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

Smartphone Ownership. The Pew Internet \& American Life Project conducted a survey of smartphone ownership. One aspect of the study involved the gender of smartphone owners. Of 1029 sampled men, 607 owned a smartphone; and of 1223 sampled women, 648 owned a smartphone. At the 1% significance level, do the data provide sufficient evidence to conclude that a difference exists in the percentages of smartphone owners between men and women?

Short Answer

Expert verified

The data does not support the conclusion that there is a difference in the percentages of smartphone owners between men and women.

Step by step solution

01

Given Information

Sample size is n=1029.

Number of success is x=607.

Sample proportion would be p^=xn

=6071029

=0.590

02

Explanation

For men:

A significance level of α,

To find zα/2using table II.

The significance level is 1%. i.e. α=0.01.

Using table II,

zα/2=z0.01/2

=z0.005

=2.58.

The confidence interval for pis

p^±zα/2·p^(1-p^)n

CI=0.590±2.580.590(1-0.590)1029

CI=0.590±0.04

CI=0.55to0.63

CI=55%to63%

03

Explanation

For women:

Sample size is n=1223.

Number of success is x=648.

Sample proportion would be p^=xn

=6481223

=0.5298

A significance level of α,

To find zα/2using table II.

Significance level is 1%.

i.e. α=0.01.

Using table II,

zα/2=z0.01/2

=z0.005

=2.58.

04

Explanation

The confidence interval for pis given as-

p^±zα/2·p^(1-p^)n

CI=0.5298±2.580.5298(1-0.5298)1223

CI=0.5298±0.0368

CI=0.4930to56.66

CI=59.30%to56.66%

One could assume that the data does not provide adequate information to indicate that men and women have different percentages of smartphone owners.

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

x1=30,n1=80,x2=15,n2=20;95%confidence interval

Racial Crossover. Refer to Exercise 11.108 and find and interpret a 95% confidence interval for the difference between the stroke incidences of white and African-American elderly.

Margin of error=0.03

Confidence level=99%

Educated guess=0.5

(a) Obtain a sample size that will ensure a margin of error of at most the one specified (provided of course that the observed value of the sample proportion is further from 0.5that of the educated guess.

(b). Compare your answer to the corresponding one and explain the reason for the difference, if any.

Margin of error=0.02

Confidence level=90%

Educated guess=0.1

(a) Obtain a sample size that will ensure a margin of error of at most the one specified (provided of course that the observed value of the sample proportion is further from 0.5that of the educated guess.

(b). Compare your answer to the corresponding one and explain the reason for the difference, if any.

Genetic Binge Eating. According to an article in Science News, binge eating has been associated with a mutation of the gene for a brain protein called melanocortin 4 receptor (MC4R). In one study, F. Horber of the Hirslanden Clinic in Zurich and his colleagues genetically analyzed the blood of 469 obese people and found that 24 carried a mutated MCAR gene. Suppose that you want to estimate the proportion of all obese people who carry a mutated MC4R gene

a. Determine the margin of error for a 90%confidence interval.

b. Without doing any calculations, indicate whether the margin of error is larger or smaller for a 95%confidence interval. Explain your answer

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