252y0121 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 6,9
16  6 
 3  6
z
 P 1.00  z  1.11
1. P 3  x  16   P 
9
9 

 P1.00  z  0  P0  z  1.11  .3413  .3665  .7078
5 6
0  6
z
 P 0.67  z  0.11
2. P0  x  5  P 
9 
 9
 P0.67  z  0  P0.11  z  0  .2486  .0438  .2048
14  6 
  24  6
z
 P 3.33  z  0.89 
3. P24  x  14   P 
9 
 9
 P3.33  z  0  P0  z  0.89   .4996  .3133  .8129
4  6
 2  6
z
 P 0.89  z  0.22 
4. P2  x  4  P 
9
9 

 P 0.89  z  0  P 0.22  z  0  .3133  .0871  .2262
2  6

5. F 2  (The Cumulative probability up to 2) . Px  2  P  z 
9 

 Pz  0.44   Pz  0  P 0.44  z  0  .5  .1700  .3300
6. A symmetrical interval about the mean with 79% probability. We want two
points x .895 and x.105 , so that Px.895  x  x105   .7900 . Make a diagram showing
6 in the middle at the center of a 79% region split into two areas with probabilities
of .3950. From the diagram, if we replace x by z, P0  z  z.105   .3950 .
The closest we can come is P0  z  1.25   .3944 or P0  z  1.26   .3962 . Since
neither of these is much closer than the other, use z.105  1.255 , and
x    z.105  6  1.255 9  6  11.295 , or -5.295 to 17.295. To check this note that
17 .295  6 
  5.295  6
z
P 5.295  x  17 .295   P

9
9


 P1.295  z  1.2558   2P0  z  1.255   2.3950   .7900
x.008 We want a point x.008 , so that Px x.008   .008 . Make a diagram of z
showing zero in the middle, .4920 between 0 and z.008 and .008 above z.008 . From
the diagram, if we replace x by z, P0  z  z.008   .4920 . The Normal table says
P0  z  2.41  .4920 . So z.008  2.41 , and x    z.008  6  2.419  27.69. To check this note
7.
27 .69  6 

Px  27 .69   P  z 
  Pz  2.41  Pz  0  P0  z  2.41  .5  .4920  .008
9


252y0121 3/19/01
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 Miller's, the sample mean is
115.25 and the sample standard deviation is 1.75255.
xm
xa
Obs Miller's Albert's
1
112
119
2
115
121
3
115
122
4
117
120
5
117
122
6
117
124
7
115
122
8
114
122
a. Compute s a , the standard deviation for Albert'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 x 2 for Albert's and x1 for Miller's. x 2  115 .25 and s 2  1.75255 , so s 22  3.07143 .
Line
1
2
3
4
5
6
7
8
x1
119
121
122
120
122
124
122
122
972
x
x12
14161
14641
14884
14400
14884
15376
14884
14884
118114

x12  n1x12 118114  8121 .50 2 16
972
 121 .50 s12 


 2.2857 s1  1.51186
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
2
1
1
2
unknown,
sˆ p 
n1  n2  2
variances
DF  n1  n2  2
assumed equal)
x1 
1

2
252y0121 3/19/01
x1  121 .50, s12  2.2857 , s1  1.51186 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  121 .50  115 .25  6.25
sˆ 2p 
n1  1s12  n2  1s22 = 72.2857   73.0714   2.2857  3.0714
n1  n2  2
sd  s p
1
1


n1 n2
14
2
2.67955  1  1   2.67955 .25  
8
8
 2.67955
  .01,
14
t .005
 2.977
0.6698875  0.818467
Confidence Interval:   d  t sd  6.25  2.977 0.81847   6.25  2.4366 or 3.81 to 8.69. 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
Solution: b) From the formula table;
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
H 0 :   0
H 1:   0
  d  t 2 s d
Difference
between Two
Means(
unknown,
variances
assumed
unequal)
s12 s 22

n1 n2
sd 
DF 
 s12 s22 
  
n

 1 n2 
  1   2
2
n1
n1  1
s 22
d 0
sd
d cv   0  t  2 s d
Same as
H 0: 1   2
2
   
s12
t
2
n2
n2  1
H 1:  1   2
if  0  0
x1  121 .50, s12  2.2857 , s1  1.51186 x 2  115 .25, s 22  3.0714 , s 2  1.75255
  .01
d  x1  x 2  121 .50  115 .25  6.25
s12 2.2857

