Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card
Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card
8th Edition
ISBN: 9781464158933
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Chapter 1, Problem 163E

(a)

To determine

To find: The graphical and numerical summary of the provided data of number of users of internet per 100 people in the year 2010 (Users2010) and to find the numerical summaries of the provided data.

Solution: The obtained histogram indicates that distribution of the data is right skewed. According to the numerical summary, mean of the data is 35.64, minimum value of the data is 0.21, first quartile is 10.31, the median is 31.40, third quartile is 55.65 and the maximum value is 95.63.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given: The data of number of users of internet per 100 people (Users2010) is provided in the question.

Country Name Users2010 Users2011
Afghanistan 3.65 4.58
Albania 45.00 49.00
Algeria 12.50 14.00
Andorra 81.00 81.00
Angola 10.00 14.78
Antigua and Barbuda 80.00 82.00
Argentina 40.00 47.70
Aruba 42.00 57.07
Australia 75.89 78.95
Austria 75.20 79.75
Azerbaijan 46.68 50.75
Bahamas, The 43.00 65.00
Bahrain 55.00 77.00
Bangladesh 3.70 5.00
Barbados 70.20 71.77
Belarus 32.15 39.96
Belgium 73.74 76.20
Benin 3.13 3.50
Bermuda 85.13 88.85
Bhutan 13.60 21.00
Bolivia 22.40 30.00
Bosnia and Herzegovina 52.00 60.00
Botswana 6.00 7.00
Brazil 40.65 45.00
Brunei Darussalam 53.00 56.00
Bulgaria 45.98 50.80
Burkina Faso 2.40 3.00
Burundi 1.00 1.11
Cambodia 1.26 3.10
Cameroon 4.30 5.00
Canada 80.04 82.68
Cape Verde 30.00 32.00
Cayman Islands 66.00 69.47
Central African Republic 2.00 2.20
Chad 1.70 1.90
Chile 45.00 53.89
China 34.39 38.40
Colombia 36.50 40.40
Comoros 5.10 5.50
Congo, Dem. Rep. 0.72 1.20
Congo, Rep. 5.00 5.60
Costa Rica 36.50 42.12
Cote d'Ivoire 2.10 2.20
Croatia 60.12 70.53
Cuba 15.90 23.23
Cyprus 52.99 57.68
Czech Republic 68.64 72.89
Denmark 88.76 89.98
Djibouti 6.50 7.00
Dominica 47.45 51.31
Dominican Republic 31.40 35.50
Ecuador 29.03 31.40
Egypt, Arab Rep. 30.20 35.62
El Salvador 15.90 17.69
Eritrea 5.40 6.20
Estonia 74.15 76.53
Ethiopia 0.75 1.10
Faeroe Islands 75.20 80.73
Fiji 20.00 28.00
Finland 86.91 89.33
France 77.28 76.77
French Polynesia 49.00 49.00
Gabon 7.23 8.00
Gambia, The 9.20 10.87
Georgia 26.29 35.28
Germany 82.53 83.44
Ghana 9.55 14.11
Greece 44.57 53.40
Greenland 63.85 64.62
Guatemala 10.50 11.73
Guinea 1.00 1.30
Guinea-Bissau 2.45 2.67
Guyana 29.90 32.00
Honduras 11.09 15.90
Hong Kong SAR, China 71.85 75.03
Hungary 52.91 58.97
Iceland 95.63 96.62
India 7.50 10.07
Indonesia 10.92 18.00
Iran, Islamic Rep. 16.00 21.00
Iraq 2.47 4.95
Ireland 69.78 77.48
Israel 65.68 68.17
Italy 53.74 56.82
Jamaica 28.07 31.99
Japan 77.65 78.71
Jordan 27.83 35.74
Kazakhstan 31.03 44.04
Kenya 14.00 28.00
Kiribati 9.07 10.00
Korea, Rep. 81.62 81.46
Kuwait 61.40 74.20
Kyrgyz Republic 18.02 19.58
Lao PDR 7.00 9.00
Latvia 68.82 72.43
Lebanon 43.68 52.00
Lesotho 3.86 4.22
Liberia 2.30 3.00
Libya 14.00 17.00
Liechtenstein 80.00 85.00
Lithuania 62.82 67.17
Luxembourg 90.71 90.70
Macao SAR, China 53.80 58.00
Macedonia, FYR 51.90 56.70
Madagascar 1.70 1.90
Malawi 2.26 3.33
Malaysia 56.30 61.00
Maldives 28.30 34.00
Mali 1.90 2.00
Malta 63.08 69.03
Mauritania 4.00 4.50
Mauritius 28.73 35.51
Mexico 31.05 36.15
Moldova 32.40 37.85
Mongolia 12.90 20.00
Montenegro 37.50 40.00
Morocco 49.00 51.00
Mozambique 4.17 4.30
Myanmar 0.25 0.98
Namibia 11.60 12.00
Nepal 7.93 9.00
Netherlands 90.71 92.13
New Caledonia 42.66 51.17
New Zealand 83.01 86.18
Nicaragua 10.00 10.60
Niger 0.83 1.30
Nigeria 24.00 28.43
Norway 93.27 93.45
Oman 62.00 68.00
Pakistan 8.00 9.00
Panama 40.10 42.70
Papua New Guinea 1.28 2.00
Paraguay 19.80 23.90
Peru 34.77 36.50
Philippines 25.00 29.00
Poland 62.47 65.02
Portugal 51.28 55.57
Puerto Rico 45.63 48.50
Qatar 81.60 86.20
Romania 40.02 44.12
Russian Federation 43.31 49.31
Rwanda 8.00 7.00
Sao Tome and Principe 18.75 20.16
Saudi Arabia 41.00 47.50
Senegal 16.00 17.50
Serbia 45.56 47.19
Seychelles 41.00 43.61
Singapore 71.14 75.06
Slovak Republic 76.16 74.87
Slovenia 69.35 71.40
Solomon Islands 5.00 6.00
South Africa 18.05 20.95
Spain 65.81 67.92
Sri Lanka 12.12 15.13
St. Lucia 40.06 42.01
St. Vincent and the Grenadines 38.50 43.01
Suriname 31.59 32.00
Swaziland 12.41 20.43
Sweden 90.01 90.88
Switzerland 82.17 82.99
Syrian Arab Republic 20.66 22.44
Tajikistan 11.55 13.03
Tanzania 11.00 12.00
Thailand 22.40 23.70
Timor-Leste 0.21 0.88
Togo 3.00 3.50
Tonga 16.00 25.00
Trinidad and Tobago 48.50 55.20
Tunisia 36.56 38.81
Turkey 39.82 42.10
Turkmenistan 3.00 5.00
Tuvalu 25.00 30.00
Uganda 12.50 13.01
Ukraine 23.09 30.25
United Arab Emirates 68.00 70.00
United Kingdom 77.75 81.71
United States 74.25 78.24
Uruguay 46.57 51.57
Uzbekistan 19.22 28.57
Venezuela, RB 37.56 40.44
Vietnam 30.97 35.45
West Bank and Gaza 43.44 56.82
Yemen, Rep. 12.35 14.91
Zambia 10.13 11.50
Zimbabwe 11.50 15.70

