Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
9th Edition
ISBN: 9781319013387
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Chapter 2.2, Problem 36E

(a)

To determine

To find: A new data set that has Internet users as users per 10,000 people and births as births per 10,000 people.

(a)

Expert Solution
Check Mark

Answer to Problem 36E

Solution: The partial data set is shown below:

Country Name

Country Code

Birth rate

Users Per 100

Users per 10000

Birth per 10000

Aruba

ABW

11.323

57.0684

5706.84

113.23

Angola

AGO

40.838

14.776

1477.6

408.38

Albania

ALB

12.748

49.00

4900

127.48

United Arab Emirates

ARE

12.755

70.00

7000

127.55

Argentina

ARG

17.006

47.7040

4770.4

170.06

Burundi

BDI

33.722

1.11

111

337.22

Benin

BEN

39.073

3.50

350

390.73

Explanation of Solution

Given: The data of birthrate (births per 1000 people) versus Internet use (users per 100 people) for 106 countries is given below.

Introduction to the Practice of Statistics, Chapter 2.2, Problem 36E , additional homework tip  1

Introduction to the Practice of Statistics, Chapter 2.2, Problem 36E , additional homework tip  2

Introduction to the Practice of Statistics, Chapter 2.2, Problem 36E , additional homework tip  3

Introduction to the Practice of Statistics, Chapter 2.2, Problem 36E , additional homework tip  4

Calculation: To obtain a new data set that has Internet users as users per 10,000 people and births as births per 10,000 people use Minitab. Follow the steps below:

Step 1: Open the Minitab worksheet which contains the data.

Step 2: Go to Calc > Calculator.

Step 3: Write “Users per 10000” in the column for Store result in variable. Select Users in the column for Expression and select “&*#8221; from the calculator and write (10000/100) in the column for Expression.

Step 4: Click OK.

The new data set for users for 10,000 people is obtained. To obtain the new data set for Birthrate 2011 for 10,000 people follow the similar steps.

Step 1: Open the Minitab worksheet which contains the data.

Step 2: Go to Calc > Calculator.

Step 3: Write “Birth rate per 10000”in the column for Store result in variable. Select Birthrate 2011 in the column for Expression and select “&*#8221; from the calculator and write (10000/1000) in the column for Expression.

Step 4: Click OK.

The new data set for Birthrate 2011 for 10,000 people is obtained. The partial data set is shown below:

Country Name

Country Code

Birth rate

Users

Users per 10000

Birth per 10000

Aruba

ABW

11.323

57.0684

5706.84

113.23

Angola

AGO

40.838

14.7760

1477.60

408.38

Albania

ALB

12.748

49.0000

4900.00

127.48

United Arab Emirates

ARE

12.755

70.0000

7000.00

127.55

Argentina

ARG

17.006

47.7040

4770.40

170.06

Burundi

BDI

33.722

1.1100

111.00

337.22

Benin

BEN

39.073

3.5000

350.00

390.73

(b)

To determine

To explain: The reason for these transformations to be linear transformations.

(b)

Expert Solution
Check Mark

Answer to Problem 36E

Solution: These transformations are linear transformations because every variable is multiplied by a numerical value.

Explanation of Solution

Consider the results of part (a). The transformations provide the linearity for the new variables. These transformations can be considered as a linear transformations because every variable is multiplied by a number.

(c)

To determine

To graph: A scatterplot by using transformed variables.

(c)

Expert Solution
Check Mark

Explanation of Solution

Graph: To obtain the scatterplot use Minitab. Follow the steps below:

Step 1: Go to Stat > Graph > Scatterplot > Simple.

Step 2: Select “Birth per 10,000” in Y variables and “Users per 10,000” in X variables.

Step 3: Click OK.

The required scatterplot is obtained as:

Introduction to the Practice of Statistics, Chapter 2.2, Problem 36E , additional homework tip  5

Interpretation: The scatterplot represents that the relationship between numbers of births per10,000 versus Internet users per 10,000 is negative.

(d)

To determine

To explain: The comparison between obtained scatterplot and figure 2.13.

