252y0122 3/26/01
ECO252 QBA2
Name
SECOND HOUR EXAM Hour of Class Registered (Circle)
March 20, 2001
MWF 10 11 TR 12:30 2:00
I. (14 points) Do all the following. (Make diagrams!!!)
x ~ N 7,9
16  7 
3 7
z
 P 1.11  z  1.00 
1. P 3  x  16   P 
9
9 

 P1.11  z  0  P0  z  1.00   .3665  .3413  .7078
67
0  7
z
 P 0.78  z  0.11
2. P0  x  6  P 
9 
 9
 P0.78  z  0  P0.11  z  0  .2823  .0438  .2385
14  7 
  24  7
z
 P 3.44  z  0.78 
3. P24  x  14   P 
9 
 9
 P3.44  z  0  P0  z  0.78   .4997  .2823  .7820
57
 2  7
z
 P 1.00  z  0.22 
4. P2  x  5  P 
9 
 9
 P1.00  z  0  P0.22  z  0  .3413  .0871  .2542
27

5. F 2  (The Cumulative probability up to 2) . Px  2  P  z 
9 

 Pz  0.56   Pz  0  P0.56  z  0  .5  .2123  .2877
6. A symmetrical interval about the mean with 81% probability. We want two
points x .905 and x.095 , so that Px.905  x  x .095   .8100 . Make a diagram showing
6 in the middle at the center of a 81% region split into two areas with probabilities
of .4050. From the diagram, if we replace x by z, P0  z  z.095   .4050 .
The closest we can come is P0  z  1.31  .4049 . Use z .095  1.31 , and
x    z.105  7  1.319  7  11.79 , or -4.79 to 18.79. To check this note
18 .795  7 
  4.79  7
z
that P4.79  x  18.79   P

9
9


 P 1.31  z  1.31  2P0  z  1.31  2.4049   .8098  .8100
7.
x.006 . We want a point x.006 , so that Px x .006   .006 . Make a diagram of z
showing zero in the middle, .4940 between 0 and z .006 , and .006 above z .006 . From
the diagram, if we replace x by z, P0  z  z.006   .4940 . The Normal table says
P0  z  2.51  .4940 . So z .006  2.51 , and x    z.006  7  2.519  29.59.
29 .59  7 

To check this note Px  29 .59   P  z 
  Pz  2.51
9


 Pz  0  P0  z  2.51  .5  .4940  .006 .
252y0122 3/20/01
II. (6 points-2 point penalty for not trying part a.) Show your work!
A test shopper goes to Miller's Supermarkets 8 times and to Albert's Supermarkets 8 times. The
amount that was spent using a standardized shopping list is shown below. You can regard these data as two
independent random samples from populations with a normal distribution. For Albert's, the sample mean is
115.25 and the sample standard deviation is 1.75255.
xm
xa
Obs Miller's Albert's
1
118
112
2
119
115
3
120
115
4
118
117
5
120
117
6
122
117
7
120
115
8
120
114
a. Compute s m , the standard deviation for Miller's. (3)
b. Compute a 99% confidence interval for the difference between the population means 1 and  2
assuming that the variances are equal for the two parent populations. According to your confidence interval,
is there a significant difference between the population means (You must tell why!)? (3)
c. (Extra Credit) (i) Redo the confidence interval on the assumption that the variances are not equal. (6)
(ii) Use an F-test to tell whether you should have assumed that the variances were or were not equal. (2)
Solution: a) Use x1 for Miller's and x 2 for Albert's. x 2  115 .25 and s 2  1.75255 , so s 22  3.07143 .
Line
1
2
3
4
5
6
7
8
x
x1
118
119
120
118
120
122
120
120
957
x12
13924
14161
14400
13924
14400
14884
14400
14400
114493

x12  n1x12 114493  8119 .625 2 11 .875
957
 119 .625 s12 


 1.696429 s1  1.30247
n1
8
n1  1
7
7
b) From Table 3 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Difference
H 0:   0
  d  t  2 sd
d cv   0  t  2 sd
d  0
t
between Two
H 1:   0
s
1
1
d
Means (
sd  s p







n

1s12  n2  1s 22
1
2
n
n
1
2
unknown,
sˆ 2p  1
n1  n2  2
variances
DF  n1  n2  2
assumed equal)
x1 
1

