252y0333 11/25/03
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ECO252 QBA2
THIRD HOUR EXAM
Nov 25, 2003
Name KEY
Hour of Class Registered (Circle)
I. (30+ points) Do all the following (2points each unless noted otherwise).
TABLE 11-0
Shiffler and Adams present the partially complete ANOVA table below that resulted from the analysis of a problem with 3 rows
and 3 columns.
ANOVA
Source of Variation
SS
Columns
18
Rows
40
df
MS
F
F
Interaction
Within (Error)
208
Total
296
62
1.
Complete the table. Assume a 5% significance level. You may not be able to get exactly the
degrees of freedom you are looking for, but you should be able to come close. (4)
Solution: The SS column must add up, which accounts for the 30. For c columns the DF must be
c  1  2 . For r rows, DF = r 1  2 . For interaction, DF = r  1c  1  22  4. DF for within is 62
– 2 – 2 – 4 =54. MS column is SS column divided by df column. F column is MS column divided by
MSW  3.852 . F  is from F table.
ANOVA
F
Source of Variation
SS
df
MS
F
Columns
Rows
Interaction
Within (Error)
18
40
30
208
2
2
4
54
9
20
7.5
3.852
2.336ns
5.192 s
1.947ns
3.17
3.17
2.54
Total
296
62
2. Is there significant interaction? Explain your answer. Solution: Since the computed F is less than
the table F, we do not reject the null hypothesis of no interaction – this is shown by ns for ‘not
significant,’ though I need to know why it is not significant. Remember SS and df columns must
add up.
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TABLE 13-6
The following Minitab table (with many parts deleted) was obtained when "Score received on an exam
(measured in percentage points)" (Y) is regressed on "percentage attendance" (X) for 22 students in a
Statistics for Business and Economics course.
Regression Analysis: Orders versus Weight
The regression equation is
Score = …… + ….. Attendance
Predictor
Constant
Attendance
Coef
39.3927
0.34058
S = 20.2598
SE Coef
37.2435
0.52852
R-Sq = 2.034%
T
1.0576
0.6444
P
0.3028
0.5266
R-Sq(adj) = -2.864%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
20
21
SS
MS
F
P
0.523
3.
Referring to Table 13-6, which of the following statements is true?
a) -2.86% of the total variability in score received can be explained by percentage
attendance.
b) -2.86% of the total variability in percentage attendance can be explained by score
received.
c) *2% of the total variability in score received can be explained by percentage attendance.
d) 2% of the total variability in percentage attendance can be explained by score received.
Explanation: R 2 is the total variability in Y that is explained by the regression line.
4.
Referring to Table 13-6, which of the following statements is true?
a) If attendance increases by 0.341%, the estimated average score received will increase by
1 percentage point.
b) If attendance increases by 1%, the estimated average score received will increase by
39.39 percentage points.
c) *If attendance increases by 1%, the estimated average score received will increase by
0.341 percentage points.
d) If the score received increases by 39.39%, the estimated average attendance will go up by
1%.
Explanation: The equation is Score  39.3927  .34058  Attendance . If attendance goes up by 1,
Score goes up by .34058.
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5.
(Text CD problem 12.51)The manager of a commercial mortgage department has collected data
over 104 weeks concerning the number of mortgages approved. The data is the x and O columns
below ( x is the number of mortgages approved and O is the number of weeks that happened, for
example there were 32 weeks in which 2 mortgages were approved) and the problem asks if it
follows a Poisson distribution.
x
Row
O
E
1
2
3
4
5
6
7
8
9
10
11
12
13
0
1
2
3
4
5
6
7
8
9
10
11
12
13
25
32
17
9
6
1
1
0
0
0
0
0
12.7355
26.7445
28.0817
19.6572
10.3200
4.3344
1.5170
0.4551
0.1195
0.0279
0.0059
0.0011
0.0002
Since we have no guide as to what the parameter of the distribution is, the x and O columns
were multiplied together to tell us that there were 219 mortgages approved over 104 weeks to
give us an average of 2.1 mortgages per week. The E above is the computer – generated Poisson
distribution multiplied by 104 .
In a Kolmogorov – Smirnov procedure we make the O and E into cumulative distributions and
compare them as is done below.
Row
Fo
Fe
D
1
2
3
4
5
6
7
8
9
10
11
12
13
0.12500
0.36538
0.67308
0.83654
0.92308
0.98077
0.99038
1.00000
1.00000
1.00000
1.00000
1.00000
1.00000
0.12246
0.37962
0.64963
0.83864
0.93787
0.97955
0.99414
0.99851
0.99966
0.99993
0.99999
1.00000
1.00000
0.0025435
0.0142304
0.0234453
0.0021047
0.0147973
0.0012180
0.0037536
0.0014857
0.0003369
0.0000689
0.0000126
0.0000019
0.0000000
Assume this is correct and explain how you would finish this analysis and why you would or
would not reject the null hypothesis. (4) Solution: The null hypotheses is H 0 : Poisson .
n
 O  104 . We use the K-S table and find that the critical value for the maximum
1.36