(d)

Expert Solution
Check Mark

Answer to Problem 36E

Solution: The obtained scatterplot in part (c) represents that the relationship between variables is more linear than the relationship which is represented by the scatterplot of births per 1000 people and internet users per 100 people. Both the scatterplots shows moderate relationship between the variables.

Explanation of Solution

The scatterplot which is obtained in part (c) represents that the relationship between numbers of births per 10,000 versus Internet users per 10,000 is negative.

The scatterplot of births per 1000 people and internet users per 100 people is shown below:

Introduction to the Practice of Statistics, Chapter 2.2, Problem 36E , additional homework tip  6

The scatterplot represents that the relationship between births per 1000 people versus Internet users per 100 people for 106 countries is quite linear and negative. The scatterplot which is obtained in part (c) represents that the relationship between birth rate per 10000 people and internet users per 10000 people is more linear than the provided scatterplot. Both the scatterplot show a moderate relationship between the variables.

(e)

To determine

To explain: The reason why World Bank selected Internet users as users per 100 people and births as births per 1000 people.

(e)

Expert Solution
Check Mark

Answer to Problem 36E

Solution: To make the proper ratio, the analyst of World Bank selected Internet users as users per 100 people and births as births per 1000 people.

Explanation of Solution

The users of internet can be male or woman both but births can be given by females only. To make the ratio accurate, the analyst of World Bank selected Internet users as users per 100 people and births as births per 1000 people.

