This is the second exam from Spring 1988. Unfortunately the take-home exam that accompanied it
is not available since the disk that contains it is unreadable by this version of Word.
3/23/98 252y9822
ECO252 QBA2
Name
SECOND HOUR EXAM Hour of Class Registered (Circle)
February 18, 1998
MWF 10 11 TR 12:30 2:00
Hour of Class Attended (If Different)
______________
I. (14 points) Do all the following.
x ~ N 5,3
1.
16  5
 3  5
z
P 3  x  16  P
  P 2.67  z  3.67
 3
3 
 P 2.67  z  0  P 0  z  367
.  .4962.4999 .9961
2.
3.
4.
5.
3  5
 05
z
.  z  0.67
P 0  x  3  P
  P 167
 3
3 
 P 167
.  z  0  P 0.67  z  0 .4525.2486 .2039
0  5
 2  5
z
. 
P 2  x  0  P
  P 2.33  z  167
 3
3 
 P 2.33  z  0  P 167
.  z  0 .4901.4525 .0376
0  5

. 
P x  0  P z 
  P z  167

3 
 P z  0  P 167
.  z  0 .5000.4525 .0475
2  5

. 
F  2 (The Cumulative probability) P x  2  P z 
  P z  100

3 
 P z  0  P 100
.  z  0 .5000.3413 .1587
6. A symmetrical interval about the mean with 82% probability.
We want two points x .09 and x.91 , so that P x.91  x  x.09  .8200 . From the
diagram, if we replace x by z, P 0  z  z.09  .4100 . The closest we can come is
. , and x    z.09  5  134
P 0  z  134
.  .4099 . So z.09  134
.  3  5  4.02 , or 0.98 to 9.02.
check:
7.
9.02  5
 0.98  5
z
.  z  134
.   2.4099 .8198
P 0.98  x  9.02  P
  P 134
 3
3 
x .18
We want a point x .18 , so that P x  x .18  .18 . From the diagram, if we replace
x by z, P 0  z  z.18  .32 . The closest we can come is P0  z  0.91 .3186
or P0  z  0.92 .3212 . Use something between the two. So z.18  0.915 , and
x    z.18  5  0.915 3  5  2.745 , or 7.745.
7.745  5

check: P x  7.745  P z 
  P z  0.92  P z  0  P 0  z  0.92 .5000.3212 .1788

3 
2
3/20/98 252y9822
II. (6 points-2 point penalty for not trying part a.) Show your work!
I wish to decide whether to become a high school teacher or a university professor. I will do so on
the basis of earnings alone. I am interested in both a high salary and the certainty of getting it. The data that
I get in my sample is shown below. Assume that the data represent independent samples taken from
populations with the normal distribution.
x1
x2
d
High School Teachers
University Professors
Difference
38.7
39.2
0.50
42.5
68.5
-25.00
62.6
43.7
18.90
38.3
47.2
-8.90
49.2
51.3
-2.10
73.5
29.9
43.60
a. Compute s1 , the standard deviation for high school teachers. (3) Note that x 2  46.63, s2  12.99 ,
d  4.50, sd  2386
. . You will not necessarily need all of these in your computations.
b. Test to see if means for salaries of both types of teacher are equal. You may assume that they come from
populations with identical variances. Does this indicate that one of the occupations is more lucrative?(3)
c. (Extra Credit) Test to see if variances are equal. What can you say about the relative safety of the two
occupations?(2)
a. Solution:
x1
s12 
x

1
n
x
2
1
Item
304.8

 50.80
6
 nx12
n 1

16512.48  6 50.80
. s1  14.343 .
 205728
.
5
2
1
2
3
4
5
6
Total
b. Solution: From page 10 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Interval
Difference
H 0:   0
  d  t  2 sd
between Two
H 1:   0
1
1
Means (
sd  s p

  1   2
n1 n2
unknown,
variances
DF  n1  n2  2
assumed equal)
x12
1497.69
1806.25
3918.76
1466.89
2420.64
5402.25
16512.48
x1
38.7
42.5
62.6
38.3
49.2
73.5
304.8
Test Ratio
t
s p2 
d  0
sd
Critical Value
d cv   0  t  2 sd
 n1  1 s12   n2  1 s22
n1  n2  1
3
3/20/98 252y9822
H 0 :   0
H 0: 1   2
H 1 :    0 Same as H 1 :  1   2
  1   2
if  0  0
s p2 
 n1  1 s12   n2  1 s22
sd  s p
n1  n2  1
=
x1  511333
.
, s12  197.835,
d  x1  x 2  4.50
5 205.728  512.78
10
2

