EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
EBK PRACTICE OF STAT.F/AP EXAM,UPDATED
6th Edition
ISBN: 9781319287573
Author: Starnes
Publisher: MPS PUB
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 1.1, Problem 30E

a.

To determine

To find the distribution of superpower preferencefor the students in the sample from each country (the United States and the United Kingdom).To make a segmented bar graph to compare these distributions.

a.

Expert Solution
Check Mark

Answer to Problem 30E

The result for the distribution of superpower preferencefor the students in the sample from each country (the United States and the United Kingdom)is as below:

    SuperpowerCountry
    U.K.U.S.
    Fly27%20.93%
    Freeze time26%20.47%
    Invisibility15%17.21%
    Superstrength10%10.70%
    Telepathy22%30.70%

The segmented bar graph to compare these distributions is as below:

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 1.1, Problem 30E , additional homework tip  1

Explanation of Solution

Given:

The study includes 415 children from the United Kingdom and the United States who completed a survey in a recent year. Each students countryof origin was recorded along with which superpower they would most like to have: theability to fly, ability to freeze time, invisibility, superstrength, or telepathy (ability to readminds). The data are summarized in the following table.

    SuperpowerCountry
    U.K.U.S.
    Fly5445
    Freeze time5244
    Invisibility3037
    Superstrength2023
    Telepathy4466

Concept Used:

Let us first determine the column total of each column, which is the sum of all counts in the corresponding column. Then determine the distribution of superpower preference from each country, by dividing the counts by the column total. Finally, make a segmented bar graph to compare these distributions.

Calculation:

Determine the column total of each column from the given table.

    SuperpowerCountry
    U.K.U.S.
    Fly5445
    Freeze time5244
    Invisibility3037
    Superstrength2023
    Telepathy4466
    Column Total54+52+30+20+44=20045+44+37+23+66=215

Determine the distribution of superpower preference from each country,by dividing the counts by the column total as shown in the below table:

    SuperpowerCountry
    U.K.U.S.
    Fly54200=27100=0.27=27%45215=9430.2093=20.93%
    Freeze time52200=1350=0.26=26%442150.2047=20.47%
    Invisibility30200=320=0.15=15%372150.1721=17.21%
    Superstrength20200=110=0.10=10%232150.1070=10.70%
    Telepathy44200=1150=0.22=22%662150.3070=30.70%

Create a segmented bar graph, where in the width of each bars should be equal and the height should be equal to the percentage of superpowers as shown below.

  EBK PRACTICE OF STAT.F/AP EXAM,UPDATED, Chapter 1.1, Problem 30E , additional homework tip  2

b.

To determine

To describe the association between country of origin and superpower preferencefor the students in the sample from the segmented bar graph obtained in part (a).

b.

Expert Solution
Check Mark

Answer to Problem 30E

We note that the most popular superpower in the U.S. appears to be telepathy, while the most popular superpowers in the U.K. appear to be freeze time and fly.

We also note that the superstrength appears to be the least popular superpower in both the countries.

Explanation of Solution

We note that the most popular superpower in the U.S. appears to be telepathy, while the most popular superpowers in the U.K. appear to be freeze time and fly as shown from the bar graph obtained in part (a).

We also note that the superstrength appears to be the least popular superpower in both the countriesas shown from the bar graph obtained in part (a).

