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(ISOM111)[2001](f)final~2106^_10358.pdf
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ISMT111 2001 Final Examination Suggested Answer
Question 1: [16 Marks]
n=(1.96/.08)2p(1-p)
a) Case 1. choose p=0.5, hence n=150.0625
b) Case 2. An idea about p based on last years figures. Total sales =$1000000+$20,000 translates into 1000+20 cars and hence p is approximately 20/1020=0.0196 approximately .02. Hence n is approximately 11.8

Question 2: [16 Marks]
Part a) is a straightforward one-sided Z-test.
One test for H0 =60 versus Ha <60.
So z*=(52-60)sqrt(15)/10 = -3.098

For b)
Use a t with 14 degrees of freedom,
T* = (52-60) sqrt(15)/22 = -1.408


Question 3: [16 Marks]
2 (4 15.822 + 6 12.772) 1001.0896 + 978.4374
Pooled variance s = == 197.9527
p 5 + 7 . 2 10
= 8.23679 .

(a)
t10,0.005=3.169. (60.33-32.21)3.169(8.23679)=28.1226.1 = (2.013,54.227).

(b)
Step 1: t*=[(60.33-32.21)-15]/8.23679=13.12/8.239=1.59. Step 2: Sig. prob.=right tailed-area from 1.59, of a t-density (df=10) which is between 5% and 10%. Hence, 5% < p-value < 10% and do not reject the H0 at =5%.



Question 4: 20 20 [17 Marks]
xy
20 ii
0.958 190.5
i=1 i=1
Sxy = xi yi .=10.292 .=1.16705 i=1 20 20
.20 .2
.x .
20 2 .i=1 i . 0.9582
S = x .=0.061638 .=0.0157498 xxi=1 i 20 20
20
20 ..yi ..22
2 .i=1 . 190.5
Syy = y.=1978.3 .=163.7875 i=1 i 20 20
20 20
xi yi
== 190.5
i 1 0.958 i 1
x == =0.0479 y ===9.525
2020 2020
Sxy 1.16705
. ..
a) == =74.1 =y .x =5.9756
1 01
Sxx 0.0157498
The least squares regression line is: y. =5.9756 +74.1x
2
Sxy
b)SSE =Syy .=77.3098 Sxx
SSE
s =

=2.0724
n .2
c) When x = 0.07, y. =5.9756 +74.10.07 =11.1626
Degrees of freedom = n C 2 = 20 C 2 = 18 t0.05 = 1.734
90% confidence interval for E(y|x) is
1(xp .x)2
y. t0.05s
+
n Sxx

1 (0.07 .0.0479)2 =11.1626 1.734(2.0724) +
20 0.0157498 =(10.140,12.185)
d) When x = 0.07, y. = 11.1626 90% prediction interval for y is
1(xp . x)2 y. t0.05s
1 ++ n Sxx
1 (0.07 . 0.0479)2
= 11.1626 1.734(2.0724) 1 ++
20 0.0157498
= (7.426,14.899)
e) The confidence interval for E(y|x) is much narrower than the prediction interval for y.
f) Increasing the sample size n will increase Sxx, both of them will reduce the widths of the two intervals.

Question 5: [17 Marks]
a) Linearity, constant variance, normality and independence of errors.
b) R2 = SSR/SST = 586.43/589.20 = 0.995.
This means that 99.5% of the variation of the stock return can be explained by the regression.

c) Since p-value = 0.000 < , we reject H0 (1 = 0) at 1% significance level. There is sufficient evidence to indicate the market return is a useful predictor of stock return.
d) n = 7, degrees of freedom = n C 2 = 7 C 2 = 5 t0.005 = 4.032
99% Confidence Interval for 1 is .
4.032s
1 1
= 1.76214 4.032 0.05415
= (1.544,1.980)
e) The stock should be considered aggressive. Since 1 is well below the lower bound of the 99% Confidence Interva