5/01/03 252y0342
Introduction
This is long, but that’s because I gave a relatively thorough explanation of everything that I did.
When I worked it out in the classroom, it was much, much shorter.
The easiest sections were probably 1a, 2a, 2b, 2c, 7b, 7c, and 7d in Part I. There are some very
easy sections in part I if you just follow suggestions and look at p-values and R-squared.
After doing the above, I might have computed the spare parts in 3b in Part II.
Spare Parts Computation:
SSx 
x 2  nx 2  1019346  11265 .636 2
x 2922
 243158 .7
x

 265 .636
n
11
Sxy 
xy  nx y  4419959  11265 .636 1184 .36 


 y  13028  1184 .36
y
n

 959263 .8
11
SSy 
y
2
 ny 2  19891990  111184 .36 2
 4462195 .3
And not recomputed them every time I needed them in Problems 4 and 5 ! Only then would I have tried the
multiple regression.
Many of you seem to have no idea what a statistical test is. We have been doing them every day. The most
common examples of this were in part II.
8
10
 .069565 and p 2 
 .09709 . Hey look! They’re different! Whoopee! And
Problem 1b: p 3 
115
103
you think that you will get credit for this? Whether these two proportions come from the same population
or not, chances are they will be somewhat different, you need one of the three statistical tests shown in the
solution to show that they are significantly different.
Problem 2c: Many of you started with
x
1019346  11265 .636 2 243158 .7

 24315 .87 s  24315 .86  155 .90 This is
n 1
10
10
fine and you got some credit for knowing how to compute a sample variance, though you probably had
already computed the numerator somewhere else in this exam. But then you told me that this wasn’t 200.
Where was your test?:
s x2 
2
 nx 2



