=========================preview======================
(ISOM111)[2000](f)final~2106^_10356.pdf
Back to ISOM111 Login to download
======================================================
The Hong Kong University of Science and Technology
ISMT111 C Business Statistics, Fall 2000
All Section
Final Exam

1a)
Let . be the mean sales of the new computer per week.
H0 : . . 45
H1 : . . 45

1b) x . .
test statistic t . s
n
43.5 . 45
3
16
..2 degree of freedom . 16 . 1 . 15 P-value is about 0.03.
. H0 is accepted at 2.5% level of significant.
1c) 43.5 . 45test statistic z . 3
16
..2
P-value = 0.0228.
. H0 is rejected at 2.5% level of significance.
2)
Let n be the sample size that the statistician should choose. . 1.96 .2
n .. .. p(1 . p)
. 0.05 .
. 1536.64 . p(1. p)

2a)
n . 1536.64 . 0.5(1. 0.5)

. 384.16
. 385

2b)
n . 1536.64 . 0.275(1. 0.275)

. 306.37
. 307

3a)
The pooled sample variance is

( n1 . 1)s1 2 . ( n2 . 1)s2 2
s2 .
p n1 . n2 . 2

(11. 1) . 2.12 . (10 . 1) . 22
11. 10 . 2
. 2.05332
. s . 2.0533
p
3b)
Let .1 and . 2 be the mean car mileages for leaded and unleaded gas respectively.

We have to test

H: . . . . 0
01 2
H: . . . . 0
11 2
The test statistic is
x . x . ( . . . )
12 12
t .
11


S .
p
nn
12
17.2 . 18.9
Degree of freedom = 20 C 1 = 19
P-value is about 2 . 3.5% .
. H0 is accepted at 5% level of significance.

3c)
We have to test

H: . . . . 0
01 2
H: . . . . 0
11 2
The test statistic is
17.2 . 18.9

..1.895
Degree of freedom = 20 C 1 = 19
P-value is about 3.5%.
. H0 is rejected at 5% level of significance.
4a)
.
Y. 951 . 63.8X
4b)
If the bus fare increases by $1, then the number of passenger decreases by 63.8 on
average.

4c)
951, No I dont believe in this estimate because the X is out of range. Extrapolation is not
an appropriate thing to do in regression.

4d) Standard error for the slope = 7.609. Thus the 95% confident interval is .63.8 . 2.447 . 7.61 or (-82.42, -45.18)
4e)
Yes, the new slope will be smaller because the point (6.5, 300) is way below the
regression line in part (a).

4f)
951-63.8(3.75) . 711.75 , 711.75(3.75) . 2669,
951-63.8(4.25) . 679.85 , 679.85(4.25) . 2889 .

2889 C 2669 = 220. The total revenue increases by $220 when the fare increases from
$3.75 to $4.25.

5a)
.
Y. 149.056 . 50.118X
5b)
S2
r2 . XY . 82.3% SS
XX YY
82.3% of the variation in fuel usage can be explained by distance. 5c)
.
Y . 149.056 . 50.118.10 . 650.236
5d)
99% prediction interval .650.236 . 2.58. 95.09 or (405, 896)
Because 900 is outside the 99% prediction interval so we are 99% confident that there
will be no fuel shortage.

6a)

0.35

6b) SSE
(n-k-1)
Adjusted R-square . (1.

) .100%
(n -1)
SYY
2.6972
22
) .100% . 81%
. (1.
15.4816 24

6c)
If the income of the household remains the same and the size increase by 1 then on
average the food consumption increase by 354 dollars.

6d)
Because F = 52.13 with p-value essentially equals to 0. We can reject the null hypothesis
at 1% level of significant.