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Lecture 1

# MGEB12H3 Lecture 1: assignment 2 solution (2015 winter)

Department
Economics for Management Studies
Course Code
MGEB12H3
Professor
Vinh Quan
Lecture
1

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Department of Management, UTSC
MGEB12 Quantitative Methods in Economics II - Assignment #2 – Solution
Problem 1
(a) p = fraction error, H0:p .02 , Ha: p > .02 , Reject H0 if Zo Zα
120
6
=
p
= 0.05,
=
=
=
120
)02.1(02.
)1(
00
n
pp
p
σ
.01278,
01278.
02.05.
)1(
00
0
0
=
=
n
pp
pp
Z
= 2.3474
Zα = Z0.025 = 1.96, Since 2.3474 1.96, therefore reject Ho, ABC copier has more defect than standard.
(b)
Sample size n = 120, let # of defects = d, p0 = 0.02, Zα = Z0.025 = 1.96
n
pp
pp
Z)1(
00
0
0
=
=
01278.
02.0120/
d
, do not reject H0 if
01278.0
02.0120/
d
< 1.96 →
0450488.
120
<
d
so d <
5.405 or d = 5 incorrectly stapled.
(c) H0:p .02 , Ha: p > .02 , Reject H0 if Zo Zα
So reject if
96.1
0
p
pp
σ
96.1
01278.
02.0
p
Fail to reject Ho if
045.0
<
p
,
Β = P[type II error] = P[fail to reject Ho]
]
01425.0
02.0
[]
120
)025.1(025.
025.0045.0
120
)025.1(025.
025.0
[]045.0[
=
=<
zP
p
PpP
]403.1[
=
zP
=
(.9192+.9207)/2=.92
(d)
dept = population #1, n1 = 120,
1
p
= 6/120 = 0.05
co-op = population #2, n2 = 100,
2
p
= 6/100 = 0.06
100(1- α) = 99 get α = 0.01
P[z ≥ zα/2 ] = α/2 so P[z ≥ z0.005 ] = 0.005, P[z ≤ z0.005 ] = 0.995 so Z0.005 = 2.575
Confidence interval:
100
)94(.06.0
120
)95(.05.0
575.206.005.0
+±
→ -.08977 ≤ p1 - p2 ≤ .06977
(e)
Dept = population #1, n1 = 120,
1
p
= 6/120 = 0.05
Co-op = population #2, n2 = 100,
2
p
= 6/100 = 0.06
Ho: p1 - p2 = 0
Ha: p1 - p2 ≠ 0
Reject Ho if Z0 ≤ - Zα/2 or Z0 ≥ Zα/2
Case 1: D0 = 0:
21
21
nn
xx
p
+
+
=
= (6 + 6)/(120+ 100) =12/220
Test statistic
21
021
0
)1()1(
n
pp
n
pp
Dpp
Z
+
=
=
)
100
1
120
1
)(220/208)(220/12(
006.05.
0
+
=
Z
= -0.32522
α = 0.01 so get zα/2 = z0.005 = 2.575 and -zα/2 = -z0.005 = -2.575
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