/*! 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 40 The article referenced in the pr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article referenced in the previous exercise also reported that \(38 \%\) of the 1,200 social network users surveyed said it was OK to ignore a coworker's friend request. If \(p=0.38\) is used as an estimate of the proportion of all social network users who believe this, is it likely that this estimate is within 0.05 of the actual population proportion? Use what you know about the sampling distribution of \(\hat{p}\) to support your answer.

Short Answer

Expert verified
No, it's not likely that the estimate of 0.38 is within 0.05 of the actual population proportion, given the large Z-scores and assuming a 95% confidence level.

Step by step solution

01

Define the method

This problem requires hypothesis testing, and in this context, we are using the Z-score formula. The Z-score represents how many standard deviations an observation or datum is from the mean.
02

Mean and Standard Deviation

Using the given proportion \(p = 0.38\) as the population mean and standard deviation \(\sigma = \sqrt{p(1-p)/n}\), we get the mean \(\mu = 0.38\) and standard deviation \(\sigma = \sqrt{(0.38)(1-0.38)/1200} = 0.0141\). It is important to note that this only applies if the sampling distribution is approximately normal. This will be checked in the next step.
03

Checking the Normal condition

To use the Normal model, we need \(np\) and \(n(1-p)\) both to be at least 10.\nThe number expected with the event is \(1200 \times 0.38 = 456\), and the number expected without the event is \(1200 \times (1-0.38) = 744\). Since both these numbers are greater than 10, the sampling distribution is approximately normal, and we can proceed with calculating the Z-score.
04

Calculate the Z-scores

Given the interval within 0.05 of the actual proportion, this corresponds to two intervals of interest, \(p+0.05\) and \(p-0.05\).\nFor \(p+0.05 = 0.43\), \(Z = (0.43 - 0.38) / 0.0141 = 3.55\). The Z-score for \(p-0.05 = 0.33\) is \(Z = (0.33 - 0.38) / 0.0141 = -3.55\)
05

Interpret the Z-scores

The Z-scores for both upper and lower limits are beyond the usual threshold for a statistical test (±1.96 for a 95% confidence level). This means that the estimate 0.38 is not within 0.05 of the true population proportion, assuming a 95% confidence level.

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.

Sampling Distribution
When diving into hypothesis testing, the concept of a sampling distribution is essential. Imagine you take multiple samples from a population and calculate a statistic, like the mean or proportion, for each sample. Here, the focus is on the proportion \(\hat{p}\), which represents the proportion estimate from each sample.
A typical goal is to understand the distribution of these sample estimates. We consider the collection of all sample proportions, each resulting from a different sample of the same size, and this creates the sampling distribution of \(\hat{p}\).
Understanding the distribution:
  • The mean of the sampling distribution of \(\hat{p}\) is the population proportion \(p\).
  • The standard deviation, often called the standard error, is given by \(\sigma = \sqrt{\frac{p(1-p)}{n}}\), where \(n\) is the sample size.
  • If both \(np\) and \(n(1-p)\) are greater than 10, the sampling distribution is approximately normal.
This normality allows us to use the standard normal distribution for calculating probabilities and conducting hypothesis tests. This is crucial when we're trying to determine the likelihood of our sample result.
Z-score Calculation
In the world of statistics, Z-score offers a powerful tool to standardize data. The Z-score tells you how many standard deviations a data point is from the mean of the distribution. This can help you understand if your sample estimate significantly differs from what you expect.
To calculate the Z-score for a sample proportion \(\hat{p}\), you'll use the formula:\[ Z = \frac{\hat{p} - p}{\sigma} \]Here, \(\hat{p}\) is your sample proportion, \(p\) is the assumed population proportion, and \(\sigma\) is the standard deviation of the sampling distribution, which we've seen earlier.
Suppose you want to determine if your proportion estimate of 0.38 is within 0.05 of the true proportion. You'd calculate Z-scores for the boundaries \(p+0.05\) and \(p-0.05\) to interpret if these proportions are typical under a normal model:
  • If Z is beyond ±1.96, it suggests that the sample mean is statistically significantly different from the hypothesized mean at a 95% confidence level.
Z-scores thus guide decision-making in hypothesis testing by showing whether deviations from the mean are due to random sampling variation or signal a significant effect.
Confidence Interval
Confidence intervals are widely used in statistics to indicate the reliability of an estimate. An interval gives a range of values, derived from the sample, which is likely to contain the population parameter with a certain level of confidence.
To construct a confidence interval for a population proportion, we use the sample proportion \(\hat{p}\) and its standard error. A typical formula for a confidence interval is:\[ \hat{p} \pm Z^* \times \sigma \]Here, \(Z^*\) is the critical value from the normal distribution corresponding to the desired confidence level, and \(\sigma\) is the standard deviation of the sampling distribution.
Key aspects of confidence intervals:
  • The confidence level (like 95%) tells us how sure we are that the interval contains the true population parameter.
  • A 95% confidence interval means that if you took 100 different samples and computed an interval for each, approximately 95 of those intervals would contain the population parameter.
  • They provide a balance between precision (narrow intervals) and confidence (higher percentage), influenced by sample size and variability.
In the context of the exercise, verifying if 0.38 falls within a specified confidence interval helps in understanding how likely it is to estimate the true population proportion accurately.