Graph: To draw the histogram of the provided data, below mentioned steps are followed in Minitab.

Step 1: Enter the data into Minitab worksheet.

Step 2: Go to Graph, select Histogram and select simple and click OK.

Step 3: Select the data variable column of “Users2010” and click on Scale.

Step 4: Check minor ticks under Y Scale Low and X Scale Low and then click OK.

The obtained histogram is as follows:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 1, Problem 163E , additional homework tip  1

Calculation: Use the following steps in Minitab to find the numerical summary for percent users of internet per 100 people.

Step 1. Go to stat option then click on Basis statistics and then choose Display Descriptive Statistics.

Step 2. Click on statistics and check minor ticks under Mean, First quartile, Median, Third quartile, Minimum and Maximum.

The numerical summary for percent of users of internet per 100 people in the year 2010 is as follows:

Varible                   Mean         Minimum       Q1          Median        Q3      MaximumSmokeEveryDay     35.64           0.21           10.31         31.40        55.65      95.63

Interpretation: The obtained histogram of the data is plotted and it shows that the data is normal and is right skewed. The mean of the data is 35.64, minimum value of the data is 0.21, first quartile is 10.31, the median is 31.40, third quartile is 55.65 and the maximum value is 95.63.

(b)

To determine

To find: The change in number of users per 100 people from 2010 to 2011 and to analyze the changes.

(b)

Expert Solution
Check Mark

Answer to Problem 163E

Solution: The minimum value of the data is 1.285, first quartile is 0.996, the median is 2.570, third quartile is 4.811 and the maximum value is 22.000.

Explanation of Solution

Calculation:

Use the following steps in Minitab to find the change in number of users per 100 people from 2010 to 2011.

Step 1. Go to Calc then click on Calculator and enter change in Store result option and enter the formula ‘User2011’-‘User2010’ in the expression option and then click OK.

Step 2. Go to Stat option then click on Basis Statistics and then choose Display Descriptive Statistics.

Step 3. Click on Statistics and check minor ticks under First quartile, Median, Third quartile, Minimum and Maximum.