 0.285714
n1
8
s 22 3.0714

 0.383929
n2
8
sd 
s12 s 22

 0.669643  0.8183172
n1 n 2
s12 s 22

 0.669643
n1 n 2
DF 
 s12 s 22 



 n1 n 2 


2
2
2
 s12 
 s 22 
 
 
 n1 
 n2 
 
 

n1  1
n2 1

.669643 2
2.28571 2  .383929 2
4
 13 .7052 , so use 13 degrees of freedom.
7
3
252y0121 3/19/01
13
t .005
 3.012
Confidence Interval: The 2-sided interval is   d  t sd   6.25  3.012 0.669643   6.25  2.01 or
2
4.24 to 8.26
H :    2
s 2  2.2857 
c) F test.  0 1
. To test this at the 1% significance level, test 12  
  0.7442 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
 1.343 against F.005
.005
.001
.005
0.7442
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
252y0121 3/19/01
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)
Data Display
int
15.6
15.8
17.8
18.1
14.6
16.6
17.3
17.0
18.7
15.3
16.4
18.4
17.6
14.0
19.2
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
MTB > ttest mu=18.3 'int';
SUBC> alt=-1.
First Test
T-Test of the Mean
Test of mu = 18.300 vs mu < 18.300
Variable
int
N
15
Mean
16.827
StDev
1.538
SE Mean
0.397
T
-3.71
P-Value
0.0012
 H 0 :   18 .3
b) Solution: 
Since the p-value is below   .05, reject the null hypothesis of equality and
 H 1 :   18 .3
conclude that now interest rates are better (lower).
5
252y0121 3/19/01
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
16.83
1.54
int1 20
15.47
3.22
SE Mean
0.40
0.72
95% C.I. for mu int - mu int1: ( -0.33,
T-Test mu int = mu int1 (vs >): T= 1.65
3.04)
P=0.055
DF=
28
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 :   1
population,  0
or 
. Since the p-value is above   .05, do not reject the null
 H 1 :   1
 H 1 :  int   int1
hypothesis of equality. Conclude that now interest rates on the net cannot 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)
2
s12 s 22

n1 n2
sd 
DF 
 s12 s22 
  
n

 1 n2 
s12 s22 

n1 n2
  1   2
s12
2
n1
s 22
t
d 0
sd
d cv   0  t  2 sd
Same as
H 0: 1   2
2
   
n1  1
sd 
H 0 :   0
H 1:   0
  d  t  sd
Critical Value
2
n2
n2  1
.40 2  .72 2
H 1:  1   2
if  0  0
 .82365
t
d 0
16 .83  15 .47   0  1.65 Make a diagram

sd
.82365
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%.
MTB >
6
252y0121 3/19/01
2. Bowerman, O'Connell and Hand tell us that (i) of 30 investors who invested in a bond fund, 7 were
dissatisfied, (ii) of 30 investors who invested in a stock fund, 6 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 proportioned satisfied was the same for each investment (6)
Solution: To summarize the information in the problem -   .01 and
Investment
Bond Fund
Stock Fund
Annuity
23
24
30
Satisfied
7
6
10
Dissatisfied
30
30
40
Total
.7667
.8000
.7500
Proportion
satisfied
We are comparing p1  .7667 , n1  30 and p 2  .8000 , n2  30.
Interval for
Confidence
Hypotheses
Test Ratio
Interval
Difference
p  p 0
p  p  z 2 sp
H 0 : p  p0
z
between
 p
H 1 : p  p0
p  p1  p2
proportions
If p  0

p

p

p
p1q1 p2 q 2
0
01
02
q  1 p
s p 

p01q 01 p02 q 02
 p 

n1
n2
or p 0  0
n
n
1
Or use s p
s p 
2
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
p1 q1 p 2 q 2
.7667 .2333  .800 .200 



 .0059624  .0053333  .0112954  .10628
n1
n2
30
30
p  p1  p 2  .0333 , p 0 
23  24 n1 p1  n 2 p 2 30 .7667   30 .8000 


 .7833 ,
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

.7833 .2167  130  130  .011316  .10638
s p  .0333  2.576 .10628   .033  .274 or -.307 to .241. 2/3 of you were following
n1

1
n3
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

 0.313 Make a Diagram showing a 'reject' region below
Test Ratio: z 
 p
.10638
-2.327. Since -0.131 is above this, do not reject H 0 .
or Critical Value: pcv  p0  z  p  0  2.327 .10638   .2473 . Make a Diagram showing a
'reject' region below -.2473. Since -0.0333 is above this, do not reject H 0 .
or Confidence Interval: p  p  z s p  .0333  2.327 .10628   .2140 p  .2140 is an interval
that includes 0. In all cases do not reject H 0 .
7
252y0121 3/19/01
c)
DF  r  1c  1  12  2
H 0 : Homogeneousor p1  p 2  p 3 
H 1 : Not homogeneousNot all ps are equal
O
Satisfied
Not
Total
 .2012   9.2103
Bond Stock Annuity Total p r
 23
24
30 
77
.77


7
6
10 
23
.23

30
30
40
100 1.00
E
Satisfied
Not
Total
Bond Stock Annuity Total p r
 23 .1 23 .1
30 .8
77
.77


6.9
9.2
23
.23
 6.9
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
one way is needed.)
O2
O
E
E
23
23.1
22.9004
7
6.9
7.1014
24
23.1
24.9351
6
6.9
5.2174
30
30.8
29.2208
10
9.2
10.8696
100
100.0
100.2447
 2  0.2447 is computed two different ways here - only
OE
-0.10000
0.10000
0.90000
-0.90000
-0.80000
0.80000
0.00000
O  E 2
E
0.000433
0.001449
0.035065
0.117391
0.020779
0.069565
0.24468
O  E 2  100 .2337 100  0.2447
O2
n 
E
E
Since this is less than 9.2103, do not reject H 0 .
(Diagram!)


