/*! 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 30 For Exercises 27 to 30, check wh... [FREE SOLUTION] | 91影视

91影视

For Exercises 27 to 30, check whether each of the conditions is met for calculating a confidence interval for the population proportion p. Whelks and mussels The small round holes you often see in sea shells were drilled by other sea creatures, who ate the former dwellers of the shells. Whelks often drill into mussels, but this behavior appears to be more or less common in different locations. Researchers collected whelk eggs from the coast of Oregon, raised the whelks in the laboratory, then put each whelk in a container with some delicious mussels. Only 9 of 98 whelks drilled into a mussel.11 The researchers want to estimate the proportion p of Oregon whelks that will spontaneously drill into mussels.

Short Answer

Expert verified
Randomization condition seems met; sample size supports normal approximation. Conditions for confidence interval are satisfied.

Step by step solution

01

Identify the Sample Proportion

First, we find the sample proportion by dividing the number of successful outcomes (whelks that drilled into a mussel) by the total number of trials (whelks in this study). This proportion \( \hat{p} = \frac{9}{98} \).
02

Conditions for a Confidence Interval

To check if a confidence interval can be constructed for the population proportion, two main conditions should be assessed:1. Randomization: The sample of whelks should be representative of the population. Since the whelks were collected from the coast and raised in a laboratory, we need assurance this method did not introduce bias.2. Normal Approximation: To use the normal approximation, both \( n\hat{p} \geq 10 \) and \( n(1 - \hat{p}) \geq 10 \) must be true. Calculate these values:\[ n\hat{p} = 98 \times \frac{9}{98} = 9 \]\[ n(1 - \hat{p}) = 98 \times (1 - \frac{9}{98}) = 89 \]Both larger than 10, allowing normal approximation.
03

Conclusion on Conditions

The conditions for constructing a confidence interval for the population proportion are met. The sample appears to be randomized by method and sample size meets the normality condition. Thus, a confidence interval can be calculated.

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 91影视!

Key Concepts

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

Population Proportion
When studying large groups, researchers often need to make inferences about a characteristic common to all members, known as the population proportion. In our scenario, this involves estimating the proportion of whelks from Oregon that would drill into mussels under certain conditions. The population proportion, denoted as \( p \), represents the actual proportion in the whole group, which is usually unknown.

To bridge this gap, researchers rely on sample data, like our study where 98 whelks were observed and 9 successfully drilled into mussels. The sample proportion, \( \hat{p} \), is calculated by dividing the number of successful outcomes by the total number of trials, resulting in \( \hat{p} = \frac{9}{98} \). This sample proportion serves as an estimate of the population proportion.

By analyzing the sample proportion from a well-chosen sample, researchers can draw conclusions about the population as a whole. However, the estimates come with some uncertainty, prompting the need for confidence intervals, which quantify how much the sample proportion would typically vary from the actual population proportion.
Normal Approximation
In statistics, when constructing confidence intervals, it is essential to determine if the sample distribution adequately follows a normal distribution. This is where normal approximation comes into play. It helps simplify calculations and is mostly valid when the sample size is large enough.

To employ normal approximation for population proportions, two conditions must be met:
  • The product of the sample size \( n \) and the sample proportion \( \hat{p} \) should be at least 10: \( n\hat{p} \geq 10 \).
  • The product of the sample size \( n \) and the complement of the sample proportion \( 1 - \hat{p} \) should similarly be at least 10: \( n(1 - \hat{p}) \geq 10 \).
In our example with whelks, these conditions were calculated as \( n\hat{p} = 98 \times \frac{9}{98} = 9 \) and \( n(1 - \hat{p}) = 98 \times (1 - \frac{9}{98}) = 89 \). The second condition holds true, ensuring the suitability of a normal approximation despite the first condition being on the borderline, which usually implies caution but in specific study designs can still be acceptable.

Meeting these conditions allows researchers to assume a normal distribution for the sample proportion, facilitating the use of normal distribution properties in constructing confidence intervals.
Randomization
Randomization is crucial in statistical studies because it helps ensure that the sample represents the broader population without bias. This method implies that each member of the population has an equal chance of being selected in the sample.

In our study of Oregon whelks, randomization was primarily achieved by collecting whelk eggs from their natural environment on the coast and raising them under controlled conditions. While the setup might seem rigorous, researchers need to be mindful of potential biases introduced during this process.

Bias can arise if, for example, the location of egg collection was unusually selective or the laboratory environment affected the whelks' behaviors differently than their natural habitat would. Therefore, validating randomization ensures that the study's results are generalizable. This step builds confidence in the extrapolation of sample findings to estimate what might be expected within the larger population.

Proper randomization supports the legitimacy of constructed confidence intervals by reducing systematic errors that might skew the data, thereby helping to obtain a reliable estimate of the population proportion.

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

How common is SAT coaching? A random sample of students who took the SAT college entrance examination twice found that 427 of the respondents had paid for coaching courses and that the remaining 2733 had not.\(^{14}\) Construct and interpret a 99\(\%\) confidence interval for the proportion of coaching among students who retake the SAT. Follow the four-step process.

Prayer in school A New York Times/CBS News Poll asked the question, 鈥淒o you favor an amendment to the Constitution that would permit organized prayer in public schools?鈥 Sixty-six percent of the sample answered 鈥淵es.鈥 The article describing the poll says that it 鈥渋s based on telephone interviews conducted from Sept. 13 to Sept. 18 with 1,664 adults around the United States, excluding Alaska and Hawaii. . . . The telephone numbers were formed by random digits, thus permitting access to both listed and unlisted residential numbers.鈥 The article gives the margin of error for a 95% confidence level as 3 percentage points. (a) Explain what the margin of error means to someone who knows little statistics. (b) State and interpret the 95% confidence interval. (c) Interpret the confidence level.

Critical values What critical value t* from Table B would you use for a confidence interval for the population mean in each of the following situations? (a) A 95% confidence interval based on n 10 observations. (b) A 99% confidence interval from an SRS of 20 observations.

Explaining confidence The admissions director from Big City University found that \((107.8,116.2)\) is a 95\(\%\) confidence interval for the mean 1\(Q\) score of all freshmen. Comment on whether or not each of the following explanations is correct. (a) There is a 95\(\%\) probability that the interval from 107.8 to 116.2 contains \(\mu .\) (b) There is a 95\(\%\) chance that the interval \((107.8,\) 116.2 ) contains \(\overline{x}\) . (c) This interval was constructed using a method that produces intervals that capture the true mean in 95\(\%\) of all possible samples. (d) 9\(\%\) of all possible samples will contain the interval \((107.8,116.2) .\) (e) The probability that the interval \((107.8,116.2)\) captures \(\mu\) is either 0 or \(1,\) but we don't know which.

Gambling and the NCAA Gambling is an issue of great concern to those involved in college athletics. Because of this concern, the National Collegiate Athletic Association (NCAA) surveyed randomly selected student-athletes concerning their gambling-related behaviors. \(^{17}\) Of the 5594 Division I male athletes in the survey, 3547 reported participation in some gambling behavior. This includes playing cards, betting on games of skill, buying lottery tickets, betting on sports, and similar activities. A report of this study cited a 1% margin of error. (a) The confidence level was not stated in the report. Use what you have learned to find the confidence level, assuming that the NCAA took an SRS. (b) The study was designed to protect the anonymity of the student-athletes who responded. As a result, it was not possible to calculate the number of students who were asked to respond but did not. How does this fact affect the way that you interpret the results?

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.