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ECO252 QBA2
Final EXAM
December , 2006
Version 1
Name and Class hour:____KEY________________
I. (18+ points) Do all the following. Note that answers without reasons and citation of appropriate
statistical tests receive no credit. Most answers require a statistical test, that is, stating or implying a
hypothesis and showing why it is true or false by citing a table value or a p-value. If you haven’t done it
lately, take a fast look at ECO 252 - Things That You Should Never Do on a Statistics Exam (or Anywhere
Else)
Since I will use this as an example of approaches to model building I will include general rules in my
solutions.
Regression A seeks to explain the selling price of a home in terms of a group of variables explained on the
output sheet. Note that regressions 1 and 7 are identical. Look at the definitions of the variables carefully
and, in particular, notice which are interaction variables.
a) The homes in this regression are in three different areas. There are dummy variables to indicate that the
homes are in Area 1 or Area 2. Why isn’t there a dummy variable for Area 3? (1) A model with a set of
dummy variables X d .1 , X d .2 ,  X d .k 1 , X d .k is overdetermined if X d .k  1  X d .1  X d .3   X d .k 1 .
In practice this means if we have say 5 categories we must let one be a ‘control group’ and not be
identified by a dummy variable.
b) In Regression 1, what coefficients are significant at the 5% level? (2) A coefficient,  i is insignificant
if the null hypothesis H 0 :  i  0 cannot be shown to be false at a reasonable significance level
(Usually   .05 or   .01 ). In practice this means that the t-ratio t 



bi
is not between t  or the
2
s bi

pvalue  2P t  t computed or, if the t-ratio is negative, pvalue  2P t  t computed is below a reasonable
significance level.
c) What independent variables did I remove from the problem to get to Regression 2 from Regression 1?
Why? (2) The best subsets approach, according to Behrenson et. al., involves: (i) Choosing a large set
of candidate independent variables; (ii) running a regression with all the candidate variables and
using the VIF option, which tests for collinearity; (iii) eliminating variables with a VIF over 5; (iv)
continuing to run regressions and eliminate candidate variables until there are no variables with a
VIF over 5; (v) performing a best-subsets regression on the model without high VIFs and computing
C p ; (vi) shortlisting the models with a C p less than or close to k  1 , where k is the number of
independent variables in that regression; (vii) choosing from the shortlist on the basis of things like
significance of coefficients and R-squared; (viii) using residual analysis and influence analysis to
further refine the model by adding nonlinear terms, transforming variables and eliminating
suspicious observations. Note that terms like a squared term are largely exempt from the VIF rules, if
they are correlated with the untransformed variable.
d) Following the same process, I went on to remove one or more variables each time until I got to
Regression 5. When I got to Regression 5 I ran the ‘best subsets regression.’ 6. I concluded that it was time
to quit removing variables. Between the best subsets regression and the characteristics of the coefficients of
the results in Regression 5 I felt that I had gone as far as was reasonable in removing independent variables.
What are the three things that led me to think that regression 5 was the best that I could do? (3)
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e) Using Regression 5 and assuming that all homes have two baths, Regression 5 effectively becomes 3
regressions relating price to living area. Take the coefficient of bath, multiply it by two and add it to the
constant to get the effective intercept for homes with two baths. Using L or any other symbol that you find
convenient for living area, what are the equations relating living area to price in (3 points)
Area 1?
Area 2?
Area 3?
[11]
f) Continuing with Regression 5 and assuming that a home has 2(thousand) square feet of living area and 2
baths, what would it sell for in
Area 1?
Area 2?
Area 3?
What is the percent difference between the lowest and highest price? (2)
g) We have not yet dealt with the question of whether the coefficients in Regression 5 are reasonable. In
order to do this look at two homes in Area 1 that have two baths. If one has 2(thousand) square feet of
living area and the other 3, how would there prices differ? Does that seem reasonable? Try the same for a
home in area 3. (3)
[16]
h) As I warned you, I now repeated Regression 1 as Regression 7, without using the VIFs. Much to my
surprise, I ended up dropping the same variables as I did after Regression 1. Why? (1) This is a rather
outdated technique of eliminating insignificant coefficients and worrying about R-squared and
adjusted R-squared.
i) Continuing in the same way, I worked myself to Regression 9. Looking at the things I usually check, this
looked pretty good. Then I tried to check the coefficients in the same way that I did in g). Why was I very
unhappy? What is there in Regression 8 that could explain these results? (4)
j) Regression 11 is a stepwise regression. The printout, which continues on page 7, presents four different
possible regressions in column form. In each case a coefficient has a t-value under it and a p-value for a
significance test. After the fourth try, the computer refused to add any more independent variables. The only
regression here that I thought was worth looking at was the one with four independent variables. What can
you tell me about its acceptability? (3) Stepwise regression adds independent variables in sequence,
starting with the one with the highest correlation with the dependent variable. It then adds or
removes variables using an F test for significant improvement and picking, at first, independent
variables that provide the lowest p-values for the test. When it can no longer find new variables that
provide an F-ratio with a p-value below a pre-specified significance level, it quits.
k) Do an F test to compare regressions 2 and 3 and to find out if lot 1 and lot 2 have any explanatory power.
(3)
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II. Hand in your fourth computer problem. (2 to 7 points)
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III. Do at least 4 of the following 7 Problems (at least 12 each) (or do sections adding to at least 50 points –
(Anything extra you do helps, and grades wrap around). You must do parts a) and b) of problem 1. Show
your work! State H 0 and H1 where applicable. Use a significance level of 5% unless noted otherwise.
Do not answer questions without citing appropriate statistical tests – That is, explain your hypotheses
and what values from what table were used to test them. Clearly label what section of each problem
you are doing! The entire test has about 151 points, but 70 is considered a perfect score. Don’t waste
our time by telling me that two means, proportions, variances or medians don’t look the same to you.
You need statistical tests! There are two blank pages below.
1. a) If I want to test to see if the mean of x 2 is larger than the mean of x1 , my null hypotheses are:
(Note: D  1   2 )
i) 1   2 and D  0
ii) 1   2 and D  0
v) 1   2 and D  0
vi) 1   2 and D  0
iii) 1   2 and D  0
vii) 1   2 and D  0
iv) 1   2 and D  0
viii) 1   2 and D  0 (2)
Solution: Since this question will appear on future exams, it is left to the student.
The first two columns below represent times for 25 workers on an industrial task. The third column is the
difference between them
d
Row
x1
x2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
6.11
5.13
6.42
4.65
5.82
4.08
4.01
5.26
5.25
7.66
6.29
5.41
6.17
5.50
4.06
6.19
6.71
4.41
5.25
4.85
6.50
5.24
7.29
4.99
4.26
4.81
4.19
5.17
4.07
4.58
2.97
3.39
4.14
4.31
6.68
5.37
3.95
4.93
4.04
2.40
4.71
5.93
2.93
4.25
4.41
4.68
3.50
6.09
2.87
3.06
1.30
0.94
1.25
0.58
1.24
1.11
0.62
1.12
0.94
0.98
0.92
1.46
1.24
1.46
1.66
1.48
0.78
1.48
1.00
0.44
1.82
1.74
1.20
2.12
1.20
Assume that   .05 . Minitab gives us the following summary (edited).
Descriptive Statistics: x1, x2, d
Variable
x1
x2
d
N
25
25
25
N*
0
0
0
Mean
5.50
4.30
1.20
SE Mean
0.200
0.212
…………
StDev
1.00
1.06
………
Minimum
4.010
2.400
0.4400
Q1
4.750
3.445
0.9400
Median
Q3 Maximum
5.260 6.240 7.660
4.250 4.870 6.680
1.200 1.4700 2.120
In the d column, the column sum is 30.08 and the sum of the first 24 numbers squared is 38.585. Do not
recompute things that have been done for you if you want to ever get much done on this exam.
Clearly label parts b, c, d etc. The null hypothesis is the same for parts c, d and e, so state it clearly.
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b). Find the sample variance for the d column. (2) Solution: 38 .585  1.20 2  40 .025 . The last line of
the output above should read d
25
0 1.20
0.080 0.400
0.4400 etc.
d
40 .025  251.20 2
 0.16771 , s  0.409 .
n 1
24
The formula table has the following for the difference between two means.
Interval for
Confidence
Hypotheses
Test Ratio
Interval
s d2 
2
D1: Difference
between Two
Means (
known)
d 2