2
252y0122 3/20/01
x1  119 .625 , s12  1.696429 , s1  1.30247 x 2  115 .25, s 22  3.0714 , s 2  1.75255
DF  n1  n 2  2  8  8  2  14
d  x1  x 2  119 .625  115 .25  4.375
sˆ 2p 
n1  1s12  n2  1s22 = 71.696429   73.0714   1.696429  3.0714
n1  n2  2
sd  s p
1
1


n1 n2
14
2
2.38391  1  1   2.38391 .25  
8
8
 2.38391
  .01,
14
t .005
 2.977
0.595979  0.77200
Confidence Interval:   d  t sd  4.375  2.977 0.77200   4.375  2.2982 or 2.076 to 6.673. The
2
interval does not include 0, so there is a significant difference between the means.
H 0 :   0
H :    2
H :    2  0

Formally, our hypotheses are H 1 :   0 or  0 1
or  0 1
We reject H 0 .
H 1 :  1   2
H 1 :  1   2  0
  1   2
c) (i) From the formula table;
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
  d  t  sd
Difference
Between Two
Means(
Unknown,
Variances
Assumed
Unequal)
2
s12 s22

n1 n2
sd 
DF 
 s12 s22 
  
n

 1 n2 
2
   
s12
2
n1
n1  1
s 22
2
n2
n2  1
H 0 :   0
H 1:   0
  1   2
t
d  0
sd
d cv   0  t  2 sd
Same as
H 0: 1   2
H 1:  1   2
if  0  0
x1  119 .625 , s12  1.696429 , s1  1.30247 x 2  115 .25, s 22  3.0714 , s 2  1.75255
  .01
d  x1  x 2  119 .625  115 .25  4.375
s12 1.69643

 0.212054
n1
8
s 22 3.0714

 0.383929
n2
8
sd 
s12 s 22

 0.595982  0.77200
n1 n 2
s12 s 22

 0.595982
n1 n 2
DF 
 s12 s 22 



 n1 n 2 


2
2
2
 s12 
 s 22 
 
 
 n1 
 n2 
 
 

n1  1
n2 1

.595982 2
0.212054 2  0.383929 2
7
 12 .925 , so use 12 degrees of freedom.
7
3
252y0122 3/20/01
12
t .005
 3.055
Confidence Interval: The 2-sided interval is   d  t sd  4.375  3.055 0.77200   4.375  2.359 or
2
2.016 to 6.734.
H :    2
s 2  1.6964 
c) (ii) F test.  0 1
. To test this at the 1% significance level, test 12  
  0.5523 and
s 2  3.0714 
H 1 :  1   2
s 22
s12