Chapter 1 Solutions

EBK PRACTICE OF STAT.F/AP EXAM,UPDATED

Ch. 1.1 - Prob. 21ECh. 1.1 - Prob. 22ECh. 1.1 - Prob. 23ECh. 1.1 - Prob. 24ECh. 1.1 - Prob. 25ECh. 1.1 - Prob. 26ECh. 1.1 - Prob. 27ECh. 1.1 - Prob. 28ECh. 1.1 - Prob. 29ECh. 1.1 - Prob. 30ECh. 1.1 - Prob. 31ECh. 1.1 - Prob. 32ECh. 1.1 - Prob. 33ECh. 1.1 - Prob. 34ECh. 1.1 - Prob. 35ECh. 1.1 - Prob. 36ECh. 1.1 - Prob. 37ECh. 1.1 - Prob. 38ECh. 1.1 - Prob. 39ECh. 1.1 - Prob. 40ECh. 1.1 - Prob. 41ECh. 1.1 - Prob. 42ECh. 1.1 - Prob. 43ECh. 1.1 - Prob. 44ECh. 1.2 - Prob. 45ECh. 1.2 - Prob. 46ECh. 1.2 - Prob. 47ECh. 1.2 - Prob. 48ECh. 1.2 - Prob. 49ECh. 1.2 - Prob. 50ECh. 1.2 - Prob. 51ECh. 1.2 - Prob. 52ECh. 1.2 - Prob. 53ECh. 1.2 - Prob. 54ECh. 1.2 - Prob. 55ECh. 1.2 - Prob. 56ECh. 1.2 - Prob. 57ECh. 1.2 - Prob. 58ECh. 1.2 - Prob. 59ECh. 1.2 - Prob. 60ECh. 1.2 - Prob. 61ECh. 1.2 - Prob. 62ECh. 1.2 - Prob. 63ECh. 1.2 - Prob. 64ECh. 1.2 - Prob. 65ECh. 1.2 - Prob. 66ECh. 1.2 - Prob. 67ECh. 1.2 - Prob. 68ECh. 1.2 - Prob. 69ECh. 1.2 - Prob. 70ECh. 1.2 - Prob. 71ECh. 1.2 - Prob. 72ECh. 1.2 - Prob. 73ECh. 1.2 - Prob. 74ECh. 1.2 - Prob. 75ECh. 1.2 - Prob. 76ECh. 1.2 - Prob. 77ECh. 1.2 - Prob. 78ECh. 1.2 - Prob. 79ECh. 1.2 - Prob. 80ECh. 1.2 - Prob. 81ECh. 1.2 - Prob. 82ECh. 1.2 - Prob. 83ECh. 1.2 - Prob. 84ECh. 1.2 - Prob. 85ECh. 1.2 - Prob. 86ECh. 1.3 - Prob. 87ECh. 1.3 - Prob. 88ECh. 1.3 - Prob. 89ECh. 1.3 - Prob. 90ECh. 1.3 - Prob. 91ECh. 1.3 - Prob. 92ECh. 1.3 - Prob. 93ECh. 1.3 - Prob. 94ECh. 1.3 - Prob. 95ECh. 1.3 - Prob. 96ECh. 1.3 - Prob. 97ECh. 1.3 - Prob. 98ECh. 1.3 - Prob. 99ECh. 1.3 - Prob. 100ECh. 1.3 - Prob. 101ECh. 1.3 - Prob. 102ECh. 1.3 - Prob. 103ECh. 1.3 - Prob. 104ECh. 1.3 - Prob. 105ECh. 1.3 - Prob. 106ECh. 1.3 - Prob. 107ECh. 1.3 - Prob. 108ECh. 1.3 - Prob. 109ECh. 1.3 - Prob. 110ECh. 1.3 - Prob. 111ECh. 1.3 - Prob. 112ECh. 1.3 - Prob. 113ECh. 1.3 - Prob. 114ECh. 1.3 - Prob. 115ECh. 1.3 - Prob. 116ECh. 1.3 - Prob. 117ECh. 1.3 - Prob. 118ECh. 1.3 - Prob. 119ECh. 1.3 - Prob. 120ECh. 1.3 - Prob. 121ECh. 1.3 - Prob. 122ECh. 1.3 - Prob. 123ECh. 1.3 - Prob. 124ECh. 1.3 - Prob. 125ECh. 1.3 - Prob. 126ECh. 1.3 - Prob. 127ECh. 1.3 - Prob. 128ECh. 1 - Prob. 1ECh. 1 - Prob. 2ECh. 1 - Prob. 3ECh. 1 - Prob. 4ECh. 1 - Prob. 5ECh. 1 - Prob. 6ECh. 1 - Prob. 7ECh. 1 - Prob. 8ECh. 1 - Prob. 9ECh. 1 - Prob. 10ECh. 1 - Prob. R1.1RECh. 1 - Prob. R1.2RECh. 1 - Prob. R1.3RECh. 1 - Prob. R1.4RECh. 1 - Prob. R1.5RECh. 1 - Prob. R1.6RECh. 1 - Prob. R1.7RECh. 1 - Prob. R1.8RECh. 1 - Prob. R1.9RECh. 1 - Prob. R1.10RECh. 1 - Prob. T1.1SPTCh. 1 - Prob. T1.2SPTCh. 1 - Prob. T1.3SPTCh. 1 - Prob. T1.4SPTCh. 1 - Prob. T1.5SPTCh. 1 - Prob. T1.6SPTCh. 1 - Prob. T1.7SPTCh. 1 - Prob. T1.8SPTCh. 1 - Prob. T1.9SPTCh. 1 - Prob. T1.10SPTCh. 1 - Prob. T1.11SPTCh. 1 - Prob. T1.12SPTCh. 1 - Prob. T1.13SPTCh. 1 - Prob. T1.14SPT
Knowledge Booster
Background pattern image
Statistics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Text book image
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Text book image
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Text book image
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
Text book image
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Text book image
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman
Statistics 4.1 Point Estimators; Author: Dr. Jack L. Jackson II;https://www.youtube.com/watch?v=2MrI0J8XCEE;License: Standard YouTube License, CC-BY
Statistics 101: Point Estimators; Author: Brandon Foltz;https://www.youtube.com/watch?v=4v41z3HwLaM;License: Standard YouTube License, CC-BY
Central limit theorem; Author: 365 Data Science;https://www.youtube.com/watch?v=b5xQmk9veZ4;License: Standard YouTube License, CC-BY
Point Estimate Definition & Example; Author: Prof. Essa;https://www.youtube.com/watch?v=OTVwtvQmSn0;License: Standard Youtube License
Point Estimation; Author: Vamsidhar Ambatipudi;https://www.youtube.com/watch?v=flqhlM2bZWc;License: Standard Youtube License