5/06/03 252y0342
ECO252 QBA2
FINAL EXAM
May 7, 2003
Name
KEY
Hour of Class Registered (Circle)
I. (18 points) Do all the following. Note that answers without reasons receive no credit.
A researcher wishes to explain the selling price of a house in thousands on the basis of its assessed
valuation, whether it was new and the time period. New is 1 if the house is new construction, zero
otherwise. The researcher assembles the following data for a random sample of 30 home sales. Use   .10
in this problem. That’s why p-values above .10 mean that the null hypothesis of insignificance is not
rejected.
—————
4/25/2003 9:58:00 PM
————————————————————
Welcome to Minitab, press F1 for help.
MTB > Retrieve "C:\Documents and Settings\RBOVE\My Documents\Drive D\MINITAB\2x03421.MTW".
Retrieving worksheet from file: C:\Documents and Settings\RBOVE\My Documents\Drive
D\MINITAB\2x0342-1.MTW
# Worksheet was saved on Fri Apr 25 2003
Results for: 2x0342-1.MTW
MTB > print c1 - c4
Data Display
Row
Price
Value
New
Time
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
26
27
28
29
30
69.00
115.50
100.80
96.90
72.00
61.90
97.00
87.50
96.90
81.50
69.34
97.90
96.00
92.00
94.10
101.90
109.50
88.65
93.00
83.00
106.70
97.90
97.30
90.50
95.90
113.90
94.50
86.50
91.50
93.75
66.28
86.31
84.78
79.74
65.54
59.93
79.98
75.22
81.88
72.94
60.80
81.61
79.11
77.96
78.17
80.24
85.88
74.03
75.27
74.31
84.36
77.90
79.85
74.92
79.07
85.61
76.50
72.78
72.43
76.64
0
0
1
1
0
0
1
0
1
0
0
1
0
0
1
1
1
0
0
0
0
1
1
0
1
0
1
0
0
0
1
2
2
3
4
4
4
5
5
5
6
6
7
9
10
10
10
11
11
11
12
12
12
12
12
13
14
14
17
17
1. Looking for a place to start, the researcher does individual regressions of price against the individual
independent variables.
a. Explain why the researcher concludes from the regressions that valuation (‘value’) is the most
important independent variable. Consider the values of R 2 and the significance tests on the
slope of the equation (2)
b. What kind of variable is ‘new.’ Explain why the regression of ‘price’ against ‘new’ is
equivalent to a test of the equality of 2 sample means, and what the conclusion would be. (2)
5/06/03 252y0342
2
Solution: a) Note that the p-values for the first two regressions are below 10%, indicating a significant
slope. The regression against ‘Time,’ however, shows a p-value above 10% for the slope, indication that
time is a poor explanatory variable. If we compare the regression against ‘Value’ with the regression against
‘New,’ we find a much higher R-sq for ‘Value,’ which indicates that this independent variable does a much
better job of explaining ‘Price’ than ‘New.’
b) ‘New’ is a dummy variable, it is one when something is true and zero when it is false. The equation
Price = 88.5 + 9.93 New only gives us two values for ‘Price.’ If the house is not new, the
predicted price is 88.5 (thousand) and if the house is new, and we use the values from the ‘Coef’ column,
the predicted price is 88.458 + 9.926 = 98.3 (thousand) . Remember that a regression predicts the average
value of a dependent variable for a given value of the independent variable, so these are predicted means for
old and new houses respectively, and the fact that the p-value is .031, below 10% indicates that the
difference is significant.
MTB > regress c1 1 c2
Regression Analysis: Price versus Value
The regression equation is
Price = - 44.2 + 1.78 Value
Predictor
Coef
SE Coef
T
P
Constant
-44.172
7.346
-6.01
0.000
Value
1.78171
0.09546
18.66
0.000
slope is significant.
S = 3.475
R-Sq = 92.6%
R-Sq(adj) = 92.3%
The p-value below .10 shows us that the
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
28
29
SS
4206.7
338.1
4544.8
Unusual Observations
Obs
Value
Price
6
59.9
61.900
11
60.8
69.340
MS
4206.7
12.1
Fit
62.606
64.156
F
348.37
SE Fit
1.719
1.642
P
0.000
Residual
-0.706
5.184
St Resid
-0.23 X
1.69 X
X denotes an observation whose X value gives it large influence.
MTB > regress c1 1 c3
Regression Analysis: Price versus New
The regression equation is
Price = 88.5 + 9.93 New
Predictor
Constant
New
Coef
88.458
9.926
S = 11.70
SE Coef
2.759
4.362
R-Sq = 15.6%
T
32.07
2.28
P
0.000
0.031
R-Sq(adj) = 12.6%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
28
29
Unusual Observations
Obs
New
Price
2
0.00
115.50
6
0.00
61.90
26
0.00
113.90
SS
709.3
3835.5
4544.8
Fit
88.46
88.46
88.46
MS
709.3
137.0
SE Fit
2.76
2.76
2.76
F
5.18
P
0.031
Residual
27.04
-26.56
25.44
St Resid
2.38R
-2.33R
2.24R
R denotes an observation with a large standardized residual
5/06/03 252y0342
MTB > regress c1 1 c4
3
Regression Analysis: Price versus Time
The regression equation is
Price = 86.4 + 0.698 Time
Predictor
Coef
SE Coef
T
P
Constant
86.355
4.942
17.47
0.000
Time
0.6980
0.5057
1.38
0.178
slope is insignificant.
S = 12.33
R-Sq = 6.4%
R-Sq(adj) = 3.0%
The p-value above .10 shows us that the
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
28
29
SS
289.6
4255.2
4544.8
Unusual Observations
Obs
Time
Price
2
2.0
115.50
6
4.0
61.90
MS
289.6
152.0
Fit
87.75
89.15
F
1.91
SE Fit
4.07
3.27
P
0.178
Residual
27.75
-27.25
St Resid
2.38R
-2.29R
R denotes an observation with a large standardized residual
MTB > regress c1 2 c2 c4;
SUBC> dw;
SUBC> vif.
2. The researcher now adds time. Compare this regression with the regression with Value alone. Are the
coefficients significant? Does this explain the variation in Y better than the regression with value alone? .
What would the predicted selling price be for an old house with a valuation of 80 in time 17? (3)
Solution: a) This is working out beautifully. R-sq rose, as it almost always does. R-sq adjusted went up,
which is better news. The low p-value associated with the ANOVA indicates a generally successful
regression. The low p-value (.008 is much less than 1%) on the coefficient of ‘Time’ indicates that the
coefficient is highly significant. Both other coefficients are even more significant as shown by the low pvalues.
b) The regression equation is Price = - 45.0 + 1.75 Value + 0.368 Time, so, for the
values given above Price = - 45.0 + 1.75(80) + 0.368(17)= -45.0 + 140 + 6.3 =
101.3 (thousand).
Regression Analysis: Price versus Value, Time
The regression equation is
Price = - 45.0 + 1.75 Value + 0.368 Time
Predictor
Constant
Value
Time
Coef
-44.988
1.75060
0.3680
S = 3.097
SE Coef
6.553
0.08576
0.1281
R-Sq = 94.3%
T
-6.87
20.41
2.87
P
0.000
0.000
0.008
VIF
1.0
1.0
R-Sq(adj) = 93.9%
Analysis of Variance
Source
Regression
Residual Error
Total
Source
Value
Time
DF
1
1
DF
2
27
29
SS
4285.8
258.9
4544.8
MS
2142.9
9.6
F
223.46
P
0.000
Seq SS
4206.7
79.2
4
5/06/03 252y0342
Unusual Observations
Obs
Value
Price
2
86.3
115.500
11
60.8
69.340
20
74.3
83.000
Fit
106.842
63.656
89.146
SE Fit
1.385
1.474
0.680
Residual
8.658
5.684
-6.146
St Resid
3.13R
2.09R
-2.03R
R denotes an observation with a large standardized residual
Durbin-Watson statistic = 2.73
3. The researcher now adds the variable ‘new’ Remember that there is nothing wrong with a negative
coefficient unless there is some reason why it should not be negative.
a. What two reasons would I find to doubt that this regression is an improvement on the
regression with just value and time by just looking at the t tests and the sign of the coefficients?
What does the change in R 2 adjusted tell me about this regression? (3)
b. We have done 5 ANOVA’s so far. What was the null hypothesis in these ANOVA’s and what
does the one where the null hypothesis was accepted tell us? (2)
c. What selling price does this equation predict for an old home with a valuation of 80 in time
17? What percentage difference is this from the selling price predicted in the regression with just
time and value? (2)
d. The last two regressions have a Durbin-Watson statistic computed. What did this test for,
what should our conclusion be, and why is it important? (3)
e. The column marked VIF (variance inflation factor) is a test for (multi)collinearity. The rule of
thumb is that if any of these exceeds 5, we have a multicollinearity problem. None does. What is
multicollinearity and why am I worried about it? (2)
f. Do an F test to show whether the regression with ‘value’, ‘time’ and ‘new’ is an improvement
over the regression with ‘value’ alone. (3)
Solution: a) Note that the p-value of the coefficient of ‘New’ is well above 10%, indicating that the
coefficient is not significant. The negative sign on ‘New’ may also give us problems, since we usually
expect new construction to be more expensive. It is possible, however, that the assessors are systematically
giving higher valuations to newer housing. R-sq did go up, but it almost always goes up. R-sq adjusted fell,
indicating that the explanation is not really better.
b) All but one of the five ANOVAs gave us low p-values, indicating that there is a linear relation between
the independent variables and the dependent variable. The high p-value on the regression against ‘Time’
alone indicates that this variable alone is no better than the average value of ‘Price’ in predicting ‘Price.’
c) Remember that we previously predicted that a home with a valuation of 80 (thousand) in time 17 would
have Price = 101.3 (thousand). Our new equation gives Price = - 47.7 + 1.79 Value +
0.351 Time - 1.22 New = - 47.7 + 1.79(80) + 0.351(17) - 1.22 New(0)
= -47.7 + 143.2 + 6.0 = 101.5 (thousand), a change of price of less than 0.2%.
d) The Durban – Watson statistic tests for a pattern (first-order autocorrelation) in the residuals. If there is
autocorrelation, we could make a better prediction by including time patterns of price variation. In this case
n  30 and k  3, giving us, on the 5% table in the text d L  1.21 and dU  1.65, while the regression
printout says “Durbin-Watson statistic = 2.60”. The diagram in the outline reads
0
+
 0
dL
+
?
dU
+
 0
2
+
 0
4  dU
+
?
4 dL
+
 0 4
+
Since the value given by Minitab is between d U and 4  d U , we do not have significant autocorrelation. In
the previous regression, the D-W statistic was 2.73, n  30 and k  2, giving us, on the 5% table in the
text d L  1.28 and dU  1.57 and, again no significant autocorrelation.
e) Multicollinearity is close correlation between the independent variables and makes accurate values of the
coefficients hard to get.
5
5/06/03 252y0342
f) The current regression has the ANOVA:
Source
Regression
Residual Error
Total
Source
Value
Time
New
DF
3
26
29
DF
1
1
1
SS
4294.0
250.7
4544.8
MS
1431.3
9.6
F
148.42
P
0.000
MS
4206.7
12.1
F
348.37
P
0.000
Seq SS
4206.7
79.2
8.2
The regression against ‘Value’ alone gave:
Source
Regression
Residual Error
Total
DF
1
28
29
SS
4206.7
338.1
4544.8
We can itemize the regression sum of squares in the current regression by using either the sequential sum of
squares in the current regression or looking at the regression sum of squares in the ‘Value’ alone regression.
Source
Value
2 more variables
Residual Error
Total
DF
1
3
26
29
SS
4206.7
87.9
250.7
4544.8
MS
4206.7
42.95
9.6
F
438.20
4.47
F.05
2.98
3,26  2.98 from the F table. Since our computed F is larger
The appropriate F test tests 4.47 against F.05
than the table F, we reject the null hypothesis that the two new independent variables have no explanatory
value.
MTB > regress c1 3 c2 c4 c3;
SUBC> dw;
SUBC> vif.
Regression Analysis: Price versus Value, Time, New
The regression equation is
Price = - 47.7 + 1.79 Value + 0.351 Time - 1.22 New
Predictor
Constant
Value
Time
New
Coef
-47.675
1.79394
0.3508
-1.218
S = 3.105
R-Sq = 94.5%
Analysis of Variance
Source
DF
Regression
3
Residual Error
26
Total
29
Source
Value
Time
New
SE Coef
7.190
0.09804
0.1298
1.322
DF
1
1
1
SS
4294.0
250.7
4544.8
T
-6.63
18.30
2.70
-0.92
P
0.000
0.000
0.012
0.366
VIF
1.3
1.0
1.3
R-Sq(adj) = 93.8%
MS
1431.3
9.6
F
148.42
P
0.000
Seq SS
4206.7
79.2
8.2
Unusual Observations
Obs
Value
Price
Fit
SE Fit
Residual
2
86.3
115.500
107.862
1.777
7.638
11
60.8
69.340
63.502
1.487
5.838
20
74.3
83.000
89.492
0.778
-6.492
R denotes an observation with a large standardized residual
Durbin-Watson statistic = 2.60
St Resid
3.00R
2.14R
-2.16R
6
5/06/03 252y0342
II. Do at least 4 of the following 7 Problems (at least 15 each) (or do sections adding to at least 60 points Anything extra you do helps, and grades wrap around) . 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. Remember: 1) Data must be in order for Lilliefors. Make sure that you do not
cross up x and y in regressions.
1. (Berenson et. al. 1220) A firm believes that less than 15% of people remember their ads. A survey is
taken to see what recall occurs with the following results (In these problems calculating proportions won’t
help you unless you do a statistical test):
Medium
Mag
TV
Radio Total
Remembered
25
10
8
43
Forgot
73
93
107
273
Total
98
103
115
316
a. Test the hypothesis that the recall rate is less than 15% by using proportions calculated from the
‘Total’ column. Find a p-value for this result. (5)
b. Test the hypothesis that the proportion recalling was lower for Radio than TV. (4)
c. Test to see if there is a significant difference in the proportion that remembered according to the
medium. (6)
d. The Marascuilo procedure says that if (i) equality is rejected in c) and
 
