3/22/00 252y0021
ECO252 QBA2
SECOND HOUR EXAM
March 21,2000
Name _____key___________
Hour of class registered _____
Class attended if different ____
Show your work! Make Diagrams!
I. (14 points) Do all the following.
x ~ N 9,6
9 9
 1 9
z
 P 1.67  z  0  .4525
1. P1  x  9  P 
6 
 6
11  9 
  11  9
z
 P 3.33  z  0.33 
2. P11  x  11  P 
6 
 6
 P 3.33  z  0  P0  z  0.33  .4996  .1293  .6289
0 9
9 9
z
 P 3.00  z  1.50 
3. P9  x  0  P 
6
6 

 P3.00  z  0  P1.50  z  0  .49865  .4332  .06545
15  9 

4. F 15  (The cumulative probability up to 15) Px  15   P  z 
6 

 Pz  1.00   Pz  0  P0  z  1.00   .5  .3413  .8413
17 .5  9 
14  9
z
 P0.83  z  1.42 
5. P14  x  17 .5  P 
6 
 6
 P0  z  1.42   P0  z  0.83  .4222  .2967  .1255
6.
A symmetrical interval about the mean with 48% probability.
We want two points x .74 and x.26 , so that Px.74  x  x 26   .4800 . From the diagram,
if we replace x by z, P0  z  z.26   .2400 . The closest we can come is
P0  z  0.64   .2389 or P0  z  0.65   .2422 . Since these are about the same
distance from .2400 use z.26  0.645 , and x    z.26  9  0.645 6  9  3.87 ,
12 .87  9 
 5.13  9
z
or 5.13 to 12.87. To check this note that P5.13  x  12 .87   P

6
6


 P0.645  z  0.645   2P0  z  0.645   22400   .4800
x.085 We want a point x .085 , so that Px x .085   .085 . From the diagram,
if we replace x by z, P0  z  z.085   .4150 . The closest we can come is
P0  z  1.37   .4147 . So z.085  1.37 , and x    z.085  9  1.37 6  9  8.22 , or 17.22 .
7.
17 .22  9 

To check this note that Px  17 .22   P z 
  Pz  1.37 
6


 Pz  0  P0  z  1.37   .5  .4147  .0853  .085
3/22/00 252y0021
II. (6 points - 2 point penalty for not trying)
A manufacturer of mowers wants to compare the amount of time it takes to mount a mower engine
using two different processes. Two independent samples are taken and the time in minutes appears below. !
Note: Most of you didn't bother to tell me what your hypotheses were in any problem but the first
one. This means that I couldn't really tell whether your critical values etc. were right or wrong!
x1
x2
2.1
3.3
4.1
7.3
9.1
5.3
3.1
8.3
2.1
4.3
3.3
Note that x1  4.1 and s1  2.915476
s

 8.5
2
1
a. Compute the sample standard deviation for the second process, s 2 . Show your work.(3)
b. Compute a 95% 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)
a. Solution: x 2  5.3 and s 2  2.097618
x2 
x
n
x

2

s
2
2
 4.4

31 .8
 5.30
6
190 .54  65.30 2
n 1
5
 4.40 . s 2  2.097618 .
s22
2
2
 nx22
x 22
10.89
53.29
28.09
68.89
18.49
10.89
190.54
x2
3.3
7.3
5.3
8.3
4.3
3.3
31.8

b. 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)
Test Ratio
t
sˆ 2p 
sˆ2p 
n1  1s12  n2  1s22
sd  s p
2
n1  n2  2
1
1


n1 n2
=
n1  1s12  n2  1s22
n1  n2  2
DF  n1  n 2  2  5  6  2  9
48.50   54.40  34 .00  22 .00

 6.22222
9
9
6.22222  1  1   6.22222 .366667  
5
6
d cv   0  t  2 sd
d  0
sd
x1  4.1, s12  8.50, s1  2.915476 x 2  5.3, s 22  4.40, s 2  2.097618
d  x1  x 2  4.1  5.3  1.2
Critical Value
  .05,
9
t.025
 2.262
2.28148  1.51046
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Confidence Interval:   d  t sd  1.20  2.262 1.51046   1.20  3.417 or -4.62 to 2.22. The
2
interval includes 0, so there is no significant difference between the means.
H 0 :   0
H 0 :  1   2
H :    2  0

