Answers for Problem Set 7 - Welcome | Department of...
2
Answers for Problem Set 7 Economics 206 Beomsoo Kim Fall 2017 1. n1997 = 1100, x �1997 = 0.35, n1992 = 1200, x �1992 = 0.27, Δ = 0.35 - 0.27 = -0.08 H0: d = μ1997-μ1992 = 0 n1992 + n1997= 2300 so, Z0.025 = 1.96 For two population proportions using independent samples Z-test: Z = p 1 � −p 2 � (p p � �1−p p � �)^0.5� 1 n 1 + 1 n 2 = 0.08/(0.46*0.04) = 4.15 – since | 4.15 | > 1.96, reject the null (where, p � 1 = x �1997 = 0.35, p � 2 = x �1992 = 0.27, p � p= p 1 �+p 2 � n 1 +n 2 ) d = ∆ ±Z 0.025 � p 1 �(1−p 1 �) n 1 + p 2 �(1−p 2 �) n 2 = Δ ± Z0.025 se(Δ) = 0.08 ± 1.96(0.019) = [ 0.042,0.118] Since 0 is not part if the confidence interval, we reject the null. 2. H0:d=μnonunion - μunion s 2 p = (n n −1)s n 2 +(n u −1)s u 2 n n −n u −2 = 1205(43.3)+375(22.75) 1580 = 38.421 so, sp = √38.421 = 6.20 se(Δ) = s p � 1 n 1 + 1 n 2 = 6.20� 1 376 + 1 1206 = 0.366 nn+nu-2 = 1580 so, t(1580)0.025 = 1.96 Δ = x - 2 - x - 1 = 0.72 95% confidence interval d = Δ ± t(1580)0.025 s p � 1 n 1 + 1 n 2 = Δ ± t(1580)0.025 se(Δ) = 0.72 ± 1.96(0.366) = [1.44, 0.0026] reject null, but barely 3. H0: μmale –μfemale = 0 X � male = 0.48 (let p � 1) X � female= 0.40 (let p � 2) X � male – X � female = p � 1-p � 2 = 0.48-0.40 = 0.08 Z-test: Z = p 1 �−p 2 � {p p � (1−p p � )}^0.5� 1 n 1 + 1 n 2 = 0.08/(0.496*0.075)= 2.132 (Where, p � p = p 1 �+p 2 � n 1 +n 2 = 0.44 ) since | 2.132 | > 1.96 at 95% confidence level reject the null at 99% confidence level critical value is 2.58 cannot reject the null 95 % confidence interval 0.08±Z 0.025� p 1 �(1−p 1 �) n 1 + p 2 �(1−p 2 �) n 2 = 0.08 ± 1.96*0.037= [0.007, 0.1525] reject the null 99% confidence interval 0.08± Z 0.005 � p 1 �(1−p 1 �) n 1 + p 2 �(1−p 2 �) n 2 = 0.08 ± 2.58*0.037= [-0.015, 0.1754] cannot reject the null
Transcript of Answers for Problem Set 7 - Welcome | Department of...