1
7,7  . F 7,7  is not on our table but F 7,7   6.99 is and F 7,7  must be
 1.811 against F.005
.005
.001
.005
0.5523
7,7  , do not reject H . Conclude that we should have
larger than 6.99. Since neither ratio is larger than F.005
0
assumed that the variances were equal.
4
252y0122 3/20/01
III. Do at least 3 of the following 4 Problems (at least 10 each) (or do sections adding to at least 30 points Anything extra you do helps, and grades wrap around) . You must do problem 1a! Show your work! State
H 0 and H1 where applicable. Do not answer a question 'yes' or 'no' without citing a statistical test.
Use a 95% confidence level unless another level is specified.
1a. Turn in your computer output from computer problem 1 only. (3 - 2 point penalty for not handing this
in.)
b. (Bowerman et. al.) A researcher knows that last year's average credit card interest rate was 18.3% but
believes that, due to increased competition, current rates are better. She checks the interest rates on a
random sample of credit cards and comes up with the data given as 'int' on the next page. What does the first
test tell us about whether rates have improved? State the null and alternative hypotheses and, using a 5%
significance level, tell whether the researcher is right. (3)
c. She now wishes to compare these rates with rates she finds in a search of banks that advertise on the net.
She assembles another random sample as 'int1' and runs a second test. Again state the hypotheses and
conclusion. (2)
d. Using the statistics that have been computed for you, verify the value of t in the second test and show
graphically how the computer got the p-value. (3)
Worksheet size: 100000 cells
MTB > RETR 'C:\MINITAB\2X0123-2.MTW'.
Retrieving worksheet from file: C:\MINITAB\2X0123-2.MTW
Worksheet was saved on 3/13/2001
MTB > print 'int'
Data Display
int
16.4
16.6
18.6
18.9
15.4
17.4
18.1
17.8
19.5
16.1
17.2
19.2
18.4
14.8
20.0
12.3
14.1
12.9
15.7
11.4
15.3
10.2
15.3
10.5
20.5
17.0
16.5
14.1
18.3
MTB > print 'int1'
Data Display
int1
18.5
17.3
18.7
15.1
13.6
22.1
First Test
MTB > ttest mu=18.3 'int';
SUBC> alt=-1.
T-Test of the Mean
Test of mu = 18.300 vs mu < 18.300
Variable
int
N
15
Mean
17.627
StDev
1.538
SE Mean
0.397
T
-1.70
P-Value
0.056
 H 0 :   18 .3
b) Solution: 
Since the p-value is above   .05, do not reject the null hypothesis of
 H 1 :   18 .3
equality and do not conclude that now interest rates are better (lower). The researcher has not been proved
right.
5
252y0122 3/20/01
Second Test
MTB > twosample 'int''int1';
SUBC> alt=1.
Two Sample T-Test and Confidence Interval
Twosample T for int vs int1
N
Mean
StDev
int
15
17.63
1.54
int1 20
15.47
3.22
SE Mean
0.40
0.72
95% C.I. for mu int - mu int1: ( 0.47, 3.84)
T-Test mu int = mu int1 (vs >): T= 2.62 P=0.0070
DF=
28
MTB >
c) Solution: If  is the mean interest rate from the old population and 1 is the rate from the net
 H 0 :  int   int1
 H 0 :   1
population, 
or 
Since the p-value is below   .05, reject the null
 H 1 :   1
 H 1 :  int   int1
hypothesis of equality. Conclude that now interest rates on the net can be proved to be better (lower).
d) Solution: If we are comparing two samples, from the formula table;
Interval for
Confidence
Hypotheses
Test Ratio
Interval
Difference
between Two
Means(
unknown,
variances
assumed
unequal)
  d  t 2 s d
2
1
s
s

n1 n2
sd 
DF 
 s12 s22 
  
n

 1 n2 
s12 s22 

n1 n2
2
   
s12
2
n1
n1  1
sd 
2
2
s 22
2
n2
n2  1
.40 2  .72 2
H 0 :   0
H 1:   0
  1   2
t
d  0
sd
Critical Value
d cv   0  t  2 sd
Same as
H 0: 1   2
H 1:  1   2
if  0  0
 .82365
t
d 0
17 .63  15 .47   0  2.62 Make a

sd
.82365
diagram showing an almost-normal curve with a mean at zero. The p-value is the probability that t  1.65 ,
so shade the area above 1.65 and label it 5.5%.
6
252y0122 3/20/01
2. Bowerman, O'Connell and Hand tell us that (i) of 30 investors who invested in a bond fund, 6 were
dissatisfied, (ii) of 30 investors who invested in a stock fund, 5 were dissatisfied and (iii) of 40 investors
who invested in a tax-deferred annuity, 10 were dissatisfied. You may assume that those who were not
dissatisfied were satisfied. Use a 99% confidence level.
a. Do a two sided confidence interval for the difference between the proportion satisfied with the bond fund
and with the stock fund. (3)
b. Test that the proportion of investors who were satisfied with the stock fund was higher than the
proportion who invested in the bond fund. (3)
c. Test the hypothesis that the proportion satisfied was the same for each investment (6)
Solution: To summarize the information in the problem -   .01 and
Investment
Bond Fund
Stock Fund
Annuity
24
25
30
Satisfied
6
5
10
Dissatisfied
30
30
40
Total
.8000
.8333
.7500
Proportion
satisfied
We are comparing p1  .8000 , n1  30 and p 2  .8333 , n2  30.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
pcv  p0  z 2  p
Difference
p  p 0
p  p  z 2 sp
H 0 : p  p0
z
between
If p0  0
 p
H 1 : p  p0
p  p1  p2
proportions
 p  p0 q 0  1 n  1 n 
If p  0
p 0  p 01  p 02
p1q1 p2 q 2
q  1 p
s p 

p01q 01 p02 q 02
n p  n2 p2
n1
n2
 p 

or p 0  0
p0  1 1
n1
n2
n1  n 2
Or use s p
1
s p 
2
p1 q1 p 2 q 2
.800 .200  .8333 .1667 