The numerical summary for change in number of users per 100 people from 2010 to 2011 is as follows:

Varible                    Minimum             Q1           Median          Q3      MaximumSmokeEveryDay     1.285              0.996           2.570         4.811       22.000

To draw the histogram of the provided data, below mentioned steps are followed in Minitab.

Step 1: Enter the data into Minitab worksheet.

Step 2: Go to Graph, select Histogram and select simple and click OK.

Step 3: Select the data variable column of “change” and click on Scale.

Step 4: Check minor ticks under Y Scale Low and X Scale Low and then click OK.

The obtained histogram is as follows:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 1, Problem 163E , additional homework tip  2

Interpretation: The obtained histogram of the data is plotted and it shows that the data is normal and is left skewed. The minimum value of the data is 22.00, first quartile is 4.811, the median is 2.570, third quartile is 0.996 and the maximum value is 1.285.

(c)

To determine

To find: The percent change in number of users per 100 people from 2010 to 2011 and to analyze the changes.

(c)

Expert Solution
Check Mark

Answer to Problem 163E

Solution: The minimum value of the data is 12.50, first quartile is 5.58, the median is 10.75, third quartile is 20.70 and the maximum value is 327.32.

Explanation of Solution

Calculation: Use the following steps in Minitab to find the percent change in number of users per 100 people from 2010 to 2011.

Step 1. Go to Calc then click on Calculator and enter % change in Store result option and enter the formula (User2011User2010)*100/User2010 in the expression option and then click OK.

Step 2. Go to stat option then click on Basis statistics and then choose Display Descriptive Statistics.

Step 3. Click on statistics and check minor ticks under First quartile, Median, Third quartile, Minimum and Maximum.

The numerical summary for change in number of users per 100 people from 2010 to 2011 is as follows:

Varible                    Minimum      Q1       Median      Q3       MaximumSmokeEveryDay     12.50        5.58        10.75      20.70      327.32

To draw the histogram of the provided data, below mentioned steps are followed in Minitab.

Step 1: Enter the data into Minitab worksheet.

Step 2: Go to Graph, select Histogram and select simple and click OK.

Step 3: Select the data variable column of “%change” and click on Scale.

Step 4: Check minor ticks under Y Scale Low and X Scale Low and then click OK.

The obtained histogram is as follows:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 1, Problem 163E , additional homework tip  3

Interpretation: The obtained histogram of the data is plotted and it shows that the data is right skewed. The minimum value of the data is 12.50, first quartile is 5.58, the median is 10.75, third quartile is 20.70 and the maximum value is 327.32.

(d)

To determine

The summary to the analysis from part (a) to (c) and to compare between the change and percent change in number of users per 100 people from 2010 to 2011

(d)

Expert Solution
Check Mark

Answer to Problem 163E

Solution: Internet utilization has considerably expanded from 2010 to 2011 and the data for internet use in the year 2010 is right skewed.

Explanation of Solution

The data for internet use in the year 2010 is right skewed. The change in internet use from 2010 to 2011 and the percent change are also right skewed. Moreover, there is a gap between the percentage change and it is obtained from the histogram. For every one of the data, numerical summary appears the most ideal approach to depict the distribution. Internet utilization has considerably expanded from 2010 to 2011.