(ii) p 2  p3   2 s p , where the chi – squared is what you used in c) and the standard deviation is
2
what you would use in a confidence interval solution to b), you can say that you have a significant
difference between TV and Radio. Try it! (5)
Solution: I have never seen so many people lose their common sense as did on this problem. Many of you
seemed to think that the answer to c) was the answer to a) in spite of the fact that .15 appeared nowhere in
your answer. An A student tried at one point to compare the total fraction that forgot with the total fraction
that remembered, even though the method she used was intended to compare fractions of two different
groups and the two fractions she compared could only have been the same if they were both .5, since they
had to add to one.
a) From the formula table.
Interval for
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
Proportion
p  p0
p  p  z 2 s p
pcv  p0  z 2  p
H 0 : p  p0
z

H
:
p

p
p
1
0
pq
p0 q0
sp 
p 
n
n
q  1 p
q0  1  p0
H 1 : p  .15 It is an alternate hypothesis because it does not contain an equality. The null
hypothesis is thus H 0 : p  .15. Initially, assume   .05 and note than n  316 , x  43 so that
p0 q0
x
43
.15.85 

 .1361 .  p 

 .0004075  .02019 . This is a one-sided test and
n 316
n
316
z  z .05  1.645 . This problem can be done in one of three ways.
p
7
5/06/03 252y0342
(i) The test ratio is z 
p  p0
p

.1361  .15
 0.6885 . Make a diagram of a normal curve with a
.02019
mean at zero and a reject zone below - z   z.05  1.645 . Since z  0.6885 is not in the
'reject' zone, do not reject H 0 . We cannot say that the proportion who do not recall is
significantly below 15%. We can use this to get a p-value. Since our alternate hypothesis is
.1361  .15 

H 1 : p  .15 , we want a down-side value, i.e. P p  .1361   P z 

.02019 

 Pz  .68846   Pz  0  P0.69  z  0  .5  .2549  .2451 . Since the p-value is above the
significance level, do not reject H 0 . Make a diagram. Draw a Normal curve with a mean at .15
and represent the p-value by the area below .1361, or draw a Normal curve with a mean at zero and
represent the p-value by the area below -0.69.
(ii) Since the alternative hypothesis says H 1 : p  .15 we need a critical value that is below .15.
We use pcv  p0  z  p  .15 1.645.02019  .1168. Make a diagram of a normal curve with a
mean at .15 and a ‘reject’ zone below .1168. Since p  .1361 is not in the 'reject' zone, do not
reject H 0 . We cannot say that the proportion is significantly below 15%.
pq
. To make the 2-sided confidence interval,
n
p  p  z 2 s p , into a 1-sided interval, go in the same direction as H 1 : p  .15 We get
(iii) To do a confidence interval we need s p 
pq
.1361 .8639 

 .000372  .01928 . Thus the interval is
n
316
sp 
 .1361  1.645 .01928   .1678 .
p  p  z s p
p  1678 does not contradict the null hypothesis.
8
10
 .069565 , n1  115 and p 2 
 .09709 , n 2  103 .
115
103
Confidence
Hypotheses
Test Ratio
Critical Value
Interval
pcv  p0  z 2  p
p  p 0
p  p  z 2 sp
H 0 : p  p0
z
If p0  0
 p
H 1 : p  p0
p  p1  p2
b) We are comparing p 3 
Interval for
Difference
between
proportions
q  1 p
s p 
p1q1 p2 q 2

n1
n2
p 0  p 01  p 02
or p 0  0
If p  0
p 
 p 
p01q01 p02q02

n1
n2
Or use s p
s p 
p0 q 0  1 n1 
1
n2

n p  n2 p2
p0  1 1
n1  n 2
p3 q3 p 2 q 2
.069565 .930435  .09709 .90291 



 .00056283  .00085108  .0014139  .037602
n3
n2
115
103
p  p3  p 2  .02752 , p 0 
n p  n 2 p 2 115 .069565   103 .09709 
8  10
 1 1

 .08257 ,
115  103
n1  n 2
115  103
  .05, z  z.05  1.645 . Note that q  1  p and that q and p are between 0 and 1.
 p  p 0 q 0

1
n1

1
n3

.08257 .91743  1115  1103 
.00139415  .037338
H 0 : p 3  p 2
H 0 : p 3  p 2  0
H0 : p  0
Our hypotheses are 
or 
or 
H1 : p 0
H 1 : p 3  p 2
H1 : p 3  p 2  0
There are three ways to do this problem. Only one is needed
8
5/06/03 252y0342
(i) Test Ratio: z 
p  p 0
 p

 .02752  0
 0.7371
.037338
Make a Diagram showing a 'reject'
region below -1.645. Since -0.7371 is above this value, do not reject H 0 .
(ii) Critical Value: pcv  p0  z  p becomes pcv  p0  z p
2
 0  1.645 .037338   .061421 . Make a Diagram showing a 'reject' region below - 0.06142.
Since p  .02752is not below this value, do not reject H 0 .
(iii) Confidence Interval:: p  p  z s p becomes p  p  z sp
2
 .02752  1.645 .037338   0.03390 . Since
not reject H 0 .
c)
DF  r  1c  1  12  2
H 0 : Homogeneousor p1  p 2  p 3 
H 1 : Not homogeneousNot all ps are equal
O
R
F
Total
M
 25

 73
98
T
010
R Total
008  043

093 107  273
103 115
316
pr
.13608
.86392
1.0000
p  .03390 does not contradict p  0 , do
.2052   5.9915
E
On
1
2
3
Total
pr
13 .3358 14 .0162 15 .6492  43 .000
.13608


Oft
84 .6642 89 .9838 99 .3508  273 .000 .86392
Total 98 .0000 103 .000 115 .000 316 .000 1.00000
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  2  17.4689  333.469  316 is computed two different
ways here - only one way is needed.)
O2
O  E 2
Row
E  O2
E O
O
E
E
E
1
2
3
4
5
6
25
73
10
93
8
107
316
13.3358
84.6642
14.0162
88.9838
15.6492
99.3508
316.000
-11.6642
11.6642
4.0162
-4.0162
7.6492
-7.6492
0.0000
136.053
136.053
16.130
16.130
58.510
58.510
10.2020
1.6070
1.1508
0.1813
3.7389
0.5889
17.4689
46.866
62.943
7.135
97.198
4.090
115.238
333.469
Since the  2 computed here is greater than the  2 from the table, we reject H 0 .
d) The Marascuilo procedure says that if (i) equality is rejected in c) and
 
(ii) p 2  p3   2 s p , where the chi – squared is what you used in c) and the standard deviation is
2
what you would use in a confidence interval solution to b), you can say that you have a significant
difference between TV and Radio.
OK – We already have DF  r  1c  1  12  2,  .2052  5.9915
s p 
p3 q3 p 2 q 2
.069565 .930435  .09709 .90291 