1.36
 0.1334 . since the maximum of the D column is less than the
n
104
critical value, we do not reject the null hypothesis.
discrepancy is
6.
Referring to the previous problem, a more direct method of comparing the observed and expected
data is below. Answer the following questions.
a) What method is being used? (1)
Answer: Chi-squared.
b) How many degrees of freedom do we have? (1)
Answer: Normally, DF would be the number of rows minus 1, but here we have
estimated a parameter from the data and lost a degree of freedom. 7 – 1 – 1 = 5.
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3
c)
Why are the columns shorter here than in Problem 5? (1)
Answer: The text says that E cannot be below 1. The E s are that are below 1 and the
corresponding O s are added together to form a new 7th line.
d) Do we reject our null hypothesis? Why? (3) Solution: To get our computed chi –
2
squared, take the O
column and subtract n  104 . The result, 1.8415, must be
E
compared with the 5% value of chi-squared from the table. The value from the table is
11.00705, so we do not reject the null hypothesis.
O2
O
E
E
Row
1
2
3
4
5
6
7
13
25
32
17
9
6
2
104
12.7355
26.7445
28.0817
19.6572
10.3200
4.3344
2.1267
104.0000
13.2700
23.3693
36.4650
14.7020
7.8488
8.3056
1.8808
105.8415
7.
In problems 5 and 6, one of the methods was used improperly. Why? Answer: The K-S method
was wrong because we did not completely know the distribution we were testing for and had to
estimate the mean from the data.
8.
Random samples of salaries (in thousands) for lawyers in 3 cities are presented by Dummeldinger.
They are repeated in the three left columns.
1
2
3
4
5
6
7
Atlanta
45.5
47.9
43.1
42.0
49.0
52.0
39.0
DC
41.5
40.1
39.0
56.5
37.0
49.0
43.0
LA
52.0
72.0
41.0
54.0
33.0
42.0
50.0
rank-At
12.0
13.0
11.0
8.5
14.5
17.5
3.5
80.0
rank-DC
7.0
5.0
3.5
20.0
2.0
14.5
10.0
62.0
rank-LA
17.5
21.0
6.0
19.0
1.0
8.5
16.0
89.0
You are asked to analyze them, which you do using a Kruskal – Wallis procedure. You are aware
that the tables you have are only appropriate for columns with 5 or fewer items in them, so you
drop the last two items in each column and after ranking the items from 1 to 15 get a Kruskal –
Wallis H of 1.82. If you use the tables, What did you test and what is the conclusion? (3)
Answer: The table gives a number of p-values, but none for 1.82. However, it should be plain
from the values given, that 1.82 has a p-value above .102 We cannot reject the null hypothesis,
which is equal medians, because the p-value is above any significance level we are likely to use.
Another way to do this is to note from the Friedman table that the p-value for H  5.66 is .057
and the p-value for H  5.78 is .049, so that H .05  5.8. Since 1.83 is cousiderably below 5.8,
do not reject the null hypothesis.
9.
You remember how to work with column sizes that are too large for the table. You rank the data as
appears in the three right columns above. Compute the Kruskal – Wallis H and use it to test your
null hypothesis at the 5% significance level.(3)
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 12
 SRi 2 

  3n  1
Solution: Compute the Kruskal-Wallis statistic H  
 nn  1 i  ni 
 12  80 2 62 2 89 2 

