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Exercises 12.6 Chapter Test

1.

Suppose State A has a population of 935,000 and five representatives. State B has a population of 2,343,000 and 11 representatives.

  1. Find the average constituency of each state.

  2. Which state is more poorly represented?

Answer.

  1. A = 187,000 ppl/rep; B = 213,000 ppl/rep

  2. B

Solution.
\begin{equation*} A = \frac{935,000}{5}=187,000 \end{equation*}
\begin{equation*} B = \frac{2,343,000}{11}=213,000 \end{equation*}

B is more poorly represented, because they have a higher average constituency.

2.

A county task force has 15 members from four different regions. The population of each region is listed below.

  • Region 1: 60,400

  • Region 2: 125,800

  • Region 3: 82,750

  • Region 4: 315,200

Apportion the 15 members of the task force using:

  1. Hamilton's method.

  2. Jefferson's method.

  3. Webster's method.

  4. Huntington-Hill method.

Answer.

  1. Region 1 = 2 seats, Region 2 = 3 seats, Region 3 = 2 seats, Region 4 = 8 seats

  2. Region 1 = 1 seat, Region 2 = 3 seats, Region 3 = 2 seats, Region 4 = 9 seats

  3. Region 1 = 2 seats, Region 2 = 3 seats, Region 3 = 2 seats, Region 4 = 8 seats

  4. Region 1 = 2 seats, Region 2 = 3 seats, Region 3 = 2 seats, Region 4 = 8 seats

Solution.

  1. The standard divisor is:

    \begin{equation*} 584,000 \div 15 = 38,943.33 \end{equation*}

    Table 12.6.1.
    Region Quota Initial Final
    1 1.55 1 2
    2 3.23 3 3
    3 2.12 2 2
    4 8.09 8 8
    Total 14 15
    Region 1 recieved the extra seat because they have the largest decimal part.

    Region 1 = 2 seats, Region 2 = 3 seats, Region 3 = 2 seats, Region 4 = 8 seats

  2. Table 12.6.2.
    Region Quota Initial
    1 1.55 1
    2 3.23 3
    3 2.12 2
    4 8.09 8
    Total 14
    This is too small, so pick a smaller divisor. Using a modified divisor of \(35,000\text{,}\) the results are:
    Table 12.6.3.
    Region Quota Initial
    1 1.73 1
    2 3.59 3
    3 2.36 2
    4 9.01 9
    Total 15
    Region 1 = 1 seat, Region 2 = 3 seats, Region 3 = 2 seats, Region 4 = 9 seats

  3. Table 12.6.4.
    Region Quota Initial
    1 1.55 2
    2 3.23 3
    3 2.12 2
    4 8.09 8
    Total 15
    This gives the required total, so we are done.

    Region 1 = 2 seats, Region 2 = 3 seats, Region 3 = 2 seats, Region 4 = 8 seats

  4. Table 12.6.5.
    Region Quota Lower Quota Geo Mean Initial
    1 1.55 1 \(\sqrt{1\cdot 2}=1.41\) 2
    2 3.23 3 \(\sqrt{3\cdot 4}=3.46\) 3
    3 2.12 2 \(\sqrt{2 \cdot 3}=2.45\) 2
    4 8.09 8 \(\sqrt{8 \cdot 9}=8.49\) 8
    Total 15
    This gives the required total, so we are done.

    Region 1 = 2 seats, Region 2 = 3 seats, Region 3 = 2 seats, Region 4 = 8 seats

3.

A regional transportation department has three bus lines: A, B and C. The number of passengers that use each line per day are listed below.

  • A: 23,530

  • B: 5,550

  • C: 70,920

  1. Use Hamilton’s method to apportion 100 buses.

  2. One year later, the passengers per day was reported to be: A = 24,030, B = 5,650, and C = 71,100. Re-apportion the 100 buses using Hamilton’s method.

  3. Did a paradox occur? If so, which one? If not, why not?

Answer.

  1. A = 23, B = 6, C = 71

  2. A = 24, B = 6, C = 70

  3. If C grew at a faster rate than A, it would have been the population paradox. However, there wasn’t a paradox. Although A gained a bus and C lost as bus, that is expected because the number of passengers on A grew by about 2% (500 people), but the number of passengers on C grew by less than a 1% (180 people).

Solution.
  1. The standard divisor is \(100,000 \div 100 = 1000\)
    Table 12.6.6.
    Bus Quota Initial Final
    A 23.53 23 23
    B 5.55 5 6
    C 70.92 70 71
    Total 98 100
  2. The standard divisor is \(100,680 \div 100 = 1006.8\)
    Table 12.6.7.
    Bus Quota Initial Final
    A 23.87 23 24
    B 5.61 5 6
    C 70.52 70 70
    Total 98 100

  3. If C grew at a faster rate than A, it would have been the population paradox. However, there wasn’t a paradox. Although A gained a bus and C lost as bus, that is expected because the number of passengers on A grew by about 2% (500 people), but the number of passengers on C grew by less than a 1% (180 people).

4.

An instructor at a Fitness Center can teach eight classes. A pre-registration survey indicates the following interest:

  • 66 want to take yoga

  • 39 want to take karate

  • 18 want to take weight training

  • 23 want to take meditation

Assume the instructor will teach at least one class for each area. Use the Huntington-Hill method to apportion the classes.

Answer.

Yoga = 4, Karate = 2, Weight Lifting = 1, Meditation = 1

Solution.

The standard divisor is \(146 \div 8 = 18.25\)

Table 12.6.8.
Class Quota Lower Quota Geo Mean Initial
Yoga 3.62 3 \(\sqrt{3\cdot 4}=3.46\) 4
Karate 2.14 2 \(\sqrt{2\cdot 3}=2.45\) 2
Weight Lifting 0.99 0 \(\sqrt{0 \cdot 1}=0\) 1
Meditation 1.26 1 \(\sqrt{1 \cdot 2}=1.41\) 1
Total 15
This gives the required total, so we are done.