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Chapter 1 Solutions

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card

Ch. 1.1 - Prob. 11ECh. 1.1 - Prob. 12ECh. 1.1 - Prob. 13ECh. 1.1 - Prob. 14ECh. 1.1 - Prob. 15ECh. 1.2 - Prob. 16UYKCh. 1.2 - Prob. 17UYKCh. 1.2 - Prob. 18UYKCh. 1.2 - Prob. 19UYKCh. 1.2 - Prob. 20UYKCh. 1.2 - Prob. 21UYKCh. 1.2 - Prob. 22UYKCh. 1.2 - Prob. 23UYKCh. 1.2 - Prob. 24UYKCh. 1.2 - Prob. 25ECh. 1.2 - Prob. 26ECh. 1.2 - Prob. 27ECh. 1.2 - Prob. 28ECh. 1.2 - Prob. 29ECh. 1.2 - Prob. 30ECh. 1.2 - Prob. 31ECh. 1.2 - Prob. 32ECh. 1.2 - Prob. 33ECh. 1.2 - Prob. 34ECh. 1.2 - Prob. 35ECh. 1.2 - Prob. 42ECh. 1.2 - Prob. 36ECh. 1.2 - Prob. 43ECh. 1.2 - Prob. 37ECh. 1.2 - Prob. 44ECh. 1.2 - Prob. 38ECh. 1.2 - Prob. 45ECh. 1.2 - Prob. 46ECh. 1.2 - Prob. 39ECh. 1.2 - Prob. 40ECh. 1.2 - Prob. 41ECh. 1.3 - Prob. 47UYKCh. 1.3 - Prob. 48UYKCh. 1.3 - Prob. 49UYKCh. 1.3 - Prob. 50UYKCh. 1.3 - Prob. 51UYKCh. 1.3 - Prob. 52UYKCh. 1.3 - Prob. 53UYKCh. 1.3 - Prob. 54UYKCh. 1.3 - Prob. 55UYKCh. 1.3 - Prob. 56UYKCh. 1.3 - Prob. 57UYKCh. 1.3 - Prob. 58UYKCh. 1.3 - Prob. 59UYKCh. 1.3 - Prob. 60UYKCh. 1.3 - Prob. 67ECh. 1.3 - Prob. 69ECh. 1.3 - Prob. 61ECh. 1.3 - Prob. 62ECh. 1.3 - Prob. 63ECh. 1.3 - Prob. 64ECh. 1.3 - Prob. 65ECh. 1.3 - Prob. 66ECh. 1.3 - Prob. 74ECh. 1.3 - Prob. 75ECh. 1.3 - Prob. 76ECh. 1.3 - Prob. 71ECh. 1.3 - Prob. 68ECh. 1.3 - Prob. 70ECh. 1.3 - Prob. 77ECh. 1.3 - Prob. 78ECh. 1.3 - Prob. 79ECh. 1.3 - Prob. 80ECh. 1.3 - Prob. 81ECh. 1.3 - Prob. 82ECh. 1.3 - Prob. 83ECh. 1.3 - Prob. 84ECh. 1.3 - Prob. 85ECh. 1.3 - 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. 72ECh. 1.3 - Prob. 97ECh. 1.3 - Prob. 98ECh. 1.3 - Prob. 99ECh. 1.3 - Prob. 100ECh. 1.3 - Prob. 73ECh. 1.4 - Prob. 101UYKCh. 1.4 - Prob. 102UYKCh. 1.4 - Prob. 103UYKCh. 1.4 - Prob. 104UYKCh. 1.4 - Prob. 105UYKCh. 1.4 - Prob. 106UYKCh. 1.4 - Prob. 107UYKCh. 1.4 - Prob. 108UYKCh. 1.4 - Prob. 109ECh. 1.4 - Prob. 110ECh. 1.4 - Prob. 111ECh. 1.4 - Prob. 112ECh. 1.4 - Prob. 113ECh. 1.4 - Prob. 114ECh. 1.4 - Prob. 115ECh. 1.4 - Prob. 116ECh. 1.4 - Prob. 117ECh. 1.4 - Prob. 118ECh. 1.4 - Prob. 119ECh. 1.4 - Prob. 120ECh. 1.4 - Prob. 121ECh. 1.4 - Prob. 122ECh. 1.4 - Prob. 123ECh. 1.4 - Prob. 124ECh. 1.4 - Prob. 125ECh. 1.4 - Prob. 126ECh. 1.4 - Prob. 127ECh. 1.4 - Prob. 128ECh. 1.4 - Prob. 129ECh. 1.4 - Prob. 130ECh. 1.4 - Prob. 131ECh. 1.4 - Prob. 132ECh. 1.4 - Prob. 133ECh. 1.4 - Prob. 134ECh. 1.4 - Prob. 135ECh. 1.4 - Prob. 136ECh. 1.4 - Prob. 137ECh. 1.4 - Prob. 138ECh. 1.4 - Prob. 139ECh. 1.4 - Prob. 140ECh. 1.4 - Prob. 141ECh. 1.4 - Prob. 142ECh. 1.4 - Prob. 143ECh. 1.4 - Prob. 144ECh. 1.4 - Prob. 145ECh. 1.4 - Prob. 146ECh. 1.4 - Prob. 147ECh. 1.4 - Prob. 148ECh. 1.4 - Prob. 149ECh. 1.4 - Prob. 150ECh. 1.4 - Prob. 151ECh. 1.4 - Prob. 152ECh. 1.4 - Prob. 153ECh. 1.4 - Prob. 154ECh. 1.4 - Prob. 155ECh. 1 - Prob. 156ECh. 1 - Prob. 157ECh. 1 - Prob. 158ECh. 1 - Prob. 159ECh. 1 - Prob. 160ECh. 1 - Prob. 161ECh. 1 - Prob. 162ECh. 1 - Prob. 163ECh. 1 - Prob. 164ECh. 1 - Prob. 165ECh. 1 - Prob. 166ECh. 1 - Prob. 167ECh. 1 - Prob. 168ECh. 1 - Prob. 169ECh. 1 - Prob. 170ECh. 1 - Prob. 171ECh. 1 - Prob. 172ECh. 1 - Prob. 173ECh. 1 - Prob. 174ECh. 1 - Prob. 175ECh. 1 - Prob. 176ECh. 1 - Prob. 177E
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