  322   .02597  18165   66  1.4025 . Since the 5% chi-square



7
7 
 7 
 2122   7
for 2 degrees of freedom is 5.9915, we do not reject our null hypothesis. Note: In spite of my
warnings, most of you changed the 12 to something else, most likely n . If this worked the formula

 1
would be H  ?
 n  1
 SRi 2 
  3n  1 . Unless I was lying, this is not the case.

n
i


 
i
10. The Kruskal – Wallis test above was done on the assumption that the underlying data did not
follow the Normal distribution. Let’s assume that you found out that the underlying distributions
were Normal and had a common variance. The method to use would be.
a) Friedman Test
b) Chi – squared test.
c) *One way ANOVA
d) Two – way ANOVA
TABLE 13-8
The regression equation is
GPA = 0.5681 + .1021 ACT
Predictor
Constant
ACT
S = 0.2691
Coef
.5681
.1021
SE Coef
0.9284
0.0356
R-Sq = 0.5774%
T
0.6119
2.8633
P
0.5630
0.0286
R-Sq(adj) = 0.5069%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
6
7
SS
0.5940
0.4347
1.0287
MS
0.5940
0.0724
F
8.1986
P
.0287
It is believed that GPA (grade point average, based on a four point scale) should have a positive linear
relationship with ACT scores. Given below is the Excel output from regressing GPA on ACT scores
using a data set of 8 randomly chosen students from a Big Ten university.
11. Referring to Table 13-8, the interpretation of the coefficient of determination in this regression is
that
a) *57.74% of the total variation of ACT scores can be explained by GPA.
b) ACT scores account for 57.74% of the total fluctuation in GPA.
c) GPA accounts for 57.74% of the variability of ACT scores.
d) *none of the above (error)
12. Referring to Table 13-8, the value of the measured test statistic to test whether there is any linear
relationship between GPA and ACT is
a) 0.0356.
b) 0.1021.
c) 0.7598.
d) *2.8633.
The t statistic.
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13. Referring to Table 13-8, what is the predicted average value of GPA when ACT = 20?
a) *2.61
.5681 + .1021(20) = 2.6101
b) 2.66
c) 2.80
d) 3.12
14. Referring to Table 13-8, what are the decision and conclusion on testing whether there is any
linear relationship at the 1% level of significance between GPA and ACT scores?
a) *Do not reject the null hypothesis; hence, there is not sufficient evidence to show that
ACT scores and GPA are linearly related.
b) Reject the null hypothesis; hence, there is not sufficient evidence to show that ACT scores
and GPA are linearly related.
c) Do not reject the null hypothesis; hence, there is sufficient evidence to show that ACT
scores and GPA are linearly related.
d) Reject the null hypothesis; hence, there is sufficient evidence to show that ACT scores
and GPA are linearly related.
Answer: The p-value is above the significance level, so do not reject the null hypothesis.
The null hypothesis is usually that a relationship is not significant.
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ECO252 QBA2
Third EXAM
Nov 25 2003
TAKE HOME SECTION
Name: _________________________
Social Security Number: _________________________
Please Note: computer problems 2 and 3 should be turned in with the exam. In problem 2, the 2 way
ANOVA table should be completed. The three F tests should be done with a 5% significance level and you
should note whether there was (i) a significant difference between drivers, (ii) a significant difference
between cars and (iii) significant interaction. In problem 3, you should show on your third graph where the
regression line is.
II. Do the following: (23+ points). Assume a 5% significance level. Show your work!
1. Assume that each column below represents a random sample of sales of the popular cereal brand,
‘Whee!’ As it was moved from shelf 1 (lowest) to shelf 4 (highest) of a group of supermarkets. Assume that
the underlying distribution is Normal and test the hypothesis 1   2   3   4 .
a) Before you start add the second to last digit of your social security number to the 451 in column 4 and
find the sample variance of sales from shelf 4. For example, Seymour Butz’s SS number is 123456789 and
he will change 451 to 459. This should not change the results by much. (2)
b) Test the hypothesis (6) Show your work – it is legitimate to check your results by running these problems
on the computer, but I expect to see hand computations for every part of them.
c) Compare means two by two, using any one appropriate statistical method, to find out which shelves are
significantly better than others. (3)
d) (Extra Credit) What if you found out that each row represented one store? If this changes your analysis,
redo the analysis. (5)
e) (Extra Credit) What if you found out that each row represented one store and that the underlying
distribution was not Normal? If this changes your analysis, redo the analysis. (5)
f) I did some subsequent analysis on theist problem. The output, in part said
Levene's Test (any continuous distribution)
Test Statistic: 0.609
P-Value
: 0.613
What was I testing for and what should my conclusion be? (2)
Row
1
1
2
3
4
5
6
7
8
9
10
336
417
208
420
366
227
357
353
518
388
Sales of ‘Whee’ Cereal
Shelf
2
3
4
440
277
374
421
481
349
328
449
462
373
464
479
492
456
338
413
383
554
497
510
354
423
321
424
518
451
311
462
339
202
Sum of
Sum of
1362860
Sum of
Sum of
1602366
Sum of
Sum of
2140264
shelf 1 = 3590.0
squares of shelf 1 =
shelf 2 = 3954.0
squares of shelf 2 =
shelf 3 = 4586.0
squares of shelf 3 =
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252y0333 11/25/03
The original data is presented in my format for 1-way ANOVA
sum
x1
x2
x3
x4
1
2
3
4
5
6
7
8
9
10
Sum
nj
336
417
208
420
366
227
357
353
518
388
3590
10
440
277
374
421
481
349
328
449
462
373
+3954
+10
464
479
492
456
338
413
383
554
497
510
+4586
+10
354
423
321
424
518
451
311
462
339
202
+ 3805
+10
x j
359.0
395.4
458.6
380.5
SS 13628460 + 1602366 +2140264 +1524677
x 2j
398.375  x
=6630167
Most of this line was done for you.
128881 +156341.16+210313.96+144780.25 =640316.37
 x  15935 ,
From the above
x
=15935 40 398.375 6630167 1602754.8125
= 40
n  40 ,
 x  15935  398 .375 .
n
SST 
2
ij
 6630167 ,
x
2
i.
 6630167 ,
and
40
 x
x
 x
2
ij
 n x  6630167  40 398 .375 2  6630167  6348105 .625  282061 .375 .
2
1524677  10 380 .52
 8541 .6111
n 1
9
b) We are testing H o : 1   2   3   4
a) s 2 
SSB 
2
 nx 2
 n
2
j x j

