ECON 2121B: Methods of Economic Statistics

Spring 2012

Answer Key to Problem Set 1

Exercise 1 The scores of the final exam of 20 students in class X is {88, 91, 88, 85,82, 82, 94, 76, 71, 85, 86, 69, 72, 68, 67, 66, 61, 69, 63, 54}.

1. Calculate the mean, median, mode, variance, standard deviation, range,

and the inter-quartile range. Also calculate the z-score of the student with

score 82.

2. Make the stem-and-leaf plot. Also draw a histogram with a class width

of 10 and with the lowest value being 50. That is, 50-59 is a class, and

the next class is 60-69, and so on. Make a table of cumulative relative

frequency based on the histogram and draw a corresponding ogive.

3. Calculate the weighted mean using the histogram (or, in fact, the relative

frequeny table behind the histogram). Is this weighted mean the same as

the mean?

Solutions

1. (We treat the data set as population here.)

(population) mean µ = 75.85

median = (72 + 76)/2 = 74

mode = 69, 82, 85, 88

(population) variance σ2 = 121.6275

(population) standard deviation σ = 11.0285

range = 94− 54 = 40

inter-quartile range = 3rd quartile− 1st quartile = 85.5− 67.5 = 18

z-score of the student with score 82 = (82− µ)/σ = 0.5576

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Figure 1: Histogram

2. steam-and-leaf plot:

5 | 4

6 | 1 3 6 7 8 9 9

7 | 1 2 6

8 | 2 2 5 5 6 8 8

9 | 1 4

Score Cumulative Relative Frequency

50-59 0.05

60-69 0.40

70-79 0.55

80-89 0.90

90-100 1.00

3. Weighted mean = 55×0.05+65×0.35+75×0.15+85×0.35+95×0.1 = 76

The weighted mean is NOT the same as the mean.

Exercise 2 Continued from the above, suppose you also know the scores of the

midterm. They are {107, 103, 107, 109, 103, 91, 108, 110, 95, 94, 105, 98, 101, 96,95, 95, 87, 85, 72, 69}. The sequence of the midterm and final scores are made

according the same roster. Use Excel to make a scatter diagram. Calculate the

covariance and correlation coefficient.

Solutions

covariance = 94.775

correlation coefficient = 0.7697

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Figure 2: Scatter Diagram

Exercise 3 A ≡ {x|1 ≤ x ≤ 20 & x is an integer}?B ≡ {x|1 ≤ x ≤ 20 & x is an even number}?C ≡ {x|1 ≤ x ≤ 20 & x is an odd number}?D ≡ {x|1 ≤ x ≤ 10 & x is an integer}?E ≡ {x|10 ≤ x ≤ 20 & x is an integer}?F ≡ {x|1 ≤ x ≤ 20 & x is a multiple of 3}?G ≡ {x|1 ≤ x ≤ 20 & x is a multiple of 5}?Suppose A is the universal set.

1. Find B’(B’ for the complement of B)?

2. Find B∩

D and B∪D.

3. Find (B∩F )′ and B′ ∪F ′ and check if the two are equal?

4. Find (B∪F )′ and B′ ∩F ′ and check if the two are equal?

5. Find G∩(C

∪F ) and (G

∩C)

∪(G

∩F ) and check if the two are equal?

6. Find G∪(C

∩F ) and (G

∪C)

∩(G

∪F ) and check if the two are equal?

7. Describe a partition and give an example.

Solutions

1. B′ = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} = C

3

2. B ∩D = {2, 4, 6, 8, 10}

B ∪D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20}

3. B ∩ F = {6, 12, 18}

(B ∩ F )′ = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20}

B′ = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

F ′ = {1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20}

B′ ∪ F ′ = {1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20}

So (B ∩ F )′ = B′ ∪ F ′

4. B ∪ F = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20}

(B ∪ F )′ = {1, 5, 7, 11, 13, 17, 19}

B′ ∩ F ′ = {1, 5, 7, 11, 13, 17, 19}

So (B ∪ F )′ = B′ ∩ F ′

5. C ∪ F = {1, 3, 5, 6, 7, 9, 11, 12, 13, 15, 17, 18, 19}

G ∩ (C ∪ F ) = {5, 15}

G ∩ C = {5, 15}

G ∩ F = {15}

(G ∩ C) ∪ (G ∩ F ) = {5, 15}

So G ∩ (C ∪ F ) = (G ∩ C) ∪ (G ∩ F )

6. C ∩ F = {3, 9, 15}

G ∪ (C ∩ F ) = {3, 5, 9, 10, 15}

G ∪ C = {1, 3, 5, 7, 9, 10, 11, 13, 15, 17, 19, 20}

G ∪ F = {3, 5, 6, 9, 10, 12, 15, 18, 20}

(G ∪ C) ∩ (G ∪ F ) = {3, 5, 9, 10, 15, 20}

So G ∪ (C ∩ F ) = (G ∪ C) ∩ (G ∪ F )

7. A partition of a set X is a set of nonempty subsets of X such that (i) the

subsets are mutually exclusive, and (ii) the union of the subsets is equal

to X.

For example, a partition of A can be {B,C}, or {D,E}, or

{{1, 3, 15}, {2, 6, 8, 10, 11, 17}, {4}, {5, 7, 14, 20}, {9, 12, 13, 19}, {16, 18}}, etc.