 .0053333  .0046297  .009963  .09982
n1
n2
30
30
p  p1  p 2  .0333 , p 0 
24  25 n1 p1  n 2 p 2 30 .8000   30 .8333 


 .8167 ,
30  30
n1  n 2
30  30
  .01, z  z.01  2.327, z 2  z.005  2.576. Note that q  1  p and that q and p are between 0 and 1.
 p  p 0 q 0

1
a) p  p  z
2

.8167 .1833  130  130  .0099815  .099907
s p  .0333  2.576 .09982   .033  .257 or -.290 to .224.
n1

1
n3
2/3 of you were following the old exam so slavishly that you never realized that I had asked for a
confidence interval!
H 0 : p  0
H 0 : p1  p 2
H 0 : p1  p 2  0
b) 
Same as 
or 
(Only one of the following methods is
H 1 : p 0
H 1 : p1  p 2
H 1 : p1  p 2  0
needed!)
p  p 0  .0333  0

 0.333 Make a Diagram showing a 'reject' region below
Test Ratio: z 
 p
.099907
-2.327. Since -0.033 is above this, do not reject H 0 .
or Critical Value: pcv  p0  z  p  0  2.327 .099907   .2324 . Make a Diagram showing a
'reject' region below -.2324. Since -0.0333 is above this, do not reject H 0 .
or Confidence Interval: p  p  z s p  .0333  2.327 .09982   .1990 p  .1990 is an interval
that includes 0. In all cases do not reject H 0 .
7
252y0122 3/20/01
b)
H 0 : Homogeneousor p1  p 2  p 3 
H 1 : Not homogeneousNot all ps are equal
O
Satisfied
Not
Total
Bond Stock Annuity Total p r
 24
25
30 
79
.79


6
5
10 
21
.21

30
30
40
100 1.00
DF  r  1c  1  12  2
 .2012   9.2103
E
Satisfied
Not
Total
Bond Stock Annuity Total p r
 23 .7
23 .7
31 .6
79
.79


6.3
8.4
21
.21
 6.3
30 .0
30 .0
40 .0
100 1.00
The proportions in rows, p r , are used with column totals to get the items in E . Note that row and column
sums in E are the same as in O . (Note that  2  0.7434 is computed two different ways here - only
one way is needed.)
O2
O  E 2
OE
O  E 2
O
E
E
E
24
23.7 -0.30000
0.09000
0.003797
24.3038
6
6.3
0.30000
0.09000
0.014286
5.7143
25
23.7 -1.30000
1.69000
0.071308
26.3713
5
6.3
1.30000
1.69000
0.268254
3.9683
30
31.6
1.60000
2.56000
0.081013
28.4810
10
8.4 -1.60000
2.56000
0.304762
11.9048
100
100.0
0.00000
0.743420
100.7435
O  E 2  100 .7435 100  0.7435 or .743420
O2
n 
E
E
Since this is less than 9.2103, do not reject H 0 .
(Diagram!)