 n x  10 640316 .37   40 398 .375 2  6403163 .7  6348105 .625  55058 .075
2
ANOVA
Source of Variation
SS
df
Between
Within (Error)
55058
227003
3
36
Total
282061
39
MS
18353
6306
F
F
3,36
2.91s F.05  2.87
Since Our computed F exceeds the table F, we reject the null hypothesis.
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252y0333 11/25/03
c) Recall the following from the outline.
Scheffe’ Confidence Interval: If we desire intervals that will simultaneously be valid for a given
confidence level for all possible intervals between column means, use
 1
1 
.
1   2  x1  x2   m  1Fm 1, n  m   s

 n
n2 
1

Tukey Confidence Interval: This also applies to all possible differences.
1   2  x1  x2   q m,n  m 
s
2
1
1
. This gives rise to Tukey’s HSD (Honestly Significant

n1 n 2
Difference) procedure. Two sample means x .1 and x .2 are significantly different if x.1  x.2 is greater
than 1  2  qm, n  m 
s
2
1
1

n1 n2
From the solution to grass 3, If you used a Scheffe’ interval, there are m  4 columns. The error side is
1
1 
1 1

 32.87 6306     104 .206

n
n
 10 10 
1
2 


For the Tukey comparison, we have an error side with q m, n  m  q 4,36
.05  3.81 .


