3/22/99 252y9921 ECO252 QBA2
SECOND HOUR EXAM
March 23, 1999
Name
Hour of Class Registered (Circle)
MWF TR 10 12 12:30 2:00 night
I. (14 points) Do all the following. Use diagrams! Show your work! x ~ N (Why do too many of you still believe that a probability can be negative?)
1. P

7
 x

16


.
3413

P



7

9
7

2.
z

16

9
7



P

0
 z

1 .
00

P

0 x 3



P


0 .
78

P


 z
0


9
0
7

  z

3

9

0 .
44
7


 z


P


0

0 .
78


.
2823 z
 
0 .
44

.
1700


.
1123
3.
P


0

P


1 .
00


P


 z


0
2
9

7
 
 z


0 .
78
0


9 z
7




0

P


1 .
00

.
3413
 z


.
2823

0 .
78


.
0590
4. P
 x


3


P

P


0 .
44 z

 z
3


7

P
P
 z z

9
0
 

0



0 .
44

.
1700

.
5000

.
6700
5. F
 

(The Cumulative probability)
P
 z
 
0 .
33


P
 z

0
 
P
 x

4


0 .
33
 z


0

P


 z

.
5000
4

9
7



.
1293

.
3707
6. A symmetrical interval about the mean with 68% probability.
We want two points diagram, if we replace x by z,
P

0
 z

0 .
99

 x
.
3389
.
16 and x
.
84
. So between even better), and x

P z
.
16


0

, so that

 z
 z
0 .
99 z
.
16

P
.
16

 x
.
84

 x
 x
.
16
.
3400


.
6800 . From the
. The closest we can come is
(1.00 is acceptable and something in

7

0 .
99
 

7

8 .
91 , or –1.91 to 15.91. check: P


1 .
91
 x

15 .
91


P




1 .
91
9

7
 z

15 .
91

9
7



P


0 .
99
 z

0 .
99


2

.
3389


.
6778
7. x
.
045 x by z, P
We want a point

0
 z
 z
.
045

 x
.
045
, so that
.
4550
P
 x
 x
.
045


.
045 . From the diagram, if we replace
. The closest we can come is P

0
 z

1 .
70


.
4554
(though 1.69 or something between the two is acceptable). So x
   z
.
045
 
7

1 .
70
 

7

15 .
3 , or 22.3. z
.
045

1 .
70 , and check: P
 x

22 .
3


P


 z

22 .
3

9
7



P
 z

1 .
70


P
 z

0
 
0
 z

1 .
70


.
5000

.
4554

.
0446

3/22/99 252y9921
II. (6 points-2 point penalty for not trying part a.) Show your work!
A new software package has been designed to help system analysts design information systems.
We wish to see if there is a significant difference between the time required to develop a system using the new technology and using the current technology. Two independent samples are taken. The results are shown below. Assume that the data represent independent samples taken from populations with the normal distribution. x
1 x
2
Current Technology
300
280
344
385
372
New Software
276
222
310
338
200
360 302
288
I will be happy to nominate everyone who had trouble dealing with n
1
 n
2 for the Bourbon prize for inflexibility. Samples only need to be the same size when data is paired and there is no reason to believe that this data is paired!
a.
Compute
Note that s
2
, the standard deviation for time required to design a system using the new software. (3) x
1

332 .
714 , s
1

42 .
812 . b.
Do a two-sided confidence interval for the difference between the means for the two technologies
  
.
01

. (2) Indicate the following: What assumptions did you make about the variances of the
populations from which the samples were taken? Is there a significant difference between the means
(You must tell why)? (1)
Solution: x
2

 x
2
 n
2 x
2
2 s
2
2  n
2

2671 .
467

1648
6
 n
2 x
2
2

1 s
2


274 .
667
467008

53 .
586

6

274 .
667

2
5
Item
1
2
3
4
5 x
2
276
222
310
338
200 x
2
2
76176
49284
96100
114244
40000
6
Total
302
1668
91204
477168 b. Solution: Assume equal variances.
From Table 3 of the Syllabus Supplement:
Interval for Confidence Hypotheses Test Ratio
Interval
   t
 s
2 d
H
0:

0
Difference
between Two
Means (

unknown,
variances
assumed equal) s d
 s
 p
1
1
 n n
1
2
DF
 n
1
 n
2

2
0
 
1
 
2 t

2 p
 d
 
0 s d
 n
1

1
 s
1
2 
 n
2

1
 s
2
2 n
1
 n
2

1 x
1
DF

332 .
714 , s
1
2
 n
1
 n
2


42 .
812

2 
1832 .
867 , x
2

274 .
667 , s
2
2

2

7

6

2

11 , t
.
11
005

3 .
355 .
3/22/99 252y9921

2471 .
467 , d
 x
1
 x
2
Critical Value d cv
 
0
 t
 s
2 d

58 .
047 ,
 
.
01 ,
2

s

2 p

 n
1

1
 n
1 s
2
1

 n
2
 n

2
2

1
 s
2
2

6

1832 .
867
 
2671 .
467

11

10997 .
202

11
13357 .
335

2214 .
049
 s d


 d s
 p

1 n
1 t

2 s d


1 n
2

58 .
047
 s
2 p


1

1



 n
1 n
2

3 .
106

26 .
178
2214 .
049


1
7


58 .
0

1
6



685 .
301

26 .
178
87 .
3 or –23.3 to 139.3. This interval includes 0. Since the interval includes zero, we can say that there is no significant difference between the means. Or if the null hypothesis is H
0
:
 
0 or H
0
:

1
 
2

0 or H
0
:

