close solutoin list

S O C In problem 3.6, you computed mean and median income for the 13 Canadian provinces and territories and for 13 U.S. states in two separate years. Now compute the standard deviation and range for each year, and, taking account of the two measures of central tendency and the two measures of dispersion, write a paragraph summarizing the distributions. What do the measures of dispersion add to what you already knew about central tendency? Did the median income of the provinces and states become more or less variable over the period? The scores are reproduced here. Median Income for Canadian Provinces and Territories, 2000 and 2011 (Canadian dollars) Province or Territory 2000 2011 Newfoundland and Labrador 38,800 67,200 Prince Edward Island 44,200 66,500 Nova Scotia 44,500 66,300 New Brunswick 43,200 63,930 Quebec 47,700 68,170 Ontario 55,700 73,290 Manitoba 47,300 68,710 Saskatchewan 45,800 77,300 Alberta 55,200 89,930 British Columbia 49,100 69,150 Yukon Columbia 56,000 90,090 Northwest Territories 61,000 105,560 Nunavut 37,600 65,280 Mean = Median = Range = Standard Deviation = Median Income for Thirteen States, 1999 and 2012 (U.S dollars) State 1999 2012 Alabama 36,213 43,464 Alaska 51,509 63,348 Arkansas 29,762 39,018 California 43,744 57,020 Connecticut 50,798 64,247 Illinois 46,392 51,738 Kansas 37,476 50,003 Maryland 52,310 71,836 Michigan 46,238 50,015 New York 40,058 47,680 Ohio 39,617 44,375 South Dakota 35,962 49,415 Texas 38,978 51,926 Mean = Median = Range = Standard Deviation = 3.6 S O C The following tables list the median family incomes for the 13 Canadian provinces and territories in 2000 and 2011 and for 13 states of the United States in 1999 and 2012. For the provinces and the states, compute the mean and median family income for each year and compare the two measures of central tendency. Which measure of central tendency is greater for each year? Are the distributions skewed? In which direction? Median Income for Canadian Provinces and Territories, 2000 and 2011 (Canadian dollars) Province or Territory 2000 2011 Newfoundland and Labrador 38,800 67,200 Prince Edward Island 44,200 66,500 Nova Scotia 44,500 66,300 New Brunswick 43,200 63,930 Quebec 47,700 68,170 Ontario 55,700 73,290 Manitoba 47,300 68,710 Saskatchewan 45,800 77,300 Alberta 55,200 89,930 British Columbia 49,100 69,150 Yukon Columbia 56,000 90,090 Northwest Territories 61,000 105,560 Nunavut 37,600 65,280 Mean = Median = Median Income for Thirteen States, 1999 and 2012 (U.S dollars) State 1999 2012 Alabama 36,213 43,464 Alaska 51,509 63,348 Arkansas 29,762 39,018 California 43,744 57,020 Connecticut 50,798 64,247 Illinois 46,392 51,738 Kansas 37,476 50,003 Maryland 52,310 71,836 Michigan 46,238 50,015 New York 40,058 47,680 Ohio 39,617 44,375 South Dakota 35,962 49,415 Texas 38,978 51,926 Mean = Median =

BuyFind

Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836
BuyFind

Essentials Of Statistics

4th Edition
HEALEY + 1 other
Publisher: Cengage Learning,
ISBN: 9781305093836

Solutions

Chapter
Section
Chapter 4, Problem 4.5P
Textbook Problem
75 views

S O C In problem 3.6, you computed mean and median income for the 13 Canadian provinces and territories and for 13 U.S. states in two separate years. Now compute the standard deviation and range for each year, and, taking account of the two measures of central tendency and the two measures of dispersion, write a paragraph summarizing the distributions. What do the measures of dispersion add to what you already knew about central tendency? Did the median income of the provinces and states become more or less variable over the period? The scores are reproduced here.

Median Income for Canadian Provinces and Territories, 2000 and 2011 (Canadian dollars)

Province or Territory 2000 2011
Newfoundland and Labrador 38,800 67,200
Prince Edward Island 44,200 66,500
Nova Scotia 44,500 66,300
New Brunswick 43,200 63,930
Quebec 47,700 68,170
Ontario 55,700 73,290
Manitoba 47,300 68,710
Saskatchewan 45,800 77,300
Alberta 55,200 89,930
British Columbia 49,100 69,150
Yukon Columbia 56,000 90,090
Northwest Territories 61,000 105,560
Nunavut 37,600 65,280
Mean =
Median =
Range =
Standard Deviation =

Median Income for Thirteen States, 1999 and 2012 (U.S dollars)

State 1999 2012
Alabama 36,213 43,464
Alaska 51,509 63,348
Arkansas 29,762 39,018
California 43,744 57,020
Connecticut 50,798 64,247
Illinois 46,392 51,738
Kansas 37,476 50,003
Maryland 52,310 71,836
Michigan 46,238 50,015
New York 40,058 47,680
Ohio 39,617 44,375
South Dakota 35,962 49,415
Texas 38,978 51,926
Mean =
Median =
Range =
Standard Deviation =

3.6 S O C The following tables list the median family incomes for the 13 Canadian provinces and territories in 2000 and 2011 and for 13 states of the United States in 1999 and 2012. For the provinces and the states, compute the mean and median family income for each year and compare the two measures of central tendency. Which measure of central tendency is greater for each year? Are the distributions skewed? In which direction?

