4/14/98 252y9831
ECO252 QBA2
THIRD HOUR EXAM
April 16, 1998
Name
Hour of Class Registered (Circle)
MWF 10 11 TR 12:30 2:00
Hour of Class Attended (If Different)
______________
I. (16 points) Do all the following.
1. Hand in your computer printouts.(6)
2. Do not do the following unless you handed in all three outputs:
year
1
2
3
4
5
6
7
8
visitors
4068
3934
3931
4144
4780
5760
6457
5126
Fper$
4.2566
4.2250
5.4396
6.5793
7.6203
8.7355
8.9799
6.9256
The material at the left is the number of visitors
(in thousands) and the francs per dollar exchange
rate for a sample of years. Two Minitab
procedures were run on the data.
There is no need for you to do any calculations
from the original data.
a)My first test read as follows:
MTB  ttest mu=5000 'visitors'
SUBC alt=-1.
TEST OF MU=5000 VS MU L.T. 5000
N
MEAN STDEV SE MEAN
visitors 8
4775
944
334
T
-0.67
P VALUE
0.26
In the language used in class:
(i) What are H 0 and H1 (1)
(Do not do any unnecessary calculations)
(ii) If  .05 do we accept H 0 ? Why? (1.5)
(iii) Sketch a diagram of the t-distribution to show the
meaning of the p-value (2)
 H 0 :   5000
Solution: (i) 
 H1 :   5000
(ii) Accept H 0 if pval   so accept H 0 or accept H 0 if t lies above t  , so accept H 0 .
(iii) Note that a p-value is a probability, not just a point on the x axis.
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b)My next test read as follows:
MTB  regress 'visitors' on 1 'Fper$' 'resid' 'pred'
The regression equation is
visitors = 1768 + 456 Fper$
Predictor
Constant
Fper$
s = 462.6
Coef Stdev
1768.5 645.7
455.86 94.72
R-sq = 79.4%
t-ratio
p
2.74 0.034
4.81 0.003
R-sq (adj) = 76.0%
Analysis of Variance
SOURCE
Regression
Error
Total
DF
1
6
7
SS
4956347
1283854
6240202
MS
4956347
213976
F
p
23.16 0.003
i) If the exchange rate is 5 Francs per dollar, how many visitors can Europe expect? (1.5)
ii) Is the regression coefficient of Fper$ significant at the 1% level? Why? (2)
iii) What two items on the printout could be used to compute R-sq? (2)
Solution: (i) visitors = 1768 + 456(5)=4048.
H0 :  1  0
 6
1,6
(ii) 
Accept H 0 if t lies between t .005 or if p-value is above .01 or if F lies below F.01 .
H
:


0
1
1

Since these are all false we reject H 0 and say that  1 is significant.
RSS 4956347
(iii) R 2 

.794
TSS 6240202
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4/14/98 252y9831
III. Do at least 2 of the following 3 Problems (at least 12 each) (or do sections adding to at least 24 points Anything extra you do helps, and grades wrap around) . Show your work! State H 0 and H1 where
applicable. You must do 1a!
1.
A researcher interviews 3 independent samples of 5 individuals of different ages and asks them to
name the seven dwarfs. The amount of time in seconds for the three samples is below.
For your convenience the following facts are given:
x
3
 350,
x
2
3
x
1
 208,
x
2
1
 ?,
x
2
 443,
x
2
2
 47029,
 26300 .
a) Find the sample variance of x1 (2)
b)Do a test of the equality of the means at a significance level of 5% . (Assume that the underlying
distributions are normal) (10)
c) Do a confidence interval for the difference between the means of the second and third populations on the
assumption that this interval is one of the three possible contrasts. (3)
2
x12  nx12 9194  5 414
. 
a)
s x21 

 135.3
n 1
4
x1
x2
x3
x12
people
people
people
under 12
12-25
over 25
60
120
75
3600
45
65
45
2025
38
60
75
1444
30
48
55
900
35
150
100
1225

Sum
 x  1001
208
443
350
b)
nj
5
5
5
x. j
41.6
88.6
70.0
x
SS
9194
47029
26300
82523 
1730.56
7849.96
4900
x 
2
.j
2
9194 
x
n  15
x
n  66.7333
 x
14480.47    x 
2
2
.j
  x  n x  82523  15 66.7333  15722.9333
SSB   n  x   n x  5  x   15 x  514480.47  15 66.7333
SST 
2
2
j
2
2
.j
2
2
.j
H 0 :  1   2   3 H 1: Not all means equal.
Source
Between
Within
Total
2
1
SS
5602.282
10120.65
15722.933
 2 ,12 
Since F  F.05 , accept H 0 .
DF
2
12
14
MS
2801.1415
843.3875
F
3.321
2
 5602.283
 2 ,12 
F.05
3.89 NS
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c)  2   3  x 2  x 3 
 1
1
 1 1
    88.6  70.0  2 389
.  843.3875   
 5 5
 n1 n2 
 m  1 F m1,nm s
 8.6  2624.6219  8.6  5123
.
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2. a) Data from 1) is repeated below. Assuming that the underlying distributions are not normal, do a test of
the equality of medians. (6)
x1
x2
x3
r1
r2
r3
people
people
people
under 12
12-25
over 25
60
120
75
9
14
11
45
65
45
4
10
5
38
60
75
3
8
12
30
48
55
1
6
7
35
150
100
2
15
13
19
53.5
47.5
b) Complete the following ANOVA table for three speeds, four temperatures and three measurements per
cell. Name the three tests and their conclusions (5).
SOURCE
speed
temperature
interaction
within
total
SS
DF
227.06
113.44
142.06
MS
F
683.22
c) The author of the text mentions a randomized block experiment with 5 treatments and six blocks for
which SSTR = 287, SST = 1446, and SSE = 180. Is there a difference between treatment means? (4)
Solution: a) H 0 :  1   2   3
H 1: Not all medians equal.
check: 19  535
.  47.5 
 120
H
1516
2
12  19 2 535.  2  47.5 2 
 5479.5