D  d z 2  d
 12
d 

n1
 22
n2
H 0 : D  D0 *
z
H 1 : D  D0 ,
D  1   2
d  D0
d
Critical Value
d cv  D0  z d
d  x1  x 2
D2: Difference
between Two
Means (
unknown,
variances
assumed equal)
D  d  t 2 s d
D3: Difference
between Two
Means(
unknown,
variances
assumed
unequal)
D  d  t 2 s d
sd  s p
1 1

n1 n2
t
H 1 : D  D0 ,
D  1   2
sˆ 2p 
d  D0
sd
d cv  D0  t  2 s d
n1  1s12  n2  1s22
n1  n2  2
DF  n1  n2  2
DF 
H 0 : D  D0 *
s12 s22

n1 n2
sd 
 s12 s22 
  
n

 1 n2 
H 1 : D  D0 ,
t
d  D0
sd
d cv  D0  t  2 s d
t
d  D0
sd
d cv  D0 t  2 s d
D  1   2
2
   
s12
2
n1
n1  1
D4: Difference
between Two
Means (paired
data.)
H 0 : D  D0 *
s 22
2
n2
n2  1
D  d t  2 s d
H 0 : D  D0 *
H 1 : D  D0 ,
d  x1  x 2
D  1   2
s
df  n  1 where
sd  d
n
n1  n 2  n
We have the following facts - n  n1  n2  25 , x1  5.50, s1  1.00 , s x1  0.200 , x 2  4.30, s 2  1.06 ,
s d2
0.16771

 0.08190
n
25
n
c) On the assumption that the underlying distributions are Normal and that the first two columns represent
independent samples from populations that represent plants 1 and 2 and come from populations with similar
variances, can we conclude that average workers in plant 2 complete the task faster than those in plant 1?
(4)
 H 0 : 1   2
H 0 : D  0
Solution: Our hypotheses are 
or if D  1   2 
. This is a right side test. If we
H
:



2
 1 1
H 1 : D  0
s x2  0.212 , d  5.50  4.30  1.20, s d  0.409 , s d 
sd

do a critical value for d , we need a d cv above zero. If we use a test ratio it will be on the right side of the t
distribution, above zero. df  n1  n 2  2  25  25  2  48
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Because we have assumed equal variances, we use method D2. df  n1  n 2  2  25  25  2  48
sˆ 2p 
n1  1s12  n2  1s 22
s d  sˆ p
t
n1  n 2  2
 
24 1.00 2  24 1.06 2
24  24  2

1.0000  1.1236
 1.0618
2
1
1
2

 1.0618
 0.084944  0.2915 . If we use a test ratio, we compute
n1 n 2
25
d  D0 1.20  0
48
48
 1.677 . and t .025
 2.011. Because this is a 1-sided test, we reject

 4.1166 . t .05
sd
0.2915
the null hypothesis if the t ratio is above 1.677. It is, so we reject H 0 and conclude that workers in plant 2
are faster. If we use a critical value for d , we use d cv  D0  t s d  0  1.677 0.2915   0.489 . Since
d  1.20 is above 0.489, we reject the null hypothesis.
d) (Extra credit) Repeat part c) after dropping the assumption that the variances are similar. (5)


  s2 s2 2
  1  2 
  n1 n 2 
Solution: If we drop the equal variance assumption, we use Method D3. Here df  
2
2
  s2 
 s 22 
 
 1 
 n2 
  n1 
 


n2 1
 n1  1
We have s x1  0.200 and s x2  0.212 , 20. So





 .





s12
s2
 0.200 2  0.04000 , 1  0.212 2  0.04494 and
n1
n1
s12 s 22
0.08494 2

 0.200 2  0.212 2  0.04000  0.04494  0.08494 . So df 
n1 n 2
0.04000 2 0.04494 2

24
24
s2 s2
0.0072148
0.1731552

 47 .838 s d  1  2  0.08494  0.2915 . If we use a test
0.0016000  0.0020196
n1 n 2
0.0036196
24
d  D0 1.20  0
47 
47 
 1.678 . and t .025
 2.012. Because this is a 1ratio, we compute t 

 4.1166 . t .05
sd
0.2915

sided test, we reject the null hypothesis if the t ratio is above 1.678. It is, so we reject H 0 and conclude
that workers in plant 2 are faster. If we use a critical value for d , we want
d cv  D0  t s d  0  1.678 0.2915   0.489 . Since d  1.20 is above 0.489, we reject the null
hypothesis.
e) Actually, these data supposedly represent performance of a single sample of 25 workers on two
administrations of a standard test of manual dexterity. The question was ‘Did the time for the test improve
between the first and second administration?’ (3) These are now paired data and we use method D4.
d  1.20, s d  0.409 , s d 
sd
n
use a test ratio, we compute t 

s d2
0.16771
24 
24 
 2.064 and t .05
 1.711.

 .08190 t .025
n
25
If we
d  D0 1.20  0

 14 .6520 . Because this is a 1-sided test, we reject the
sd
0.08190
null hypothesis if the t ratio is above 1.711. It is, so we reject H 0 and conclude that workers in plant 2 are
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252y0641 2/23/07
faster. If we use a critical value for d , we want d cv  D0  t s d  0  1.7110.0819   0.140 . Since
[11]
d  1.20 is above 0.140, we reject the null hypothesis.
f) Assume that the means above come from independent samples, but that the data represent samples for
populations with known population variances of 1.00 and 1.06. Test the null hypothesis that you used in
part c) and find an exact p-value. (3) We can now say that s d 
z
d  D0
d