m  1Fm1,nm  s
1
1
6306  1
1

 3.81
    95 .676
2  10 10 
2 n1 n 2
We have the following differences
x.1  x.2  359 .0  395 .4  36.4 x.2  x.3  395 .4  458 .6  63 .2
q m, n  m 
s
x.1  x.3  359 .0  458 .6  99.6
x.2  x.4  395 .4  380 .5  14.90
x.1  x.4  359 .0  380 .5  21.5
x.1  x.4  359 .0  380 .5  21.5
For the Tukey comparison only, x.1  x.3  359 .0  458 .6  99.6 is larger than 95.676, so we say that
there is a significant difference between those shelves. The more conservative Scheffe interval gives us
nothing.
d) 2-way ANOVA You have already done SSC.
x3
x4
x
440
277
374
421
481
349
328
449
462
373
+3954
+10
464
479
492
456
338
413
383
554
497
510
+4586
+10
354
423
321
424
518
451
311
462
339
202
+ 3805
+10
1594
1596
1395
1721
1703
1440
1379
1818
1816
1473
=15935
= 40
395.4
458.6
380.5
x1
x2
nj
336
417
208
420
366
227
357
353
518
388
3590
10
x j
359.0
1
2
3
4
5
6
7
8
9
10
Sum
SS 13628460 + 1602366 +2140264 +1524677
x 2j
i..
ni
x i.
4 398.50
4 399.00
4 348.75
4 430.25
4 425.75
4 360.00
4 344.75
4 454.50
4 454.00
4 398.25
40 398.375
SS
647108
658988
528245
741353
747885
547300
478443
846570
843898
590577
6630167
x i2.
158802.2500
159201.0000
121626.5625
185115.0625
181263.0625
129600.0000
118852.5625
206570.2500
206116.0000
135608.0625
1602754.8125
398.375  x
=6630167
128881 +156341.16+210313.96+144780.25 =640316.37
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252y0333 11/25/03
 x  15935 , n  40 ,  x  6630167 ,  x
 x  15935  398 .375 .
 640316 .37 and x 
2
ij
From the above
x
2
.j
n
2
i.
 6630167
40
  n x  6630167  40398 .375   6630167  6348105 .625  282061 .375
SSC   n x  n x  10 640316 .37   40 398 .375   6403163 .7  6348105 .625  55058 .075
SSR   n x  n x  41602754 .8125   40 398 .375   6411019  6348105  62914
SST 
2
x ij2
2
j j
2
i i.
2
2
2
2
2
ANOVA
Source of Variation
SS
Columns (shelves)
df
MS
F
55058
3
18353
Rows (stores)
Within (Error)
62914
164090
9
27
6990
6077
Total
282061
39
F
3, 27
3.02s F.05  2.96
9, 27
1.15ns F.05  2.25
The null hypotheses are no significant difference between store means and no significant difference between
shelf means. We reject the hypothesis that store means are equal.
e) In general if the parent distribution is Normal use ANOVA, if it's not Normal, use Friedman or
Kruskal-Wallis. If the samples are independent random samples use 1-way ANOVA or Kruskal
Wallis. If they are cross-classified, use Friedman or 2-way ANOVA. So the other method that allows for
cross-classification is Friedman and we use it if the underlying distribution is not Normal.
The null hypothesis is H 0 : Columns from same distribution or H 0 : 1   2   3   4 . We use a
Friedman test because the data is cross-classified by store. This time we rank our data only within rows.
There are c  4 columns and r  10 rows.
x1
x2
x3
x4
r1
r2
r3
r4
1
2
3
4
5
6
7
8
9
10
Sum
336
417
208
420
366
227
357
353
518
388
3590
440
277
374
421
481
349
328
449
462
373
+3954
464
479
492
456
338
413
383
554
497
510
+4586
354
423
321
424
518
451
311
462
339
202
+ 3805
1
2
1
1
2
1
3
1
4
3
19
3
1
3
2
3
2
2
2
2
2
22
4
4
4
4
1
3
4
4
3
4
35
2
3
2
3
4
4
1
3
1
1
24
To check the ranking, note that the sum of the three rank sums is 19 + 22 + 35 +24 = 100, and that
rcc  1 10 45
SRi 

 100 .
the sum of the rank sums should be
2
2

10
252y0333 11/25/03
 12
Now compute the Friedman statistic  F2  
 rc c  1


 SR   3r c  1
2
i
i

 12
19 2  22 2  35 2  24 2   310 5  0.06 361  484  1225  576   250  158 .76  150  8.76

