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S O C Labor force participation rates (percent employed), percent high school graduates, and mean income for males and females in ten states are reported here. Calculate the mean and the standard for both groups for each variable and describe the differences. Are males and females unequal on any of these variables? How great is the gender inequality? Labor Force Participation Rate Percentage HS Graduates Mean Income State Male Female Male Female Male Female A 65.8 54.3 81.0 81.9 55,623 50,012 B 76.7 63.0 88.4 89.7 52,345 51,556 C 71.8 57.2 82.4 84.6 55,789 48,231 D 76.1 66.6 89.5 90.9 48,907 46,289 E 75.1 63.1 86.9 88.7 62,023 58,034 F 69.9 61.1 86.3 86.4 55,000 53,897 G 73.6 59.6 87.1 87.6 49,145 47,148 H 70.5 60.3 87.0 87.6 51,897 50,659 I 66.3 55.2 81.7 84.1 51,238 45,289 J 74.5 67.1 89.1 91.6 60,746 56,489

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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 4.7P
Textbook Problem
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S O C Labor force participation rates (percent employed), percent high school graduates, and mean income for males and females in ten states are reported here. Calculate the mean and the standard for both groups for each variable and describe the differences. Are males and females unequal on any of these variables? How great is the gender inequality?

Labor Force Participation Rate Percentage HS Graduates Mean Income
State Male Female Male Female Male Female
A 65.8 54.3 81.0 81.9 55,623 50,012
B 76.7 63.0 88.4 89.7 52,345 51,556
C 71.8 57.2 82.4 84.6 55,789 48,231
D 76.1 66.6 89.5 90.9 48,907 46,289
E 75.1 63.1 86.9 88.7 62,023 58,034
F 69.9 61.1 86.3 86.4 55,000 53,897
G 73.6 59.6 87.1 87.6 49,145 47,148
H 70.5 60.3 87.0 87.6 51,897 50,659
I 66.3 55.2 81.7 84.1 51,238 45,289
J 74.5 67.1 89.1 91.6 60,746 56,489

Expert Solution
To determine

To find:

The mean and standard deviation of male and female for the given variable

Explanation of Solution

Given:

The following table shows the data of 3 different variables of 10 states.

Labor Force Participation Rate Percentage HS Graduates Mean Income
State Male Female Male Female Male Female
A 65.8 54.3 81.0 81.9 55, 623 50, 012
B 76.7 63.0 88.4 89.7 52, 345 51, 556
C 71.8 57.2 82.4 84.6 55, 789 48, 231
D 76.1 66.6 89.5 90.9 48, 907 46, 289
E 75.1 63.1 86.9 88.7 62, 023 58, 034
F 69.9 61.1 86.3 86.4 55, 000 53, 897
G 73.6 59.6 87.1 87.6 49, 145 47, 148
H 70.5 60.3 87.0 87.6 51, 897 50, 659
I 66.3 55.2 81.7 84.1 51, 238 45, 289
J 74.5 67.1 89.1 91.6 60, 746 56, 489

Formula used:

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:

Consider the labor force participation rate for males.

The size of the population is 10.

The mean is given by,

X¯=i=1NXiN

Substitute 10 for N, 65.8 for X1, 76.7 for X2 and so on in the above mentioned formula,

X¯=65.8+76.7+...+74.510=720.310=72.03 ……(1)

Consider the following table of sum of squares,

(Xi) (XiX¯) (XiX¯)2
65.8 6.23 38.8129
76.7 4.67 21.8089
71.8 0.23 0.0529
76.1 4.07 16.5649
75.1 3.07 9.4249
69.9 2.13 4.5369
73.6 1.57 2.4649
70.5 1.53 2.3409
66.3 5.73 32.8329
74.5 2.47 6.1009
(XiX¯)=0 (XiX¯)2=134.941

From equation (1), substitute 65.8 for X1 and 72.03 for X¯ in (X1X¯).

(X1X¯)=(65.872.03)(X1X¯)=6.23

Square the both sides of the equation.

(X1X¯)2=(6.23)2(X1X¯)2=38.8129

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=38.8129+21.8089+...+6.1009=134.941 ……(2)

The standard deviation is given by,

s=(XiX¯)2N

From equation (2), substitute 134.941 for (XiX¯)2 and 10 for N in the above mentioned formula,

s=134.94110=13.49413.67

Thus, standard deviation for the labor force participation rate for males is 3.67.

Consider the labor force participation rate for females.

The mean is given by,

X¯=i=1NXiN

Substitute 10 for N, 54.3 for X1, 63.0 for X2 and so on in the above mentioned formula,

X¯=54.3+63+...+67.110=607.510=60.75 ……(3)

Consider the following table of sum of squares,

(Xi) (XiX¯) (XiX¯)2
54.3 6.45 41.6025
63 2.25 5.0625
57.2 3.55 12.6025
66.6 5.85 34.2225
63.1 2.35 5.5225
61.1 0.35 0.1225
59.6 1.15 1.3225
60.3 0.45 0.2025
55.2 5.55 30.8025
67.1 6.35 40.3225
(XiX¯)=0 (XiX¯)2=171.785

From equation (3), substitute 54.3 for X1 and 60.75 for X¯ in (X1X¯).

(X1X¯)=(54.360.75)(X1X¯)=6.45

Square the both sides of the equation.

(X1X¯)2=(6.45)2(X1X¯)2=41.6025

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=41.6025+5.0625+...+40.3225=171.785 ……(4)

The standard deviation is given by,

s=(XiX¯)2N

From equation (4), substitute 171.785 for (XiX¯)2 and 10 for N in the above mentioned formula,

s=171.78510=17.17854.144

Thus, standard deviation for the labor force participation rate for females is 4.144.

Consider the percentage HS graduates of males.

The mean is given by,

X¯=i=1NXiN

Substitute 10 for N, 81 for X1, 88.4 for X2 and so on in the above mentioned formula,

X¯=81+88.4+...+89.110=859.410=85.94 ……(5)

Consider the following table of sum of squares,

(Xi) (XiX¯) (XiX¯)2
81 4.94 24.4036
88.4 2.46 6.0516
82.4 3.54 12.5316
89.5 3.56 12.6736
86.9 0.96 0.9216
86.3 0.36 0.1296
87.1 1.16 1.3456
87 1.06 1.1236
81.7 4.24 17.9776
89.1 3.16 9.9856
(XiX¯)=0 (XiX¯)2=87.144

From equation (5), substitute 81 for X1 and 85.94 for X¯ in (X1X¯).

(X1X¯)=(8185.94)(X1X¯)=4.94

Square the both sides of the equation.

(X1X¯)2=(4.94)2(X1X¯)2=24.4036

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=24.4036+6.0516+...+9.9856=87.144 ……(6)

The standard deviation is given by,

s=(XiX¯)2N

From equation (6), substitute 87.144 for (XiX¯)2 and 10 for N in the above mentioned formula,

s=87

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