Formally, our hypotheses are H 1 :   0 or 
or  0 1
We do not reject H 0 .
H 1 :  1   2
H 1 :  1   2  0
  1   2
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3/22/00 252y0021
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) . Show your work! State H 0 and H 1 where
appropriate. Use a 95% confidence level unless another level is specified.
1. For your convenience, the data from the previous page, giving samples of the time to mount a mower
engine, is copied below. Test the hypothesis that the mean time to mount a mower is greater for method two
than method one.
x1
2.1
4.1
9.1
3.1
2.1
x2
3.3
7.3
5.3
8.3
4.3
3.3
Note that x1  4.1 and s1  2.915476
s
2
1

 8.5
a) Choose the correct null hypothesis from the list below and write the alternative hypothesis. (2)
H 0 :  2  4.1
H 0 : 1   2
H 0 : x1  x 2
H 0 :  2  4.1
H 0 : 1   2
H 0 : x1  x 2
H 0 :  2  4.1
H 0 : 1   2
H 0 : x1  x 2
H 0 :  2  4.1
H 0 : 1   2
H 0 : x1  x 2
H 0 :  2  4.1
H 0 : 1   2
H 0 : x1  x 2
b) Assume that  1   2 and find a critical value appropriate for this problem, using a confidence
level of 90%(3)
c) Use your critical value to test the hypothesis. State clearly whether you reject the null
hypothesis. (2)
d) Repeat the test using (i) a test ratio (2) and (ii) a confidence interval. (2)
find an approximate p-value for your null hypothesis. (1)
e) Do a 90% confidence interval for the ratio of the standard deviations of the populations from
which x1 and x 2 come. On the basis of this comparison, what assumption should we have made
about the equality of the variances? (3)
Solution: a) The hypothesis says that 1   2 . Since this does not include an equality, it is the alternate
 H 0 : 1   2
hypothesis. The null hypothesis is the opposite, 1   2 , so that our hypotheses are 
or
 H 1 : 1   2
 H 0 : 1   2  0
H 0 :   0
or 
where   1   2 . Did you make the same mistake on this exam as

H
:




0
2
 1 1
H 1 :   0
you did in the first one?
4
3/22/00 252y0021
s
b)   .10 , x1  4.1 and s1  2.915476
We have already computed sˆ2p 
2
1

 8.5 , x 2  5.3 and s 2  2.097618
n1  1
 n2  1
n1  n2  2
s12
s22
1
1


n1 n2
DF  n1  n2  2  5  6  2  9 sd  s p
=
s
2
2

 4.4 .
48.50   54.40 
 6.22222
9
6.22222  1  1   1.51046
5
6
9
 1.383
d  x1  x 2  4.1  5.3  1.2 . t .10
Critical Value: d cv   0  t  2 sd is the formula for a 2-sided critical value, but
since our alternative hypothesis is H 1 :   0 or H1 : 1  2  0 , we want a
critical value below zero and to use t 9  t 9. So d    t s  0  1.383 1.51046   2.089

cv
.10
0
 d
c) d  1.20 is above this value, so we do not reject H 0 . Unless you gave me a
clear indication of what the rejection region was, e.g. a diagram, and unless
the reject region did not include d  0, you did not get full credit on c) or d) .
d) (i) Test Ratio: t 
d   0  1.20  0
9

 0.794 . Since this is above  t .10
 1.383 ,
sd
1.51046
we do not reject H 0 .
(ii) Confidence Interval: The 2-sided interval is   d  t sd , but since our alternative
2
hypothesis is H 1 :   0 , we use   d  t s d  1.20  1.383 1.51046   1.20  2.089  0.889 .
Since   0.889 does not contradict H 0 :   0 , we do not reject H 0 .
9
9
(iii) t  0.749 . This value of t is between t .25
 0.703 and t .20
 0.883 ,
so we can say that .20  p  value  .25.
e) A 2-sided confidence interval is
s12
s 22


F DF2 , DF1 
2
 12
 22

s12
s 22
s 22
s12


F DF1 , DF2 
2
 22
 12

s 22
1
s12 FDF2 , DF1 
or
2
1
2
2
DF1 , DF2  , and s1  8.5, s 2  4.4, n1  5 and n 2  6 . So DF1  n1  1  4
F
2


4,5  5.19 and
and DF2  n2  1  5 .   .10 , which means that F DF1 , DF2  F.05
2


5, 4   6.26 . If we use the first formula,
F DF2 , DF1  F.05
2
2.6866 
 22
 12
2
4.4
5.19   22  4.4 1 reduces to
8.5
 1 8.5 6.26
 0.08269 and, if we take square roots, 1.639 
2
1
 0.288 . If we use the second formula,