x 2  46.63, s2  12.99
 .05 or as you specified.
DF  n1  n2  2  6  6  2  10
205.728  163.3284
10
 184.5542 t .025  2.228
2
1
1
 1 1
.
 7.8423

 s 184.5542     615181
 6 6
n1 n2
Test Ratio: t 
d   0 4.50  0

 0.5737 This is between 2.228 .
sd
7.8433
or Critical Value: d cv   0  t  2 sd  0   2.228 7.8433  17.75
d  4.50 is between these values.
or Confidence Interval:   d  t  2 sd  4.50   2.228 7.8433  4.50  17.75 or -13.25 to 22.25. The
interval includes 0. In all cases accept H 0 . There is no difference in salaries.
c. Solution: From page 11 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Interval
Ratio of
 22 s22 DF1 , DF2
H 0 :  12   22
 2 F.5 .5 
Variances
2
2
1 s1
H 1 :  12   22
1
DF1 , DF2
F1  2
 DF1 , DF2
DF1  n1  1
F 2
DF2  n2  1
 2

.5  .5   2    or
1  
2

s2
205.728
5,5
F    12 
 1260
.
s2 12.78 2
There is no difference in risk.
Test Ratio
F DF1 , DF2 
Critical Value
s12
s22
and
F DF2 , DF1 
s22
s12
s2
1
5,5
5,5
. Since both are below F.025  7.15 , accept H 0 .
F    22 
.
s1 1260
4
3/23/98 252y9822
III. Do at least 2 of the following 4 Problems (at least 10 each) (or do sections adding to at least 20 points Anything extra you do helps, and grades wrap around) . Show your work! State H 0 and H1 where
applicable.
1. In a random sample of 600 television sets made between Tuesday and Thursday from a production line
80 were defective. In a random sample of 200 sets made on Monday 40 were defective.
a. Using a 5% test, test the hypothesis that the defect rate is larger on Monday than during the middle of the
week. (5)
b. If you conclude that the defect rate is higher on Monday, test the Hypothesis that it is more than 3%
higher (3).
c. If you found out that of 200 sets made on Friday 34 were defective, test for a difference between
Monday, Tuesday-Thursday, and Friday. (6)
On Monday
On Tuesday-Thursday
On Friday
x1  40, n1  200, p1 
40
.2000
200
x 2  80, n2  600, p2 
80
.1333
600
x 3  34, n2  200, p1 
34
.1700
200
a. Solution: From page 11 of the Syllabus Supplement:
Interval for
Confidence
Interval
p  p  z 2 sp
Difference
between
proportions
q  1 p
p  p1  p2
p1q1 p2 q 2

n1
n2
s p 
Hypotheses
Test Ratio
H 0 : p  p0
H 1 : p  p0
p 0  p 01  p 02
or p 0  0
z
p  p 0
 p
If p  0
 p 
p01q 01 p02 q 02

n1
n2
Or use s p
Critical Value
pcv  p0  z 2  p
If p0  0
 p 
p0 q 0  1 n1 
1
n2

n p  n2 p2
p0  1 1
n1  n 2
H 0 : p  0
H 0 : p1  p2 p  p1  p2 .0667.
Same as
. .
H 1 : p  0
H 1 : p1  p2  .05 or as you specified. z  1645
p0 
.20.80 .1333.8677
p1q1 p2 q 2


n1
n2
sp 
200

600
 .00099 .03151
n1 p1  n2 p2 200.20  600.1333
40  80


.15
n1  n2
200  600
200  600
 p 
p0 q 0

Test Ratio: z 
1
n1

1
n2

p  p 0
 p

.15.85 1 200  1 600 
.000850 .02915
.0667  0
 2.287 This is above 1.645.
.02915
or Critical Value: pcv  p0  z  p  0  1645
. .02915 .0480
p .0667. is above this value.
or Confidence Interval: p  p  z sp or p .0733  1645
. .02915 .0187 . The
interval does not include 0. In all cases reject H 0 .
5
3/23/98 252y9822
b. Solution:
H 0 : p .03
H 1 : p  .03
Test Ratio: z 
p  p 0 .0667 .03

 1165
.
This is below 1.645.
s p
.03151
or Critical Value: pcv  p0  z  p .03  1645
. .03151 .0818
p .0733. is below this value.
or Confidence Interval: p  p  z sp or p .0733  1645
. .02915 .0187 . The
interval does not contradict H 0 : p .03 In all cases accept H 0 .
c. Solution:
H 0 : Homogeneous
H 1 : Not homogeneous
DF   r  1 c  1  1 2  2
2 2
 .05   5.9915
O
pr
40
80
34
154
E
.154
160 520 166 846 .846
200 600 200 1000 1000
.
O
E
40
80
34
160
520
166
1000
30.8
92.4
30.8
169.2
507.6
169.2
1000.0
O2
E
51.9481
69.2641
37.5325
151.3002
532.7029
162.8605
1005.6083
pr
30 .8 92 .4 30 .8 .154
169 .2 507 .6 169 .2 .846
200 .0 600 .0 200 .0 1.000
O  E 
O2
n 
 1005.6083  1000  5.6083
E
E
Since this is less than 5.9915. accept H 0 .