 12
n1

 22
n2
 0.08494  0.2915 and that
1.20  0
 4.1166 . The p-value to four places is Pz  4.11  .5  .5  0 .
0.2915
[14]
g) Using the value of s d that you used in e), make a confidence interval with a confidence level of 92%.
You must find the value of z  needed to do this first. Of course, it is not on the t-table. (2) [16]
2
Solution: The significance level is   1  .92  .08 , z   z.04 so for z make a diagram. Draw a Normal
2
curve with a mean at 0. z .04 is the value of z with 4% of the distribution above it. Since 100 – 4 = 96, it
is also the 96th percentile. Since 50% of the standardized Normal distribution is below zero, your diagram
should show that the probability between z .04 and zero is 96% - 50% = 46% or P0  z  z.04   .4600 .
The relevant part of the Normal table appears below.
0.00
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
z
1.5
1.6
1.7
1.8
1.9
0.4332
0.4452
0.4554
0.4641
0.4713
0.4345
0.4463
0.4564
0.4649
0.4719
0.4357
0.4474
0.4573
0.4656
0.4726
0.4370
0.4484
0.4582
0.4664
0.4732
0.4382
0.4495
0.4591
0.4671
0.4738
0.4394
0.4505
0.4599
0.4678
0.4744
0.4406
0.4515
0.4608
0.4686
0.4750
0.4418
0.4525
0.4616
0.4693
0.4756
0.4429
0.4535
0.4625
0.4699
0.4761
0.4441
0.4545
0.4633
0.4706
0.4767
The closest we can come to this is P0  z  1.75   .4599 . (1.76 is also acceptable here.) So z .04  1.75 . A
2-sided confidence interval for the difference between the means is
D  d  z  2  d  1.20  1.750.2915   1.20  0.51 or 0.69 to 1.71.
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2. Let us expand the problem of question 1 by adding another column. The full data set with lots done for
you looks like this. The first three columns represent the given data. In the next three columns I have take
the first three columns and squared them. I have added the first three rows to get the seventh column. I have
computed row means in the 9th column. The tenth column is a row sum of squares. In the 11 th to the 13th
columns the numbers in the first three columns are ranked from 1 to 75. Sums are provided for all 13
columns.
(1)
(2)
(3)
(4)
(5)
(6)
Row
x 1
x 2
x 3
x21
x22
x 23
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
x1
6.11
5.13
6.42
4.65
5.82
4.08
4.01
5.26
5.25
7.66
6.29
5.41
6.17
5.50
4.06
6.19
6.71
4.41
5.25
4.85
6.50
5.24
7.29
4.99
4.26
x2
4.81
4.19
5.17
4.07
4.58
2.97
3.39
4.14
4.31
6.68
5.37
3.95
4.93
4.04
2.40
4.71
5.93
2.93
4.25
4.41
4.68
3.50
6.09
2.87
3.06
x3
5.46
4.66
5.80
4.36
5.20
3.53
3.70
4.70
4.78
7.17
5.83
4.68
5.55
4.77
3.23
5.45
6.32
3.67
4.75
4.63
5.59
4.37
6.69
3.93
3.66
x1sq
37.3321
26.3169
41.2164
21.6225
33.8724
16.6464
16.0801
27.6676
27.5625
58.6756
39.5641
29.2681
38.0689
30.2500
16.4836
38.3161
45.0241
19.4481
27.5625
23.5225
42.2500
27.4576
53.1441
24.9001
18.1476
x2sq
23.1361
17.5561
26.7289
16.5649
20.9764
8.8209
11.4921
17.1396
18.5761
44.6224
28.8369
15.6025
24.3049
16.3216
5.7600
22.1841
35.1649
8.5849
18.0625
19.4481
21.9024
12.2500
37.0881
8.2369
9.3636
x3sq
29.8116
21.7156
33.6400
19.0096
27.0400
12.4609
13.6900
22.0900
22.8484
51.4089
33.9889
21.9024
30.8025
22.7529
10.4329
29.7025
39.9424
13.4689
22.5625
21.4369
31.2481
19.0969
44.7561
15.4449
13.3956
(2)
(3)
Sum (1)
(4)
(5)
(6)
(7)
x
i
rsum
16.38
13.98
17.39
13.08
15.60
10.58
11.10
14.10
14.34
21.51
17.49
14.04
16.65
14.31
9.69
16.35
18.96
11.01
14.25
13.89
16.77
13.11
20.07
11.79
10.98
(7)
(8)
(9)
(10)
x
2
i
x i
x i2
rmean
5.46
4.66
5.80
4.36
5.20
3.53
3.70
4.70
4.78
7.17
5.83
4.68
5.55
4.77
3.23
5.45
6.32
3.67
4.75
4.63
5.59
4.37
6.69
3.93
3.66
rmsq
29.8116
21.7156
33.6013
19.0096
27.0400
12.4374
13.6900
22.0900
22.8484
51.4089
33.9889
21.9024
30.8025
22.7529
10.4329
29.7025
39.9424
13.4689
22.5625
21.4369
31.2481
19.0969
44.7561
15.4449
13.3956
rssq
90.280
65.589
101.585
57.197
81.889
37.928
41.262
66.897
68.987
154.707
102.390
66.773
93.176
69.325
32.676
90.203
120.131
41.502
68.188
64.408
95.401
58.805
134.988
48.582
40.907
(8)
(9)
(10)
(11)
r1
rank1
63.0
44.0
68.0
31.0
59.0
19.0
15.0
50.0
48.5
75.0
66.0
52.0
64.0
55.0
17.0
65.0
72.0
27.5
48.5
41.0
69.0
47.0
74.0
43.0
23.0
(11)
(12) (13)
r2
r3
rank2
40.0
21.0
45.0
18.0
29.0
4.0
7.0
20.0
24.0
70.0
51.0
14.0
42.0
16.0
1.0
36.0
61.0
3.0
22.0
27.5
33.5
8.0
62.0
2.0
5.0
rank3
54.0
32.0
58.0
25.0
46.0
9.0
12.0
35.0
39.0
73.0
60.0
33.5
56.0
38.0
6.0
53.0
67.0
11.0
37.0
30.0
57.0
26.0
71.0
13.0
10.0
(12) (13)
The sums of the columns will not fit on the table so they are printed here.
(1)Sum of x1 = 137.51; (2)Sum of x2 = 107.43; (3)Sum of x3 = 122.48;
(4)Sum of x1sq = 780.400; (5)Sum of x2sq = 488.725; (6)Sum of x3sq = 624.649;
(7)rsum = 367.42; (8)Sum of rmean = 122.473; (9)Sum of rmsq = 624.587; (10)Sum of rssq = 1893.77;
(11)Sum of rank 1 = 1236.5; (12)Sum of rank 2 = 662; (13)Sum of rank 3 = 951.5. You are left to find
column means and the grand mean. Please avoid Recomputing stuff that I have done for you. Life is not
that long. You will need to get column and overall means. Almost everything else is done for you.
a) Consider the first three columns to be three independent random samples from Normal distributions with
similar variances. Compare the means using an appropriate statistical test or tests. (6)
Solution: Scary looking isn’t it. But almost all the real work has been done for you. In 252ANOVAex3 we
have the display on the next page that we can bend to our will.
8
252y0641 2/23/07
Tableau presented in 252ANOVAex3.
Individual
Display 1
Sum
ni
x i 
45
35
20
40
40
180
5
140
110
75
120
135
580
15
3
3
3
3
3
15
46.6667
36.6667
25.0000
40.0000
45.0000
(38.6667)
35
36
(38.6667)
x
10475
6375
6850
23700
2
 xijk
2025
1225
1296
4546
1
2
3
4
5
Sum
nj
Display 3
50
45
30
45
55
225
5
Display
2
45
30
25
35
40
175
5
45
x j 
SS
x j  2
n
x
SS
xi 2
6550
4150
1925
4850
6225
23700
2177.7778
1344.4444
625.0000
1600.0000
2025.0000
7772.2222
 x ij2
 x i 2
 x .2j .
Version for this problem. Items from the last page have been added.
Row
Column 1
Column 2
Column 3
Sum
1
2
…….
24
25
Sum
6.11
5.13
…………
4.99
4.26
137.51
25
4.81
4.19
…………
2.87
3.06
107.43
25
5.46
4.66
………….
3.93
3.66
122.48
25
16.38
13.98
............
11.79
10.98
367.42
75
3
3
………..
3
3
75
nj
ni
n
x j 
5.5004
4.2972
4.8992
(4.8989)
x
SS
780.400
488.725
624.649
1893.774
2
 xijk
18.4659
24.0022
72.7225
30.2544
x j  2
x i 
5.46
4.66
…………..
3.93
3.66
(4.8989)
x
SS
xi 2
90.280
65.598
……………
48.582
40.907
1893.77
29.8116
21.7156
……………..
15.4449
13.3956
624.587
 x ij2
 x i 2
 x .2j .
Almost everything was done for you on the last page. The new calculations are in boldface. Note that
5.5004  4.2972  4.8992 367 .42
x