 .00056283  .00085108  .0014139  .037602
n3
n2
115
103
p  p3  p 2  .02752 . I guess we really should use
 .2025  7.3778  2.7162 and
 22 s p   2.7162 .037602   .10213 . Since p 2  p3 is obviously smaller than this, we do not have a
significant difference in these 2 proportions.
9
5/06/03 252y0342
2. (Berenson et. al. 1142) A manager is inspecting a new type of battery. These are subjected to 4 different
pressure levels and their time to failure is recorded. The manager knows from experience that such data is
not normally distributed. Ranks are provided.
PRESSURE
Use
low
1
2
3
4
5
8.2
8.3
9.4
9.6
11.9
rank normal
11
12
15
16
19
7.9
8.4
10.0
11.1
12.5
rank
high
rank
whee!
rank
9
13
17
18
20
6.2
6.5
7.3
7.8
9.1
4
5
7
8
14
5.3
5.8
6.1
6.9
8.0
1
2
3
6
10
a. At the 5% level analyze the data on the assumption that each column represents a random
sample. Do the column medians differ? (5)
b. Rerank the data appropriately and repeat a) on the assumption that the data is non-normal but
cross classified by use. (5)
c. This time I want to compare high pressure (H) against low - moderate pressure (L). I will write
out the numbers 1-20 and label them according to pressure.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
H H H H H H H H L
H L L L H L L L L L L
Do a runs test to see if the H’s and L’s appear randomly. This is called a Wald-Wolfowitz test for the
equality of means in two nonnormal samples. Null hypothesis is that the sequence is random and the means
are equal. What is your conclusion? (5)
Solution: a) This is a Kruskal – Wallis Test, Equivalent to one-way ANOVA when the underlying
distribution is non-normal.
H 0 : Columns come from same distribution or medians equal. I am basically copying the outline. There are
n  20 data items, so rank them from 1 to 20. Let n i be the number of items in column i and SRi be the
rank sum of column i . n 
1
2
3
4
5
8.2
8.3
9.4
9.6
11.9
11
12
15
16
19
SR1  73
n
i
7.9
8.4
10.0
11.1
12.5
.
9
13
17
18
20
SR2  77
6.2
6.5
7.3
7.8
9.1
4
5
7
8
14
SR3  38
5.3
5.8
6.1
6.9
8.0
1
2
3
6
10
SR4  22
To check the ranking, note that the sum of the four rank sums is 73.0 + 77.0 + 38.0 + 22.0 = 210.0, and
nn  1 20 21

 210 .
that the sum of the first n numbers is
2
2
 12
 SRi 2 

  3n  1
Now, compute the Kruskal-Wallis statistic H  
 nn  1 i  ni 
 12  73 .02 77 .02 38 2 22 2 

  321  12 1 5329  5929  1444  484   63  12 .3486 .








20
21
5
5
5
5
420 5




If the size of the problem is larger than those shown in Table 9, use the  2 distribution, with df  m  1  3 ,
where m is the number of columns. Compare H with  .2053  7.81475 . Since H is larger than  .205 , reject
the null hypothesis.
10
5/06/03 252y0342
b) This is a Friedman test ,equivalent to two-way ANOVA with one observation per cell when the
underlying distribution is non-normal.
H 0 : Columns come from same distribution or medians equal. Note that the only difference between this
and the Kruskal-Wallis test is that the data is cross-classified in the Friedman test.
1
2
3
4
5
8.2
8.3
9.4
9.6
11.9
4
3
3
3
3
SR1  16
7.9
8.4
10.0
11.1
12.5
3
4
4
4
4
SR2  19
6.2
6.5
7.3
7.8
9.1
2
2
2
2
2
SR3  10
5.3
5.3
6.1
6.9
8.0
1
1
1
1
1
SR4  5
Assume that   .05 . In the data, the pressures are represented by c  4 columns, and the uses by
r  5 rows.. In each row the numbers are ranked from 1 to c  4 . For each column, compute SRi , the rank
sum of column i .
To check the ranking, note that the sum of the four rank sums is 16 + 19 + 10 + 5 = 50, and that the sum of
cc  1
the c numbers in a row is
. However, there are r rows, so we must multiply the expression by r .
2
rcc  1 545
SRi 

 50 .
So we have
2
2
 12

SRi2   3r c  1
Now compute the Friedman statistic  F2  
 rc c  1 i


 


 12
16 2  19 2  10 2  52   355   12 256  361  100  25   75  14 .04 . Since the

100

 545

size of the problem is larger than those shown in Table 10, use the  2 distribution, with df  c  1 ,
where c is the number of columns. Again,  .2053  7.81475 . Since our statistic is larger than  .205 , reject the
null hypothesis.
c) . This time I want to compare high pressure (H) against low - moderate pressure (L). I will write out the
numbers 1-20 and label them according to pressure.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
H H H H H H H H L
H L L L H L L L L L L
If we do a runs test, n  20 , r  6, n1  10 and n 2  10 . Can you see why r  6 ? I have underlined
alternate runs to help you. The runs test table gives us critical values of 6 and 16. The directions on the table
say to reject randomness if r  6 or r  16 . Since the sequence is not random, the means are not equal.
11
5/06/03 252y0342
3. A researcher studies the relationship of numbers of subsidiaries and numbers of parent companies in 11
metropolitan areas and finds the following:
Area
parents
subsidiaries
1
2
3
4
5
6
7
8
9
10
11
x2
y
x
658
396
357
266
231
223
207
156
146
143
139
2922
2602
1709
1852
1223
875
666
1519
884
477
564
657
13028
xy
432964
156816
127449
70756
53361
49729
42849
24336
21316
20449
19321
1019346
1712116
676764
661164
325318
202125
148518
314433
137904
69642
80652
91323
4419959
y2
6770404
2920681
3429904
1495729
765625
443556
2307361
781456
227529
318096
431649
19891990
a. Do Spearman’s rank correlation between x and y and test it for significance (6)
b. Compute the sample correlation between x and y and test it for significance (6)
c. Compute the sample standard deviation of x and test to see if it equals 200 (4)
Solution: a) First rank x and y into rx and r y respectively. (As usual, people tried to compute rank
correlations without ranking, I warned you!) Compute d  rx  ry and d 2 .
Area
parents
1
2
3
4
5
6
7
8
9
10
11
subsidiaries
x
y
rx
r y d  rx  ry d 2
658
396
357
266
231
223
207
156
146
143
139
2602
1709
1852
1223
875
666
1519
884
477
564
657
11
10
9
8
7
6
5
4
3
2
1
66
11
9
10
7
5
4
8
6
1
2
3
66
0
1
-1
1
2
2
-3
-2
2
0
-2
0
0
1
1
1
4
4
9
4
4
0
4
32
 d  0 is a check on the correctness of the ranking.
6 d
192
632 
From the outline: r  1 
 1
 1  0.1455  0.8545
 1
11121 
nn  1
1111  1
Note that
2
s
2
2
The 11 line from the rank correlation table has
n
  .050
  .025
  .010
  .005
11
.5273
.6091
.7000
.7818
H 0 :  s  0
If you tested 
at the 5% level, reject the null hypothesis of no relationship if rs is above .5273
H 1 :  s  0
H 0 :  s  0
or, if you tested 
at the 5% level, reject the null hypothesis if rs is above .6091. So, in this
H 1 :  s  0
case we have a significant rank correlation.
12
5/06/03 252y0342
 x  2922,
b) We need the usual spare parts. From above n  11,
 y  13028,  x
2
 1019346,
 xy  4419959
and
y
2
 19891990. (In spite of the
fact that most column computations were done for you, many of you wasted time and energy doing them
over again.)
Spare Parts Computation:
SSx 
x 2  nx 2  1019346  11265 .636 2
x 2922
 243158 .7
x

 265 .636
n
11
Sxy 
xy  nx y  4419959  11265 .636 1184 .36 
y 13028
 959263 .8
y

 1184 .36
n
11




SSy 
y
2
 ny 2  19891990  111184 .36 2
 4462195 .3
The simple sample correlation coefficient is r 
 XY  nXY
 X  nX  Y
2
R
2
 nY 2
square root of
 XY  nXY 
Sxy 
959263 .8