8
252y0122 3/20/01
3. You have a sample of earned incomes for 9 couples, both of whom are teachers. You wish to test if the
women make more than the men. (Look at problem 4 before you do this problem - the data are identical,
what you do with it is different.)
a. You decide to compare means. Test to see if the women make more than the men assuming that the data
comes from a Normal distribution. (5)
b. You are reminded that the income data is usually highly skewed, so you ought to compare medians
instead of means. Repeat the test. (5)
For your convenience, not only the raw data are provided, but the data is ranked from 1 to 20 in the
columns r1 and r2 . If you need the ranks of the differences, r , you will have to do it yourself. Minitab
gives us the following results: x1  61.64, s1  12.23 (These are the sample mean and sample standard
deviation of x1 ), x 2  55.40, s 2  8.40, d  6.24, s d  17.19. You do not need all this information in
every part of the problem. Data is in thousands.
Row
1
2
3
4
5
6
7
8
9
women
x1
47.5945
58.5687
43.4502
62.3263
60.3484
65.3845
80.1389
78.6558
58.2903
rank
r1
4
10
1
13
12
15
18
17
9
men
x2
64.0374
57.6019
57.3682
50.1957
44.6027
47.4112
48.4382
58.9646
69.9989
rank
difference
d
r2
14 -16.4428
8
0.9668
7 -13.9180
6
12.1306
2
15.7457
3
17.9734
5
31.7007
11
19.6912
16 -11.7085
rank
r
Solution: This is paired data. a) Assume   .05 . All tests of the mean or median in Problems three and
four are one sided. The following table may help you choose a method. Save it for the final exam!
Comparing 2 Samples
Paired Samples
Independent Samples
Location - Normal distribution.
Method D4
Methods D1- D3
Compare means.
Location - Distribution not
Normal. Compare medians.
Method D5b
Proportions
Method D5a
Method D6
Variability - Normal distribution.
Method D7
Compare variances.
From the outline, there are three ways of approaching a problem involving two means. We know that
s
H :    2
16 .57
 5.73 . We are testing  0 1
or
d  x1  x 2  6.24 ,   1   2 , s d  17.19, s d  d 
n
9
 H 1 : 1   2
 H 0 : 1   2  0
H 0 :   0
8
 1.860 .
or 
, df  n  1  8 , tn 1  t .05

 H 1 : 1   2  0
H 1 :   0
(i) . Confidence Interval:   d  t  2 s d or 1   2   x1  x 2   t  2 s d . This interval becomes
  d  t s d  6.24 - 1.860 5.73   6.24 - 10.66  -4.42 . Since this interval includes zero, we
cannot reject H 0 .
9
252y0122 3/20/01
(ii). Test Ratio: t 
x  x 2   10   20 
d  0
d   0 6.24  0
or t  1

 1.089 . Make
. t
sd
sd
sd
5.73
a diagram showing an almost Normal curve with a mean at zero and a 'reject' region above
8
tn 1  t .05
 1.860 . Since 0.684 is not in this region, we cannot reject H 0 .
(iii). Critical Value: d CV   0  t  2 s d or x1  x 2 CV  10   20   t  2 s d . Because this is a onesided test, we want one critical value above zero. The critical value formula becomes
d CV   0  t s d  0  1.860 5.73   10 .658 Make a diagram showing an almost Normal curve
with a mean at zero and a 'reject' region above 10.658. Since 6.24 is not in this region, we cannot
reject H 0 .
Note that only one of the methods used above was expected.
b) If the underlying distributions are not Normal and the two samples are paired, we should use a Wilcoxon
Signed Rank test.
difference
rank
H :   2
Our hypotheses are  0 1
. The signed
d
r
H 1 : 1   2
-16.4428
-6
ranks of the differences are at right. If we add
0.9668
1
-13.9180
-4
ranks with like signs, we get T   12 and
12.1306
3
T   33 (Check: Their total is the sum of the
15.7457
5
910 
17.9734
7
 45 . )
numbers 1 through 9, which is
2
31.7007
9
According to the Wilcoxon table for a 1-sided
19.6912
8
5% test, reject the null hypothesis if the smaller
-11.7085
-2
of these totals is less than 8.
Since neither of the totals is below 8, do not reject H 0 .
10
252y0122 3/20/01
4. You have two independent random samples of 9 incomes from each of two towns. You wish to test if
the people in town 1 make more than people in town 2. (The numbers just happen to be the same as in the
last problem)
a. You decide to compare means. Test to see if the people in town 1 make more than the people in town 2
assuming that the data comes from a Normal distribution. (5)
b. You are reminded that the income data is usually highly skewed, so you ought to compare medians
instead of means. Repeat the test. (5)
For your convenience, not only the raw data are provided, but the data is ranked from 1 to 20 in the
columns r1 and r2 . If you need the ranks of the differences, r , you will have to do it yourself. Minitab
gives us the following results: x1  61.64, s1  12.23, x 2  55.40, s 2  8.40, d  6.24, s d  17.19. You
do not need all this information in every part of the problem. Data is in thousands.
c. In part a of this question, what assumption did you make about the variances of x1 and x 2 ? Test it here.
(3).
d. Using the means and variances given above, but assuming that n1  n 2  150 do a 2-sided 98.8%
confidence interval for 1   2 . (4)
Row
1
2
3
4
5
6
7
8
9
town 1
x1
47.5945
58.5687
43.4502
62.3263
60.3484
65.3845
80.1389
78.6558
58.2903
rank
r1
4
10
1
13
12
15
18
17
9
town 2
x2
64.0374
57.6019
57.3682
50.1957
44.6027
47.4112
48.4382
58.9646
69.9989
rank
difference
d
r2
14 -16.4428
8
0.9668
7 -13.9180
6
12.1306
2
15.7457
3
17.9734
5
31.7007
11
19.6912
16 -11.7085
rank
r
Solution: a) Assume   .05 . All tests of the mean or median in Problems three and four are one sided.
From the outline, there are three ways of approaching a problem involving two means. We know that
 H 0 : 1   2
 H 0 : 1   2  0
H 0 :   0
or 
or 
. It is most
d  x1  x 2  6.24 . We are testing 
 H 1 : 1   2
 H 1 : 1   2  0
H 1 :   0
convenient to assume that  1   2 , though we really ought to test it in part c). It is not wrong to assume
that variances differ, but the solution will only be provided to people who did so. If we assume that
16
 1.746 ,
variances are equal we find n1  n2  9 , df  n1  n 2  2  16, t .05
n  1s12  n2  1s 22 812.232  88.40 2 149 .5729  70.5600

