Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Downloading “apps” to your cell phone. Refer toExercise 4.173 (p. 282) and the August 2011 survey by thePew Internet & American Life Project. The study foundthat 40% of adult cell phone owners have downloadedan application (“app”) to their cell phone. Assume thispercentage applies to the population of all adult cell phoneowners.

  1. In a random sample of 50 adult cell phone owners, howlikely is it to find that more than 60% have downloadedan “app” to their cell phone?
  2. Refer to part a. Suppose you observe a sample proportionof .62. What inference can you make about the trueproportion of adult cell phone owners who have downloadedan “app”?
  3. Suppose the sample of 50 cell phone owners is obtainedat a convention for the International Association forthe Wireless Telecommunications Industry. How willyour answer to part b change, if at all?

Short Answer

Expert verified
  1. There is 0.2% chance that more than 60% have downloaded the application.
  2. The sample proportion is too small. So, there can assume that either the sample is taken randomly or there is a problem in the population mean proportion.

c. The population mean can’t be rejected and there is no problem with that. So, the sample is taken randomly

Step by step solution

01

Given information

Referring to exercise 4.173 (p. 282), the study discovers that the application was downloaded by 40% of adults on their cell phones. The percentage applied to the population of adult cell phone owners.

So, the probabilityp=0.40

02

Calculate the Probability

a.

Consider the standard error of the 50 random sample,σ

So,

σ=p1-pn=0.40×0.6050=0.692

Now, the probability that more than 60% have downloaded the application is,

PrX>0.60=Prp^-pσ>0.60-0.400.692=Prz>2.89=1-Prz<2.89=1-0.99807=0.0020

Therefore, there is a very low chance of 0.2% that more than 60% have downloaded the application.

03

Determine the inference

b.

Referring to part a., the sample proportion is too small. So, there can assume that either the sample is taken randomly or there is a problem in the population mean proportion.

04

Justify the inference

c.

Hence the sample is from the International Association for the Wireless Telecommunication industry. So, the sample is biased and categorizes the adults from the same industry who are more interested to download the application.

Therefore, the population mean can’t be rejected and there is no problem with that. So, the sample is taken randomly.

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 Vaia!

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

Producing machine bearings. To determine whether a metal lathe that produces machine bearings is properly adjusted, a random sample of 25 bearings is collected and the diameter of each is measured.

  1. If the standard deviation of the diameters of the bearings measured over a long period of time is .001 inch, what is the approximate probability that the mean diameter xof the sample of 25 bearings will lie within.0001 inch of the population mean diameter of the bearings?
  2. If the population of diameters has an extremely skewed distribution, how will your approximation in part a be affected?

Refer to Exercise 5.18. Find the probability that

  1. x¯is less than 16.
  2. x¯is greater than 23.
  3. x¯is greater than 25.
  4. x¯falls between 16 and 22.
  5. x¯ is less than 14.

Requests to a Web server. In Exercise 4.175 (p. 282) youlearned that Brighton Webs LTD modeled the arrivaltime of requests to a Web server within each hour, using auniform distribution. Specifically, the number of seconds xfrom the start of the hour that the request is made is uniformly

distributed between 0 and 3,600 seconds. In a randomsample of n= 60 Web server requests, letrepresentthe sample mean number of seconds from the start of thehour that the request is made.

  1. Find Ex¯and interpret its value.
  2. Find Varx¯.
  3. Describe the shape of the sampling distribution of x¯.
  4. Find the probability that x¯is between 1,700 and 1,900seconds.
  5. Find the probability that x¯exceeds 2,000 seconds.

A random sample of n = 80 measurements is drawn from a binomial population with a probability of success .3.

  1. Give the mean and standard deviation of the sampling distribution of the sample proportion,P^
  2. Describe the shape of the sampling distribution ofP
  3. Calculate the standard normal z-score corresponding to a value ofP=.35.
  4. FindP(P=.35.)

Motivation of drug dealers. Refer to the Applied Psychology in Criminal Justice (September 2009) investigation of the personality characteristics of drug dealers, Exercise 2.80 (p. 111). Convicted drug dealers were scored on the Wanting Recognition (WR) Scale. This scale provides a quantitative measure of a person’s level of need for approval and sensitivity to social situations. (Higher scores indicate a greater need for approval.) Based on the study results, we can assume that the WR scores for the population of convicted drug dealers have a mean of 40 and a standard deviation of 5. Suppose that in a sample of 100 people, the mean WR scale score is x = 42. Is this sample likely selected from the population of convicted drug dealers? Explain.

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.

Sign-up for free