Answers for Problem Set 7 Economics 206
Beomsoo Kim Fall 2017
1. n1997 = 1100, x�1997 = 0.35, n1992 = 1200, x�1992 = 0.27, Δ = 0.35 - 0.27 = -0.08 H0: d = μ1997-μ1992 = 0 n1992 + n1997= 2300 so, Z0.025 = 1.96
For two population proportions using independent samples
Z-test: Z = p1�−p2�
(pp��1−pp��)^0.5� 1n1+ 1n2
= 0.08/(0.46*0.04) = 4.15 – since | 4.15 | > 1.96, reject the null
(where, p�1 = x�1997 = 0.35, p�2 = x�1992 = 0.27, p�p= p1�+p2�n1+n2
)
d = ∆ ± Z0.025�p1�(1−p1�)
n1+ p2�(1−p2�)
n2
= Δ ± Z0.025 se(Δ) = 0.08 ± 1.96(0.019) = [ 0.042,0.118]
Since 0 is not part if the confidence interval, we reject the null. 2. H0:d=μnonunion - μunion
s2
p = (nn−1)sn2+(nu−1)su2
nn−nu−2 = 1205(43.3)+375(22.75)
1580 = 38.421 so, sp = √38.421 = 6.20
se(Δ) = sp�1n1
+ 1n2
= 6.20� 1376
+ 11206
= 0.366
nn+nu-2 = 1580 so, t(1580)0.025 = 1.96 Δ = x-2 - x-1 = 0.72
95% confidence interval
d = Δ ± t(1580)0.025 sp�1n1
+ 1n2
= Δ ± t(1580)0.025 se(Δ) = 0.72 ± 1.96(0.366) = [1.44, 0.0026]
reject null, but barely 3. H0: μmale –μfemale = 0
X�male = 0.48 (let p�1) X�female= 0.40 (let p�2) X�male – X�female = p�1-p�2 = 0.48-0.40 = 0.08
Z-test: Z = p1�−p2�
{pp� (1−pp� )}^0.5� 1n1+ 1n2
= 0.08/(0.496*0.075)= 2.132 (Where, p�p = p1�+p2�n1+n2
= 0.44 )
since | 2.132 | > 1.96 at 95% confidence level reject the null at 99% confidence level critical value is 2.58 cannot reject the null 95 % confidence interval
0.08±Z 0.025�p1�(1−p1�)
n1+ p2�(1−p2�)
n2 = 0.08 ± 1.96*0.037= [0.007, 0.1525] reject the null
99% confidence interval
0.08± Z 0.005 �p1�(1−p1�)
n1+ p2�(1−p2�)
n2 = 0.08 ± 2.58*0.037= [-0.015, 0.1754] cannot reject the null
4. A) The 95% confidence interval is
d = Δ ± t(nn+ng-2) sp�1nn
+ 1ng
t(16+14-2)0.025 = t(28)0.025 = 2.048 s2
p = (nn−1)sn2+(ng−1)su2
nn−ng−2 = 15(0.5)2+13(0.4)2
28 = 0.208
sp = 0.456
�1nn
+ 1ng
= � 116
+ 114
= 0.366
x-n - x-g = 6.8 - 6.4 = 0.4
d = Δ ± t(nn+ng-2) sp�1nn
+ 1ng
= 0.4 ± 2.048*(0.456)(0.366) = 0.4 ± 0.342 = [0.058, 0.742]
B) The t-ratio for this problem is t = � ∆−dse(∆)
� . Recall that the se(Δ) = sp�1nn
+ 1ng
=0.456(0.366)=0.167
so t = � ∆−dse(∆)
� = �0.4− 00.167
� = 2.395, which falls above the 95% critical value of 2.048 so in this case, we can reject the null.
C) With a 99% confidence level, α=0.1 and α/2 = 0.05 so t(28)0.005 = 2.763 so the t-statistic (2.395) is smaller than the 99%
critical value (2.76), so we cannot reject the null hypothesis of no difference at the 99% confidence level. 5. nw=22, x�w =1.04 sw = 0.71 nwo=76, x�wo =1.82, swo =1.32 95 confidence interval for d=μw – μwo
A) d= 1.04-1.82 = -0.78
sp2 = �21(0.71)2+75(1.32)2
22+76−2� = 1.472
SE (d) = sp�122
+ 176
= 1.21� 122
+ 176
= 0.29
95 % confidence interval -0.78 ±t(22+76-2) 0.025 *SE(d)=-0.78 ± 1.96*0.29 = [-1.35, -0.21]
B) t test: -0.78/0.296 = -2.635 since |-2.635| > 1.96 then reject the null
C) if we change the confidence level to 99 % then Z0.005=2.57 then still reject the null 6. In this problem, N=25, X�=18 and s=8. With σ unknown, the general form for the confidence interval for one mean is
X� ± t(n-1)α/2�sn, so for a 99% confidence interval, t(n-1)0.005 =t(24)0.005 = 2.797 and
μ = X� ± t(n-1)0.005�sn = 18 ± 2.797*1.6 = 18 ± 4.47= [13.52, 22.48]
t test: 18-21/1.6=1.87 99 % t value 2.797 can’t reject the null
7. Facts of the problem: N=16, X�=44 and s=8.
90% confidence level t(n-1)0.025 =t(15)0.05 = 1.753 diff=84.5-90= -5.5
SE(diff) = �sn=12/4=3
t value = -5.5/3=-1.83 reject the null since |-1.83|>1.763