s p2  1


 110 .06645 and
n1  n 2  2
16
2
1 
  1
1 1
  110 .06645     24 .4592  4.9456 .
s d  s p2  
n
n
9 9
2 
 1
Only one of the methods below is expected.
(i) . Confidence Interval:   d  t  2 s d or 1   2   x1  x 2   t  2 s d . This interval
becomes   d  t s d  6.24 - 1.746 4.9456   6.24 - 8.635  -2.3950 . Since this interval
includes zero, we cannot reject H 0 .
11
252y0122 3/20/01
(ii). Test Ratio: t 
x  x 2   10   20 
d  0
or t  1
.
sd
sd
d   0 6.24  0

 1.262 . Make a diagram showing an almost Normal curve with a
sd
4.9456
mean at zero and a 'reject' region above t 16  1.746 . Since 0.7436 is not in this region,
t
.05
we cannot reject H 0 .
(iii). Critical Value: d CV   0  t  2 s d or x1  x 2 CV  10   20   t  2 s d . Because this
is a one-sided test, we want one critical value above zero. The critical value formula
becomes d CV   0  t sd  0  1.7464.9456  8.6350 Make a diagram showing an
almost Normal curve with a mean at zero and a 'reject' region above 8.874. Since 6.24 is
not in this region we cannot reject H 0 .
b) If the underlying distributions are not Normal and the two samples are independent, we should use a
H 0 : 1   2
Mann-Whitney-Wilcoxon Rank test. Our hypotheses are 
. The rank total for Town 1 (gotten
H 1 : 1   2
from adding the r1 column) is T1  99 and, for Town 2, T2  72 . (Check: Their total is the sum of the
18 19 
 171 . ) According to Table 6 (for a 1-sided 5% test), we do not
2
reject the null hypothesis if the smaller of these totals ( W  72 ) is between 54 and 90. Since both of the
totals are in that interval, do not reject H 0 .
numbers 1 through 18, which is
2
H 0 :  1   2
s 2  12 .23 
c) 
. To test this at the 5% significance level, test 12  
  2.120 and
s 2  8.40 
H 1 :  1   2
s 22
s12

1
8,8  4.43 . Since neither ratio is larger than 4.43, do not reject H .
 0.4717 against F.025
0
2.120
d) If the confidence level is 98.8%, the significance level is   1  .988  .012 . Since our samples have a
total number of degrees of freedom of df  n1  n 2  2  160  160  2  318 , we can use z instead of t .
 s2 s2 
 12 .23 2 8.40 2
sd   1  2   

 160
 n1 n 2 
160




  149 .5729  70 .5600  1.21143 . z  z.006 , and we found
2

160

that it was 2.51 on page 1. Our confidence interval is   d  t  2 s d  6.24  2.511.21143   6.24  3.04,
or 3.20 to 9.28.
12