UNIVERSITI MALAYSIA PERLIS TEST 1 09 February 2011 EQT 373

UNIVERSITI MALAYSIA PERLIS
TEST 1
09 February 2011
EQT 373 – Statistics for Engineers
Time: 1 Hour and 30 Minutes
INSTRUCTIONS
Please make sure that this question booklet has FOUR (4) printed
pages including the front page.
This question booklet has THREE questions. Answer ALL
questions.
You are required to answer all the questions in this question
booklet.
All necessary working must be shown clearly.
NAME: ___________________________________________
MATRIC NUMBER: ________________________________
GROUP / COURSE: _________________________________
LECTURER: _______________________________________
Question 1:
The breaking strength X of a certain rivet used in a machine engine has normally
distributed with a mean 5000 psi and standard deviation 400 psi. A random sample of 36
rivets is taken.
(i)
What is the probability that the sample mean falls between 4800 psi and 5200 psi?
(5 Marks)
(ii)
What sample n of rivets would be necessary in order to have the probability for
sample mean between 4900 and 5100 is 0.99?
(5 Marks)
Solution
X

4002 
N  5000,

36 

4800  5000
5200  5000
Z
)
400 / 6
400 / 6
 P (3  Z  3)
P(4800  X  5200)  P(
 0.4987  0.4987  0.9974
 0.9974
X

4002 
N  5000,

n 

P(4900  X  5100)  0.99
P(
4900  5000
Z
5100  5000
400 / n
400 / n
P( Z  z  Z )  0.99
)  0.99
  1  0.99  0.01

 0.005
2
Z 0.005  2.5758
5100  5000
 2.5758
400 / n
n  106.1559  107
2
Question 2:
A scientist interested to know whether there is a difference in the average daily intakes of
dairy product between men and women. He took a sample of n  50 adult women and
recorded their daily intakes of dairy products in grams per day. He did the same for adult
men. A summary of his sample results is listed in table below:
Sample size
Sample mean
Sample standard deviation
Men
50
756
35
Women
50
762
30
(i) Construct a 95% confidence interval for the difference in the average daily intakes of
dairy products for men and women.
(4 Marks)
(ii) Can you conclude that there is a difference in the average daily intakes for men and
women? Test the hypothesis using   0.05 .
(6 Marks)
Solution
95% CI for the difference in the average daily intakes of dairy products for
men and women:
X

1  X 2  Z / 2
S 21 S 2 2

n1
n2
Z / 2  1.96
 756  762   1.96
352 30 2

50
50
6  12.7776
 18.7776, 6.7776
The hypothesis to be tested are:
H o : 1   2  0
H1 : 1   2  0
Rejection region: Z   z  or Z  z  .
2
2
 z    z0.025  1.96  from normal distribution table 
2
3
Therefore, test statistic is:
Z
 x  x       
1
2
1
2
756  762
s12 s2 2

n1 n2
352 302

50 50
 0.9204
Since -0.9204 does not less than -1.96 and also not exceed than 1.96, we
fail to reject H 0 (accept H 0 ) and that is, there is not sufficient evidence
to declare that there is a difference in the average daily intakes of dairy
products for men and women:
Question 3:
A Human Resource manager of a large firm believes there may be a relationship between
absenteeism and age of workers. A random sample of 10 workers were chosen and the
results below were obtained.
Age (years), x
27
61
37
23
46
58
29
36
64
40
Day absent in a year, y
15
6
10
18
9
7
14
11
5
8
(i) Find the regression line of day absent in a year (Y) on age (X).
Given:
 x  421;  y  103 ;  xy  3817 ;  x2  19661;  y2  1221
(7 Marks)
(ii) Predict how many days a 45 year-old worker would be absent from work.
(2 Marks)
(iii) Interpret the meaning of ˆ value in part (i).
(1 Marks)
4
Solution
i)
x
 x  421  42.1
n
y
10
n
SS xy   xi yi 
i 1
 y  103  10.3
n
10
n
 421103  519.3
1 n
xi  yi  3817 

n i 1 i 1
10
 421  1936.9
1 n 
SS xx   x    xi   19661 
n  i 1 
10
i 1
2
n
2
2
i

SS xy
SS xx

519.3
 0.27
1936.9
  y   x  10.3   0.27  42.1  21.667
The regression model is y  21.667  0.27 x
ii)
y  21.667  0.27  45   9.517  10 days
iii)
ˆ  0.27 . This means if age increase by one year, the number of days
absent would decrease by o.27 days.
5
LIST OF FORMULAS
For 1 - 2:
Binomial distribution

2 2 
( X 1  X 2 ) ~ N  1  2 , 1  2  ,
n1 n2 

X
B  n, p  , P( X  x)  nCx p x q n  x
ztest 
Normal distribution
x
N   , 2  , z 
( X 1  X 2 )   1  2 
 12
x
n1

ttest 
Sampling distributions
 2 
X ~ N  , 
n 



 
X 1  X 2 ~ N  1   2 , 1  2 
n1
n2 

2
2
Where

 22
n2
( X 1  X 2 )  ( 1  2 )
, with   n1  n2  2
1 1
sp

n1 n2
Sp 
n1  1s12  n2  1s2 2
n1  n2  2
For P:
 pq 
pˆ ~ N  p,

n 

P~
 pq 
N  p,
 and ztest 
n 

pˆ  p 0
p 0 (1  p 0 )
n
Confidence interval
For P1 -P2 :
  2

 x  z 2
 ,  is known
n


P1  P2 ~ N  p1  p2 ,
s  2

 x  z 2
 ,  is unknown, n  30
n


ztest 
s  2

 x  t 2,n1
 ,  is unknown, n  30
n


.  p  z


2
pq
n

x
 where pˆ 
n

ˆ ˆ pq
ˆˆ
pq

n1 n2
S xx   x 
2
S yy   y 2 
For :

 , z test  x  



n
( pˆ1  pˆ 2 )   p1  p2 
with
pˆ 
Simple Linear Regression
Hypothesis testing:
 2
X ~ N  ,

n

p1 (1  p1 ) p2 (1  p2 ) 

 and
n1
n2

or , t test 
x
s
n
S xy   xy 
ˆ  y  ˆ x
( x ) 2
n
( y ) 2
n
( x)(  y )
n
ˆ 
S xy
S xx
r
S xy
S yy S xx
x1  x2
n1  n2

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