 10 45

.
If we check the Friedman Table for c  4 and r  10 , we find that the problem is too large for the table. So
we look up the 5% value of  2 with c  1  3 degrees of freedom.. It is 7.8147. Since the computed
statistic exceeds that value we reject the null hypothesis.
f) The Levene Test is a test for equality of variances. A high p-value indicates that the null hypothesis of the
columns’ coming from populations with equal variance cannot be rejected.
Your Solution: Go to the document 252y0333TH pp 2-32 and find Results for: 2x0333-0j, where j is the
second to last digit of your SS number. The first things that appear are the sums and sums of squares for all
the columns, followed by a printout of column 4. The one-way ANOVA had to be run twice, the first time
in the form that you would use and the second time in the form that you used in Computer Assignment 2.
The second version prints out Tukey confidence intervals. Minitab then prints the results of the 2-way
ANOVA and the Friedman test.
11
252y0333 11/25/03
2. A company, operating in 12 regions, gives us its advertising expenses as a percent of those of its leading
competitor, and its sales as a percent of those of its leading competitor.
Row
1
2
3
4
5
6
7
8
9
10
11
12
Ad
77
110
110
93
90
95
100
85
96
83
100
95
Sales
85
103
102
109
85
103
110
86
92
87
98
108
Sum
Sum
Sum
Sum
of
of
of
of
Ad = 1134.0
squares of Ad = 108258
Sales = 1168.0
squares of Sales = 114750
Note that the sum and sum of squares
of sales can’t be used directly, but
they should help you to get the
corrected numbers.
Change the 103 in the ‘sales’ column by adding the second-to-last digit of your Social Security number to it.
For example, Seymour Butz’s SS number is 123456789 and he will change 103 to 112. This should not
change the results by much. The question is whether our relative advertising expenses affect our relative
sales, so ‘Sales’ should be your dependent variable and ‘Ad’ should be your independent variable.
Show your work – it is legitimate to check your results by running the problem on the computer, but I
expect to see hand computations that show clearly where you got your numbers for every part of this
problem.
a. Compute the regression equation Y  b0  b1 x to predict the ‘Sales’ on the basis of ‘Ad’. (2)
b. Compute R 2 . (2)
c. Compute s e . (2)
d. Compute s b0 and do a significance test on b0 (2)
e. Do an ANOVA table for the regression. What conclusion can you draw from this table about the
relationship between advertising expenditures and sales? Why? (2)
e. It is proposed to raise our expenditures to 110% of our competitors’ in every region. Use this
to find a predicted value for sales and to create a confidence interval for sales. Explain the
difference between this and a prediction interval and when the prediction interval would be more
useful. (3)
12
252y0333 11/25/03
Solution: Working with the original data, we get the following table. The x and x 2 columns and their
sums were not actually needed since they were done for you.
Row
Ad
i
x
Sales crprice
y
x2
xy
y2
1
2
3
4
5
6
7
8
9
10
11
12
77
110
110
93
90
95
100
85
96
83
100
95
1134
85
103
102
109
85
103
110
86
92
87
98
108
1168
7225
10609
10404
11881
7225
10609
12100
7396
8464
7569
9604
11664
108258
6545
11330
11220
10137
7650
9785
11000
7310
8832
7221
9800
10260
111090
5929
12100
12100
8649
8100
9025
10000
7225
9216
6889
10000
9025
124750
 x  1134 ,  y  1168 ,  x
n  12 ,
2
 108258 ,
Spare Parts Computation:
 x y  111090 and  y  114750 .
SSx   x 2  nx 2  108258 1294.52
2
 1095 .00
x
 x  1134  94.5
n
Sxy 
 714 .00
12
 y  1168  97.3333
y
n
a) b1 
SSy 
y
2
 ny  114750  12 97 .3333 2
2
 1064 .6667  SST
12
Sxy

SSx
 xy  nx y  111090  1294.597.3333 
 xy  nxy  714  0.652055
 x  nx 1095
2
2
b0  y  b1 x  97.3333  0.652055  94.5  35.7141
So Y  b  b x becomes. Yˆ  35.7141  0.6521 x
0
1
SSR
465
 xy  nxy   0.6521 714 .00   465 .60 So R  SST

 .4368
1064 .6667
 xy  nxy 
Sxy 
714 


 .4372
SSxSSy  x  nx  y  ny  1095 1064 .6667 
b) SSR  b1 Sxy  b1
2
2
2
R
2
2
2
2
2
2
c) d) SSE  SST  SSR  1064 .6667  465  599 .6667
1  R SST  1  R  y

2
2
or
s e2
n2

 ny 2
n2
So s e  59 .9194  7.7408
1
d) s b20  s e2  
n

2
(
s e2


s e2 
SSE 599 .6667

 59 .96667
n2
10
1  .4372 1064 .6667   59.9194
10
is always positive!)

