/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Determine the point estimator yo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the point estimator you would use and calculate the value of the point estimate. What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: "You witness two students cheating on a quiz. Do you go to the professor?" Only 19 answered "Yes."

Short Answer

Expert verified
The point estimate is 0.1105, meaning 11.05% of students are likely to report cheating.

Step by step solution

01

Understanding the Problem

We need to determine the point estimate for the proportion of students who are willing to report cheating. This involves analyzing the responses from the sample provided in the problem.
02

Identify the Point Estimator

The point estimator for a population proportion in a sample is the sample proportion. In this context, the point estimator will be calculated by dividing the number of students who answered 'Yes' by the total number of students surveyed.
03

Calculate the Sample Proportion

The sample proportion \(\hat{p}\) is calculated as follows: \(\hat{p} = \frac{x}{n}\), where \(x = 19\) and \(n = 172\).
04

Perform the Calculation

Substituting the values into the formula, we calculate the sample proportion: \(\hat{p} = \frac{19}{172} \approx 0.1105\).
05

Conclusion

The point estimator for the proportion of students willing to report cheating is \(\hat{p} = 0.1105\). This means an estimated 11.05% of students are willing to report cheating.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Sample Proportion
When conducting a study or survey, we often want to find an estimate of a particular characteristic within a population. The sample proportion, denoted by \( \hat{p} \), is a key metric used in statistical analysis to estimate the proportion of a population that has a certain attribute. To compute the sample proportion, you need two pieces of information:
  • \( x \): the number of individuals in your sample who have the particular attribute,
  • \( n \): the total number of individuals in your sample.
The formula to calculate the sample proportion is:\[\hat{p} = \frac{x}{n}\]In the given exercise, \( x \) is 19 (students who said "Yes" to reporting cheating), and \( n \) is 172 (total students surveyed). By plugging these into the formula, we find \( \hat{p} = \frac{19}{172} \approx 0.1105 \), indicating that about 11.05% of students in the sample are willing to report cheating.
Population Proportion
Understanding the population proportion is crucial when analyzing data from a sample. The population proportion refers to the percentage of the entire population that has a particular characteristic. However, since it is often impractical to survey every individual in a population, we use sample data to estimate this proportion. The sample proportion serves as a point estimator for the population proportion, meaning it provides an estimated value for the true proportion within the entire population. In statistical terms, the larger and more representative the sample, the more reliable the estimate of the population proportion will be. In our exercise, we estimate that approximately 11.05% of all students (our population of interest) would report cheating on a quiz. This estimate is drawn directly from the sample data.
Statistical Analysis
Statistical analysis is the process of collecting and interpreting data to uncover patterns and trends. In the context of this exercise, statistical analysis helps us estimate the proportion of students who would potentially report cheating.Using statistical tools such as the sample proportion allows us to make inferences about larger populations. Through computations like \( \hat{p} = \frac{19}{172} \approx 0.1105 \), we can derive meaningful estimates from sample data. Statistical analysis can include various methods such as descriptive statistics, inferential statistics, and various modeling techniques. It's a powerful tool in fields like biology, psychology, and business, as it helps translate sample data into actionable insights.
Simple Random Sample
A simple random sample (SRS) is a fundamental concept in statistics that ensures each individual in a population has an equal chance of being selected for the sample. An SRS is crucial for minimizing bias and providing a representative subset of the entire population. In this context, an SRS of 172 students was used to gauge their willingness to report cheating. Since the sample is random, it helps in painting an accurate picture of the larger university population's attitudes toward cheating. The quality of results obtained from a statistical analysis often hinges on the sampling method used. A well-executed SRS provides a solid foundation for making generalizations about the entire population, making it an indispensable tool in research and data collection.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A New York Times/CBS News Poll asked a random sample of U.S. adults the question, "Do you favor an amendment to the Constitution that would permit organized prayer in public schools?" Based on this poll, the \(95 \%\) confidence interval for the population proportion who favor such an amendment is (0.63,0.69) (a) Interpret the confidence interval. (b) What is the point estimate that was used to create the interval? What is the margin of error? (c) Based on this poll, a reporter claims that more than two-thirds of U.S. adults favor such an amendment. Use the confidence interval to evaluate this claim.

When two lights close together blink alternately, we "see" one light moving back and forth if the time between blinks is short. What is the longest interval of time between blinks that preserves the illusion of motion? Ask subjects to turn a knob that slows the blinking until they "see" two lights rather than one light moving. A report gives the results in the form "mean plus or minus the standard error of the mean." \({ }^{22}\) Data for 12 subjects are summarized as \(251 \pm 45\) (in milliseconds). (a) Find the sample standard deviation \(s_{x}\) for these measurements. Show your work. (b) A hasty reader believes that the interval given in the report is a \(95 \%\) confidence interval for the population mean. Find the actual confidence level for the given interval.

A bunion on the big toe is fairly uncommon in youth and often requires surgery. Doctors used X-rays to measure the angle (in degrees) of deformity on the big toe in a random sample of 37 patients under the age of 21 who came to a medical center for surgery to correct a bunion. The angle is a measure of the seriousness of the deformity. For these 37 patients, the mean angle of deformity was 24.76 degrees and the standard deviation was 6.34 degrees. A dotplot of the data revealed no outliers or strong skewness. \({ }^{26}\) (a) Construct and interpret a \(90 \%\) confidence interval for the mean angle of deformity in the population of all such patients. (b) Researchers omitted one patient with a deformity angle of 50 degrees from the analysis due to a measurement issue. What effect would including this outlier have on the confidence interval in part (a)? Justify your answer without doing any calculations.

Determine the point estimator you would use and calculate the value of the point estimate. How many pairs of shoes, on average, do female teens have? To find out, an AP Statistics class conducted a survey. They selected an SRS of 20 female students from their school. Then they recorded the number of pairs of shoes that each student reported having. Here are the data: $$\begin{array}{rrrrrrrrrr}50 & 26 & 26 & 31 & 57 & 19 & 24 & 22 & 23 & 38 \\\13 & 50 & 13 & 34 & 23 & 30 & 49 & 13 & 15 & 51\end{array}$$

Scientists collect data on the blood cholesterol levels (milligrams per deciliter of blood) of a random sample of 24 laboratory rats. A \(95 \%\) confidence interval for the mean blood cholesterol level \(\mu\) is 80.2 to 89.8. Which of the following would cause the most worry about the validity of this interval? (a) There is a clear outlier in the data. (b) A stemplot of the data shows a mild right skew. (c) You do not know the population standard deviation \(\sigma\). (d) The population distribution is not exactly Normal. (e) None of these are a problem when using a \(t\) interval.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.