Download HW – 1 due - UNC

ECON 570
Fall, 2010
Solution to
HW - 1
Instructor:
Saraswata Chaudhuri
Problem 1. [Points = 3] In September, Chapel Hill’s daily high temperature has a mean
of 81 degree F and a standard deviation of 10 degree F. What is the mean, standard deviation and variance in terms of Celsius?
Answer 1: Note that C = (5/9) × (F − 32) = a + b × F where a = −160/9 and b = 5/9.
Therefore, the mean temperature is a + b × 81 ≈ 27.3 degree Celsius. The standard deviation
is b × 10 ≈ 5.56 degree Celsius.
Problem 2. [Points = 3] In a given population of two-earner male/female couples, male
earnings have a mean of 40,000 USD per year and a standard deviation of 12,000 USD.
Female earnings have a mean of 45,000 USD per year and a standard deviation of 18,000
USD. The correlation between male and female earnings for a couple is 0.80. Let C denote
the combined earnings for a randomly selected couple.
(a) What is the mean of C?
(b) What is the covariance between male and female earnings?
(c) What is the standard deviation of C?
Answer 2: Let us denote male earnings by m and female earnings by f .
(a) C = m + f , and hence C̄ = m̄ + f¯ = 85, 000 USD.
(b) Cov(m, f ) = Corr(m, f ) × sd(m) × sd(f ) = .8 × 12, 000 × 18, 000 = 172800000.
(c) sd(C) =
p
sd(m)2 + sd(f )2 + 2 × Cov(m, f ) = 28523.67 USD.
Problem 3. [Points = 3] X and Y are discrete random variables with the following joint
distribution:
That is, P r[X = 1, Y = 14] = 0.02, and so forth.
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ECON 570
Fall, 2010
Solution to
HW - 1
Joint Probability
Distribution
14
1
0.02
5
0.17
Value of X
8
0.02
Value of
22
30
0.05 0.10
0.15 0.05
0.03 0.15
Instructor:
Saraswata Chaudhuri
Y
40
0.03
0.02
0.10
65
0.01
0.01
0.09
(a) Calculate the probability distribution, mean, and variance of Y .
(b) Calculate the probability distribution, mean, and variance of Y given X = 8.
(c) Calculate the covariance and correlation between X and Y .
Answer 3:
(a) The probability distribution of Y is given by
– P [Y = 14] = P [X = 1, Y = 14] + P [X = 5, Y = 14] + P [X = 8, Y = 14] = .21
– P [Y = 22] = P [X = 1, Y = 22] + P [X = 5, Y = 22] + P [X = 8, Y = 22] = .23
– P [Y = 30] = P [X = 1, Y = 30] + P [X = 5, Y = 30] + P [X = 8, Y = 30] = .30
– P [Y = 40] = P [X = 1, Y = 40] + P [X = 5, Y = 40] + P [X = 8, Y = 40] = .15
– P [Y = 65] = P [X = 1, Y = 65] + P [X = 5, Y = 65] + P [X = 8, Y = 65] = .11.
Now note that E[Y ] = 14 × P [Y = 14] + 22 × P [Y = 22] + 30 × P [Y = 30] + 40 × P [Y =
40] + 65 × P [Y = 65] = 30.15 and that E[Y 2 ] = 142 × P [Y = 14] + 222 × P [Y =
22] + 302 × P [Y = 30] + 402 × P [Y = 40] + 652 × P [Y = 65] = 1127.23. Therefore,
V (Y ) = E[Y 2 ] − E 2 [Y ] = 218.21.
(b) The probability distribution of Y conditional on X = 8 is given by
– P [Y = 14|X = 8] = P [X = 8, Y = 14]/P [X = 8] = .05
– P [Y = 22|X = 8] = P [X = 8, Y = 22]/P [X = 8] = .08
2
ECON 570
Fall, 2010
Solution to
HW - 1
Instructor:
Saraswata Chaudhuri
– P [Y = 30|X = 8] = P [X = 8, Y = 30]/P [X = 8] = .38
– P [Y = 40|X = 8] = P [X = 8, Y = 40]/P [X = 8] = .26
– P [Y = 65|X = 8] = P [X = 8, Y = 65]/P [X = 8] = .23,
where I use P [X = 8] from the following computations
– P [X = 1] = P [X = 1, Y = 14] + P [X = 1, Y = 22] + P [X = 1, Y = 30] + P [X =
1, Y = 40] + P [X = 1, Y = 65] = .21
– P [X = 5] = P [X = 5, Y = 14] + P [X = 5, Y = 22] + P [X = 5, Y = 30] + P [X =
5, Y = 40] + P [X = 5, Y = 65] = .4
– P [X = 8] = P [X = 8, Y = 14] + P [X = 8, Y = 22] + P [X = 8, Y = 30] + P [X =
8, Y = 40] + P [X = 8, Y = 65] = .39.
