Download Suggested answers, Problem Set 1

Suggested answers, Problem Set 1
ECON 30331
Bill Evans
Fall 2013
1.
The probability the tire will fail after 30 thousand miles is the Pr(x>30) which is evaluated as one minus the
CDF or 1-Pr(x≤30)=1-F(30). From the handout, we know that the CDF for the exponential distribution is
F(a) = 1-exp(-λa) so 1-F(30) = exp(-λa) = exp(-0.025*30) = 0.47. Pr(x>30)=0.47.
2.
F (a) 
a
ex
x
x
 (1  e x )2 dx . Let (1+e )=u, so e dx=du
a
ex
du
1
1
1
1
1
1  ea  1
ea
F (a)  
dx









1



 u 2 u  (1  e x )
(1  e x )2
1  ea 1  e
1  ea
1  ea
1  ea



a
3.
a.
b.
c.
4.
T
a.
a
a
Prob(Z ≤ -0.87) = Φ(-0.87) = 0.1922
Prob(Z > 1.15) = 1- Prob(Z≤1.15) = 1 - Φ(1.15) = 1 – 0.8749 = 0.1250
Prob(-1.10< z ≤ 1.67) = Φ(1.67) - Φ(-1.10) = 0.9525- 0.1357 = 0.8169
N ( ,  2 )  N (87,6.52 )
The family only turns their air conditioning on when T>92. So what is Prob(T>92)?
Pr(T>92) = 1-Pr(T≤92).
Pr(T  92)  Pr[ z  (T   ) /   (92  87) / 6.5]  Pr( z  0.77)  0.7794
Pr(T>92) = 1-Pr(T≤92) =1- 0.7794 = 0.2206
b.
John only golfs when T≤85
Pr(T  85)  Pr[ z  (T   ) /   (85  87) / 6.5]  Pr( z  0.31)  0.3783
5.
We want to calculate the threshold below which 90% of the GPAs lie, or Prob(GPA≤K) = 0.9.
In this case, we want to know K. GPA is a normal but not a standard normal distribution so we need to
standardize it by subtracting off the mean and diving by the standard deviation.
Prob[Z=(GPA-u)/σ ≤(K-u)/σ=a] = 0.90. Looking at the standard normal table, Pr(Z≤a)=0.90 when a=1.28.
a=(K-u)/σ=1.28 and K=aσ+u = 1.28(0.4)+2.9=3.41. Kids with 3.41 and above will be in the top 10% of
their class.
6.
If you plot out the points, you will see that the points lie along a negatively sloped straight line, so X
perfectly predicts Y and vice versa. This means that the estimated correlation coefficient ̂ must equal -1.
7.
I=a + bGPA + cLSAT =15GPA + LSAT
a=0, b=800, c=1
μg = 3.2 and μlsat = 150
σg = 1.2 and σs=5
1
ρgs = 0.5
σgs = ρgs σg σs
E[I] = 15 μg + μlast = 15(3.2) + 150 = 198
Var(I) = b2 σg 2 + c2 σlast2 + 2bcσgl
Because I did not give you σgl but instead, I provide ρgl, you must use the fact that σgs = ρgs σg σs
Var(I) = 152 (1.2)2 + (12)(5)2 + 2(15)(1)(1.2)(5)(0.5) = 439
8.
9.
a)
What is the Prob(Have a second child) = 0.694.
b)
What is the Prob(Have a second child | 1st child is a boy) = Prob(2nd | 1st Boy)=
Prob(2nd  1st Boy)/Prob(1st Boy) = 0.339/0.488 = 0.695
c)
What is the Prob(Have a second child | 1st child is a girl) =Prob(2nd | Girl)=
Prob(2nd  Girl)/Prob(Girl) = 0.355/.512=0.693
d)
In this case, Prob(2nd) ≈ Prob(2nd | Boy) ≈ Prob(2nd | Girl). Realization of the sex of the first child
conveys no information about the subsequent probability of having a second child. These events
are independent.
In this case, x  18 and s  8 and n=25. The 95% confidence interval for μ is:
μ = x ± t(n-1)0.025 s/n0.5
t(n-1)0.025 =t(15)0.025 = 2.064
μ = x ± t(n-1)0.025 s/n0.5 = -18 ± 2.064(8/250.5) = -18 ± 3.30= [-21.03,-14.7]
Ho: μ = -21
Using the confidence interval, we see that -21 is within the confidence interval so we cannot
reject the null that the company’s statement is correct.
b)
You could also use a t-test to answer this question. The t-ratio for this problem is
t = | (x - μ )/(s/n0.5)| = | (44-40)/(8/4) | = 2. Because the t-statistic is smaller than the critical value
of the t-distribution (2.131), in this case, we cannot reject the null hypothesis that μ = 40
At the 90% critical value, the t-ratio stays the same but the critical value of the t-distribution falls to
1.753, so in this instance, we can reject the null hypothesis.
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10.
A.
The 95% confidence interval is
d = Δ ± t(n1+n2-2)sp[(1/n1) + (1/n2)]0.5
t(20+20-2)0.025 = t(38)0.025 = 2.024
s2p = [(n1-1)s2w + (n2-1)s2l]/(n1+n2-1) = [19*122 + 19*82]/38 = 104
sp = 10.2
((1/n1) + (1/n2))0.5 = ((1/20) + (1/20))0.5 = 0.316
x1 - x2 = -16 - -7 = -9
d = Δ ± t(n1+n2-2)sp[(1/n1) + (1/n2)]0.5 = -9 ± 2.024(10.2)(0.316) = -9 ± 6.3 = [-15.53,-2.47]
B.
Since 0 is not in the confidence interval, we can reject the null that the weight loss is the same in the
two progrms.
C.
The t-ratio for this problem is t = | (Δ - d )/(se(Δ)| . Recall that the se(Δ) = sp ((1/n1) + (1/n2))0.5 =
10.2*.316= 3.22, so t = | (Δ - d )/(se(Δ)| = | (-16 - -7)/3.22| = 2.795 which falls above the 99%
critical value of 2.712. So in this case, we can still reject the null.
11.
Given a linear combination of random variables, Z = a + bY1 + cY2, the handout demonstrates that Var(Z)
= b2Var(Y1) + c2Var(Y2) + 2bcCov(Y1,Y2). In this case Z = Y1 + Y2 so a=0, b=1 and c=1. We are also
given the fact that Y1 and Y2 have the same variance, σ2y, plus Y1 and Y2 are independent so we also know
that Cov(Y1,Y2) = σ12 = 0. Plugging all these values into the equation above, Var(Z) = 2 σ2y
12.
Given the results above, if Z = Y1 + Y2 + Y3 + ..... Yn and each value Yi is independent of Yj, then Var(Z)
= Var(Y1 + Y2 + Y3 + ..... Yn) = Var(Y1) + Var(Y2) + ..... Var(Yn) = nσ2y
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