 .8480828




SSx
SSy
243158
.7 4462195 .3
 X  nX  Y  nY 
2
2
2
2
2
2
2
2
2
r  .8480828  .9209 . From the outline, if we want to test H 0 : xy  0 against H1 : xy  0 and
x and y are normally distributed, we use t n  2  
r

1 r 2
n2
.9209
1  .9209 2
11  2

.9209
.01688

.9209
 7.088 .
.12993
9
 2.262 , we reject H 0 . Note that R-squared is always between 0 and 1 and that correlations are
Since t .025
always between -1 and +1.
c) From the formula table (but the outline is better)
Interval for
Confidence
Hypotheses
Interval
VarianceH 0 :  2   02
n  1s 2
2  2
Small Sample
 .5 .5 2 
H1: :  2   02
s 2DF 
VarianceH 0 :  2   02
 
Large Sample
 z  2DF 
H1 :  2   02
2
We already know n  11 and SSx 
x
2
 nx 2
x
2
Test Ratio
2 
Critical Value
n  1s 2
 cv2 
 02
n  1s 2
 .25.5 2 
z 
2  2  2DF   1
 nx 2  1019346 11265.6362  243158.7 . Assume   .05 .


SSx 243158 .7

 24315 .87 s  24315 .87  155 .93
n 1
10
H 0 :   200
n  1s 2  1024315 .87   6.0790 .
2 
a) Only the test ratio method is normally used. 
 02
200 2
H 1 :   200
then s x2 
n 1

10  20 .4832
DF  n  1  10.  .2025
Make a diagram. Show a curve with a mean at 10 and rejection zones
10  20 .4832
10  3.2470
(shaded) above  .2025
and below  .2975
. Since your value of  2 is between the two
critical values, do not reject H 0 .
13
5/06/03 252y0342
4. Data from the previous page is repeated:
Area
parents
1
2
3
4
5
6
7
8
9
10
11
658
396
357
266
231
223
207
156
146
143
139
2922
subsidiaries
x2
y
x
2602
1709
1852
1223
875
666
1519
884
477
564
657
13028
432964
156816
127449
70756
53361
49729
42849
24336
21316
20449
19321
1019346
y2
xy
1712116
676764
661164
325318
202125
148518
314433
137904
69642
80652
91323
4419959
6770404
2920681
3429904
1495729
765625
443556
2307361
781456
227529
318096
431649
19891990
a. Test the hypothesis that the correlation between x and y is .7 (5)
b. Test the hypothesis that x has the Normal distribution. (9)
c. Test the hypothesis that x and y have equal variances. (4)
Solution: a) From the previous page we know n  11 and
 XY  nXY 
Sxy 
959263 .8


 X  nX  Y  nY  SSxSSy 243158 .74462195 .3  .8480828
2
R
2
2
2
2
2
2
2
r  .8480828  .9209 . From the Correlation section of the outline:
If we are testing H 0 : xy   0 against H 1 : xy   0 , and  0  0 , the test is quite different.
1 1 r 
z  ln 
We need to use Fisher's z-transformation. Let ~
 . This has an approximate mean
2  1 r 
~
n 2 
z  z
1
1  1 0 
 and a standard deviation of s z 

of  z  ln 
, so that t
.

n3
sz
2  1 0 
(Note: To get ln , the natural log, compute the log to the base 10 and divide by .434294482. )
H 0 : xy  0.7
Test 
when n  11, r  .9209 and r 2  .8481   .05  .
H 1 : xy  0.7
1  1  r  1  1  .9209  1  1.9209  1
1
~
z  ln 
  ln 
  ln 
  ln 24 .28445   3.18984   1.5949
2  1  r  2  1  .9209  2  0.0791  2
2
1 1 0
 z  ln 
2  1 0
sz 
 1  1  .7  1  1.7  1
1
  ln 
 2  1  .7   2 ln  0.3   2 ln 5.66670   2 1.73460   0.86730

~
z   z 1.5949  0.86730
1
1
1


 0.35355 .

 2.058 . Compare this
Finally t 
n3
11  3
8
sz
0.35355
with  t n2 2  t .9025  2.262 . Since 2.058 lies between these two values, do not reject the null hypothesis.
14
5/06/03 252y0342
1 1 r 
~
z10  log 
 . This has an approximate mean of
2  1 r 
~
n  2 
z   z 10
0.18861
and a standard deviation of s z 10 
, so that t
.
 10
n3
s z 10
Note: To do the above with logarithms to the base 10, try
1 1 0
 z 10  log
2  1 0




b) From the previous page we know n  11 , x 
 x  nx
 x  nx

SSx 
2
2
2
 x  2922  265 .636 ,
n
11
 1019346 11265.636  243158.7 and
2

2

SSx 243158 .7


 24315 .87 s  24315 .87  155 .93
n 1
n 1
10
Use the setup in Problem E9 or E10. The best method to use here is Lilliefors because the data is
not stated by intervals, the distribution for which we are testing is Normal, and the parameters of the
xx
distribution are unknown. We begin by putting the data in order and computing z 
(actually t ) and
s
proceed as in the Kolmogorov-Smirnov method. For example, in the second row
143  265 .636
O  11 - so
z
 0.79 , O  1 because there is only one number in each interval, n 
155 .93
each value of O is 1  .0909 . Since the highest number in the interval represented by Row 2 is
n
11
z  0.79 , Fe  F 0.79   Pz  0.79   Pz  0  P0.79  z  0  .5  .2852  .2146 .
s x2

Row
1
2
3
4
5
6
7
8
9
10
11
x
139
143
146
156
207
223
231
266
357
396
658
z
-0.81
-0.79
-0.77
-0.70
-0.38
-0.27
-0.22
0.00
0.59
0.84
2.52
O
O
Fo
n
D  Fe  Fo
Fe
1 .0909 .0909
1 .0909 .1818
1 .0909 .2727
1 .0909 .3636
1 .0909 .4545
1 .0909 .5455
1 .0909 .6364
1 .0909 .7273
1 .0909 .8192
1 .0909 .9091
1 .0909 1.0000
11 0.9999
.5
.5
.5
.5
.5
.5
.5
.5
.5
.5
.5
-
.2910
.2852
.2794
.2580
.1480
.1064
.0871
=
=
=
=
=
=
=
=
+ .2224 =
+ .2995 =
+ .4941 =
.2090
.2146
.2206
.2420
.3520
.3936
.4129
.5000
.7224
.7995
.9941
.1181
.0328
.0521
.1216
.1025
.1519
.2235
.2273
.0968
.1096
.0059
From the Lilliefors table for   .05 and n  11 , the critical value is .249. Since the maximum
deviation (.2273) is below the critical value, we do not reject H 0 .
c) From the previous page SSy 
y
2
 ny 2
y
2
 ny 2  19891990 111184.362  4462195.3
SSy 4462195 .3

 446219 .53 s x2 
n 1
10
x
2
 nx 2
SSx 243158 .7


 24315 .87
n 1
n 1
10
If we follow 252meanx4 in the outline: Our Hypotheses are H 0 :  x2   y2 and H 1 :  x2   y2 .
s 2y 
n 1