2
, we cannot reject it. b. Alternate Solution: Assume unequal variances. From Table 3 of the Syllabus Supplement:
Interval for Confidence Hypotheses Test Ratio Critical Value
Interval
Difference
between Two
Means(

unknown,
variances
assumed
unequal)
   t s s d
DF
 s
1
 s
1
2 n
1
2 s 2
2
 s 2
2 n
2
    n
1
2

 s
2
1 n
1 n
1

1

 s
2
2 n
2 n
2


2 n
2 d

2
1
H
0
:
0
0
 
1
 
2
Same as
H
0
:

1
 

1
 
2
2 if

0

0 t
 d
 
0 s d d cv
 
0
 t s d
2 s
1
2 n
1 s
2
2 n
2

1832 .
867
7

261 .
8381

2671 .
467 s n
1
2
1
6
 s
2
2 n
2


445 .
2445
707 .
0826 s d
 s
1
2 n
1
 s
2
2 n
2

707 .
0826

26 .
591
DF t
9
.
005
 s
1
2 s
1
2 n
1
2
 s
2
2 n
2 s
2
2
2
2

 n
1 n
1

1

3 .
250 , so n
2 n
2
 

1
 d
 t

2 s d

261
6

707
.
8381
2
.
0826


445
2
.
2445
5

2

58 .
047


3 .
250

26

9 .
79 , so use 9 degrees of freedom.
.
591


58 .
0

86 .
4 or –28.4 to 144.4. Conclusion is the same as for a.
3

3/23/99 252y9921
III. Do at least 3 of the following 5 Problems (at least 10 each) (or do sections adding to at least 30 points -
Anything extra you do helps, and grades wrap around) . Show your work! State H
0
and H
1 where applicable.
1.
A firm is currently contracting with firm 1 for delivery of raw materials. It will stay with firm 1 unless it can demonstrate that delivery times for firm 2 are faster. The statistics from the two samples are below.
Assume that the parent populations are Normal.
Firm 1 n
1

50 x
1

14 days s
1

3 days
Firm 2 n
2

30 x
2

12 .
5 days s
2

2 days a.
Test to see if the mean for population 1 is larger than the mean for population 2.
(It’s incredible that 2/3 of you couldn’t write down “
(i) State your hypotheses in terms of

1
and
  
2
” here!)

2
. (1)


(ii)
(iii)
Do the test assuming that
 2
1
  2
2
. (3)
Do the test assuming that
 2
1
  2
2
. (4)
(iv) On the basis of these tests would you change supplier? (why?) (1) b.
Test the variances for equality.

 
.
02

(3)

.
01

Solution: See previous pages for formulas a.
(i)
(ii)


H
 H
1
0 x
1
:
:



14 ,


0
0
where s
1
2  2



9 ,

1 x
2
 
2
or

12 .
5 , s
2
2

H

 H
1
0

:
:
 
2

1

1



2

2

4 , d



0
0
or x
1


H

 H
1
0 x
2
:
:

1

1

1 .
5 ,





2

2
.
.
01 ,
DF s

2 p


 n
1 n
1
 n
2

1 n
1
 s
1
2


2
 n
2

 n
2

50
2


1

30 s
2
2


2

78 , t
.
78
01

49
 

29
 
78
2 .
37 .

7 .
14103 s d
 s
 p
1 n
1

1 n
2
 s

2 p


1 n
1

1 n
2



7 .
14103


1
50

1
30



0 .
38085

0 .
61713
(You still can’t do a 2-sided test for a 1-sided Hypothesis!)
Test Ratio: t
 d
 s d

0 
1 .
50

0
0 .
61713

2 .
431 This is above 2.37. or Critical Value: d cv
 
0
 t

2 s d

0


2 .
37

0 .
61713


1 .
46 .
1.5 lies above this value.
or Confidence Interval:
  d
 t
 s d

1 .
5


2 .
37

0 .
61713


0 .
037
This interval does not include 0. In all cases reject H
0
.
4

3/22/99 252y9921
(iii) s
1
2 n
1 s
2
2 n
2 s
1
2


9
50
4
30

0 .
18000


0
0
.
13333
.
31333 n
1
 s
2
2 n
2 s d
 s
1
2 n
1
 s
2
2 n
2

0 .
31333

0 .
559762
DF t

77
.
01
 s
1
2 n
1 n
1 s
1
2 n
1
2 .
37 .
2

1

 s
2
2 n
2
, so
Test Ratio: t
 s
2
2 n
2 n
2
 d
2
2


0

0 .
18000
.
31333

49

2
0 .
13333
29

2


1
 d
 t

2 s d
 
0 s d or Critical Value: d cv


1 .
5

1 .
50

0
0 .
559762
 
0
 t

2 s d

2 .
680


77 .
0474

3 .
250

26 .
591


0


2 .
37

0 .
559762
, so use 77 degrees of freedom.
58 .
0

86 .
4 or –28.4 to 144.4.
This is above 2.37.


1 .
37 .
1.5 lies above this value.
or Confidence Interval:
  d
 t
 s d

1 .
5


2 .
37

0 .
559762


0 .
173 b.
This interval does not include 0. In all cases reject H
0
.
(iv) As usual, rejection of a hypothesis calls for action. Since
  
2
, we have shown that firm 2 is faster.
H

H
1
0
:

1
2
:

1
2
 
  2
2
2
2 s
1
2 s
2
2

9
4

2 .
25 , s
2
2 s
1
2

4
9

1 . Since the second ratio is below 1, the only one we need check is s
1
2 s
2
2
. Since F

2
Solution continues in252z9921

F

49 ,
.
01
29


2 .
27 is larger than our ratio, do not reject H
0
.
5