 4.8989 and 754.8989 2  1799 .941591 .
3
75
a) This is a one-way ANOVA. SST 
SSB 
n x
2
j .j
 x
2
 nx 2  1893.774  754.89892  93.8324
 nx 2  2572.7225  754.89892  18.1209 . SSW  SST  SSB = 93.8324 – 18.1209
= 75.7115.
Source
SS
Between
18.1208
DF
2
MS
9.0604
F
F.05
8.86s
F 2,72  3.13
H0
Column means equal
Within
75.7115
72
1.0231
Total
93.8324
74
The difference between the column means is significant because the computed F exceeds the column F. It
looks like there is plentiful rounding error, since the computer gets the following.
One-way ANOVA: x1, x2, x3
Source DF
Factor
2
Error
72
Total
74
S = 1.025
SS
MS
F
P
18.10 9.05 8.60 0.000
75.72 1.05
93.81
R-Sq = 19.29%
R-Sq(adj) = 17.05%
b) Actually as in 1e) these data represent three tests of a single random sample of 25 workers. Consider the
data blocked by worker and compare means. (4) Solution: If the data is cross classified, we need a 2-way
ANOVA. SST 
ANOVA.
x
SSC  R
2
.j
 x
2
ij
 nx 2  1893.774  754.89892  93.8324 . This is the same as 1-way
 nx 2  2572.7225  754.89892  18.1209. This is the same as SSB in one-way
ANOVA.
9
252y0641 2/23/07
SSR  C
x
2
i.
 nx 2  3624.587  1538.66672  73.8194
SSW  SST  SSR  SSC = 93.8324 – 73.8194 – 18.1209 = 1.8921
Source
SS
DF
MS
F
Rows
73.8194
24
3.0755
120.28s
Columns
18.1208
2
9.0604
8.86s
F.05
F 24,48  1.75
F 2, 48  3.19
H0
Row means equal
Column means equal
Within
1.8921
48
0.02557
Total
93.8324
74
This time, as indicated by the ‘s’s in the F column, both thee difference between the column means and the
difference between the row means are significant. It looks like there is plentiful rounding error, since the
computer gets the following.
Two-way ANOVA: C11 versus C13, C12
Source DF
C13
24
C12
2
Error
48
Total
74
S = 0.1998
SS
MS
F
P
73.7956 3.07482
77.01 0.000
18.0961 9.04807 226.63 0.000
1.9164 0.03993
93.8081
R-Sq = 97.96%
R-Sq(adj) = 96.85
c) Consider the first three columns to be three independent random samples from a distribution that is not
Normal. Compare the medians using an appropriate statistical test or tests. (5) [31] Solution: If the
columns are independent random samples and the distribution is not Normal, we have a Kruskal-Wallis test.
The null hypothesis is that we have equal medians. The table on the previous page has (11)Sum of rank 1 =
1236.5, (12)Sum of rank 2 = 662 and (13)Sum of rank 3 = 951.5. As in the previous sections, the number of
items in all the columns is n  75 , the number of items in each column is ni  25 and we have 3 columns,
but too many items in a column to use the K-W table, so we say that we will consider the K-W statistic to
have a Chi-squared distribution with 3 – 1 = 2 degrees of freedom. We must compute the Kruskal-Wallis
 12
 SRi 2 

  3n  1
statistic H  
 nn  1 i  ni 
 12  1236 .52 662 2 951 .52 

  376   12  2872528 .5   228  241 .8971  228  13 .8971




25
25
25 
5700 
25

 7576  

Since this is larger than  .205  5.9915, we reject our null hypothesis.
10
252y0641 2/23/07
3. A sales manager wishes to predict newspaper circulation on Sunday ( y ) on the basis of weekday
morning circulation ( x1 ) weekday evening circulation ( x 2 ) and time ( x3 ). The data is below (Use
  .01) . All circulation data is in millions sold. Note: The product
computed in bold face.
y
x1
Row
1
2
3
4
5
6
7
8
9
S
54
55
55
56
56
57
58
59
60
AM
27
29
30
33
34
35
36
37
39
x2
x3
PM
34
33
31
29
29
28
26
25
24
T
1
2
3
4
5
6
7
8
9
The quantities below are given:
y  510,
x  300,
n 9,
x


2
2
 7549,
compute
1
 x y  ?,  x
1
2y
x
2
 259,
 14623 and
 x y was not given and is
1
x1 y
1458
1595
1650
1848
1904
1995
2088
2183
2340
17061
 y  28932,  x  10126,
 x x  8525. Yes, you will have to
2
2
1
1 2
 x y as part of a). You do not need all of these.
1
Spare Parts Computation:
510
300
259
x1 y  17061 , Y 
 56 .6667 , X 1 
 33 .3333 , X 2 
 28 .7778
9
9
9

 X 12  nX 12  SSX1  10126 933.33332  126.0000*
†Needed only in next problem
 X  nX  SSX  7549  928.7778  = 95.5556*†
 Y  nY  SST  SSY  28932  956.6667 2 = 32.0000*
 X Y  nX Y  SX Y  17061  933.3333 56.6667  = 61.0000
 X Y  nX Y  SX Y  14623  928.7778 56.6667  = -53.6667†
 X X  nX X  SX X  8525  933.3333 28.7778  = -108.333†
2
2
2
2
2
2
2
2
1
1
2
2
1
2
1
2
1
2
1
2
*Must be positive. The rest may well be negative.
a) Compute a simple regression of Sunday circulation against morning circulation.(8) Solution: The
coefficients are b1 
S xy
SS x

 XY  nXY
 X  nX
2
2

61 .000
 0.4841 and
126 .000
b0  Y  b1 X  56 .6667  0.4841  33 .3333   40 .530 . So Yˆ  40 .530  0.4841 X .
b) Compute R 2 (4) SSR  b1 S xy  0.4841 (61)  29.5301
Solution: R 2 
S xy 2
612  .9229
SSR b1 S xy 0.4841 61


 .9228 or

SST
SSy
32 .0000
SS x SS y 126 32 
c) Compute s e (3) Solution: s e2 
SSE SST  SSR SS y  b1 S xy 32  29 .5301



 0.35284
n2
n2
n2
7
s e  0.35284  0.5940
11
252y0641 2/23/07
d) Compute s b1 ( the standard deviation of the slope) and do a confidence interval for 1 .(3)
1 X 2 
2
2 1 
Solution: s b20  s e2  
 and s b1  s e 
 . We, of course, want the second formula.
 n SS x 
 SS x 
0.35284
s b21 
 0.0028003 s b  0.0529 To test for significance, H 0 : 1  0 , we compute
1
126
b  0 0.4841
t 1

 9.1512 . Since   .01 , t.7005 = 3.499. The ‘do not reject’ zone is between ±2.499.
sb1
0.0529
Since our computed t does not lie between these values, reject the null hypothesis and declare b1
significant.
e) Do a prediction interval for units when morning circulation rises to 45 million. (3) Why is this interval
likely to be larger than other prediction intervals we might compute for morning circulation we have
actually observed? (1)
[53]
1 X X 2

Solution: The Prediction Interval is Y0  Yˆ0  t sY , where sY2  s e2   0
 1 and
n

SS x


ˆ
ˆ
X

33
.
3333
.
.
=
45.
We
have
already
found
, s 2  .35284 ,
Y

40
.
530

0
.
4841
X
X
Y b b X

0
0
1
0
0
1
SS x  126 . So Yˆ0  40.530  0.484145 = 62.3145.

e
 1 45  33 .3333 2

 10

sY2  0.35284  
 1  0.35284   1.08025   0.7732 and sY  0.8793 . t.7005 = 3.499,
9

126
9




so Y0  Yˆ0  t sY  62.3145  3.449 0.8793   62.3  3.0 . Because 45 is relatively high compared to the
mean and the variance of X, it yields a relatively large prediction interval.
12
252y0641 2/23/07
4. Data from problem 3 is repeated. (Use   .01) .
A sales manager wishes to predict newspaper circulation on Sunday ( y ) on the basis of weekday morning
circulation ( x1 ) weekday evening circulation ( x 2 ) and time ( x3 ). The data is below (Use   .01) . All
circulation data is in millions sold.
Row
1
2
3
4
5
6
7
8
9
y
x1
x2
x3
S
54
55
55
56
56
57
58
59
60
AM
27
29
30
33
34
35
36
37
39
PM
34
33
31
29
29
28
26
25
24
T
1
2
3
4
5
6
7
8
9
The quantities below are given:
y  510,
x  300,
n 9,



x 22
 x y  ?,  x
 7549,
x
1
1
2y
2
 259,
 14623 and
 y  28932,  x
 x x  8525.
2
 10126,
2
1
1 2
a) Do a multiple regression of Sunday circulation against morning and evening circulation. (12)
Solution: We have already computed the spare parts.
Y  56.6667 , X 1  33 .3333 , X 2  28 .7778 .
X
2
1
 nX 12  SSX 1  126.0000*,
32.0000*,
X
1X 2
 X Y  nX Y  SX Y
1
1
1
X
2
2
 nX 22  SSX 2 = 95.5556* ,
= 61.0000,
X
2Y
Y
2
 nY 2  SST  SSY =
 nX 2 Y  SX 2 Y = -53.6667
 nX 1 X 2  SX 1 X 2 = -108.333
Note: * These spare parts must be positive. The rest may well be negative.
We substitute these numbers into the Simplified Normal Equations:
X 1Y  nX 1Y  b1
X 12  nX 12  b2
X 1 X 2  nX 1 X 2


