/*! 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 35 The article "Career Expert Provi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article "Career Expert Provides DOs and DON'Ts for Job Seekers on Social Networking" (CareerBuilder.com, August 19,2009 ) included data from a survey of 2,667 hiring managers and human resource professionals. The article noted that many employers are using social networks to screen job applicants and that this practice is becoming more common. Of the 2,667 people who participated in the survey, 1,200 indicated that they use social networking sites (such as Facebook, MySpace, and LinkedIn) to research job applicants. For the purposes of this exercise, assume that the sample can be regarded as a random sample of hiring managers and human resource professionals. a. Suppose you are interested in learning about the value of \(p,\) the proportion of all hiring managers and human resource managers who use social networking sites to research job applicants. This proportion can be estimated using the sample proportion, \(p .\) What is the value of \(p\) for this sample? b. Based on what you know about the sampling distribution of \(p,\) is it reasonable to think that this estimate is within 0.02 of the actual value of the population proportion? Explain why or why not.

Short Answer

Expert verified
a) Approximately 0.45 or 45%. b) Assuming the Central Limit Theorem, and given that the sample size is large enough, it is reasonable to think that this estimate could be within 0.02 of the actual population proportion.

Step by step solution

01

Calculate the Sample Proportion

You can calculate sample proportion (denoted as \( p \)) by dividing the number of successes (in this case, the number of hiring managers and human resource professionals that use social networking sites for applicant research, which is 1,200) by the total number of trials (which in this case is the number of all participating professionals, which is 2,667). So, the formula would be: \( p = \frac{1200}{2667} \).
02

Calculate the Result

By doing the calculation from Step 1, the value of the sample proportion, \( p \), comes out to be approximately 0.45. This interpretation of this value signifies that about 45% of hiring managers and human resource professionals use social networking sites for applicant research.
03

Making Inferences about the Population Proportion

Knowing that the sample is a random sample from the population of interest, by the Central Limit Theorem, we can assume the sampling distribution of \( p \) will be approximately normal if the sample size is large enough (usually if np and n(1-p) are both greater than 5). Here \( np = 0.45*2667 = 1200.15\) and \( n(1-p) = 1466.85 \), which are both greater than 5. Hence, it's reasonable to think that the estimate could be within 0.02 of the actual 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.

Sampling Distribution
When you gather data from a sample and want to understand how a proportion behaves across multiple samples, you look at the sampling distribution. This is essentially the distribution of all possible sample proportions you could obtain from taking multiple samples of the same size from the population. The key characteristics to remember about the sampling distribution of the sample proportion are:
  • It is approximately normally distributed if the sample size is large enough. Specifically, if both the number of successes (p imes n ext{)} and failures ((1-p) imes n ext{)} are greater than 5, you can assume normality.
  • The mean of the sampling distribution is the true population proportion (p ext{)}.
  • The standard deviation, known as the standard error, measures how much the sample proportion may vary from sample to sample, calculated as \( \sqrt{\frac{p(1-p)}{n}} \).
By knowing these properties, you can make inferences about how close your sample proportion is likely to be to the true population proportion.
Sample Proportion
The sample proportion is a simple yet powerful statistical concept. It provides an estimate of the proportion in a larger population based on data from a sample. In easy terms, it tells you how many subjects in your sample meet a particular criterion.To calculate it, you divide the number of subjects that meet the criterion by the total number of subjects in the sample. For example, if 1,200 hiring managers use social networks to screen applicants out of a total sample of 2,667, the sample proportion is calculated as:\[ p = \frac{1200}{2667} \approx 0.45 \]This means approximately 45% of the surveyed hiring managers have used social networking in their hiring process. A point to note is that the sample proportion is our best estimate of the true population proportion, but it is just an estimate, so there is always some level of uncertainty. By using the sampling distribution, we can understand and calculate this uncertainty more precisely.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that when you take a sufficiently large random sample from a population, the sampling distribution of the sample mean (or proportion) will be approximately normally distributed, regardless of the original population distribution. Here is why the CLT is so important:
  • It allows us to use normal probability theory to make inferences about population parameters.
  • It applies as long as the sample size is large enough (a common rule of thumb is that a sample size of at least 30 is sufficient, though with proportions, the criteria that np > 5 and n(1-p) > 5 is used).
In the context of the exercise, the CLT provides assurance that even if not every hiring manager was sampled, our estimate of the proportion using networking sites (our sample proportion) is reliable. The theorem helps justify using the normal distribution to assess how close the sample proportion might be to the population proportion. This is why it's reasonable to expect the estimate to be within a certain range (like 0.02) of the actual 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

For which of the following combinations of sample size and population proportion would the standard deviation of \(\hat{p}\) be smallest? $$ \begin{array}{ll} n=40 & p=0.3 \\ n=60 & p=0.4 \\ n=100 & p=0.5 \end{array} $$

The report "California's Education Skills Gap: Modest Improvements Could Yield Big Gains" (Public Policy Institute of California, April \(16,2008,\) www.ppic.org) states that nationwide, \(61 \%\) of high school graduates go on to attend a two-year or four-year college the year after graduation. The proportion of high school graduates in California who go on to college was estimated to be \(0.55 .\) Suppose that this estimate was based on a random sample of 1,500 California high school graduates. Is it reasonable to conclude that the proportion of California high school graduates who attend college the year after graduation is different from the national figure? (Hint: Use what you know about the sampling distribution of \(\hat{p}\). You might also refer to Example \(8.5 .)\)

In a study of pet owners, it was reported that \(24 \%\) celebrate their pet's birthday (Pet Statistics, Bissell Homecare, Inc., 2010 ). Suppose that this estimate was from a random sample of 200 pet owners. Is it reasonable to conclude that the proportion of all pet owners who celebrate their pet's birthday is less than \(0.25 ?\) Use what you know about the sampling distribution of \(\hat{p}\) to support your answer.

A random sample of 50 registered voters in a particular city included 32 who favored using city funds for the construction of a new recreational facility. For this sample, \(\hat{p}=\frac{32}{50}=\) 0.64 . If a second random sample of 50 registered voters was selected, would it surprise you if \(\hat{p}\) for that sample was not equal to 0.64 ? Why or why not?

A certain chromosome defect occurs in only 1 in 200 adult Caucasian males. A random sample of 100 adult Caucasian males will be selected. The proportion of men in this sample who have the defect, \(\hat{p},\) will be calculated. a. What are the mean and standard deviation of the sampling distribution of \(\hat{p}\) ? b. Is the sampling distribution of \(\hat{p}\) approximately normal? Explain. c. What is the smallest value of \(n\) for which the sampling distribution of \(\hat{p}\) is approximately normal?

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.