be around 1.96.
  • In this exercise, a 99% confidence level requires a Z-score of 2.576.
This means for a 99% confidence interval, you extend 2.576 standard errors on either side of the sample proportion.
Demographic Information
Demographic information encompasses data relating to the characteristics of a population. In this study, researchers collected data about 17-year-olds who are still in school.
Demographics can include variables such as age, gender, education level, and in this study, parental education level.
Understanding these demographics is key to interpreting the findings correctly. Only students still in school were included, which influences the results.
This exclusion of dropouts is significant. Demographic factors such as parental education could impact dropout rates, thus affecting the overall estimation of how many 17-year-olds have a parent who graduated college.

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Most popular questions from this chapter

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

\(\hat{p}\). 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.

A high-school teacher in a lowincome urban school in Worcester, Massachusetts, used cash incentives to encourage student learning in his AP statistics class. \({ }^{28}\) In 2010,15 of the 61 students enrolled in his class scored a 5 on the AP statistics exam. Worldwide, the proportion of students who scored a 5 in 2010 was \(0.15\). Is this evidence that the proportion of students who would score a 5 on the AP statistics exam when taught by the teacher in Worcester using cash incentives is higher than the worldwide proportion of \(0.15\) ? (a) State hypotheses, find the \(P\)-value, and give your conclusions in the context of the problem. Do you have any reservations about using the \(z\) test for proportions for this data? (b) Does this study provide evidence that cash incentives cause an increase in the proportion of \(5 \mathrm{~s}\) on the AP statistics exam? Explain your answer.

Explain whether we can use the \(z\) test for a proportion in these situations. (a) You toss a coin 10 times in order to test the hypothesis \(H_{0}: p=0.5\) that the coin is balanced. (b) A local candidate contacts an SRS of 900 of the registered voters in his district to see if there is evidence that more than half support the bill he is sponsoring. (c) A college president says, " \(99 \%\) of the alumni support my firing of Coach Boggs." You contact an SRS of 200 of the college's 15,000 living alumni to test the hypothesis \(H_{0}: p=0.99\).

The Internal Revenue Service plans to examine an SRS of individual federal income tax returns from each state. One variable of interest is the proportion of returns claiming itemized deductions. The total number of tax returns in a state varies from almost 30 million in California to approximately 500,000 in Wyoming. (a) Will the margin of error for estimating the population proportion change from state to state if an SRS of 2000 tax returns is selected in each state? Explain your answer. (b) Will the margin of error change from state to state if an SRS of \(1 \%\) of all tax returns is selected in each state? Explain your answer.
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