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Essentials Of Statistics

4th Edition
HEALEY + 1 other
ISBN: 9781305093836

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BuyFindarrow_forward

Essentials Of Statistics

4th Edition
HEALEY + 1 other
ISBN: 9781305093836
Textbook Problem

S O C Listed here are the rates of abortion per 100,000 women for 20 states in 1973 and 1975. Compute the median, mean, standard deviation, range and interquartile range for both years. Describe what happened to these distributions over the two-year period. Did the average rate increase or decrease? What happened to the dispersion of this distribution? Did the measures of dispersion increase or decrease? What happened between 1973 and 1975 that might explain these changes in central tendency and dispersion? (HINT: It was a Supreme Court decision.)

State 1973 1975
1 Mississippi 0.2 0.6
2 Arkansas 2.9 6.3
3 Montana 3.1 9.9
4 Maine 3.5 9.5
5 South California 3.8 10.3
6 Tennessee 4.2 19.2
7 Texas 6.8 19.1
8 Arizona 6.9 15.8
9 Ohio 7.3 17.9
10 Nebraska 7.3 14.3
11 Virginia 7.8 18.0
12 Iowa 8.8 14.7
13 Massachusetts 10.0 25.7
14 Pennsylvania 12.1 18.5
15 Colorado 14.4 24.6
16 Florida 15.8 30.5
71 Michigan 18.7 20.3
18 Hawaii 26.3 31.6
19 California 30.8 33.6
20 New York 53.5 40.7

Source: U.S. Bureau of the Census. 1977. Statistical Abstract of the United States: 1977. Washington, DC: Government Printing Office, 1977.

To determine

To find:

The mean, median, range, interquartile range and standard deviation for the given years.

Explanation

Given:

The following table shows the rates of abortion per 100, 000 women for 20 states in 1973 and 1975.

State 1973 1975
Mississippi 0.2 0.6
Arkansas 2.9 6.3
Montana 3.1 9.9
Maine 3.5 9.5
South California 3.8 10.3
Tennessee 4.2 19.2
Texas 6.8 19.1
Arizona 6.9 15.8
Ohio 7.3 17.9
Nebraska 7.3 14.3
Virginia 7.8 18.0
Iowa 8.8 14.7
Massachusetts 10.0 25.7
Pennsylvania 12.1 18.5
Colorado 14.4 24.6
Florida 15.8 30.5
Michigan 18.7 20.3
Hawaii 26.3 31.6
California 30.8 33.6
New York 53.5 40.7

Formula used:

Let the data values be Xi’s.

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

Median=((N2)th+(N2+1)th)2

The formula to calculate interquartile range is given by,

Q=Q3Q1

Where,

Q1=N×0.25

And,

Q3=N×0.75

The formula to calculate range is given by,

Range=High ScoreLow score

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 the year 1973,

Arrange the data in the increasing order.

The data in increasing order is given by,

S. No Rate
1 0.2
2 2.9
3 3.1
4 3.5
5 3.8
6 4.2
7 6.8
8 6.9
9 7.3
10 7.3
11 7.8
12 8.8
13 10
14 12.1
15 14.4
16 15.8
17 18.7
18 26.3
19 30.8
20 53.5

The number of terms is 20, which is even.

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

Median=((N2)th+(N2+1)th)2

Substitute 20 for N in the above mentioned formula,

Median=((202)th+(202+1)th)2=(10)th+(10+1)th2=(10)th+(11)th2

The value corresponding to 10th and 11th is 7.3 and 7.8.

The required median is,

Median=7.3+7.82=15.12=7.55

The highest rate is 53.5 and the lowest cost is 0.2.

The range is given by,

Range=Highest rateLowest rate

Substitute 53.5 for highest rate and 0.2 for lowest rate in the above mentioned formula,

Range=53.50.2=53.3

The size of the population is 20.

Q1=20×0.25=5

The value at 5th position ordered sample is 3.8.

And,

Q3=20×0.75=15

The value at 15th position ordered sample is 14.4.

The formula to calculate interquartile range is given by,

Q=Q3Q1

Substitute 3.8 for Q1 and 14.4 for Q3 in the above mentioned formula,

Q=14.443.8=10.6

The mean is given by,

X¯=i=1NXiN

Substitute 20 for N, 0.2 for X1, 2.9 for X2 and so on in the above mentioned formula,

X¯=0.2+2.9+...+53.520=244.220=12.21 ……(1)

Consider the following table of sum of squares,

(Xi) (XiX¯) (XiX¯)2
0.2 12.01 144.2401
2.9 9.31 86.6761
3.1 9.11 82.9921
3.5 8.71 75.8641
3.8 8.41 70.7281
4.2 8.01 64.1601
6.8 5.41 29.2681
6.9 5.31 28.1961
7.3 4.91 24.1081
7.3 4.91 24.1081
7.8 -4.41 19.4481
8.8 3.41 11.6281
10 2.21 4.8841
12.1 0.11 0.0121
14.4 2.19 4.7961
15.8 3.59 12.8881
18.7 6.49 42.1201
26.3 14.09 198.5281
30.8 18.59 345.5881
53.5 41.29 1704.864
(XiX¯)=0 (XiX¯)2=2975.098

From equation (1), substitute 0.2 for X1 and 12.21 for X¯ in (X1X¯).

(X1X¯)=(0.212.21)(X1X¯)=12.01

Square the both sides of the equation.

(X1X¯)2=(12.01)2(X1X¯)2=144.2401

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=144.23401+86.6761+...+1704.864=2975.098 ……(2)

The standard deviation is given by,

s=(XiX¯)2N

From equation (2), substitute 2975

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