 X Y  nX Y  b  X X
2
which are
2
1
1
2
 
 nX X   b  X
1
2
2
2
2

 nX  ,
2
2
 61 .0000  126 .000 b1  108 .333 b2

 53 .6667   108 .333 b1  95 .5554 b2
and solve them as two equations in two unknowns for b1 and b2 . These are a fairly tough pair of equations
to solve until we notice that, if we multiply 95.5554 by 108 .333
95 .5554
 1.133719 we get 108.333. The
 61 .0000  126 .000 b1  108 .333 b2
equations become 
. If we add these together, we get
  60 .8430   122 .819 b1  108 .333 b2
0.1570
 0.04936 . Now remember that 61  126 b1  108 .333 b2 .
0.1570  3.181b1 . This means that b1 
3.181
This can be rewritten as 108 .333 b2  61  126 b1 . Let’s substitute b1  0.04936 .
55 .5064
 0.5124 . (It’s worth checking your
108 .333
work by substituting your values of b1 and b2 back into the normal equations.) Finally we get b0 by using
108 .333 b2  61  126 0.0436   55 .5064 . So b2  
Y  56.6667 , X 1  33 .3333 , X 2  28 .7778 in b0  Y  b1 X 1  b2 X 2
 56 .6667  0.04936 33 .3333   0.5124 28 .7778   56 .6667  1.6453  14 .7457  69 .7671 . Thus our
equation is Yˆ  b0  b1 X 1  b2 X 2  69.7671 0.0494X 1  0.5124X 2 .
13
252y0641 2/23/07
Note: My flabber is ghasted. Minitab said The regression equation is S = 69.6 + 0.049 AM 0.506 PM. This is very close for a first try.
b) Compute R 2 and R 2 adjusted for degrees of freedom for both this and the previous problem. Compare
the values of R 2 adjusted between this and the previous problem. Use an F test to compare R 2 here with
the R 2 from the previous problem.(6)
Solution: From the first regression we have SST  SSy  32.000 , R 2  RY2.1  0.9228 * and we can
compute SSR  b1 S x1 y  0.484161  29.5301 so that
From the second regression Yˆ  b0  b1 X 1  b2 X 2  69.7671 0.0494X 1  0.5124X 2
Y  56.6667 , X 1  33 .3333 , X 2  28 .7778 ,
X
2Y
 X Y  nX Y  SX Y
1
1
1
= 61.0000,
 nX 2 Y  SX 2 Y = -53.6667. This means that
SSR  b1 Sx1 y  b2 Sx2 y  0.0494 61  0.5124 53.6667   3.0134  27.4988  30.5122 .
SSR 30 .5122

 0.9535 *. If we use R 2 , which is R 2 adjusted for degrees of freedom, for
SST 32 .0000
the first regression, the number of independent variables was k  1 and
R 2  RY2.12 
R2 
n  1R 2  k  80.9228  1  .9117
R2 
n  1R 2  k  80.9535  2  .9380 .
n  k 1
7
and for the second regression k  2 and
R-squared adjusted is supposed to rise if our new variable has
n  k 1
6
any explanatory power. Note: * These numbers must be positive. The rest may well be negative.
There are two ways to do the F test. We can use the second regression to give us SSE  SST  SSR2
 32.000  30.5122  1.4878 . In the second regression, the explained sum of squares rises by 30.5122 –
29.5301 = 0.9821. We can make an ANOVA table for looking at a new variable as follows. Assume that we
have SSR1 for the first regression on k independent variables and add r new independent variables and
get a new SSR2
Source
SS
First Regression SSR1
2nd Regression
SSR 2  SSR1
Error
Total
SSE
SST
Source
2 Regression
MS
Fcalc
k
MSR1
MSR1 MSE
r
MSR2
MSR2 MSE
F
F k , nk r 1
F r , nk r 1
n  k  r  1 MSE
n 1
SS
First Regression 29.5301
nd
DF
0.9821
DF
Fcalc
MS
1
29.5301
1
0.9821
F
1,6   13 .75
119s F.01
3.96ns F 1,6   13.75
.01
Error
1.4878
6
0.24797
Total
32
8
We can get the same results using R 2 . Remember RY2.12  0.9535 and RY2.1  0.9228 .
Fcalc
Source
SS
DF
MS
First Regression RY2.1  0.9228
F
1,6   13 .75
F.01
1
.9228
119s
1,6   13 .75
3.96ns F.01
2nd Regression
RY2.12  RY2.1  0.9535  .9228  .0307
1
.0307
Error
Total
1  RY2.12  1  0.9535  .0465
1
8
6
.00775
14
252y0641 2/23/07
c) Compute the regression sum of squares and use it in an F test to test the usefulness of this regression. (5)
Solution: Remember SSR  30.5122 and SSE  SST  SSR2  32.000  30.5122  1.4878 .
Source
SS
nd
First and 2 Regression
DF
30.5122
2
MS
Fcalc
15.25361
61.524
F
2,6   10 .92
F.01
Error
1.4878
6
0.24797
Total
32
8
The null hypothesis is no connection between Y and the X’s. It is rejected.
d) Use your regression to predict the Sunday circulation when AM circulation is 40 and PM circulation is
23.(2) Solution: From the second regression Yˆ  b0  b1 X 1  b2 X 2  69.7671 0.0494X 1  0.5124X 2
 69.7671  0.0494 40   0.5124 23  69.7671  1.9760  11.7852  83.5283
e) Use the directions in the outline to make this estimate into a confidence interval and a prediction interval.
6
(4) Solution: The error mean square is 0.24797 and has 6 degrees of freedom. So use t .005
 3.707 and
s e  0.24797  0.4979 .
The outline says that an approximate confidence interval is  Y0  Yˆ0  t
se
3.707
n
 0.4979 
  83 .53  0.61 and an approximate prediction interval is
 83 .5283  3.707 