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Chapter 2 Solutions

Introduction to the Practice of Statistics

Ch. 2.2 - Prob. 11UYKCh. 2.2 - Prob. 12UYKCh. 2.2 - Prob. 13UYKCh. 2.2 - Prob. 14UYKCh. 2.2 - Prob. 15UYKCh. 2.2 - Prob. 16ECh. 2.2 - Prob. 17ECh. 2.2 - Prob. 18ECh. 2.2 - Prob. 19ECh. 2.2 - Prob. 20ECh. 2.2 - Prob. 21ECh. 2.2 - Prob. 22ECh. 2.2 - Prob. 23ECh. 2.2 - Prob. 24ECh. 2.2 - Prob. 25ECh. 2.2 - Prob. 26ECh. 2.2 - Prob. 27ECh. 2.2 - Prob. 28ECh. 2.2 - Prob. 29ECh. 2.2 - Prob. 30ECh. 2.2 - Prob. 31ECh. 2.2 - Prob. 32ECh. 2.2 - Prob. 33ECh. 2.2 - Prob. 34ECh. 2.2 - Prob. 35ECh. 2.2 - Prob. 36ECh. 2.2 - Prob. 37ECh. 2.3 - Prob. 38UYKCh. 2.3 - Prob. 39UYKCh. 2.3 - Prob. 40ECh. 2.3 - Prob. 41ECh. 2.3 - Prob. 42ECh. 2.3 - Prob. 43ECh. 2.3 - Prob. 44ECh. 2.3 - Prob. 45ECh. 2.3 - Prob. 46ECh. 2.3 - Prob. 47ECh. 2.3 - Prob. 48ECh. 2.3 - Prob. 49ECh. 2.3 - Prob. 50ECh. 2.3 - Prob. 51ECh. 2.3 - Prob. 52ECh. 2.3 - Prob. 53ECh. 2.3 - Prob. 54ECh. 2.3 - Prob. 55ECh. 2.3 - Prob. 56ECh. 2.3 - Prob. 57ECh. 2.3 - Prob. 58ECh. 2.3 - Prob. 59ECh. 2.3 - Prob. 60ECh. 2.4 - Prob. 61UYKCh. 2.4 - Prob. 62UYKCh. 2.4 - Prob. 63UYKCh. 2.4 - Prob. 64UYKCh. 2.4 - Prob. 65ECh. 2.4 - Prob. 66ECh. 2.4 - Prob. 67ECh. 2.4 - Prob. 68ECh. 2.4 - Prob. 69ECh. 2.4 - Prob. 70ECh. 2.4 - Prob. 71ECh. 2.4 - Prob. 72ECh. 2.4 - Prob. 73ECh. 2.4 - Prob. 74ECh. 2.4 - Prob. 75ECh. 2.4 - Prob. 76ECh. 2.4 - Prob. 77ECh. 2.4 - Prob. 78ECh. 2.4 - Prob. 79ECh. 2.4 - Prob. 80ECh. 2.4 - Prob. 81ECh. 2.4 - Prob. 82ECh. 2.4 - Prob. 83ECh. 2.4 - Prob. 84ECh. 2.4 - Prob. 85ECh. 2.4 - Prob. 86ECh. 2.4 - Prob. 87ECh. 2.4 - Prob. 88ECh. 2.4 - Prob. 89ECh. 2.4 - Prob. 90ECh. 2.4 - Prob. 91ECh. 2.5 - Prob. 92UYKCh. 2.5 - Prob. 93UYKCh. 2.5 - Prob. 94ECh. 2.5 - Prob. 95ECh. 2.5 - Prob. 96ECh. 2.5 - Prob. 97ECh. 2.5 - Prob. 98ECh. 2.5 - Prob. 99ECh. 2.5 - Prob. 100ECh. 2.5 - Prob. 101ECh. 2.5 - Prob. 102ECh. 2.5 - Prob. 103ECh. 2.5 - Prob. 104ECh. 2.5 - Prob. 105ECh. 2.5 - Prob. 106ECh. 2.5 - Prob. 107ECh. 2.5 - Prob. 108ECh. 2.5 - Prob. 110ECh. 2.5 - Prob. 111ECh. 2.5 - Prob. 112ECh. 2.6 - Prob. 113UYKCh. 2.6 - Prob. 114UYKCh. 2.6 - Prob. 115UYKCh. 2.6 - Prob. 116UYKCh. 2.6 - Prob. 117UYKCh. 2.6 - Prob. 118UYKCh. 2.6 - Prob. 119ECh. 2.6 - Prob. 120ECh. 2.6 - Prob. 121ECh. 2.6 - Prob. 122ECh. 2.6 - Prob. 123ECh. 2.6 - Prob. 124ECh. 2.6 - Prob. 125ECh. 2.6 - Prob. 126ECh. 2.6 - Prob. 127ECh. 2.6 - Prob. 128ECh. 2.6 - Prob. 129ECh. 2.6 - Prob. 130ECh. 2.7 - Prob. 131ECh. 2.7 - Prob. 132ECh. 2.7 - Prob. 133ECh. 2.7 - Prob. 134ECh. 2.7 - Prob. 135ECh. 2.7 - Prob. 136ECh. 2.7 - Prob. 137ECh. 2.7 - Prob. 138ECh. 2.7 - Prob. 139ECh. 2.7 - Prob. 140ECh. 2.7 - Prob. 141ECh. 2.7 - Prob. 142ECh. 2.7 - Prob. 143ECh. 2 - Prob. 144ECh. 2 - Prob. 145ECh. 2 - Prob. 146ECh. 2 - Prob. 147ECh. 2 - Prob. 148ECh. 2 - Prob. 149ECh. 2 - Prob. 150ECh. 2 - Prob. 151ECh. 2 - Prob. 152ECh. 2 - Prob. 153ECh. 2 - Prob. 154ECh. 2 - Prob. 155ECh. 2 - Prob. 156ECh. 2 - Prob. 157ECh. 2 - Prob. 158ECh. 2 - Prob. 159ECh. 2 - Prob. 160ECh. 2 - Prob. 161ECh. 2 - Prob. 162ECh. 2 - Prob. 163ECh. 2 - Prob. 164ECh. 2 - Prob. 165ECh. 2 - Prob. 166ECh. 2 - Prob. 167ECh. 2 - Prob. 168ECh. 2 - Prob. 169ECh. 2 - Prob. 170ECh. 2 - Prob. 171ECh. 2 - Prob. 172ECh. 2 - Prob. 173ECh. 2 - Prob. 174ECh. 2 - Prob. 175ECh. 2 - Prob. 176E
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