Median Income for Canadian Provinces and Territories, 2000 and 2011 (Canadian dollars)

Province or Territory 2000 2011
Newfoundland and Labrador 38,800 67,200
Prince Edward Island 44,200 66,500
Nova Scotia 44,500 66,300
New Brunswick 43,200 63,930
Quebec 47,700 68,170
Ontario 55,700 73,290
Manitoba 47,300 68,710
Saskatchewan 45,800 77,300
Alberta 55,200 89,930
British Columbia 49,100 69,150
Yukon Columbia 56,000 90,090
Northwest Territories 61,000 105,560
Nunavut 37,600 65,280
Mean =
Median =

Median Income for Thirteen States, 1999 and 2012 (U.S dollars)

State 1999 2012
Alabama 36,213 43,464
Alaska 51,509 63,348
Arkansas 29,762 39,018
California 43,744 57,020
Connecticut 50,798 64,247
Illinois 46,392 51,738
Kansas 37,476 50,003
Maryland 52,310 71,836
Michigan 46,238 50,015
New York 40,058 47,680
Ohio 39,617 44,375
South Dakota 35,962 49,415
Texas 38,978 51,926
Mean =
Median =

Expert Solution
To determine

To find:

The median, mean, range and standard deviation of 13 Canadian provinces and U.S states in two separate years

Explanation of Solution

Given:

The following table gives the data of median income for Canadian Provinces and Territories of the year 2000 and 2011.

Province or Territory 2000 2011
Newfoundland and Labrador 38, 800 67, 200
Prince Edward Island 44, 200 66, 500
Nova Scotia 44, 500 66, 300
New Brunswick 43, 200 63, 930
Quebec 47, 700 68, 170
Ontario 55, 700 73, 290
Manitoba 47, 300 68, 710
Saskatchewan 45, 800 77, 300
Alberta 55, 200 89, 930
British Columbia 49, 100 69, 150
Yukon Columbia 56, 000 90, 090
Northwest Territories 61, 000 105, 560
Nunavut 37, 600 65, 280

The following table gives the data of median income for U.S states of the year 1999 and 2012.

State 1999 2012
Alabama 36, 213 43, 464
Alaska 51, 509 63, 348
Arkansas 29, 762 39, 018
California 43, 744 57, 020
Connecticut 50, 798 64, 247
Illinois 46, 392 51, 738
Kansas 37, 476 50, 003
Maryland 52, 310 71, 836
Michigan 46, 238 50, 015
New York 40, 058 47, 680
Ohio 39, 617 44, 375
South Dakota 35, 962 49, 415
Texas 38, 978 51, 926

Formula used:

The formula to calculate median for odd number of terms is given by,

Median=(Number of terms+1)2

The formula to calculate range is given by,

Range=Highest valueLowest value

Let the data values be Xi’s.

The formula to calculate mean is given by,

X¯=i=1NXiN

The formula to calculate standard deviation is given by,

s=(XiX¯)2N

Where, N is the population size.

Calculation:

For 13 Canadian provinces and Territories

For the year 2000 - arrange the data in the increasing order.

The data in increasing order is given by,

S.No Income
1 37, 600
2 38, 800
3 43, 200
4 44, 200
5 44, 500
6 45, 800
7 47, 300
8 47, 700
9 49, 100
10 55, 200
11 55, 700
12 56, 000
13 61, 000

The number of terms is 13, which is odd.

The median for the odd number of terms is given by,

Median=(Number of terms+1)2

Substitute 13 for number of terms in the above mentioned formula,

Median=(13+1)2=142=7

The median income corresponding to 7th number is 47, 300.

The highest income is 61, 000 and the lowest income is 37, 600.

The range is given by,

Range=Highest incomeLowest income

Substitute 61, 000 for highest income and 37, 600 for lowest income in the above mentioned formula,

Range=61,00037,600=23400

The size of the population is 13.

The mean is given by,

X¯=i=1NXiN

Substitute 13 for N, 38, 800 for X1, 44, 200 for X2 and so on in the above mentioned formula,

X¯=38,800+44,200+.............+61,000+37,60013=626,10013=48,161.5 ……(1)

Consider the following table of sum of squares,

Scores (Xi) (XiX¯) (XiX¯)2
38, 800 9,361.5 87637682
44, 200 3,961.5 15693482
44, 500 3,661.5 13406582
43, 200 4,961.5 24616482
47, 700 461.5 212982.25
55, 700 7, 538.5 56828982
47, 300 861.5 742182.25
45, 800 2,361.5 5576682.3
55, 200 7, 038.5 49540482
49, 100 938.5 880782.25
56, 000 7, 838.5 61442082
61, 000 12, 838.5 164827082
37, 600 10,561.5 111545282
(XiX¯)=0 (XiX¯)2=592950769

From equation (1), substitute 38,800 for X1 and 48161.5 for X¯ in (X1X¯).