9 

[82]
Y  Yˆ  t s  83.5283  3.707 0.4979   83.53  1.85 .
0
0
e
For the record, here is what the computer got. It looks like I had a fair amount of error in much of this
question.
Regression Analysis: S versus AM
The regression equation is
S = 40.5 + 0.484 AM
Predictor
Coef SE Coef
Constant
40.529
1.774
AM
0.48413 0.05290
S = 0.593808
R-Sq = 92.3%
Analysis of Variance
Source
DF
SS
Regression
1 29.532
Residual Error
7
2.468
Total
8 32.000
T
P
22.84 0.000
9.15 0.000
R-Sq(adj) = 91.2%
MS
29.532
0.353
F
83.75
P
0.000
Regression Analysis: S versus AM, PM
The regression equation is
S = 69.6 + 0.049 AM - 0.506 PM
Predictor
Coef SE Coef
Constant
69.57
20.61
AM
0.0494
0.3115
PM
-0.5057
0.3577
S = 0.555511
R-Sq = 94.2%
Analysis of Variance
Source
DF
SS
Regression
2 30.148
Residual Error
6
1.852
Total
8 32.000
Source
AM
PM
DF
1
1
T
P
3.38 0.015
0.16 0.879
-1.41 0.207
R-Sq(adj) = 92.3%
MS
15.074
0.309
F
48.85
P
0.000
Seq SS
29.532
0.617
15
252y0641 2/23/07
5. Data from problem 3 is repeated. (Use   .01) .
A sales manager wishes to predict newspaper circulation on Sunday ( y ) on the basis of weekday morning
circulation ( x1 ) weekday evening circulation ( x 2 ) and time ( x3 ). The data is below (Use   .01) . All
circulation data is in millions sold.
The time variable is now added with the following results.
MTB >
SUBC>
SUBC>
Regress c1 3 c2 c3 c10;
VIF;
DW.
Regression Analysis: S versus AM, PM, T
The regression equation is
S = 62.2 - 0.253 AM - 0.071 PM + 0.991 T
Predictor
Constant
AM
PM
T
Coef
62.19
-0.2533
-0.0707
0.9913
S = 0.453612
SE Coef
17.23
0.2960
0.3642
0.4957
R-Sq = 96.8%
Analysis of Variance
Source
DF
SS
Regression
3 30.971
Residual Error
5
1.029
Total
8 32.000
T
3.61
-0.86
-0.19
2.00
P
0.015
0.431
0.854
0.102
VIF
53.6
61.6
71.7
R-Sq(adj) = 94.9%
MS
10.324
0.206
F
50.17
P
0.000
Durbin-Watson statistic = 2.13279
a) What do the significance tests on the coefficients reveal? Give reasons. (2)
Solution: All of the p-values are above   .01 , indicating that the null hypotheses that these coefficients
are actually zero cannot be rejected. None of the coefficients are significant in spite of a highly significant
ANOVA result.
b) Can you explain why the coefficients of AM and PM seem unreasonable? What is the apparent reason for
this? (2) Solution: Though it may make sense to have a negative coefficient for evening readership, which
is obviously declining and being replaced by morning readership, one would expect that, if Sunday
readership reflects interest in the news there would be a positive coefficient relating AM readership to
Sunday readership. The problem seems to be the high VIF indicating a substantial degree of collinearity. In
the presence of collinearity, coefficients are not reliable.
c) Do a 2% two-sided Durbin-Watson test on the result as suggested in class. What is the hypothesis tested
and what is the result? (3) Solution: The line presented in the notes is below.
0   0 dL
?
dU
  0 2   0 4  dU ? 4  d L   0 4
+
+
+
+
+
+
+
Unfortunately the 1% values given in the text start at n  15 , but we can guess that the 1% values for k  3
would be somewhere near d L = 0.5 and d U = 1.5. The null hypothesis is that there is no autocorrelation,
and we cannot reject the null hypothesis if the Durbin-Watson statistic is between d U and 4  d U . The
statistic cited above is certainly close to 2 and apparently between d U and 4  d U , so it is unlikely that we
can reject the null hypothesis.
d) Reuse your spare parts from the previous regression if possible to compute the correlation between AM
and PM circulation and test it for significance. (4) Solution: The formula for correlation is
r
 XY  nXY
 X  nX  Y
2
2
2
 nY
. We have
2
X
2
1
 nX 12  SSX 1  126.0000*
,
16
252y0641 2/23/07
X

2
2
 nX 22  SSX 2 = 95.5556* and
X
1X 2
 nX 1 X 2  SX 1 X 2 = -108.333. Because of the sign of
X 1 X 2  nX 1 X 2  SX 1 X 2 , the correlation will be negative. r  

 108 .332
126 95.5556 
t n  2  
r

sr
SX 1 X 2 2
SSX1 SSX 2 
  0.9747 = -0.9872. The test of the null hypothesis   0 is
r

.9872

.9872
 16 .421 . We would use t.7005 = 3.499. The ‘do not
0.0601189
1  0.9747
1 r
7
n2
reject’ zone is between ±2.499. Since 16.421 is not in this interval, we cannot reject the null hypothesis.
2
e) Compute a rank correlation between AM and PM circulation and test it for significance. Can you explain
why it is larger than the correlation in d)? (4)
d
Solution: y
d2
x1
r1
x 2 r2
x3
Row
1
2
3
4
5
6
7
8
9
Sum
S
54
55
55
56
56
57
58
59
60
AM
27
29
30
33
34
35
36
37
39
PM
34
33
31
29
29
28
26
25
24
9
8
7
5.5
5.5
4
3
2
1
1
2
3
4
5
6
7
8
9
T
1
2
3
4
5
6
7
8
9
8
6
4
1.5
0.5
-2
-4
-6
-8
0
64
36
16
2.25
0.25
4
16
36
64
238.5
A correlation coefficient between rx and ry can be computed as in d) above, but it is easier to compute
d  rx  ry , and then rs  1 
 d  1  6238 .5  .9875 . This can be given a t test for H
981  1
nn  1
2
6
2
0
:  0 as
in b) above, but for n between 4 and 30, a special table should be used. Part of the table from the
Supplement is repeated below. We do not reject the null hypothesis of no rank correlation if rs is between
±.8167. Since our value of the rank correlation is below -.8167, reject the null hypothesis.
n
  .100
  .050
  .025
  .010
  .005
  .001
4
5
6
7
8
9
.8000
.7000
.6000
.5357
.5000
.4667
.8000
.8000
.7714
.6786
.6190
.5833
.9000
.8286
.7450
.7143
.6833
.9000
.8857
.8571
.8095
.7667
.9429
.8929
.8571
.8167
.9643
.9286
.9000
f) Test the hypothesis that the correlation that you computed in d) is -.99. (4) [101]
Note: since almost no one did this problem, the correct values of n, r and  were never substituted in
these equations.

H 0 : xy  0.8
Solution: Test 
when n  10, r  .704 and r 2  .496   .05  .
H
:


0
.
8

 1 xy
This time compute Fisher's z-transformation (because  0 is not zero)
1  1  r  1  1  .704  1  1.704  1
1
~
z  ln 
  ln 
  ln 
  ln 5.75676   1.75037   0.87519
2  1  r  2  1  .704  2  0.296  2
2
1  1 0
 z  ln 
2  1 0
 1  1  .8  1  1.8  1
1
  ln 
 2  1  .8   2 ln  0.2   2 ln 9.0000   2 2.19722   1.09861

17
252y0641 2/23/07
sz 
1
1


n3
10  3
1
 0.37796 .
7
Finally t 
~
z   z 0.87519  1.09861

 0.591 . Compare
sz
0.37796
this with  t n2 2  t .8025  2.306 . Since –0.591 lies between these two values, do not reject the null
hypothesis.
g) (Extra credit) If AM, PM and T are x1 , x 2 and x3 , find the partial correlation coefficient (square root
of the coefficient of partial determination) rY 3.12 . (2)
Solution: We had the following results.
MTB >
SUBC>
SUBC>
Regress c1 3 c2 c3 c10;
VIF;
DW.
Regression Analysis: S versus AM, PM, T
The regression equation is
S = 62.2 - 0.253 AM - 0.071 PM + 0.991 T
Predictor
Constant
AM
PM
T
Coef
62.19
-0.2533
-0.0707
0.9913
S = 0.453612
SE Coef
17.23
0.2960
0.3642
0.4957
R-Sq = 96.8%
Analysis of Variance
Source
DF
SS
Regression
3 30.971
Residual Error
5
1.029
Total
8 32.000
T
3.61
-0.86
-0.19
2.00
P
0.015
0.431
0.854
0.102
VIF
53.6
61.6
71.7
R-Sq(adj) = 94.9%
MS
10.324
0.206
F
50.17
P
0.000
Durbin-Watson statistic = 2.13279
The outline says rY23.12 
t 32
t 32  df

2.00 2  0.4444
2.00 2  5
18
252y0641 2/23/07
6. The following times were recorded for 6 skiers on 3 slopes. In order to assess their difficulty we look at
the median time for each slope. We do not assume a Normal distribution. Do not compute the median or
mean time for any slope.
Skier
Slope 1
Slope 2
Slope 3
1
4.9
6.1
5.2
2
4.5
6.0
5.1
3
4.1
5.4
4.9
4
4.4
4.7
5.1
5
4.5
4.9
4.5
6
3.3
3.8
3.9
a) Test the hypothesis that the median time on slope 1 is 4 minutes (3 or 2 depending on method) (3)
Solution: row
1
2
3
4
5
6
d  x1  4
x1
4.9
4.5
4.1
4.4
4.5
3.3
d
0.9
0.5
0.1
0.4
0.5
-0.7
0.9
0.5
0.1
0.4
0.5
0.7
r
r*
6
3.5
1
2
3.5
5
6+
3.5+
1+
2+
3.5+
5-
We use a Wilcoxon signed rank test with the null hypothesis that the median of x1 is 4 (or the median of
d  x1  4 is zero. If we add together the numbers in r * with a + sign we get T   16 . If we do the same
for numbers with a – sign, we get T   5. To check this, note that these two numbers must sum to the sum
nn  1 67 