DF x  n  1  10 and DFy  n 1  10 , Since the table is set up for one sided tests, if we wish to test
H 0 :  x2   y2 , we must do two separate one-sided tests. First test
 DFx, DFy 
10,10  3.72
F.025
 F.025
and then test
s 2y
s x2

s x2
s 2y

24315 .87
 0.0545 against
446219 .53
1
 DFy, DFx
10,10  3.72
 F.025
. If
 18 .351 against F.025
.0545
either test is failed, we reject the null hypothesis. Since 18.351 is larger than this critical value, reject H 0 .
15
5/06/03 252y034
5. Data from the previous page is repeated:
Area
parents
subsidiaries
x2
y
x
1
2
3
4
5
6
7
8
9
10
11
658
396
357
266
231
223
207
156
146
143
139
2922
2602
1709
1852
1223
875
666
1519
884
477
564
657
13028
y2
xy
432964
156816
127449
70756
53361
49729
42849
24336
21316
20449
19321
1019346
1712116
676764
661164
325318
202125
148518
314433
137904
69642
80652
91323
4419959
6770404
2920681
3429904
1495729
765625
443556
2307361
781456
227529
318096
431649
19891990
a. Compute a simple regression of subsidiaries against parents as the independent variable. (5)
b. Compute s e . (3)
c. Predict how many subsidiaries will appear in a city with 60 parent corporations. (1)
d. Make your prediction in c) into a confidence interval. (3)
e. Compute s b0 and make it into a confidence interval for  0 . (3)
f. Do an ANOVA for this regression and explain what it says about 1 . (3)
Solution:
We need the usual spare parts. From above n  11,
 xy  4419959 and  y
2
 x  2922,  y  13028,  x
SSx 
11
n
11
x
2
 nx 2  1019346  11265 .636 2
 243158 .7
 x  2922  265 .636
n
 1019346,
 19891990.
Spare Parts Computation:
(Repeated from previous page)
x
2
Sxy 
 xy  nxy  4419959 11265 .636 1184 .36 
 959263 .8
 y  13028  1184 .36
y
SSy 
y
2
 ny 2  19891990  111184 .36 2
 4462195 .3
b1 
Sxy

SSx
 xy  nxy  959263 .8  3.9450
 x  nx 243158 .7
2
b0  y  b1x  1184 .36  3.9450 265 .636  136 .42
2
Yˆ  b0  b1 x becomes Yˆ  136.42  3.9450 x .
b) We already know that SST  SSy 
y
2
 ny 2  19891990 111184.362  4462195.3 and that
 XY  nXY 
Sxy 
959263 .8
R 

 .8480828




SSx
SSy
243158
.7 4462195 .3
 X  nX  Y  nY 
SSR  b Sxy  b  xy  nx y   3.9450 959263 .8  3784295 .7 or
2
2
2
2
2
1
2
2
2
1
SSR  b1 Sxy  R 2 SST   .84808284462195.3  3784311
SSE  SST  SSR  4462195 .3  3784311  677884 .3
s e2 
SSE 677884 .3

 75320 .5
n2
9
16
5/06/03 252y0342
or
s e2
 y

2
  x
 ny 2  b12
2
 nx 2

n2

4462195 .3  3.9450 2 243158 .7 
9

677910
 75323 .3
9
So s e  75320 .5  274 .446
(
is always positive!)
ˆ
c) If Y  136.42  3.9450 x and x  60 , the prediction is Yˆ  136.42  3.945060  373.1.
s e2


2
1

X0  X
 . In
d) From the outline, the Confidence Interval is  Y0  Yˆ0  t sYˆ , where sY2ˆ  s e2  
n
X 2  nX 2 


ˆ
ˆ
this formula, for some specific X 0 , Y0  b0  b1 X 0 . Here X 0  60 , Y0  373.1 , X  265.636 and
n  11 . Then

X 0  X 2
2

  75320 .5 1  60  265 .636    75320 .5.26481   19945 .6 and
 11
243158 .7 
n
X 2  nX 2 



9
sYˆ  19945.6  141.229 , so that, if tn  2  t.025
 2.262 the confidence interval is
sY2ˆ



1
s e2 



2
 Y0
 Yˆ0  t sYˆ  373 .1  2.262 141 .2  373  319 . This represents a confidence interval for the average
value that Y will take when x  50 and is proportionally rather gigantic because we have picked a point
fairly far from the data that was actually experienced.
e) The outline says
2
1


X2
  75320 .5 1  265 .636    75320 .50.38110   28704 .64  169 .42 .
s b20  s e2  
n
X 2  nX 2 
11 243158 .7 


So the interval is  0  b0  t  sb0  136.42  2.262169.42  136  383. This indicates that the intercept is

2
not significant.
f) We can do a ANOVA table as follows:
Source
SS
DF
MS
SSR
MSR
Regression
1
Error
Total
SSE
SST
n2
n 1
F
MSR MSE
MSE
Source
SS
DF MS
F
Regression
3784311
1 3784311 50.243
Error
677884
9
75320
Total
4462195 11
1,9   5.12 so we reject the null hypothesis that x and
Note that F.05
y are unrelated. This is the same as
saying that H 0 : 1  0 is false.
17
5/06/03 252y0342
The Minitab output for this problem follows:
The regression equation is
subno = 136 + 3.94 parno
Predictor
Constant
parno
S = 274.4
Coef
136.4
3.9450
SE Coef
169.4
0.5566
R-Sq = 84.8%
T
0.81
7.09
P
0.441
0.000
R-Sq(adj) = 83.1%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
9
10
SS
3784219
677882
4462101
Unusual Observations
Obs
parno
subno
1
658
2602.0
7
207
1519.0
MS
3784219
75320
Fit
2732.2
953.0
F
50.24
SE Fit
233.5
89.0
P
0.000
Residual
-130.2
566.0
St Resid
-0.90 X
2.18R
R denotes an observation with a large standardized residual
X denotes an observation whose X value gives it large influence.
18
5/06/03 252y0342
6. A chain has the following data on prices, promotion expenses and sales of one product. (You can do
x x
1 2
100.
Store
1
2
3
4
5
6
7
8
9
10
11
12
): Computation of this sum is in red. It might help to divide y , x1 y and x 2 y by 10 and y 2 by
sales
promotion
y
x1
x2
x12
3842
3754
5000
1916
3224
2618
3746
3825
1096
1882
2159
2927
35989
59
59
59
79
79
79
79
79
99
99
99
99
968
200
400
600
200
200
400
600
600
200
400
400
600
4800
3481
3481
3481
6241
6241
6241
6241
6241
9801
9801
9801
9801
80852
y2
x 22
Store
1
2
3
4
5
6
7
8
9
10
11
12
price
40000
160000
360000
40000
40000
160000
360000
360000
40000
160000
160000
360000
2240000
x1 y
14760964
14092516
25000000
3671056
10394176
6853924
14032516
14630625
1201216
3541924
4661281
8567329
121407527
y  2999.08, x1  80.6667
226678
221486
295000
151364
254696
206822
295934
302175
108504
186318
213741
289773
2752491
x1 x 2
11800
23600
35400
15800
15800
31600
47400
47400
19800
39600
39600
59400
387200
x2 y
768400
1501600
3000000
383200
644800
1047200
2247600
2295000
219200
752800
863600
1756200
15479600
x 2  400.000.
and
a. Do a multiple regression of sales against x1 and x 2 . (10)
b. Compute R 2 and R 2 adjusted for degrees of freedom. Use a regression ANOVA to test the usefulness
of this regression. (6)
d. Use your regression to predict sales when price is 79 cents and promotion expenses are $200. (2)
e. Use the directions in the outline to make this estimate into a confidence interval and a prediction interval.
(4)
f. If the regression of Price alone had the following output: The regression equation is
sales = 7391 - 54.4 price
Predictor
Constant
price
Coef
7391
-54.44
S = 726.2
SE Coef
1133
13.81
R-Sq = 60.9%
T
6.52
-3.94
P
0.000
0.003
R-Sq(adj) = 56.9%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
10
11
SS
8200079
5273437
13473517
MS
8200079
527344
F
15.55
P
0.003
Do an F-test to see if adding x 2 helped. (4). The next page is blank – please show your work. I suggested
that we divide y values by 10 and y squared by 100 to make computations more tractable. No one did, so I
have redone this on the following page. After a year of statistics, too many of you decided that
x1 y  ?
x1
x 2 and did something similar for some of the other sums. Where have you been?