(X1X¯)=(38,80048161.5)(X1X¯)=9,361.5

Square the both sides of the equation.

(X1X¯)2=(9.361.5)2(X1X¯)2=87637682

Proceed in the same manner to calculate (XiX¯)2 for all the 1iN for the rest data and refer table for the rest of the (XiX¯)2 values calculated. Then the value of (XiX¯)2 is calculated as,

(XiX¯)2=87637682+15693482+...........+111545282=592950769 ……(2)

The standard deviation is given by,

s=(XiX¯)2N

From equation (2), substitute 592950769 for (XiX¯)2 and 13 for N in the above mentioned formula,

s=59295076913=45611598=6753.64

Thus, range is 23, 400 and standard deviation is 6753.64.

For 13 Canadian provinces and Territories

For the year 2011 - arrange the data in the increasing order.

The data in increasing order is given by,

S.No Income
1 63, 930
2 65, 280
3 66, 300
4 66, 500
5 67, 200
6 68, 170
7 68, 710
8 69, 150
9 73, 290
10 77, 300
11 89, 930
12 90, 090
13 105, 560

The number of terms is 13, which is odd.

The median for the odd number of terms is given by,

Median=(Number of terms+1)2

Substitute 13 for number of terms in the above mentioned formula,

Median=(13+1)2=142=7

The median income corresponding to 7th number is 68, 710.

The highest income is 105, 560 and the lowest income is 63.930.

The range is given by,

Range=Highest incomeLowest income

Substitute 105, 560 for highest income and 63, 930 for lowest income in the above mentioned formula,

Range=105,56063,930=41,630

The size of the population is 13.

The mean is given by,

X¯=i=1NXiN

Substitute 13 for N, 67, 200 for X1, 66, 500 for X2 and so on in the above mentioned formula,

X¯=67,200+66,500+............+65,28013=971,41013=74723.8 ……(3)

Consider the following table of sum of squares,

Scores (Xi) (XiX¯) (XiX¯)2
67, 200 7,523.8 56607566
66, 500 8,223.8 67630886
66, 300 8,423.8 70960406
63, 930 10,793.8 116506118
68, 170 6,553.8 42952294
73, 290 1,433.8 2055782.4
68, 710 6,013.8 36165790
77, 300 2, 576.2 6636806.4
89, 930 15, 206.2 231228518
69, 150 5,573.8 31067246
90, 090 15, 366.2 236120102
105, 560 30, 836.2 950871230
65, 280 9,443.8 89185358
(XiX¯)=0 (XiX¯)2=1937988108

From equation (3), substitute 67,200 for X1 and 74723.8 for X¯ in (X1X¯).

(X1X¯)=(67,20074723.8)(X1X¯)=7,523.8

Square the both sides of the equation.

(X1X¯)2=(7.523.8)2(X1X¯)2=56607566

Proceed in the same manner to calculate (XiX¯)2 for all the 1iN for the rest data and refer table for the rest of the (XiX¯)2 values calculated. Then the value of (XiX¯)2 is calculated as,

(XiX¯)2=56607566+67630886+.....................+89185358=1937988108 ……(4)

The standard deviation is given by,

s=(XiX¯)2N

From equation (4), substitute 1937988108 for (XiX¯)2 and 13 for N in the above mentioned formula,

Want to see this answer and more?

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

See solution

Additional Math Textbook Solutions

Find more solutions based on key concepts
Show solutions
21-47 Differentiate. y=(cosx)x

Calculus (MindTap Course List)

In Exercises 47-52, find and simplify f(a+h)f(a)h(h0) for each function. 51. f(x)=1x

Applied Calculus for the Managerial, Life, and Social Sciences: A Brief Approach

For Review Exercises 21 to 23, DEAC AD=4,BD=8,DE=3,AC=?

Elementary Geometry For College Students, 7e

Solve each equation and check: 6x+2=2(17x)

Elementary Technical Mathematics

0 2 −2 does not exist

Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th

True or False:

Study Guide for Stewart's Multivariable Calculus, 8th

In part (d) of exercise 9, data contained in the DATAfile named PGADrivingDist (PGA Tour website, November 1, 2...

Modern Business Statistics with Microsoft Office Excel (with XLSTAT Education Edition Printed Access Card) (MindTap Course List)

In Problems 11 and 12 put the given differential equation into form (3) for each regular singular point of the ...

A First Course in Differential Equations with Modeling Applications (MindTap Course List)

Weight on the Moon This is a continuation of Exercise 21. Acceleration due to gravity near the surface of the E...

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)