 21 , and that T    T   16  5  21 . We check
of the first n numbers, and that this is
2
2
5, the smaller of the two rank sums against the numbers in table 7. For a two-sided 5% test, we use the
  .025 column. For n  6 , the critical value is 1, and we reject the null hypothesis only if our test statistic
is below this critical value. Since our test statistic is 5, we do not reject the null hypothesis.
We could also use a sign test on this. We have 5 positive outcomes and 1 negative outcome. This is a 2sided test so pvalue  2P x  1 p  .5, n  6  2.10938  .2198 . Since this is above   .05 , we cannot


reject the null hypothesis.
b) Test the hypothesis that slope 1 and slope 2 have the same median times. (4)
Solution: row
1
2
3
4
5
6
x1
x2
4.9
4.5
4.1
4.4
4.5
3.3
6.1
6.0
5.4
4.7
4.9
3.8
d  x 2  x1
-1.2
-1.5
-1.3
-0.3
-0.4
-0.5
d
r
r*
1.2
1.5
1.3
0.3
0.4
0.5
4
5
6
1
2
3
456123-
We use a Wilcoxon signed rank test with the null hypothesis that the medians of x1 and x 2 are the same (or
the median of d  x 2  x1 is zero. If we add together the numbers in r * with a + sign we get T   0 . If
we do the same for numbers with a – sign, we get T   21 . To check this, note that these two numbers
nn  1 67 

 21 , and that
must sum to the sum of the first n numbers, and that this is
2
2
T   T   0  21  21 . We check 0, the smaller of the two rank sums against the numbers in table 7. For
a two-sided 5% test, we use the   .025 column. For n  6 , the critical value is 1, and we reject the null
hypothesis only if our test statistic is below this critical value. Since our test statistic is 0, we reject the null
hypothesis.
We could also use a sign test on this. We have 0 positive outcomes and 6 negative outcomes. This is a 2sided test so pvalue  2P x  6 p  .5, n  6  2P x  0 p  .5, n  6  2.01563  .03126 . Since this is




below   .05 , we reject the null hypothesis. 0.01563
19
252y0641 2/23/07
c) Test the hypothesis that the slopes all have the same median time. (4)
Solution: This is a Friedman test.
H0:  1   2   3   4 Where 1 is A, 2 is B, 3 is C and 4 is D.
H1: At least one of the medians differs.
First we rank the data within rows. The data appears below in columns marked x1 to x 4 and the
ranks are in columns marked r1 to r4 .
row
x1 Slope 1 r1
x 2 Slope 2 r2
x3 Slope 3 r3
1
2
3
4
5
6
Sum SRi
4.9
4.5
4.1
4.4
4.5
3.3
1
1
1
1
1.5
1
6.5
6.1
6.0
5.4
4.7
4.9
3.8
3
3
3
2
3
2
16
5.2
5.1
4.9
5.1
4.5
3.9
2
2
2
3
1.5
3
13.5
To check the ranking, note that the sum of the three rank sums is 6.5 + 16 + 13.5 = 36, and that
rcc  1 634
SRi 

 36 . Now compute the Friedman statistic
the sum of the rank sums should be
2
2
 12

 12
6.52  16 2  13 .52   364
 F2  
SRi2   3r c  1  
 634

 rc c  1 i



 

 12

  480 .5  72  8.0833 . The relevant part of the Friedman table appears below and since our value is
 72

between 8.333 and 9.000, the p-value is between .008 and .012. Our null hypothesis is equal column
medians and since the p-value is below .05, we reject it.
c  3,
r 6
 F2
0.000
0.333
1.000
1.333
2.333
3.000
4.000
4.333
5.333
6.333
7.000
8.333
9.000
9.333
10.333
12.000
p  value
1.000
.956
.740
.570
.430
.252
.184
.142
.072
.052
.029
.012
.008
.006
.002
.000
d) Explain what methods you would use in b) and c) if the columns were independent random samples. (1)
Solution: The Wilcoxon signed rank test for paired data corresponds to the Mann-Whitney-Wilcoxon test
for two independent samples. The Friedman test for cross-classified data corresponds to the Kruskal-Wallis
test for independent samples.
20
252y0641 2/23/07
e) Rank the skiers times on each slope from 1 (fastest) to 6. Use these as rankings of the skiers and test to
see if the ranks agree between slopes. (4) Solution: This calls for Kendall's Coefficient of Concordance.
Take k columns with n items in each and rank each column from 1 to n . The null hypothesis is
 H : Disagreement
that the rankings disagree.  0
 H 1 : Agreement
Compute a sum of ranks SRi for each row. Then S 
 SR
2
 
 n SR
2
, where SR 
n  1k
2
the mean of the SRi s. If H 0 is disagreement, S can be checked against a table for this test. If
S
, where
S  S reject H 0 . For n too large for the table use  2n1  k n  1W 
1 knn  1
12
W
S
1 k2
12
n
row
n
 SR
k 3, n 6 S 
 is the Kendall Coefficient of Concordance and must be between 0 and 1.
x1 Slope 1
4.9
4.5
4.1
4.4
4.5
3.3
1
2
3
4
5
6
SR 
3
is
2
r1
6
4.5
2
3
4.5
1
x 2 Slope 2
6.1
6.0
5.4
4.7
4.9
3.8
r2
6
5
4
2
3
1
x3 Slope 3
5.2
5.1
4.9
5.1
4.5
3.9
r3
6
4.5
3
4.5
2
1
SR
18
14
9
9.5
9.5
3
63
SR 2
324
196
81
90.25
90.25
9
790.50
 
 n SR = 790 .5  610 .52  129 . To check this note that
2
n  1k  73  10 .5
2
2
Note that if we had complete disagreement, every applicant would have a rank sum of 10.5. The
Kendall Coefficient of Concordance says that the degree of agreement on a zero to one scale is
S
12 129  0.819048 . To do a test of the null hypothesis of disagreement
W
=
2
3
1 k n n
9 63  6
12

 
  .05  , look up

S  in the table giving ‘Critical values of Kendall’s s ,’ part of which is
reproduced below. Our computed value of S is larger than S .05  103 .9, so that we reject the null
hypothesis of disagreement.
  .05
m
3
4
5
6
8
10
15
20
TABLE 12: Critical values of Kendall's s
n3
48.1
60.0
89.8
119.7
n4
n5
n6
n7
49.5
62.6
75.7
101.7
127.8
192.9
258.0
64.4
88.4
112.3
136.1
183.7
231.2
349.8
468.5
103.9
143.3
182.4
221.4
299.0
376.7
570.5
764.4
157.3
217.0
276.2
335.2
453.1
571.0
864.9
1158.7
21
252y0641 2/23/07
7. Clarence Sales is a marketing major and knows that national soft drink market shares are as below.
Classic Coke
15.6%
Pepsi
13.2%
Diet Coke
5.1%
Diet Pepsi
3.5%
Other brands
62.6%
He gets in a bit of trouble here and is sentenced to 20 hours of public service. After he finishes his public
service he takes off for Maine, gets caught littering and is sentenced to another 20 hours of public service.
During his public service, he picks up 100 cans in each state. The cans are as below.
Brand
PA
ME
Classic Coke
22
17
Pepsi
15
11
Diet Coke
13
10
Diet Pepsi
6
5
Other brands
44
57
Use a 1% significance level throughout this problem. Don’t waste our time by just computing percents and
saying that they are different. Each problem requires a statistical test or the equivalent. State your null and
alternative hypotheses in each problem.
a) Regard the cans picked up as a random sample of sales in the two states. Can we say that the proportions
of soft drink cans discarded in Pennsylvania are the same as the national market shares? (5)
Solution: A Kolmogorov-Smirnov test might work here, but I think that chi-squared is the more likely
choice. The PA data is our O . I got E by multiplying the market share proportions by 100.
H 0 :Proportions Apply
Row
1
2
3
4
5
Sum
p
E
.156
.132
.051
.035
.626
1.000
15.6
13.2
5.1
3.5
62.6
100.0
O
22
15
13
6
44
100
O  E 2
OE
6.4
1.8
7.9
2.5
-18.6
0.0
O  E 2
E
2.6256
0.2455
12.2373
1.7857
5.5265
22.4206
O2
E
31.0256
17.0455
33.1373
10.2857
30.9265
122.4206
O2
 n = 122.4206 – 100 = 22.4206, but don’t
E
E
use both. Both of these two formulas are shown above. Actually the longer calculation is better in this case
because we have a value of E that is below 5, and its contribution of 1.7857 is not what pushes us into
O  E 2  16.1029 . The relevant part of the chi-squared
rejection. DF  r  1  4 . So we have  2 
E
table is shown below. The 5% value for 4 degrees of freedom is 9.4877. Since the computed chi-square
exceeds the table chi-squared, we reject the null hypothesis.
You can use  2 