Now note that E[Y |X = 8] = 14 × P [Y = 14|X = 8] + 22 × P [Y = 22|X = 8] + 30 ×
P [Y = 30|X = 8] + 40 × P [Y = 40|X = 8] + 65 × P [Y = 65|X = 8] = 39.21 and that
E[Y 2 |X = 8] = 142 × P [Y = 14|X = 8] + 222 × P [Y = 22|X = 8] + 302 × P [Y =
30|X = 8] + 402 × P [Y = 40|X = 8] + 652 × P [Y = 65|X = 8] = 1778.69. Therefore,
V (Y |X = 8) = E[Y 2 |X = 8] − E 2 [Y |X = 8] = 241.65.
p
(c) Cov(X, Y ) = E[XY ] − E[X]E[Y ] and Corr(X, Y ) = Cov(X, Y )/ V (X)V (Y ). So we
will further need to compute E[X], V (X) and E[XY ]. Note that E[X] = 1 × P [X =
1] + 5 × P [X = 5] + 8 × P [X = 8] = 5.33, E[X 2 ] = 12 × P [X = 1] + 52 × P [X =
5] + 82 × P [X = 8] = 35.7 and hence V (X) = E[X 2 ] − E 2 [X] = 6.76. Also, by a similar
computation we get E[XY ] = 171.7. Therefore, Cov(X, Y ) = E[XY ] − E[X]E[Y ] =
11.01 and Corr(X, Y ) = .29.
Problem 4. [Points = 4] Let X and Z be independently distributed standard normal
random variables, and let Y = X 2 + Z.
3
ECON 570
Fall, 2010
Solution to
HW - 1
Instructor:
Saraswata Chaudhuri
(a) Show that E[Y |X] = X 2 .
(b) Show that E[Y ] = 1.
(c) Show that E[XY ] = 0. (Hint: Use the fact that the odd moments of a standard normal
random variable are all zero.)
(d) Show that Cov(X, Y ) = 0.
Answer 4: X ∼ N (0, 1) ⇒ E[X] = 0, V (X) = E[X 2 ]−E 2 [X] = 1, and the same for Z. Also
X and Z independent implies E[Z|X] = E[Z] = 0 and Cov(X, Z) = E[XZ] − E[X]E[Z] =
0 − 0 = 0.
(a) E[Y |X] = E[X 2 + Z|X] = E[X 2 |X] + E[Z|X] = X 2 + 0 = X 2 .
(b) By the law of iterated expectations, we have E[Y ] = EX [E[Y |X]] = EX [X 2 ] = 1.
(c) Cov(X, Y ) = Cov(X, X 2 + Z) = Cov(X, X 2 ) + Cov(X, Z) = Cov(X, X 2 ) = E[X 3 ] −
E[X]E[X 2 ] = E[X 3 ] = 0 (because the odd moments of N (0, 1) are zero).
Problem 5. [Points = 7] Grades on a standardized test are known to have a mean of
1000 for students in the United States. The test is administered to 453 randomly selected
students in Florida. In this sample, the mean is 1013 and the standard deviation is 108.
(a) Construct a 95 % confidence interval for the average test score for Florida students.
(b) Is there statistically significant evidence that Florida students perform differently than
other students in the United States?
(c) Another 503 students are selected at random from Florida. They are given a threehour preparation course before the test is administered. Their average test score is
1019 with a standard deviation of 95.
4
ECON 570
Fall, 2010
Solution to
HW - 1
Instructor:
Saraswata Chaudhuri
(i) Construct a 95 % confidence interval for the change in average test score associated
with the prep course.
(ii) Is there a statistically significant evidence that the prep course helped?
(d) The original 453 students are given the prep course and then asked to take the test
a second time. The average change in their test scores is 9 points and the standard
deviation of the change is 60 points.
(i) Construct a 95 % confidence interval for the change in average test scores.
(ii) Is there statistically significant evidence that students will perform better on their
second attempt after taking the prep course?
(iii) Students may have performed better in their second attempt because of the prep
course or because they gained test-taking experience in their first attempt. Describe an experiment that would quantify these two effects.
Answer 5:
(a) A 95 % confidence interval will be
√
[(Florida mean) ± 1.96 × (Florida sd)/ n] = [1003.05, 1022.94].
(b) Note that the National average 1000 is not included in the 95 % confidence for Florida.
So we can say that we reject at the 5 percent level the hypothesis that the Florida
students perform the same as the students from all over the nation.
(c)
(i) Note that the students with the prep course are different from the original Florida
students and it is hence reasonable to assume that there is no correlation between
the two groups. Let E[old: score w/o prep] = µwo and E[new: score with prep] =
µw . We want to compute a 95 % confidence interval for ∆ = µw − µwo , i.e. the
5
ECON 570
Fall, 2010
Solution to
HW - 1
Instructor:
Saraswata Chaudhuri
b = X̄new:w −X̄old:wo = 1019−1013 = 6
change in the average test score. Note that ∆
and since we can assume the scores of the two groups to be uncorrelated,
b =
sd(∆)
q
p
sd2new:w /nnew:w + sd2old:wo /nold:wo = 1082 /453 + 952 /503 ≈ 6.61
b ± 1.96 × sd(∆)]
b =
Therefore, a 95 % confidence interval for the change is [∆
[−6.96, 18.96].
(ii) Note that 0 in inside this confidence interval (and so are some negative values).
However, the confidence interval also contains positive values, i.e. an increase in
average score. Hence we cannot reject at the 5 % level the null hypothesis that
the prep course had no effect. But we also cannot reject at the 5 % level that the
increase in average score is equal to some positive value ranging from 0 to 18.96.
(d)
(i) A 95 % confidence interval will be
√
[(change mean) ± 1.96 × (change sd)/ n] = [3.47, 14.53].
(ii) Of course. Note that no-negative or zero value is included in the confidence
interval.
(iii) It will be more effective to divide the 453 students randomly into two groups (of
roughly equal size) and ask one group to take the test the second time without
the prep course and the other group to take the test after a prep course. We will
discuss more on this in the class.
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