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Homework 1
A.2 (i)
This is just a standard linear equation with intercept equal to 3 and slope equal to .2. The intercept is the number of classes missed by a student who lives on campus.
(ii) The average number of classes missed for someone who lives five miles away is 3 + .2(5) = 4 classes.
(iii) The difference in the average number of classes missed for someone who lives 10 miles away and someone who lives 20 miles away is 10(.2) = 2 classes.
A.4 (i) The percentage point change is 5.6 – 6.4 = – .8, or an eight-tenths of a percentage point decrease in the unemployment rate.
(ii) The percentage change in the unemployment rate is 100[(5.6 – 6.4)/6.4] = – 12.5%.
A.6 (i) The exact percentage by which Person B’s salary exceeds Person A’s is 100[42,000 – 35,000)/35,000] = 20%.
(ii) The approximate proportionate change is log(42,000) – log(35,000) .182, so the
approximate percentage change is 18.2%. [Note: log() denotes the natural log.]
A.8 From the given equation, grthemp = – .78( salestax ). Since both variables are in proportion form, we can multiply the equation through by 100 to turn each variable into percentage form. This leaves the slope as – .78. So, a one percentage point increase in the sales tax rate (say, from 4% to 5%) reduces employment growth by – .78 percentage points.
A.10 (i) The value 45.6 is the intercept in the equation, so it literally means that if class = 0, then the score is 45.6. Of course, class = 0 can never happen. And values close to zero would rarely if ever occur, except in very small schools. So, by itself, 45.6 is not of much interest. But it must be accounted for to use the equation to obtain score for sensible values of class.
(ii) We use calculus to obtain the optimal class size:
class *^ .082 / [2(.000147)] 278..
Rounded to the nearest integer, the optimal class size is 279 students. When we plug this into the equation for score , we get the largest achievable score:
score *^ 45.6 .082(279) .000147(279 )^2 57..
(iii) The following graph shows the solution rounded to the nearest integer:
(iv) It is not at all realistic to think that a student’s score is determined only by the size of his or her graduating class. For one thing, that would imply that all students graduating in the same year from the same high school would have the same score. A student’s ability, motivation, classes taken, family background, and many other factors – including some that are totally random, such as not feeling well while taking the test – would come into play. In fact, it is safe to say that, realistically, the size of the graduating class would have a small effect, if any effect at all. Multiple regression analysis allows for many observed factors to affect a variable like score , and it also recognizes that there are unobserved factors that are important and that we can never directly account for.
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Bora Kim (2924798272) (Class HW1 Econ 318 )
Problem 1
(a) E(X) = 0 × 0 .1 + 1 × 0 .2 + 2 × 0 .4 + 4 × 0 .15 + 7 × 0 .1 + 8 × 0. 05.
(b) E(3X) = 3 × E(X).
(c) Yes. For any constants a and b, E(aX + b) = aE(X) + b.
(d) V (X) = E(X^2 ) − E(X)^2 where E(X^2 ) =
x x (^2) f (x) = 0 (^2) × 0 .1 + 1 (^2) × 0 .2 + 2 (^2) × 0 .4 +
42 × 0 .15 + 7^2 × 0 .1 + 8^2 × 0 .05.
(e) V (X + 2) = V (X).
(f ) V (3X) = 9V (X).
(g) No. For any constants a and b, V (aX + b) = a^2 V (X).
(h) Std(X) =
V (X).
(i) Yes. (g) shows that for any constants a and b, Std(aX + b) = |a| × Std(X).
Problem 2
(a)
E(X) = 1 × (0.05 + 0.1 + 0.15) + 2 × (0.15 + 0.2 + 0.2) + 5 × (0.1 + 0.05 + 0) E(X^2 ) = 12 × (0.05 + 0.1 + 0.15) + 2^2 × (0.15 + 0.2 + 0.2) + 5^2 × (0.1 + 0.05 + 0) V (X) = E(X^2 ) − E(X)^2
(b)
E(X|Y = 3) =
x
x × f (x|Y = 3) = 1 ×
+ 2 ×
+ 5 ×
(c)
E(Y |X = 2) = 2 ×
+ 3 ×
+ 4 ×
(d)
V (X|Y = 3) = E(X^2 |Y = 3) − E(X|Y = 3)^2 E(X^2 |Y = 3) =
x
x^2 × f (x|Y = 3) = 1^2 ×
+ 2^2 ×
+ 5^2 ×
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Bora Kim (2924798272) (Class HW1 Econ 318 )
(e) Similarly, use
V (Y |X = 2) = E(Y 2 |X = 2) − E(Y |X = 2)^2.
(f ) Use Cov(X, Y ) = E(XY ) − E(X)E(Y )
where
E(XY ) =
x,y
xyf (x, y)
= 1 × 2 × 0 .05 + 1 × 3 × 0 .1 + 1 × 4 × 0 .15 + 2 × 2 × 0 .15 + 2 × 3 × 0. 2 +2 × 4 × 0 .2 + 5 × 2 × 0 .1 + 5 × 3 × 0 .5 + 5 × 4 × 0.
(g) Corr(X, Y ) = √CovV (X(X,Y)V (^ )Y ).
(h) If W = 3X − 2 Y ,
E(W ) = 3E(X) − 2 E(Y ), V (W ) = 9V (X) + 4V (Y ) − 12 Cov(X, Y ).
(i) Use Cov(3X + 1, 2 Y − 2) = 6Cov(X, Y ).
(j) Use
Corr(3X + 1, 2 Y − 2) = Cov(3X + 1, 2 Y − 2) √ V (3X + 1)V (2Y − 2)
6 Cov(X, Y ) 6
V (X)V (Y )
= Corr(X, Y ).
Problem 3
X ∼ N (5, 4), Y ∼ N (3, 16)
(a) Pr(X > 4) = Pr(Z > 4 √− 45 ) where Z ∼ N (0, 1). Use normal table to find the value.
(b) Use
Pr(5 < Y < 10) = Pr(
< Z <
(c) Use X + 0. 5 Y ∼ N (5 + 0. 5 × 3 , 4 + 0. 52 × 16).
Refer to Problem 2 (h).
(d) Z X^2 + Z Y^2 + Z W^2 is the sum of independent normal distributions. By the property of normal distribution, this follows χ^2 (3). Use χ^2 with degree of freedom 3 table to find the exact value.
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