2
2
8.5
6.26   12  8.5 1 , 12.0932  12  0.37222 and 3.477  2  0.610 . Since either of these
4.4
 2 4.4 5.19
2
1
intervals includes 1, we can say that the variances are not significantly different.
5
3/22/00 252y0021
2. The New York Times reported on exit polls taken of voters in 6 Republican primaries. The numbers and
per cent that voted for George W. Bush in four of these are shown below. (Use a 99% confidence level.)
State
Massachusetts
California
Connecticut
New York
485
1361
968
1060
Sample Size
262
857
542
604
Number for Bush
.540
.630
.560
.570
Proportion
a)Test the hypothesis that the proportion for Bush in New York was higher than the proportion in
Connecticut. (4)
b) Test the hypothesis that the proportion for Bush was the same in each state. (7)
Solution: a) From Table 3 of the Syllabus Supplement: Let p3 be the proportion in CT and p4 be the
proportion in NY.
Interval for
Confidence
Interval
p  p  z 2 sp
Difference
between
proportions
q  1 p
p  p1  p2
s p 
p1q1 p2 q 2

n1
n2
Hypotheses
H 0 : p  p0
H 1 : p  p0
p 0  p 01  p 02
or p 0  0
Test Ratio
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 
H 0 : p 3  p 4
H 0 : p 3  p 4  0
542
604
p3 
 .560 , p 4 
 .570 ,
or 

H
:
p

p
H
:
p

p

0
968
1060
4
4
 1 3
 1 3
542  604
n p  n4 p4 968 .560   1060 .570 
 3 3

 .565089 ,
p  p3  p 4  .010 , p0 
968  1060
n3  n4
968  1060
 p  p 0 q 0

1
n1

1
n3

.565089 .434911  1968  11060 
.00048674  .0220395
(Only one of the following methods is needed!)
p  p0 .010  0

 0.454 This is above -2.327. (Diagram!)
Test Ratio: z 
 p
.0220395
or Critical Value: pcv  p0  z  p  0  2.327 .0220395   .0513
p  .010 is above this value.
or Confidence Interval: p  p  z s p where
s p 
p3 q3 p 4 q 4
.560 .440  .570 .430 



 .00055568  0.02357 . So
n3
n4
968
1060
p  .010  2.327 0.02357   0.0448 . The interval includes 0. In all cases do not reject H 0 .
6
n2

n p  n2 p2
p0  1 1
n1  n 2
H 0 : p  0
Same as

H 1 : p 0
  .01, z  2.327 .
1
3/22/00 252y0021
b)
DF  r  1c  1  13  3
H 0 : Homogeneousor p1  p2  p3  p4 
H1 : Not homogeneousNot all ps equal
O
For
MA CA
262 857

Against  223 504
Total
485 1361
E
For
 .2013  11 .3449
CT NY
Total
pr
542 604  2265 .5847

426 456  2609 .4153
968 1060 3874 1.0000
MA CA
 283 .58 795 .78

Against 201 .42 565 .22
Total
485 .00 1361 .00
CT NY
Total
pr

565 .99 619 .78 2265 .5847

402 .01 440 .22  2609 .4153
968 .00 1060 .00 3874 1.0000
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
way is needed.)
O2
O
E
E
262
283.58
242.062
223
201.42
246.892
857
795.78
922.930
504
565.22
449.411
542
565.99
519.027
426
402.01
451.422
604
619.78
588.622
456
440.22
472.346
3874 3874.00
3892.712
 2  18.722 is computed two different ways here - only one
OE
-21.5800
21.5800
61.2200
-61.2200
-23.9900
23.9900
-15.7800
15.7800
0.0000
O  E 2  3892 .712  3874  18.712
O2
n 
E
E
Since this is less than 11.3449, reject H 0 .
(Diagram!)

