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ECO252 QBA2
THIRD HOUR EXAM
April 20, 1999
Name
Hour of Class Registered (Circle)
MWF TR 10 12 12:30 2:00 night
I. (10+ points) Do all the following;
1. Hand in your computer printouts for problems 2 and 3.(4 points – 3 point penalty for not handing in)
2. Do not do the following unless you handed in both outputs.
On the next few pages there are problems very much like the ones you did.
a. The regression relates automobile accidents (‘deaths’) to tobacco use in pounds per year.
Identify the coefficient of ’t-cons’ in the equation and explain whether it is significant at the 5% level.(2)
Does this show that tobacco consumption causes automobile accidents? Explain! (1)
b. Complete the table in the ANOVA that compares the effect of packaging and advertising plans
on the sales of a product. Is there a difference between mean sales for the advertising plans at the 5% level?
Show what numbers brought you to your conclusion. (4)
Solution: a. On the previous page 4 it says
“The regression equation is
Deaths = -42027 + 11317 t-cons
Predictor
Coef
Constant -42027
t-cons
11317
s = 3227
Stdev
19970
2603
R-sq=79.1%
t-ratio
-2.10
4.35
p
0.087
0.007
Rsq(adj) = 74.9% ”
H 0 :  1  0
To test the significance of 11317,the coefficient of ’t-cons,’ we test the hypotheses 
. If H 0 is
H 1 :  1  0
false we say that 1 is significant. We can test this coefficient three ways: (i) Note that its p-value is 0.007
5
which is less than our significance level of 5% ;(ii) note that the t-ratio is outside the range  t n2 k 1  t .025
 2.571 (Degrees of freedom can also be read from the Error or Within line in the regression analysis of
variance.) ; (iii) Look at the F-test. In each case we reject H 0 and say that 1 is significant. A regression
does not prove cause so that we cannot say that significance proves that tobacco consumption causes
automobile accidents.
b. On the previous page 3 it says (with the numbers I added in boldface)
“ Source DF
SS
MS
F
F.05
package
2
71.30
35.65
adplan
3
323.90
107.97
Interaction 6
Error
36
Total
47
8.01
45.93
449.13
1.33
1.28
F 2,36  3.26
3,36  2.87
84.35 F
6,36  2.36
1.04 F
27.85
“
H 0 : No difference between adplan means
In this case we are testing 
. If we look at the F in the adplan
H 1 : Some difference between adplan means
row, we find that it is larger than the table F and we reject H 0 .
The rule on p-value:
If the p-value is less than the significance level   reject the null
hypothesis; if the p-value is greater or equal than the significance
level, do not reject the null hypothesis.
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II. Do at least 4 of the following 6 Problems (at least 10 each) (or do sections adding to at least 40 points Anything extra you do helps, and grades wrap around) . Show your work! State H 0 and H1 where
applicable.
1. a. In the regression output supplied with this exam.
(i)
Add a regression line to the graph. (1)
(ii)
Do a 90% confidence interval for the constant in the equation. (2)
(iii)
Assuming that there is some sort of valid relationship, what automobile accident rate
would you predict for a year in which per capita tobacco consumption was 10 pounds? (2)
b. In the analysis of variance supplied with this exam.
(i)
Test for significant interaction – explain your conclusion. Use a 90% confidence level. (2)
(ii)
Do a 95% confidence interval for the difference between the means of package 1 and 3
that is
  Valid when used alone. (2)
  Valid when used with other possible differences between means. (2)
Solution: a. (i) just connect the x’s on former page 5.
5
 2.015 ( k  1 is the number of independent variables.)
(ii)   .10 , t n2 k 1  t n2 2  t .05
 0  b0  t n2 k 1 s b0  42027  2.015 19970   42027  40240 or –82267 to –1787.
b.
(iii) Deaths = -42027 + 11317 t-cons = -42027 + 11317(10) = 71143.
(i)   .10 From the previous page. (New F’s provided)
“ Source
DF
SS
MS
F
F.10
package
2
adplan
3
Interaction 6
Error
36
Total
47
F 2,36  2.47
3,36  2.26
323.90 107.97 84.35 F
6,36  1.96
8.01
1.33
1.04 F
71.30
35.65
45.93
449.13
1.28
27.85
“
All F values are approximate and come from the table on page 1020 of the text. Since the F for
interaction is less than the table value, accept H 0 : No interaction .
(ii) We have C  4 columns, R  3 rows, and P  4 observations per cell. From the outline, for
individual row means use Bonferroni intervals (with m  1 )
1   3  x1  x 3   t RC P 1
2MSW
. As explained in class, the degrees of freedom for the t
PC
statistic are the Error (Within) degrees of freedom, so we want t 36  2.02 . From the table of
2m
.025
means that appeared above the ANOVA table, the mean for package 1 is 28.700 and the mean for
package 3 is 25.737. From the table above MSW  MSE  1.28 . Putting this together
21.28 
 2.963  2.02 0.400   3.0  0.8
44
For simultaneously valid row means, use the Scheffe’ interval
2MSW
1   3  x1  x 3   R  1FR 1, RC P 1
. As explained in class, this amounts to
PC
1   3  28.700  25.737   2.02 
2,36  23.26   2.553
36
 2.02 by 2 F.05
replacing t .025
, where the degrees of freedom are those
.
used in the F-test for rows above. So 1   3  2.963  2.553 0.400   3.0  1.0 .
2
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2. Three new employees are to be evaluated by the partners in an accounting firm for the number of errors
that they make in each of six statements of varying difficulty. The data appears below. Assuming that the
underlying distribution is normal, and noting that it is blocked (classified) by the statement number,
x1  10 ,
x12  26 ,
x 2  19,
x 22  75,
compare mean error rates. (13) Note:
x
3
 23,


