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ADM 2304 Fall 2008
Final Exam Solutions
1
(a)
Let p be the proportion supporting an election (ignoring the undecided).
Ho: p=.5: Ha: p>.5
phat = 430/825 = .5212
z = .0212 / sqrt(.5*.5/825) = 1.22
z(.05)= 1.645
do not reject null H, and conclude there were not more people supporting an election
over coalition
Optional alternative approach to compare two sample proportions from the same sample.
If Ha: p(election) > p(coalition), then must calculate standard error accounting for
non
independent sample proportions:
p1hat  p2hat = .5212  .4788 = .0424
std err is sqrt( .52*.48 / 825 + .52*.48 / 825 +
2 * .52*.48 / 825)
and z = 1.22
We do not need to ignore the undecideds if we compare the nonindependent proportions:
p1hat = 430/992 = .433, p2hat = 395/992 = .398,
std err is sqrt(p1hat * q1hat /992 + p2hat*q2hat/992 + 2*p1hat*p2hat/992) = .028933
z = (.433468  .398185 ) / .028933 = 1.219457 = 1.22.
For some strange reason, many students calculated std err as
Sqrt( pbar * qbar * ( 1/430 + 1/395 ).
This is obviously incorrect.
(b)
Ho: p(W) = .30, p(Ont)=.39, p(Que)=.24, p(Mar)=.07
Observed Freq
Exp.Freq = 992*p(i)
Cont. To chisquare
West
301
297.6
.0388
Ont
390
386.88
.0252
Que
240
238.08
.0155
Mar
61
69.44
1.0258
Total
992
992
1.1
Df. = 3
Critical value is 7.81
We do not reject the null H since 1.1 not > 7.81
Conclude the observed distribution is consistent with the population distribution.
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View Full Document (c)
Test and CI for Two Proportions
Sample
X
N
Sample p
1
39
135
0.288889
2
152
390
0.389744
Difference = p (1)  p (2)
Estimate for difference:
0.100855
95% CI for difference:
(0.191344, 0.0103652)
Test for difference = 0 (vs not = 0):
Z = 2.10
PValue = 0.036
Ho: p1=p2; Ha: p1
≠
p2
P1hat = 39/135 = .29, p2hat = 152/390 = .39
Pooled proportion phat is (39+152)/(135+390) = .36
Std err is sqrt( .36*.64*(1/135 + 1/390)) = .048
Z = .10 / .048 = 2.1 or 2.1
Reject null H since
z > 1.96 for alpha = .05
Conclude that there is a difference between the proportions of BC voters and Ontario
voters who supported the coalition.
(d)
ChiSquare Test: C1, C2, C3, C4, C5, C6
Expected counts are printed below observed counts
ChiSquare contributions are printed below expected counts
C1
C2
C3
C4
C5
C6
Total
1
73
75
35
168
60
19
430
58.52
45.95
26.01
169.05
104.03
26.44
3.584
18.370
3.109
0.007
18.637
2.094
2
39
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This note was uploaded on 01/10/2010 for the course MANAGEMENT ADM 2304 taught by Professor Phansalker during the Fall '05 term at University of Ottawa.
 Fall '05
 Phansalker

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