2
6
3/23/98 252y9822
2. A group of students with insomnia are asked to test a new sleeping pill. The number of hours of sleep
each gets is summarized by the following data: x1 refers to the sleeping pill, x 2 refers to the placebo (sugar
pill) and d  x1  x2
x1  6.88, s1  0.52 , n1  11 , x 2  6.40, s2  0.75 , n2  11 , d  0.48, sd  0.33
a. Assume that the variances of the parent populations are not equal and that the samples are independent,
but that the normal distribution applies. Test the hypothesis that the mean hours for the sleeping pill is
greater than the placebo.
(i) Compute the number of degrees of freedom for the test (5).
(ii) Do the test at the 95% level using either a critical value or a test ratio. (5)
(iii) Do a two -sided confidence interval for the difference between the means. (2)
b. Assume that this represents paired data and redo the test (4)
a. Solution: From page 10 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
2
s12 s22

n1 n2
sd 
DF 
 s12 s22 
  
n

 1 n2 
s12
2
n1
  1   2
s 22
t
  1   2
d cv   0  t  2 sd
d  0
sd
Same as
H 0: 1   2
2
H 1:  1   2
   
n1  1
H 0 :  0
H 1:  0
H 0 :   0
H 1:   0
  d  t  sd
Difference
between Two
Means(
unknown,
variances
assumed
unequal)
2
if  0  0
n2
n2  1
H 0: 1   2
Same as
H 1:  1   2
s12  0.52

n1
11
2
s22  0.75

n2
11
2
s12
n1

 0.02458
s22
n2
 0.05114
 0.07572
sd 
s12 s22

n1 n 2
 0.07572  0.2752
(i)
DF 
 s12 s22 
  
 n1 n 2 
2
   
s12
2
n1
n1  1
s22
2
n2
n2  1
(ii)Test Ratio: t 

 0.07572 2
.02458 2 .05114 2
11  1

 17.81
17 
. .
So use 17 degrees of freedom. t .05  1740
11  1
d   0 0.48  0

 1744
.
This is above 1740
.
.
sd
0.2752
or Critical Value: d cv   0  t  sd  0  1740
.  0.2752  0.479
d  0.48 is above this value. With both methods reject H 0 .
17 
(iii) Confidence Interval: t .025  2.110.
  d  t  2 sd  0.48   2.110 0.2752  0.48  0.58 or -0.10 to 1.06.
7
3/23/98 252y9822
b. Solution:
H 0 :  0
H 0: 1   2
Same as
H 1:  0
H 1:  1   2
  1   2
Test Ratio: t 
d   0 0.48  0

 4.824
sd
0.0995
sd 
ss
n

0.33

11
DF  n  1  10
01089
.
 0.0995
11
t .10
.
05  1812
This is above 1812
.
.
or Critical Value: d cv   0  t  sd  0  1812
.  0.0995  01803
.
d  0.48 is above this value.
or Confidence Interval:   d  t  2 sd  0.48  1812
.  0.0995  0.2997 .
With all methods reject H 0 .
8
3/23/98 252y9822
3. The failure times in thousands of hours are given for a random sample of 8 components.
1.700 1.000 1.450 1.500 1.560 1.600 1.650 1.725
a. Test the hypothesis that these numbers came from a normal distribution. Use a 5% significance level. (5)
b. Test the hypothesis that the above data came from a normal distribution with a mean of 1.700 and a
standard deviation of 2.000 (5)
c. A television set distributorship believes that television ownership in the local area is distributed
according to a Poisson distribution with a mean of 4. A sample of 100 is taken. Is this true? Use a 5%
significance level. (5)
Number of TV sets:
0
1
2
3
4
5 or more.
Number of Households: 7 27 28 18 10
10
a. Solution: H 0 : N  ?, ? H 1 : Not Normal
Because the mean and standard deviation are unknown, this is a Lilliefors problem.
xx
From the data we find that x  1523
and s  0.2314 . t 
.
.
s
x
1000
.
1450
.
1500
.
1560
.
1600
.
1650
.
1700
.
1725
.
t
Ft
O
O
n
Fo
D
2.26 0.32 010
.
016
.
0.33 0.54 0.76 0.87
.0119 .3745 .4602 .5636 .6293 .7054 .7764 .8078
1
1
1
1
1
1
1
1
Max D .1924
O  n  8
0125
.
0125
.
0125
.
0125
.
0125
.
0125
.
0125
.
0125
.
0125
.
0.250 0.375 0.500 0.625 0.750 0.825 1000
.
.1131 .1255 .0852 .0636 .0043 .0446 .0986 .1924
Since the Critical
Value for  .05
is .288, accept H 0 .
b. Solution: H 0 : N 17
. , 2 H 1 : Not N 17
. , 2
Because the mean and standard deviation are known, this is a Kolmogorov-Smirnov problem.
x
z
.