x 32



 ? . Note also: If you wish to ignore that the data is blocked by statement, indicate this
now and compare the column means assuming that the data is three random samples from a normal
distribution.(10).
x1
x2
x3
1st
employee
2
1
0
4
2
1
statement
1
2
3
4
5
6
2nd
employee
2
3
1
6
3
4
3rd
employee
3
4
4
5
4
3
Solution: a) 2-way ANOVA (Blocked by statement) ‘s’ indicates that the null hypothesis is rejected.
Statement
Employee
Sum
SS
ni
x i.
x
x1
x2
x3
2
1
0
4
2
1
10
2
3
1
6
3
4
+ 19
3
4
4
5
4
3
+ 23
7
8
5
15
9
8
= 52 
nj
6
+6
+6
= 18  n
x j
1.6667
3.1667
3.8333
SS
26
+ 75
+ 91
2.8889  x
192   xij2
x 2j
2.7778
+ 10.0278
+ 14.6944
= 27.50 
1
2
3
4
5
6
Sum
Note that x is not a sum, but is
 n x
SSR   n x
SSC 
n
 x
i..
2
ij
 x
3
3
3
3
3
3
18
2.3333
2.6667
1.6667
5.0000
3.0000
2.6667
2.8889
17
26
17
77
29
26
192

x
5.4444
7.1111
2.7778
25.0000
9.0000
7.1111
56.4444
 x
2
ij

x
2
i.
2
j
 n x  192  18 2.8889 2  41 .7778 .
2
 n x  627 .5  182.8889 2  14 .7778 . This is SSB in a one way ANOVA.
2
2
j j
2
i i.
 x . SST 
x i2.
 n x  356 .4444   182.8889 2  19 .1111 ( SSW  SST  SSC  SSR  7.8889 )
2
Source
SS
DF
MS
F
F.05
Rows (Statement)
19.1111
5
3.8222
4.845
14.7778
2
7.3889
9.367
F 5,10  3.33 s
F 2,10  4.10 s
Columns(Employees)
Within (Error)
7.8889
10
Total
41.7778
17
b) One way ANOVA (Not blocked by statement)
Source
SS
DF
H0
Row means equal
Column means equal
0.7889
( SSW  SST  SSB  7.8889 )
MS
F.05
F
Columns(Employees)
14.7778
2
7.3889
Within (Error)
Total
27.0000
41.7778
15
17
1.8000
4.105
F 2,15  3.68 s
H0
Column means equal
3
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3. Data from problem 2 is repeated below.
x1
x2
x3
1st
employee
2
1
0
4
2
1
3rd
employee
3
4
4
5
4
3
statement
1
2
3
4
5
6
2nd
employee
2
3
1
6
3
4
a) Assume that the distribution is not normal, but that it is blocked (classified) by statement number, and
again compare the distributions represented by the columns. (5)
b) Assume that the distribution is not normal, but that each column is a random sample, and again compare
the distributions represented by the columns. (5)
Solution: a) Friedman Test H 0 : Columns from same distribution . Rank within rows.
Statement
Employee
x1
r1
x2
r2
x3
r3
1
2
1.5
2
1.5
3
3
2
1
2
3
2
4
3
3
0
2
1
2
4
3
4
4
3
6
3
5
2
5
2
2
3
2
4
3
6
1
3
4
3
3
2
Sum
6.5
13.5
16
cc  1
34
6
 36 . Since
There are r  6 rows and c  3 columns. Check: the rank sums must add to r
2
2
12
6.5 + 13.5 + 16 = 36, we are all right. The Friedman Statistic is  F2 
SR 2  3r c  1
r c c  1
12
1

6.5 2  13.5 2  16 2  364  480 .5  72  8.0833 According to the Friedman Table
634
6
( k  3, n  6 ) 7 has a p-value of .029 and 8.333 has a p-value of .012, so this p-value must lie in between. If
  .05, the p-value for our null hypothesis must be below .05, so we reject H 0 .
 


b) Kruskal-Wallis Test H 0 : Columns from same distribution Rank within entire group (1 to 18). Then
resolve ties by replacing ranks with average ranks as follows: x  1 is rank 2, 3 and 4, replace with 3; x  2
is rank 5, 6 and 7, replace with 6; x  3 is rank 8, 9, 10 and 11, replace with 9.5; x  4 is rank 12, 13, 14, 15
and 16, replace with 14. These revised ranks are r *
Statement
Employee
x1
x2
x3
r1 r1*
r2 r2*
r3 r3*
1
2
5
6
2
7
6
3
10 9.5
2
1
2
3
3
8
9.5
4
14 14
3
0
1
1
1
4
3
4
15 14
4
4
12 14
6
18 18
5
17 17
5
2
6
6
3
9
9.5
4
16 14
6
1
3
3
4
13 14
3
11 9.5
Sum
33
60
78
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4
nn  1 1819 

 171 , as 33, 60
2
2
 SRi2 

  3n  1
 ni 


Check: If there are n  18 numbers, the three sums of ranks should add to
and 78 do. The Kruskal-Wallis statistic has the formula H 

12
nn  1

i
12  33
60
78 
12 10773 


 57  6 .We cannot use the Kruskall-=Wallis table

  319  
1819   6
6
6 
1819   6 
2
2
2
because the problem is too large, so use a 5% value of  2 with 2 degrees of freedom. The table value is
5.991, so (barely) reject H 0 .
Document continues in 252z9931.
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