= 22.4206 or  2 



0.005
0.010
0.025
0.050
0.100
0.900
0.950
0.975
0.990
0.995
Degrees of Freedom
1
2
3
4
5
6
7.87946 10.5966 12.8382 14.8603 16.7496 18.5476
6.63491 9.2103 11.3449 13.2767 15.0863 16.8119
5.02389 7.3778 9.3484 11.1433 12.8325 14.4494
3.84146 5.9915 7.8147 9.4877 11.0705 12.5916
2.70554 4.6052 6.2514 7.7794 9.2364 10.6446
0.01579 0.2107 0.5844 1.0636 1.6103 2.2041
0.00393 0.1026 0.3518 0.7107 1.1455 1.6354
0.00098 0.0506 0.2158 0.4844 0.8312 1.2373
0.00016 0.0201 0.1148 0.2971 0.5543 0.8721
0.00004 0.0100 0.0717 0.2070 0.4117 0.6757
7
20.2778
18.4753
16.0128
14.0671
12.0170
2.8331
2.1674
1.6899
1.2390
0.9893
8
21.9 550
20.0902
17.5346
15.5073
13.3616
3.4895
2.7326
2.1797
1.6465
1.344
9
23.5893
21.6660
19.0228
16.9190
14.6837
4.1682
3.3251
2.7004
2.0879
1.7349
22
252y0641 2/23/07
Since the rest of this question involves tests and confidence intervals for tests of one or two proportions,
here is the relevant section of the formula table.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Proportion
p  p  z 2 s p
sp 
pq
n
H 0 : p  p0
H1 : p  p0
p  p0
z
p
q  1 p
Difference
between
proportions
q  1 p
p  p  z 2 s p
p  p1  p 2
s p 
p1 q1 p 2 q 2

n1
n2
H 0 : p  p 0
H 1 : p  p 0
z
p  p 0
 p
p 0  p 01  p 02
If p  0
or p 0  0
 p 
p 01q 01 p 02 q 02

n1
n2
Or use
pcv  p0  z 2  p
p0 q0
n
q0  1  p0
p 
pcv  p0  z 2  p
If p  0
 1
1

 n1 n 2
 p  p 0 q 0 
p0 
s p



n1 p1  n 2 p 2
n1  n 2
b) Clarence knows that that Maine is Moxie country, so he believes that the proportion of other brands sold
is higher in Maine than in Pennsylvania. Is this true? (4)
Solution: Clarence’s assertion that p1  p 2 is an alternative hypothesis, so we have a left-sided test.
 H 0 : p1  p 2
H : p  p 2  0
or  0 1
or if p  p1  p 2 ,

 H 1 : p1  p 2
H 1 : p1  p 2  0
44
57
44  57
p1 
 .44 p 2 
 .57 p 
 0.13 p 0
100
100
100
You should use  p 
H 0 : p  0
are implied.

H 1 : p  0
44  57
101


 .505
100  100 200
 1
1 
 2 
  .505 .495 
p 0 q 0  
  .005  .0707 for hypothesis tests and
 100 
 n1 n 2 
pq
p q 
 .44 .56  .57 .43  

s p   1 1  2 2   
  .002464  .002451  .004915  .0701 for
100 
n2 
 100
 n1
confidence intervals, but at this point I won’t be fussy.
p  p 0
0.13

 1.838 . Since this is a left-sided test, we reject the null
For a test ratio z 
.0707
 p
hypothesis if our computed z is below  z   z .01  2.327 . Though we would reject the null
hypothesis at the 5% significance level, we would not reject it at the 1% level. Alternately
pvalue  Pz  1.84   .5  P1.84  z  0  .5  .4671  .0329 . Since this is not below 5%, we
cannot reject the null hypothesis.
If we use a critical value we want p cv  0  2.327 .070   .1645 . Since p  0.13 is not below
this value, we do not reject the null hypothesis.
c) Create a 0.2% 2-sided confidence interval for the difference between the proportions of other brands sold
in Maine. Using your Normal table, make this into a 0.1% 2-sided interval. (3)
Solution: Recall that s p  .0701 and that a two-sided confidence interval has the form p  p  z 2 s p ,
where p  0.13 . For a 2% confidence interval we already know that z  z.01  2.327. For the 0.1%
2
interval we need  2  .0012  .0005 . Make a diagram. Draw a Normal curve with a mean at 0. z .0005 is the
value of z with 0.05% of the distribution above it. Since 100 – 0.05 = 99.95, it is also the 99.95th
percentile. Since 50% of the standardized Normal distribution is below zero, your diagram should show that
23
252y0641 2/23/07
the probability between z .0005 and zero is 99.95% - 50% = 49.95% or P0  z  z.0005   .4995 . The
relevant part of the Normal table appears below. To get .4995 we need values of z between 3.27 and 3.32.
Any of these would be acceptable, but a good guess would be z.0005  3.295 .
If z 0 is between
P0  z  z 0  is
3.08
3.11
3.14
3.18
3.22
3.27
3.33
3.39
3.49
3.62
3.90
and
and
and
and
and
and
and
and
and
and
and
3.10
3.13
3.17
3.21
3.26
3.32
3.38
3.48
3.61
3.89
up
.4990
.4991
.4992
.4993
.4994
.4995
.4996
.4997
.4998
.4999
.5000
Thus a 2% confidence interval would be p  0.13  2.327 .0701   .13  0.16 or -0.03 to 0.29
and
a 0.1% confidence interval would be p  0.13  3.295 .0701   .13  0.23 or -0.10 to 0.36
d) Actually Clarence’s mother owns the Coke franchise for Maine and last year between her sales of Classic
Coke and Diet Coke accounted for 25% of the soft drink market in Maine. She tells Clarence that her sales
are now above 25%. On the basis of Clarence’s Maine sample is that true? (2) [131]
Recall that his main sample was as below.
Brand
PA
ME
Classic Coke
22
17
Pepsi
15
11
Diet Coke
13
10
Diet Pepsi
6
5
Other brands
44
57
Classic coke and diet coke together were 17% + 10 % = 27% of the sample, so p  .27 . His mother’s
 H 0 : p1  .25
assertion that p  .25 is an alternate hypothesis and we have a right-sided test. 
.
 H 1 : p1  .25
p 
p0 q0
.25 .75 

 .001875  .04330 would be used for hypothesis tests, while s p 
100
n
pq
n
.27 .73 
 .001971  .04440 would be used for confidence intervals.
100
p  p 0 .27  .25

 0.4619 . We already know that z  z.01  2.327. We
If we use the test ratio, z 
2
0.04330
p

reject the null hypothesis if the computed z-ratio exceeds 2.327. It does not, and we do not reject the null
hypothesis.
If we want to use a critical value for p we want p cv  .25  2.327 .04330   .3507 . I’m surprised that it is
so large, but since the observed proportion does not exceed 35.07%, we cannot reject the null hypothesis.
24