  
19
5/06/03 252y0342
 y  3598.90,  x  968,  x  4800,  x  80852,
 x  2240000,  y  1214075.27,  x y  275249.1,  x y  1547960.0 and you should have found
 y  3598 .90  299 .908 , x   x  968  80.6667 , and
x
x

387200.
First,
we
compute
y


n
12
n
12
 x  5200  400 .00 . Then, we compute or copy our spare parts:
x 
Solution: a)(With y divided by 10) n  12,
2
2
1
2
1
2
2
1
2
1
1 2
1
2
2
n
12
SST  SSy 
y
 ny  1214075 .27  12 299 .908 2  134738 *
2
2
 x y  nx y  275249 .1  1280.6667 299 .908   15062
Sx y   X Y  nX Y  1547960 .0  12400 299 .908   108401
SSx1   x12  nx12  80852  1280.6672  2766*
SSx2   X 22  nX 22  2240000 124002  320000*
and Sx x   X X  nX X  387200  1280.6667 400   0 .
Sx1 y 
1
2
1
2
2
1 2
1
2
1
2
* indicates quantities that must be positive.
Then 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
2
1
1
2
 
 nX X   b  X
1
2
2
2
2

 nX  ,
2
2
 15062  2766 b1  0b2

 108401  0b1  320000 b2
which are
and solve them as two equations in two unknowns for b1 and b2 . Because of the zero Sx1 x 2 term, these
are extremely unusual normal equations, so do not use them to study for exams. The first equation can be
15062
 5.445 . The second is 320000 b2  108401 . This means
written as 2766 b1  15062 so b1 
2766
108401
 0.33875 . Finally we get b0 by solving b0  Y  b1 X 1  b2 X 2
that b2 
320000
 299 .908  5.445 80 .6667   0.33875 400   299 .908  438 .230  135 .500  602 .64 . Thus our
equation is Yˆ  b  b X  b X  604.64  5.445X  0.33875X .
0
1
1
2
2
1
2

1
 x 2  4800,  x  80852,
 x  2240000,  y  121407527,  x y  2752491,  x y  15479600 and you should have found
 y  35989  2999 .08 , x   x  968  80.6667 , and
x
x

y

387200.
First,
we
compute

n
12
n
12
 x  5200  400 .00 . Then, we compute or copy our spare parts:
x 
a)(The way most of you did it) n  12,
2
2
y  35989,
x  968,
2
1
2
1
2
1
1 2
1
2
2
n
12
20
5/06/03 252y0342
SST  SSy 
y
 ny  121407527  12 2999 .08 2  13473757 *
2
2
 x y  nx y  2752491  1280.6667 2999 .08   150620
Sx y   X Y  nX Y  15479600  12400 2999 .08   1084016
SSx1   x12  nx12  80852  1280.6672  2766*
SSx2   X 22  nX 22  2240000 124002  320000*
and Sx x   X X  nX X  387200  1280.6667 400   0 .
Sx1 y 
1
2
1
2
2
1 2
1
2
1
2
* indicates quantities that must be positive. Before we got to this point some of you decided to compute the coefficients as
if this were simple regression. Unless I saw some evidence that you knew that Sx1x2 was zero, you got no credit. For a
more conventional solution see exam 252y0341.
Then 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
2
1
1
2
 
 nX X   b  X
1
2
2
2
2

 nX  ,
2
2
 150620  2766 b1  0b2

 1084016  0b1  320000 b2
which are
and solve them as two equations in two unknowns for b1 and b2 . Because of the zero Sx1 x 2 term, these
are extremely unusual normal equations, so do not use them to study for exams. The first equation can be
15062
 54 .454 . The second is 320000 b2  1084016 . This means
written as 2766 b1  150620 so b1 
2766
108401
 3.38755 . Finally we get b0 by solving b0  Y  b1 X 1  b2 X 2
that b2 
320000
 2999 .08  54 .454 80.6667   3.38755 400   2999 .08  4392 .6245  1355 .02  6036 .68 . Thus our
equation is Yˆ  b  b X  b X  6036.68  54.454X  3.38755X .
0
1
1
2
2
1
2
b) (The way I did it) SSE  SST  SSR and that
SSR  b1 Sx1 y  b2 Sx2 y  5.445 15062   0.33875 108401   82013  38721  118734
SST  SSy 
R2 
y
2
2
 ny  134738 so SSE  134738  118734  16004
SSR 118734

 0.881 . If we use R 2 , which is R 2 adjusted for degrees of
SST 134737
freedom R 2 
n  1R 2  k  110.881  2  .855 .
n  k 1
Source
Regression
Residual Error
Total
DF
2
9
12
9
SS
118734
16004
134737
the ANOVA reads:
MS
59367.0
1778.2
F
33.39
F.05
4.26
Since our computed F is larger that the table F, we reject the hypothesis that X and Y are unrelated.
b) (The way most of you did it) SSE  SST  SSR and that
SSR  b1 Sx1 y  b2 Sx2 y  54.454 150620   3.38755 1084016   8201861 .5  3672158 .4  11874020
SST  SSy 
y
2
2
 ny  13473757 so SSE  13473757  11874020  1599737
21
5/06/03 252y0342
R2 
SSR 11874020

 0.881 . If we use R 2 , which is R 2 adjusted for degrees of
SST 13473757
freedom R 2 
n  1R 2  k  110.881  2  .855 .
Source
Regression
Residual Error
Total
n  k 1
9
DF
2
9
12
SS
11874020
1599737
13473757
the ANOVA reads:
MS
5937010
177748.56
F.05
F
33.40
4.26
Since our computed F is larger that the table F, we reject the hypothesis that X and Y are unrelated.
c) X 1  79, X 2  200
(My way) Yˆ  604.64  5.44579  0.33875200  606.64  430.155  67.75  244.235
(Your way) Yˆ  6036.68  54.45479  3.38755200  6036.68  4301.87  677.51  2412.22
d) We need to find s e . The best way to do this is to do an ANOVA or remember that s e2 
SSE
, and
n  k 1
that you got SSE  13473757  11874020  1599737 .
SSE 1599737
9
 2.262 . The outline says that an approximate
s e2 