7

O  E 2
E
1.64220
2.31207
4.70970
6.63084
1.01684
1.43161
0.40177
0.56565
18.711
3/22/00 252y0021
3. We have the following data:
x  970 ,   60 , n  225
a) If we test the hypothesis that the population mean is 978, what is the p-value for this hypothesis? (2)
b) If we use a significance value of 10% and the p-value in a), would we reject the null hypothesis? Why?
(1)
c) Find the critical values of x necessary to test the hypothesis in a) if the significance level is 10%. (2)
d) Do a power curve for the test in c). (6)
e) A table of solid-waste generation rates for countries classified by level of development was assembled.
We are hypothesizing that as countries' incomes rise they become more alike in their tendency to generate
solid waste. For 5 industrial countries the standard deviation for a measure of solid waste generation was
0.0652 and, for 7 middle-income countries, the standard deviation was 0.1828. State an appropriate
hypothesis and test it. (4)
Solution: a) From Table 3 of the Syllabus Supplement:
Interval for
Confidence
Hypotheses
Interval
Mean (
  x  z 2  x
H0 :   0
Known)
H :  
1
Test Ratio
z
0
Critical Value
x  0
x
xcv   0  z 2  x
x   0 970  978
H 0 :   978

60
x 

4 z

 2.00

x
4
n
225
H 1 :   978
p  value  2Pz  2.00   2.5  .4772   2.0228   .0456
b) If   .10 , p  value   , so reject H 0 .
c)   .10
z 2  z.05  1.645 xcv  0  z  x  978  1.6454  978  6.58 or 971.42 to 984.58.
2
d) The table below is the same as used in Problem C2 and summarizes our results and further calculations.
x cvL  971 .42 means a lower critical value and x cvU  984 .58 means an upper critical value. These become
z L and zU . Note that when 1   0 power   , and when 1  x cv Power  50 % . These do not have
to be computed.
Power
1
zL
zU
x cvL  1
x cvU  1

1  


x
978
971 .42  978
4
971 .42  981 .3
4
x
984 .58  978
4
984 .58  981 .3
4
-1.645
1.645
.4500 + .4500
= .9000
10.00%
-2.47
0.82
.4932 + .2939
= .7871
21.29%
984.58
(Same as
971.42)
971 .42  984 .58 984 .58  984 .58 -3.29
4
4
0.00
.4995
50.00%
987.9 (Same
as 968.1)
971 .42  987 .9
4
984 .58  987 .9
4
-4.12
-0.83
.5000  .2967
 .2033
79.67%
991.2 (Same
as 964.8)
971 .42  991 .2
4
984 .58  991 .2
4
-4.945
-1.655
.5000  .4400
 .0600
94.00%
981.3 (Same
as 974.7)
8
3/22/00 252y0021
Diagram:
H 0 :  12   22
e) 
H 1 :  12   22
s 22
s12

0.1828 2
0.0652 2
 7.8606 . DF2  n 2  1  7  1  6 and DF1  n1  1  5  1  4 .
6,4   4.01 , F 6,4   6.16 , F 6,4   9.20 and F 6,4   15 .21 , if you used a significance level of .10
Since F.10
.05
.025
.01
or .05 do not reject H 0 . But if you used a significance level of .025 or .01, reject H 0 .
9
3/22/00 252y0021
4. A hospital administrator wishes to compare the distribution of unoccupied beds in two hospitals.
Because she believes that the distributions are quite badly skewed she does not use a method based on
means but instead ranks the entire sample and uses a test based on these ranks. Data is below. (Source W890)
Hosp 1
rank
Hosp 2
rank
Difference
Beds
Beds
Day
d
x1
r1
x2
r2
1
6
3
34
16
-28
2
38
17
28
12
10
3
3
1
42
20
-39
4
17
9
13
6
4
5
11
5
40
19
-29
6
30
13
31
14
-1
7
15
7
5
2
10
8
16
8
32
15
-16
9
25
10
39
18
-14
10
9
4
27
11
-18
Sum
77
133
Note: The two rank sums were not in the original problem.
a. On the basis of the rank sums (and assuming that the data represents two independent samples,) test the
hypothesis that the median number of beds unoccupied differs for the two hospitals. (5)
b. A statistician claims that her method is inappropriate because she has ignored the fact that each row
corresponds to a specific day, even if the days were chosen randomly. Rerank the data to account for the
fact that it is cross-classified and repeat the analysis. (5)
c. Assume instead that this data is normally distributed paired data, test the hypothesis that the mean
number of beds unoccupied differs. You may need some of the following data: x1  17 .0, s1  11 .04 ,
x 2  29 .1, s 2  11 .86 , d  12.1 and s d  17 .24 . (4)
Solution: a) Wilcoxon-Mann-Whitney Method H 0 :  1   2
we get TL  77 and TU  133 . Check: 77  133  210 
H 1 :  1   2 . If we sum the two rankings,
20 21
. For a 5% two-tailed test with
2
n1  n 2  10 , Table 6 says that the lower critical value is 79. The lower of the two rank sums, W  77 is
less than this value, so reject H 0 .
b. Wilcoxon Signed rank test for paired data. H 0 :  1   2 H 1 :  1   2 .
difference rank
-28
8 If we add items with + and – signs separately, we
10
3.5+
find T   46, T   9 . To check this, compute
-39
10 4
2 +
T   T   9  46  55  10 11 . From
-29
9 2
-1
1 Table 7 with n  10 , TL   TL .025  8 , and
2
10
3.5+
since 9, the smaller T is above the critical value,
-16
6 -14
5 do not reject H 0 .
-18
7 -
10
3/22/00 252y0021
c. Test of equality of means for paired data.
H 0 :   0
H :    2
H :    2  0
or  0 1
  1   2 or  0 1