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 report "New Study Shows Need for Americans to Focus on Securing Online Accounts and Backing Up Critical Data" (PRNewswire, October 29,2009 ) reported that only \(25 \%\) of Americans change computer passwords quarterly, in spite of a recommendation from the National Cyber Security Alliance that passwords be changed at least once every 90 days. For purposes of this exercise, assume that the \(25 \%\) figure is correct for the population of adult Americans. a. A random sample of size \(n=200\) will be selected from this population and \(\hat{p}\), the proportion who change passwords quarterly, will be calculated. What are the mean and standard deviation of the sampling distribution of \(\hat{p} ?\) b. Is the sampling distribution of \(\hat{p}\) approximately normal for random samples of size \(n=200 ?\) Explain. c. Suppose that the sample size is \(n=50\) rather than \(n=200 .\) Does the change in sample size affect the mean and standard deviation of the sampling distribution of \(\hat{p} ?\) If so, what are the new values of the mean and standard deviation? If not, explain why not. d. Is the sampling distribution of \(\hat{p}\) approximately normal for random samples of size \(n=50 ?\) Explain.

The article "Should Pregnant Women Move? Linking Risks for Birth Defects with Proximity to Toxic Waste Sites" (Chance [1992]: \(40-45)\) reported that in a large study carried out in the state of New York, approximately \(30 \%\) of the study subjects lived within 1 mile of a hazardous waste site. Let \(p\) denote the proportion of all New York residents who live within 1 mile of such a site, and suppose that \(p=0.3\). a. Would \(\hat{p}\) based on a random sample of only 10 residents have a sampling distribution that is approximately normal? Explain why or why not. b. What are the mean and standard deviation of the sampling distribution of \(\hat{p}\) if the sample size is \(400 ?\) c. Suppose that the sample size is \(n=200\) rather than \(n=\) \(400 .\) Does the change in sample size affect the mean and

Consider the following statement: A county tax assessor reported that the proportion of property owners who paid 2012 property taxes on time was 0.93 . a. Is the number that appears in boldface in this statement a sample proportion or a population proportion? b. Which of the following use of notation is correct, \(p=0.93\) or \(p=0.93 ?\)

A random sample of 100 employees of a large company included 37 who had worked for the company for more than one year. For this sample, \(\hat{p}=\frac{37}{100}=0.37\). If a different random sample of 100 employees were selected, would you expect that \(\hat{p}\) for that sample would also be \(0.37 ?\) Explain why or why not.

Suppose that the actual proportion of students at a particular college who use public transportation to travel to campus is \(0.15 .\) In a study of parking needs at the campus, college administrators would like to estimate this proportion. They plan to take a random sample of 75 students and use the sample proportion who use public transportation, \(\hat{p},\) as an estimate of the population proportion. a. Show that the standard deviation of \(\hat{p}\) is equal to \(\sigma_{p}=0.0412\) b. If for a different sample size, \(\sigma_{p}=0.0319,\) would you expect more or less sample-to-sample variability in the sample proportions than for when \(n=75 ?\) c. Is the sample size that resulted in \(\sigma_{p}=0.0319\) larger than 75 or smaller than \(75 ?\) Explain your reasoning.

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.