 316   .05 
  48  6.795


 5 
5
5 
1516  5
Since the Kruskal-Wallis
table says that 5.78 is the 4.9% value and 7.98 is the 1% value, the p-value is below 5% and we reject H 0 .
b) Total degrees of Freedom are (3)(4)(3)-1 = 36 - 1 = 35. Degrees of freedom for speed are 3-1=2.
Degrees of Freedom for Temperature are 4 - 1 = 3. Degrees of Freedom for Interaction are (2)(3) = 6.
SOURCE
SS
DF
MS
F
F.05
 2 ,24 
F.05
 3.40 Speed means not equal
speed
227.06 2
113.53 13.57
 3,24 
F.05
 3.01 Temperature means not equal
temperature
113.44 3
37.813 4.52
 6,24 
F.05
 2.51 Interaction is present
interaction
142.06 6
23.68
2.83
within
200.66 24
8.3608
total
683.22 35
c)
F.05
SOURCE
SS
DF
MS
F
 4 ,20
F.05
 2.71 S
Treatments
287
4
71.75 7.97
5,20
F.05  2.60 S
Blocks
979
5
195.8 21.76
Error
180
20
9.0
Total
1446
29
The F test shows a difference between both block and treatment means.
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3. The following data represents gross farm income and value of land and buildings (in billions of dollars)
on farms in a random sample of 11 states.
y
income
2.6
1.8
3.9
15.4
3.7
0.4
0.5
4.9
4.1
2.6
8.1
x
value
9.3
11.1
15.0
63.3
16.2
1.4
1.2
20.9
12.3
12.0
51.0
For your convenience the following values are
given:
 x  213.7,  x  8040.53,
 y  48.0,  y  389.66,
 xy  1735.72, n  11
2
2
213.7
 19.4273
11
48.0
y
 4.3636
11
x
a. Compute the regression equation
Y  b  b x (6)
Spare Parts:
b. Compute R 2 . (4)
c. Compute s e . (3)
 3888.92
0
x
1
2
 nx 2  8040.53  1119.4273
2
 xy  nxy  1735.72  1119.4273 4.3676
d. Compute sb1 and do a significance test on b1
 803.21
y
.(4)
e. Do a prediction interval for Y when x  5 .
(5)
2
 ny 2  389.66  11 4.3636
2
 180.21  TSS
Solution:
a. b1 
 xy  nxy
 x  nx
2
2
803.21
 0.2065
3888.92

b0  y  b1 x  4.3636   0.206519.4272  0.3518
Y  b0  b1 x becomes Y  0.35180  0.2065x .
b. R 2 
b1
 xy  nxy   0.2065803.21  RSS  0.9206 or
180.21
TSS
 y  ny
 xy  nxy 
803.21


.9206
 x  nx  y  ny   3888.92180.21
 y  ny   b  xy  nxy   180.21  0.2065803.21  ESS  1594
s 
.
where
2
2
2
R
2
2
2
2
c.
2
2
2
2
2
e
ESS  TSS  RSS or
1
n2
se2
 y

n2
9
2
 ny
2
n2
1  R  or s   y
2
2
e
2
  x
 ny 2  b12
2
 nx 2

n2
6
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se  se2  1594
.
 1263
.
d. sb21 

se2
x  nx
2
2


1594
.
 0.00041
3888.92
H 0 :  1  0 H 1:  1  0
t
sb1  0.00041  0.0205
b1   10 b1  0 0.2065
. Since this is not between


 101975
.
sb1
sb1
.0205
n2
9
 t .025   t .025  2.262 , reject H 0 and conclude that  1 is significant.
e. If Y0  0.35180  0.2065x0 and x 0  5 , then Y0  0.35180  0.20655  1385
.

1
s 2y  se2  
0
n

 x0  x  2

x 2  nx 2


 1  5  19.4273 2


 1  1594
.  
 1  1824
.
 11

3888.92





s y0  1824
.
 1351
.
.
.
  2.2621351
.   138
.  3.06 .
So y 0  y 0  t s y0  1385
7