x
1000
.
1450
.
1500
.
1560
.
1600
.
1650
.
1700
.
1725
.
z
0.35 0125
.
010
.
0.07 0.05 0.025
0
0.013
F  z  .3632 .4503 .4602 .4721 .4801 .4900 .5000 .5497
O
O
n
Fo
D
1
1
0125
.
0125
.
.2382
0125
.
0.250
.2003
1
1
1
0125
.
0125
.
0125
.
0.375 0.500 0.625
.0852 .0279 .1449
1
0125
.
0.750
.2600
1
1
0125
.
0125
.
0.825 1000
.
.3750 .4502
Mas D .4502
O  n  8
Since the Critical
Value for  .05
is .457, accept H 0 .
9
3/23/98 252y9822
c. Solution: H 0 : Poisson 35
.  H 1 : Not Poisson 35
.  This can be done as a chi-square or KolmogorovSmirnov problem. f e and Fe come from the Poisson table.
x
0
O
7
O
n
.07
Fo
.07
fe
.018316
E
1832
.
Fe
.0183
D
.0517
1
27
.27
.34
.073263
7.326
.0916
.2484
2
3
28
18
.28
.18
.62
.80
.146525 14.626
.195367 19.537
.2381
.4335
.3919
.3665
4
10
.10
.90
.195367 19.537
.6288
.2712
5
10 .10 100
.
100 100
.
.37116
37.116 100000
.
97.974
0
For the Kolmogorov-Smirnov Method the 5% critical value is
O2
O
E
34
9.158
E
126.228
28 14.626
18 19.537
53.603
16.596
10 19.537
5118
.
10 38116
.
100 97.974
2.624
204.169
100
104.169
136
.
. This is less than the
 0136
.
100
2 4 
maximum value of D , which is .3919, so reject H 0 . For the Chi-Squared Method.  .05  9.4877 , this is
less than our computed  2 , which is 104.169, so reject H 0 . Note that, for the Chi-Squared method. the
first two cells must be merged, since the first has an expected value well below 5.
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3/23/98 252y9822
4. a. The lifetimes of independent samples of two brands of power equipment are given below. The
distributions are not believed to be normal, so compare the medians at the 10% significance level.(5)
Brand 1: 2.4 3.8 6.0 6.9 3.6
Brand 2: 2.0 3.9 5.0 5.7 6.5
b A sample of eight married couples are compared as to their spending on clothing. The distributions
are not believed to be normal, so that medians are compared. Use the 10% significance level to answer
whether husbands and wives spend equally in this category.(5)
Couple Number:
1
2
3
4
5 6
7
8
Husband's Spending: 310 250 240 260 240 310 481 240
Wife's Spending:
390 385 190 400 410 490 482 330
a. Solution: Wilcoxon-Mann-Whitney Method H 0 :  1   2
x1 r1 x 2
r2
2.0 1
2.4
3.6
3.8
2
3
4
Check: 29  26  55 
1011
H 1: 1   2
2
For a 10% two-tailed test, the critical values are 18 and 37. W  27
is between them, so accept H 0 .
3.9 5
5.0 6
5.7 7
6.0
8
6.5 9
6.9 10
__
27
28
b. Solution: Wilcoxon Signed rank test for paired data. H 0 :  1   2
x1 x 2 difference rank
310 390
80
3
250 385
135
5
240 190
-50
-2
260 400
140
6
240 410
170
7
310 490
180
8
481 482
1
1
240 330
90
4
T   2, T   34
H 1: 1   2 .
Check :  T    T   2  34  36 
Since  TL     TL  .05  6 , and the smaller T is below it, reject H 0 .
8 9
2
2
Note: I defined the difference as x 2  x1 . If I had defined it the other way around, it would not have
affected the results.
11