 177748 .56 , so s e  421 .60 . t .025
n2
9
s
421 .6
 2412  275 and an approximate
confidence interval is  Y0  Yˆ0  t e  2412 .22  2.262
12
n
prediction interval is Y  Yˆ  t s  2412 .22  2.262 421 .6  2412  954 .
0
0
e
e) We can copy the Analysis of Variance in the question.
Source
Regression
Residual Error
Total
DF
1
10
11
SS
8200079
5273437
13473517
MS
8200079
527344
F
15.55
P
0.003
But we just got
Source
Regression
Residual Error
Total
DF
2
9
12
SS
11874020
1599737
13473757
MS
5937010
177748.56
F
33.40
F.05
4.26
Let’s use our SST and itemize this as
Source
DF
SS
MS
X1
1
8200079
8200079
46.13
X2
1
3673941
3673941
20.67
9
12
1599737
13473757
Residual Error
Total
F
F.05
1,9   5.12
F.05
1,9   5.12
F.05
177748.56
The appropriate F to look at is opposite X2. Since 20.67 is above the table value, reject the null hypothesis
of no relationship. Yes, the added variable helped.
22
5/06/03 252y0342
I also ran this on the computer.
Regression Analysis: sales versus price, promotion
The regression equation is
sales = 6036 - 54.4 price + 3.39 promotion
Predictor
Constant
price
promotio
Coef
6035.7
-54.442
3.3875
S = 421.8
SE Coef
722.7
8.020
0.7457
R-Sq = 88.1%
T
8.35
-6.79
4.54
P
0.000
0.000
0.001
R-Sq(adj) = 85.5%
Analysis of Variance
Source
Regression
Residual Error
Total
Source
price
promotio
DF
1
1
DF
2
9
11
SS
11872129
1601387
13473517
MS
5936065
177932
F
33.36
P
0.000
Seq SS
8200079
3672050
Unusual Observations
Obs
price
sales
5
79.0
3224
Fit
2412
SE Fit
193
Residual
812
St Resid
2.16R
R denotes an observation with a large standardized residual
23
5/06/03 252y0342
7. The Lees present the following data on college students summer wages vs. years of work experience
blocked by location.
Years of Work Experience
Region
1
2
3
1
16
19
24
2
21
20
21
3
18
21
22
4
14
21
25
a. Do a 2-way ANOVA on these data and explain what hypotheses you test and what the
conclusions are. (9) (Or do a 1-way ANOVA for 6 points.) The following column sums are done for you:
x
1
 69,
x
2
 81, n1  4, n 2  4,
x
2
1
 1217 and
x
2
2
 1643. So x1  17.25,and x 2  20.25.
b. Do a test of the equality of the means in columns 2 and 3 assuming that the columns are random
samples from Normal populations with equal variances (4).
c. Assume that columns 2 and 3 do not come from a Normal distribution and are not paired data
and do a test for equal medians. (4)
d. Test the following data for uniformity. n  20.
Category
1
2
3
4
5
Numbers
0
2
0
10
8
Solution: a) This problem was on the last hour exam.
2-way ANOVA (Blocked by region) ‘s’ indicates that the null hypothesis is rejected.
Region Exper 1 Exper 2 Exper 3 sum count mean
Sum of squares
x i.. n i
SS
x1
x2
x3
x i.
x i2.
1
16.0
19.0 24.0
59.00 3 19.6667
1193
386.78
2
21.0
20.0 21.0
62.00 3 20.6667
1282
427.11
3
18.0
21.0 22.0
61.00 3 20.3333
1249
413.44
4
14.0
21.0 25.0
60.00 3 20.0000
1262
400.00
Sum
69.0
81.0 92.0
242.00 3 20.1667
4986
1627.33
4
+4
+4
12.00
nj

17.25
20.25 23.00
1217
+1643 +2126
297.56 +410.06 +529
x j
SS
x 2j
 x  242 ,
From the above
x
 x  242  20.17 .
n
SSC 
 n
SST 
12
2
j x j
n  12 ,
20.17  x
=4986
=1236.63
 x
 x
2
ij
 4986 ,
2
ij
x
2
i.
 1627 .33
x
2
.j
 1236 .63 and
 n x  4986  12 20 .17 2  4986  4881 .95  104 .05 .
2
 n x  41236 .63   12 20 .17 2  4946 .52  4881 .95  64 .57 . This is SSB in a one -
way ANOVA. SSR 
2
 n x
2
i i.
 n x  31627 .33   12 20 .17 2  4881 .99  4881 .95  0.04
2
( SSW  SST  SSC  SSR  39.44 )
Source
Rows (Regions)
SS
0.04
DF
3
Columns(Experience)
64.57
2
MS
0.0133
32.285
F
0.002
4.91
F.05
F 3,6  4.76 ns
F 2,6  5.14 s
H0
Row means equal
Column means equal
Within (Error)
39.44
6
6.573
Total
104.05
11
So the results characterized by years of experience (column means) are significantly different.
24
5/06/03 252y0342
Note that if you did this as a 1-way ANOVA, the SS and DF in the Rows line in the table would be
added to the Within line.
Computer version:
Two-way ANOVA: C40 versus C41, C42
Analysis of Variance for C40
Source
DF
SS
MS
C41
3
1.67
0.56
C42
2
66.17
33.08
Error
6
37.83
6.31
Total
11
105.67
F
0.09
5.25
P
0.964
0.048
Note that the ‘ns’ on the ANOVA on the last page seems to be due to rounding.
b) The data that we can use is repeated here.
1217  417 .25 2
2126  423 .00 2
 8.91667 , n1  4, x 3  23 .00, s 32 
 3.3333 , n3  4.
3
3
H 0 : 1   2 H 1 : 1   2
Test Ratio Method:
x  x1  x3  17.25  23.00  5.75 DF  n1  n2  2  4  4  2  6   .05,
x1  17 .25, s12 
sˆ2p 
n1  1s12  n2  1s22
sx  sˆ p
n1  n2  2
1
1


n1 n2
=
38.91667   33.3333  26 .75  10 .00

 6.1250
6
6
6.1250  1  1   6.1250 0.5 
4
4
6
t .025
 2.445
3.0625  1.750
 H 0 : 1   2
 H 0 :   0
Our hypotheses are 
or 
 H 1 : 1   2
 H 1 :   0
x   0  5.75  0
t

 3.286 so reject null hypothesis.
s x
1.750
c) The null hypothesis is H 0 : Columns come from same distribution or medians are equal.
The data are repeated in order. The second number in each column is the rank of the number among the 11
numbers in the two groups.
14 1
21 4.5
16 2
22 6
18 3
24 7
21 4.5
25 8 .
10.5
25.5
Since this refers to medians instead of means and if we assume that the underlying distribution is not
Normal, we use the nonparametric (rank test) analogue to comparison of two sample means of independent
samples, the Wilcoxon-Mann-Whitney Test. Note that data is not cross-classified so that the Wilcoxon
Signed Rank Test is not applicable.
H 0 : 1   2 H 1: 1   2 .
We get TL  10 .5 and TU  25 .5 . Check: Since the total amount of data is 4 + 4 = 8  n , 10.5 +25.5 must
nn  1 88

 36 .They do.
2
2
For a 5% two-tailed test with n1  4 and n 2  4 , Table 6 says that the critical values are 11 and 25. We
accept the null hypothesis in a 2-sided test if the smaller of thee two rank sums lies between the critical
values. The lower of the two rank sums, W  10.5 is not between these values, so reject H 0 .
equal
25
5/06/03 252y0342
d) This is basically Problem E11. This is the problem I used to use to introduce Kolmogorov-Smirnov. I
stopped because everyone seemed to assume that all K-S tests were tests of uniformity.
Five different formulas are used for a new cola. Ten tasters are asked to sample all five formulas
and to indicate which one they preferred. The sponsors of the test assume that equal numbers will prefer
each formula - this is the assumption of uniformity. Instead none prefer Formulas 1 and 3; 1 person prefers
Formula 2; 5 people prefer formula 4 and 4 people prefer Formula 5. Test the responses for uniformity.
Solution: H 0 : Uniformity or H 0 : p1  p 2  p3  p 4  p5 , where p i is the proportion that favor cola
Formula i.
Since uniformity means that we expect equal numbers in each group, we can fill the E column with fours
or just fill the next column with pi  15 .
Cola
O
1
2
3
4
5
Total
0
2
0
10
8
20
O
n
0
.1
0
.5
.4
1.0
Fo
E
0
.1
.1
.6
1.0
4
4
4
4
4
20
E
n
.2
.2
.2
.2
.2
1.0
Fe
.2
.4
.6
.8
1.0
D
.2
.3
.5
.2
0
The maximum deviation is 0.5. From the Kolmogorov-Smirnov table for n  20 , the critical values are
.20
.10
.05
.01

CV .232
.265
.294
.352
If we fit 0.5 into this pattern, it must have a p-value of less than .01, or we may simply note that if   .05 ,
0.5 is larger than .294 so we reject H 0 .
Can you see why it would be impossible to do this problem by the chi-squared method?
26