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Exercise 4 Toss two fair dice. Find the probability of the following event:

1. The sum of the two dice is 9.

2. The dice show the same number of dots.

3. The number shown by the second die is larger than the one shown by the

first.

Solutions

1. List out the 36 equally likely outcomes:

{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), . . . , (2, 6),

(3, 1), . . . , (3, 6), (4, 1), . . . , (4, 6), (5, 1), . . . , (5, 6), (6, 1), . . . , (6, 6)}

Pr(sum = 9) = Pr((3, 6) ∪ (6, 3) ∪ (5, 4) ∪ (4, 5)) =4

36= 1/9

2. Pr(same number of dots) = Pr((1, 1) ∪ (2, 2) ∪ (3, 3) ∪ (4, 4) ∪ (5, 5) ∪(6, 6)) =

6

36= 1/6

3. Pr{second larger than first} = Pr{(1, 2)∪ (1, 3)∪ (1, 4)∪ (1, 5)∪ (1, 6)∪

(2, 3)∪ (2, 4)∪ (2, 5)∪ (2, 6)∪ (3, 4)∪ (3, 5)∪ (3, 6)∪ (4, 5)∪ (4, 6)∪ (5, 6)} =15

36= 5/12

Exercise 5 If the probability is 0.54 that Stock A will increase in value during

the next month and the probability is 0.68 that Stock B will increase in value

during the next month, what is the greatest possible value for the probability

that neither of these two events will occur?

Solutions

Let A = {Stock A increases in value}, B = {Stock B increases in value}Pr(A) = 0.54, Pr(B) = 0.68

Pr(neither of these events occurs) = Pr(A′ ∩B′) = Pr((A ∪B)′)

= 1− Pr(A ∪B)

= 1− (Pr(A) + Pr(B)− Pr(A ∩B))

= 1− (0.54 + 0.68− Pr(A ∩B))

= Pr(A ∩B)− 0.22

Note that Pr(A ∩B) ≤ min{Pr(A),Pr(B)} = 0.54

Therefore, Pr(neither of these events occurs) ≤ 0.54− 0.22 = 0.32. That is,

the greatest possible value for the probability that neither of these two events

will occur is 0.32.

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Exercise 6 Questions on combination and permutation rules.

1. Five different books are on a shelf. In how many different ways could you

arrange them?

2. How many different arrangements are there of the letters of the word

“numbers”?

3. You have 5 shirts, but you will take with you only 3 for your vacation. In

how many different ways can you do this?

4. Given 6 letters {a, b, c, d, e, f}, how many different 4-letter words can be

formed from these 6 letters?

Solutions

1. P 55 =

5!

0!= 120

2. P 77 =

7!

0!= 5040

3. C35 =

5!

3!2!= 10

4. P 46 =

6!

2!= 360

Exercise 7 The Republican governor of a state is appointing a committee of

five members to consider changes in the income tax law. There are 15 state

representatives - seven Democrats and eight Republicans - available for ap-

pointment to the committee. Assume that the governor selects the committee

of five members randomly from the 15 representatives.

1. In how many different ways can the committee members be selected?

2. What is the probability that no Democrat is appointed to the committee?

3. What is the probability that the majority of the committee members are

Republican?

Solutions

1. C515 =

15!

5!10!= 3003

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2. If we use R to indicate the member from Republicans and D to indicate

the member from Democrats, the probability is

P (D = 0) =C5

8C07

C515

=8

429≈ 0.0186

3.

P (Republican is majority) = P (R = 3) + P (R = 4) + P (R = 5)

=C5

8C07

C515

+C4

8C17

C515

+C3

8C27

C515

=246

429≈ 0.573

Exercise 8 Republicans have two potential candidates (A, B) for presidential

election, while the Democrats have three (1, 2, 3). Each party must select one

to represent the party. The probability of Republicans selecting A is 0.8. If the

Republicans choose A, then the probability of the Democrats choosing 1 is 0.5,

choosing 2 is 0.3, and choosing 3 is 0.2. If the Republicans choose B, then the

probability of the Democrats choosing 1 is 0.3, choosing 2 is 0.6, and choosing

3 is 0.1.

1. State the Bayes’ rule.

2. Draw a tree diagram, and calculate all the intersection probabilities, that

is, the probability each possible pair, e.g. P (A ∩ 1), P (A ∩ 2), etc.

3. Find the conditional probability of the Republicans choosing A, given that

the Democrats choose 2.

Solutions

1. Bayes’ rule is applied to find the posterior probability that event Ai will

occur given that event B has occured:

P (Ai|B) =P (Ai)P (B|Ai)

P (A1)P (B|A1) + P (A2)P (B|A2) + ...+ P (An)P (B|An)

where A1, A2, ..., An are mutually exclusive and P (A1) + P (A2) + ... +

P (An) = 1. (i.e. A1, A2, ..., An is a partition of the event space)

2. Intersection probabilities:

P (A ∩ 1) = P (A)× P (1|A) = 0.8× 0.5 = 0.4

P (A ∩ 2) = P (A)× P (2|A) = 0.8× 0.3 = 0.24

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Figure 3: Tree Diagram

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3. Employ the Bayes’ rule:

P (A|2) =P (A)P (2|A)

P (A)P (2|A) + P (B)P (2|B)

=0.24

0.24 + 0.12=

2

3

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