Solution test 2

FCM2063
1
UNIVERSITI
TEKNOLOGI
PETRONAS
TEST 2
SEPTEMBER 2013
COURSE: FCM2063 - PROBABILITY AND STATISTICS
DATE : 04 DECEMBER 2013, Wednesday
TIME : 4.00pm – 5.00pm
PLACE : MAIN HALL
INSTRUCTIONS TO CANDIDATES
1. Answer All questions in the Answer Booklet.
2. Begin EACH answer on a new page.
3. Where applicable, show clearly steps taken in arriving at the
solutions and indicate ALL assumptions.
4. Indicate clearly answers that are cancelled, if any.
5. Do not open this Question Booklet until instructed.
Universiti Teknologi PETRONAS

FCM2063
2
Question 1:
An Ipoh council city member gave a speech in which she said that 18% of all private
homes in the city had been undervalued by the tax assessor’s office. In a follow-up
story, the local newspaper reported that it had taken random sample of 91 private
homes. Using professional evaluator to evaluate the property and checking against the
county tax records it found that 14 of the homes had been undervalued.
i. Does this data indicate that the proportion of private homes that are
undervalued by the county tax assessor is different from 18%? Use a 5%
significance level.
[7 marks]
ii. Construct a two-sided 95% on the proportion and interpret the interval.
[3 marks]
Solution:
i. (a) The parameter of interest is test about proportion, p. (1)
(b) The hypothesis testing: (1)
18.0:18.0: 10  pHvspH
(c) Test statistics is:
npp
pnXp
z
/)1(
/ˆ
00
0
0



(1)
(d) Computation:
645.0
91/)82.0)(18.0(
18.091/14
0 

z
(1)
(e) Decision:
Reject H0 if z0 > z or z0 < -z , where z0.025 = 1.96 (1)
(f) Result and conclusion:
Since Zo > -1.96, thus we fail to reject the H0 and conclude that the proportion of
private homes that undervalued is different from 18% as the claim is not true
(2)
ii. A 95% two sided CI for the proportion is
0.22ˆ0.08
91
)846.0(154.0
)96.1(154.0ˆ
91
)846.0(154.0
)96.1(154.0
96.1,154.0ˆ,91Given,
)ˆ1(ˆ
2/
ˆ)ˆ1(ˆ
2/
ˆ
2/




















p
p
Zpn
n
ppzpp
n
ppzp


(2)
Since p=0.18 falls in the interval, then we fail to reject Ho. Thus the
proportion of private homes that undervalued is different from 18% as the claim
is not true. (1)

FCM2063
3
Question 2:
An engineer investigated two different foam expanding agents that can be used
in the nozzle of fire-fighting spray equipment. A random sample of five observations
with an aqueous film-forming foam (AFFF) had a sample mean of 4.7 and a standard
deviation of 0.6. A random sample of five observations with alcohol-type concentrates
(ATC) had a sample mean of 6.9 and a standard deviation of 0.8.
i. Assume that both populations are well represented by normal distributions with
the same standard deviations draw a conclusion about difference in mean foam
expansion at 5% level of significance.
[7 marks]
ii. Find the P-value of the test in part i.
[3 marks]
Solution:
i.
(a) The parameter of interest is to test the different between the two means,
1and 2variance unknown but equal. (1)
 (b) Hypothesis testing:
0:0: 211210   HvsH
(1)
(c) Test statistics is:
21
2121
11
)()(
nn
S
xx
T
p 


 (1)
where
707.0
8
)8.0(4)6.0(4
2
)1()1( 22
21
2
22
2
11






nn
SnSn
Sp
(1)
(d) Critical value, t0.025,8 = 2.036
(e) Critical region: Reject Ho if T > 2.306 or T < -2.306
(1)
(f) Computation:
92.4
5
1
5
1
707.0
)2.714.79(
So
5,9.6,7.4 2121





T
nnxx
(1)
(g) Result and conclusion:
Since T < -2.306, then we reject Ho. It means that the mean of foam
expanding agent used for firefighting spray are not the same. (1)

FCM2063
4
iii. From a t-distribution table, for a t – distribution with 8 degree of freedom, that
T=4.92 falls between two values: 3.833 for which =0.0025 and 5.041 for which
=0.0005. So the P-value is : 2( 0.0005 < P < 0.0025) = 0.001<P<0.005. (2)
Since P < 0.05 thus we reject H0 and conclude that the means are difference at
 =0.05. (1)
Question 3:
An experiment was set up to investigate the variance of the specific heat of a certain
chemical with temperature. The data is given below.
Temperature F0
, (x) 50 60 70 80 90 100
Heat, (y) 1.60 1.63 1.67 1.70 1.71 1.71
i. Estimate the linear regression equation, xY   0 .
[6 marks]
ii. Find the coefficient of determination.
[2 marks]
iii. Find an estimate for the specific heat when the temperature is .750
F
[2 marks]
Solution:
i. = 1.4945, = 0.00234, Y = 1.4945 + 0.00234X.
(6)
ii. R2
= 90.6%.
(2)
iii. Y = 1.4945 + 0.00234(75) = 1.67
(2)
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Solution test 2

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