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(ISOM111)[2006](f)final~ac_lfyab^answer_10363.pdf
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ISMT111 F06 Final Exam Suggested Solution Question 1:
(a)
71.5 2.06 30.55 /
26 =71.5 12.34 = [59.16, 83.84]
(b)
With 95% confidence the true mean will be enclosed in the interval. That is, if we take a large number of samples and construct the intervals as in (a) then 95% of the intervals will enclose the true mean.
(c)
(80.55+59.45)/2=70 (d) (80.55-59.45)/2=10.45. SE=10.55/1.96=5.33
(1.96)2 (31.98)2
18
(e) n = 61.4 , n = 62 (f) 0.111 1.96
( ) / 36 =0.111 0.103
82
99
Question 2:
(a) H : p . p = 0 vs H : p . p > 0
02 1 a 21
(0.5 . 0.44) 0.06
(b)
== 1.087 < 1.645, accept H0. 66 + 90 156 . 11 . 0.0552
(1. ).+.
150 +180 330 .150 180 .
that the economic outlook this year is not better than last year.
(c)
p-value = 0.1379
0.5 . 0.44
(d) P{p.2 > 0.5} = P{Z >
} = P{Z > 1.62} = 0.0526 .
0.44(0.56) /180
Question 3:
(a)
Sample means have normal distribution with mean 3 and variance 1/
25 = 0.2, therefore P{2.9 x 3.3} = P{.0.5 z 1.5} = 0.6247
(b)
0.36 = margin of error 0.36/(0.2)=1.8 = critical value /2 = 0.0359 , 1. = 0.9282
(c)
t-distribution with d.f. 24, because t= 1.71 and t =1.32 so the answer is 5%.
0.05 0.1
0.2 . 0.1587
(d) P{X > 4} = P{Z > 1} = 0.1587, P{p. > 0.2} = P{Z >
} = 0.2943.
0.1587(1. 0.1587) / 25
Question 4:
(a)
-4.28+0.25430=3.34 days (b) (0.254 -0.2)/0.0285=1.895>1.771 reject H0.
1 (37.87 . 30)2
(c) 0.254 2.16(0.0285) (d) 3.34 2.16 1.11
+ =3.34 0.787
15 1511.3
Question 5:
(a)
Larger variance of X makes the SEs of b0 and b1 smaller by the corresponding formulas.
(b)
If r is positive, Y will increases by rSY . If r is negative, Y will decrease by rSY .
~~
(c)
The intercept is zero because both X and Y have sample means equal zero. The slope is
~~
r = 0.6 because X and Y have s.d. equals one.
Question 6:
2 (3.56)2 11
R = 1. 2 = 0.96
(16.57) 13
(1.5)2 + (3.56)2 = 3.863, 3.863 2.2 =8.50. The prediction interval is 75.62 8.5 0
5