/*! 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 79 Will \(\hat{p}\) from a random s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Will \(\hat{p}\) from a random sample from a population with \(60 \%\) successes tend to be closer to 0.6 for a sample size of \(n=400\) or a sample size of \(n=800 ?\) Provide an explanation for your choice.

Short Answer

Expert verified
The sample proportion \(\hat{p}\) from a random sample will tend to be closer to the population proportion of 0.6 for a sample size of \(n=800\) compared to a sample size of \(n=400\). This is because the larger sample size results in a smaller standard error, which indicates a more accurate estimate of the population proportion.

Step by step solution

01

Compute the standard error for a sample size of 400

Using the formula for the standard error of the sample proportion and the given population proportion of 60%, we compute the standard error for a sample size of 400: \[ SE(\hat{p}_{n=400}) = \sqrt{\frac{0.6(1-0.6)}{400}} = \sqrt{\frac{0.6(0.4)}{400}} \]
02

Compute the standard error for a sample size of 800

Using the formula for the standard error of the sample proportion and the given population proportion of 60%, we compute the standard error for a sample size of 800: \[ SE(\hat{p}_{n=800}) = \sqrt{\frac{0.6(1-0.6)}{800}} = \sqrt{\frac{0.6(0.4)}{800}} \]
03

Compare the standard errors

Now we will compare the standard errors to determine which sample size gives a more accurate sample proportion: \[ SE(\hat{p}_{n=400}) = \sqrt{\frac{0.6(0.4)}{400}} \approx 0.0245 \] \[ SE(\hat{p}_{n=800}) = \sqrt{\frac{0.6(0.4)}{800}} \approx 0.0173 \] The standard error of the sample proportion for a sample size of 800 is smaller (0.0173) compared to that for a sample size of 400 (0.0245).
04

Conclusion

Based on the comparison of the standard errors, we can conclude that the sample proportion \(\hat{p}\) from a random sample with a sample size of \(n=800\) will tend to be closer to the population proportion of 0.6 than a sample size of \(n=400\). This is because the larger sample size results in a smaller standard error, and a smaller standard error indicates a more accurate estimate of the population proportion.

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 survey or an experiment, researchers are often interested in the proportion of individuals within a sample that exhibit a particular characteristic, known as the sample proportion. It is usually denoted by \( \hat{p} \) and is calculated by dividing the number of individuals with the desired characteristic by the total number of individuals in the sample. For example, if we wanted to determine the proportion of left-handed students in a classroom of 30 where 9 students are left-handed, the sample proportion would be \( \hat{p} = \frac{9}{30} = 0.3 \).

The sample proportion is a critical statistic because it provides an estimate of the population proportion. However, the accuracy of this estimate depends on certain factors, such as the size of the sample. The variability of \( \hat{p} \) is higher in smaller samples, which means the estimate may be less representative of the true population proportion. This variability is measured by the standard error of \( \hat{p} \).
Population Proportion
The population proportion, often represented by the symbol \( p \), is the actual proportion of individuals in a target population who have a specified characteristic. Unlike the sample proportion, which is an estimate based on a subset of the population, the population proportion is a fixed value that we try to estimate through sampling.

For instance, if we're looking at the proportion of the general population that prefers online shopping to in-store shopping, the population proportion is the true underlying percentage of the entire population with this preference. Researchers may only have the resources to survey a sample of the population, so they will calculate the sample proportion and use it to infer the population proportion. It is this estimation process that necessitates an understanding of margin of error and standard error to gauge how much trust we can put in the sample proportion as a reflection of the population proportion.
Sample Size
The sample size, denoted as \( n \), is the total number of individuals or observations included in a sample. It's a fundamental concept in statistics because it can significantly impact the precision of our statistical estimates. A larger sample size generally leads to more reliable estimates of population parameters, such as the mean or proportion, since it reduces the effect of random fluctuations. This effect is quantified by the standard error, which tends to decrease as the sample size increases.

The relationship between standard error and sample size is inversely proportional; as one goes up, the other tends to go down. This is particularly important when considering the confidence we have in our estimates. Larger samples give us more confidence that our sample proportion is close to the actual population proportion. Understanding how sample size affects standard error is crucial for designing experiments and interpreting statistical results accurately. In our exercise example, comparing the standard errors for sample sizes of 400 and 800 directly illustrates this concept.

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

The article "Kids Digital Day: Almost 8 Hours" (USA TODAY, January 20,2010 ) summarized a national survey of 2002 Americans age 8 to \(18 .\) The sample was selected to be representative of Americans in this age group. a. Of those surveyed, 1321 reported owning a cell phone. Use this information to construct and interpret a \(90 \%\) confidence interval for the proportion of all Americans ages 8 to 18 who owned a cell phone in 2010 . b. Of those surveyed, 1522 reported owning an MP3 music player. Use this information to construct and interpret a \(90 \%\) confidence interval for the proportion of all Americans ages 8 to 18 who owned an MP3 music player in 2010 c. Explain why the confidence interval from Part (b) is narrower than the confidence interval from Part (a) even though the confidence levels and the sample sizes used to calculate the two intervals were the same.

The article "Write It by Hand to Make It Stick" (Advertising Age, July 27,2016 ) summarizes data from a survey of 1001 students age 13 to 19 years. Of the students surveyed, 851 reported that they learn best using a mix of digital and nondigital tools. Construct and interpret \(\mathrm{a}\) \(95 \%\) confidence interval for the proportion of students age 13 to 19 who would say that they learn best using a mix of digital and nondigital tools. In order for the method used to construct the interval to be valid, what assumption about the sample must be reasonable?

Use the formula for the standard error of \(\hat{p}\) to explain why a. the standard error is greater when the value of the population proportion \(p\) is near 0.5 than when it is near 1 . b. the standard error of \(\hat{p}\) is the same when the value of the population proportion is \(p=0.2\) as it is when \(p=0.8\).

The study "The Demographics of Social Media Users" (Pew Research Center, August 19,2015 ) reported that \(72 \%\) of adult American Internet users use Facebook. The \(72 \%\) figure was based on a representative sample of \(n=1602\) adult American Internet users. Suppose that you would like to use the data from this survey to estimate the proportion of all adult American Internet users who use Facebook. Answer the four key questions (QSTN) to confirm that the suggested method in this situation is a large-sample confidence interval for a population proportion.

One thousand randomly selected adult Americans participated in a survey conducted by the Associated Press (June 2006). When asked "Do you think it is sometimes justified to lie, or do you think lying is never justified?" \(52 \%\) responded that lying was never justified. When asked about lying to avoid hurting someone's feelings, 650 responded that this was often or sometimes OK. a. Construct and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who would say that lying is never justified. b. Construct and interpret a \(90 \%\) confidence interval for the proportion of adult Americans who think that it is often or sometimes \(\mathrm{OK}\) to lie to avoid hurting someone's feelings. c. Using the confidence intervals from Parts (a) and (b), comment on the apparent inconsistency in the responses.

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.