H1 :   0
H 1 :  1   2
H 1 :  1   2  0
s
17 .24
9
sd  d 
 5.4518 , DF  n  1  9, t .025
 2.262
n
10
Test Ratio: t 
d   0  12 .1  0

 2.219
sd
5.4518
d  12.1, s d  17.24,
This is on the interval
between –2.262 and +2.262.
or Critical Value: d cv   0  t s d  0  2.262 5.4518   12.3319
2
d  12.1 is on this interval.
or Confidence Interval:   d  t  2 s d  12 .1  12 .3319  or –24.43
to 0.23. This interval includes 0.
With all methods do not reject H 0 . Note that this method is more powerful than the one in c.
However, it still should not be used unless the conditions justify it.
11
3/22/00 252y0021
Extra credit. Repeat the following sections of problem III 1) assuming that  1   2 . . Test the hypothesis
that the mean time to mount a mower is greater for method two than method one.
b) Find a critical value appropriate for this problem, using a confidence level of 90%(4)
c) Use your critical value to test the hypothesis. State clearly whether you reject the null
hypothesis. (2)
d) Repeat the test using (i) a test ratio (2) and (ii) a confidence interval. (2)
Solution: b) From the formula table;
Interval for
Confidence
Interval
Hypotheses
Test Ratio
  d  t  sd
H 0 :   0
H 1:   0
t
Difference
between Two
Means(
unknown,
variances
assumed
unequal)
2
s12 s22

n1 n2
sd 
DF 
 s12 s22 
  
n

 1 n2 
  1   2
Same as
H 0: 1   2
2
2
n1
n1  1
s 22
2
n2
n2  1
  .10 , x1  4.1 and s1  2.915476
d cv   0  t  2 sd
d  0
sd
H 1:  1   2
   
s12
Critical Value
s
2
1
if  0  0

 8.5 , x 2  5.3 and s 2  2.097618
s
2
1

 4.4 . n1  5,
 H 0 : 1   2
 H 0 : 1   2  0
H :   0
or 
or  0
where   1   2 .
n 2  6. . 
 H 1 : 1   2
 H 1 : 1   2  0
H 1 :   0
s12 8.5

 1.70000
n1
5
s 22 4.4

 0.73333
n2
6
sd 
s12 s 22

 2.43333  1.55991
n1 n 2
d  x1  x 2  4.1  5.3  1.2
s12 s 22

 2.43333
n1 n 2
DF 
 s12 s 22 



 n1 n 2 


2
2
2
 
 
 
 
 n1 
 n2 
 
 

n1  1
n2 1
s12
s 22

2.43333 2
1.70000 2  0.73333 2
4
7
 7.13 , so use 7 degrees of freedom. t .10
 1.415
5
Critical Value: d cv   0  t  2 sd is the formula for a 2-sided critical value, but since our alternative
hypothesis is H 1 :   0 , we want a critical value below zero.
So d cv   0  t s d  0  1.415 1.55991   2.2073
c) d  1.20 is above this value, so we do not reject H 0 .
d) (i) Test Ratio: t 
we do not reject H 0 .
12
d   0  1.20  0
7

 0.7693 . Since this is above  t .10
 1.415 ,
sd
1.55991
3/22/00 252y0021
(ii) Confidence Interval: The 2-sided interval is   d  t sd , but since our alternative
2
hypothesis is H 1 :   0 , we use   d  t s d  1.20  1.415 1.55991   1.20  2.2073  1.0073 .
Since   1.0073 does not contradict H 0 :   0 , we do not reject H 0 .
13