8
252y0121 3/19/01
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  57.86, s 2  7.56, d  3.78, s d  16.57. You do not need all this information in
every part of the problem. Data is in thousands.
Row
women
rank
men
rank
difference
rank
d
r
x1
r1
x2
r2
1
47.5945
2
65.6336
15 -18.0391
2
58.5687
8
59.8417
10
-1.2730
3
43.4502
1
59.6314
9 -16.1812
4
62.3263
13
53.1761
6
9.1501
5
60.3484
11
48.1425
3
12.2060
6
65.3845
14
50.6700
4
14.7145
7
80.1389
18
51.5944
5
28.5445
8
78.6558
17
61.0681
12
17.5877
9
58.2903
7
70.9990
16 -12.7087
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
Method D5a
Proportions
Method D6
Variability - Normal distribution.
Compare variances.
Method D7
From the outline, there are three ways of approaching a problem involving two means. We know that
s
H :    2
16 .57
 5.52333 . We are testing  0 1
d  x1  x 2  3.78 ,   1   2 , s d  16.57, s d  d 
n
9
 H 1 : 1   2
 H 0 : 1   2  0
H 0 :   0
8
 1.860 .
or 
or 
, df  n  1  8 , tn 1  t .05
H
:




0
H
:


0
2
 1 1
 1
(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  3.78 - 1.860 5.52333   3.78 - 10.27  -6.49 . Since this interval
includes zero, we cannot reject H 0 .
9
252y0121 3/19/01
(ii). Test Ratio: t 
t
x  x 2   10   20 
d  0
or t  1
.
sd
sd
d   0 3.78  0

 0.684 . Make a diagram showing an almost Normal curve with a
sd
5.52333
8
mean at zero and a 'reject' region above 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 one-sided test, we want one critical value above zero. The critical value formula
becomes d CV   0  t s d  0  1.860 5.52333   9.734 Make a diagram showing an
almost Normal curve with a mean at zero and a 'reject' region above 9.734. Since 3.78 is
not in this region, we cannot reject H 0 .
Note that only one of the three methods above is required.
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 0 : 1   2
Our hypotheses are 
. The signed
d
r
H 1 : 1   2
-18.0391
8ranks of the differences are at right. If we add
-1.2730
1-16.1812
6ranks with like signs, we get T   19 and
9.1501
2+
T   26 (Check: Their total is the sum of the
12.2060
3+
910 
14.7145
5+
 45 . )
numbers 1 through 9, which is
2
28.5445
9+
According to the Wilcoxon table for a 1-sided
17.5877
7+
-12.7087
45% test, reject the null hypothesis if the smaller
of these totals is less than 8.
Since neither of the totals is below 8, do not reject H 0 .
10
252y0121 3/19/01
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 (These are the sample mean and sample standard
deviation of x1 ), x 2  57.86, s 2  7.56, d  3.78, s d  16.57. 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  160 do a 2-sided 98.4%
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
2
8
1
13
11
14
18
17
7
Town 2
x2
65.6336
59.8417
59.6314
53.1761
48.1425
50.6700
51.5944
61.0681
70.9990
rank
difference
d
r2
15 -18.0391
10
-1.2730
9 -16.1812
6
9.1501
3
12.2060
4
14.7145
5
28.5445
12
17.5877
16 -12.7087
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  3.78 . 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7.56 2 149 .5729  57.1536

s p2  1


 103 .36325 and
n1  n 2  2
16
2
1 
  1
1 1
  103 .36325     22 .9696  4.7927 .
s d  s p2  
9 9
 n1 n 2 
Only one of the following methods 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  3.78 - 1.746 4.7927   3.78 - 8.368  -4.588 . Since this interval
includes zero, we cannot reject H 0 .
11
252y0121 3/19/01
(ii). Test Ratio: t 
x  x 2   10   20 
d  0
or t  1
.
sd
sd
d   0 3.78  0

 0.7886 . Make a diagram showing an almost Normal curve with
sd
4.7927
a mean at zero and a 'reject' region above t 16  1.746 . Since 0.7886 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 s d  0  1.746 4.7927   8.368 Make a diagram showing an
almost Normal curve with a mean at zero and a 'reject' region above 8.368. Since 3.78 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  91 and, for Town 2, T2  80 . (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 null hypothesis if the smaller of these totals ( W  80 ) 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  
  2.617 and
s 2  7.56 
H 1 :  1   2
s 22
s12

1
8,8  4.43 . Since neither is larger than 4.43, do not reject
 0.384 against F.025
H0.
2.617
d) If the confidence level is 98.4%, the significance level is   1  .984  .016 . 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 7.56 2
sd   1  2   

 160
 n1 n 2 
160




  149 .5729  57 .1536  1.13668 . z  z.008 , and we found
2

160

that it was 2.41 on page 1. Our confidence interval is   d  t  2 s d  3.78  2.411.13668   3.78  2.74.
12