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Answer Key for Exam II Version B
1. Sample size
n = {Z
α / 2
2
* phat ( 1  phat ) / E
2
}
. n = {1.96
2
*.18*(1.18) / .06
2
} = 157.51
=
158 Ans. C
2. The number of tankers which did
not have spills
is 552. The number of tankers which had
spills is 148 ( = 700  552).
Phat = x / n
= ( 148 / 700 ) = .211
Ans. D
3.
The 90% C.I. for p is given by Phat ±
Z
α / 2
√[phat*(1phat)/(n1)] = .211 ± 1.645
√(.211*.789 / 699)
=
.1856 to .2364
Ans. B
4. Test statistic
Z = (Phat  P
o
) / √[ P
o
(1 P
o
) / n] = (.211.18) / √[(.18*.82) / 700] = 2.135
Ans. B (difference in answer due to rounding)
5. Since the oil companies’ claim is “not .18,”
we need to find the critical (table) value for a
twotailed
test corresponding to α/2 = .1/2 = .05.
We get Z
.05
= 1.645.
Ans. A
6. Given that Z* = 2.04 we can find the critical (table) value for the
twotailed
corresponding to
α/2 = .1/2 = .05, Z
.05
= 1.645.
Decision is to reject Ho since Z* (2.04) > Z
.05
(1.645).
Since we
reject Ho, there is evidence that the claim of the oil companies is correct.
Ans. B
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 Spring '08
 Dr.Jay

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