2

  59 .9194  1  94 .5
 12 1095
X 2  nX 2 


X2
or

  59 .9194 .01077  8.15548   489 .69


sb0  489.69  22.12 . Formulas for s b0 and s b1 appear in the outline.
13
252y0333 11/25/03
H 0 :  0   00
b  00
The outline says that to test 
use t  0
. Remember  00 is most often zero – and if
H
:



sb0
00
 1 0
the null hypothesis is false, we say that 1 is significant. So t 
b0  00 35 .7141

 1.615 .
sb0
22 .12
Make a diagram of an almost Normal curve with zero in the middle and, if   .05 , the rejection zones
are above t n2  t 10  2.228 and below  t n2  t 10  2.228 . Since our computed t-ratio is, at

2

.025
.025
2
1.615, nor in a rejection zone, we do not reject the null hypothesis that the coefficient is not significantly
different from zero and we say that b1 is not significant.
e) Note that the F test is the equivalent of a test on b1
Source
SS
DF
Regression
Error (Within)
Total
MS
465.0000
1
465.00
599.6667
1064.6667
10
11
59.97
F
7.75
F.05
F 1,10  4.96 s
f) Our equation says that Yˆ  35.7141  0.6521 x , so, if x  110 , Yˆ  35.7141 0.6521110  107.45 .
the Confidence Interval is   Yˆ  t s , where
Y0
1
. sY2ˆ  s e2  
n

X 0  X 
2
 X
2
 nX 2

the confidence interval is  Y0
Yˆ
0
2

  59 .9194  1  110  94 .5   18 .140 s  18.140  4.259 , so that
Yˆ

 12
1095  


 Yˆ  t s ˆ  107 .45  2.228 4.259   107 .45  9.49 or 97.96 to 116.94.
0
Y
The confidence interval for Y gives an average value for many areas in which the ad budget price was 110,
so it is not appropriate for prediction sales in one region. However, since the firm is making the ad budget
uniform over all areas, it may be quite appropriate for projecting total sales.
-------------------------------------------------------------See next page.
14
252y0333 11/25/03
Minitab output follows.
Results for: 2x0333-10.mtw
MTB > Name c15 = 'CLIM5' c16 = 'CLIM6' c17 = 'PLIM5' c18 = 'PLIM6'
MTB > Regress c2 1 c1;
SUBC>
Constant;
SUBC>
Predict c6;
SUBC>
CLimits 'CLIM5'-'CLIM6';
SUBC>
PLimits 'PLIM5'-'PLIM6';
SUBC>
Brief 2.
Regression Analysis: Sales versus Ad
The regression equation is
Sales = 35.7 + 0.652 Ad
Predictor
Constant
Ad
Coef
35.71
0.6521
S = 7.740
SE Coef
22.22
0.2339
R-Sq = 43.7%
T
1.61
2.79
P
0.139
0.019
R-Sq(adj) = 38.1%
Analysis of Variance
Source
Regression
Residual Error
Total
DF
1
10
11
SS
465.57
599.10
1064.67
MS
465.57
59.91
F
7.77
P
0.019
Predicted Values for New Observations
New Obs
1
Fit
107.44
SE Fit
4.26
95.0% CI
97.95, 116.93)
(
(
95.0% PI
87.76, 127.12)
Values of Predictors for New Observations
New Obs
1
Ad
110
Your Solution: Go to the document 252y0333TH pp 32-44 and find Results for: 2x0333-1j, where j is the
second to last digit of your SS number. The first things that appear are the results of the regression with the
correct prediction and confidence intervals. The next regression printout is wrong but useful. It computes
the regression with sales as the independent variable (X) and Ad as the dependent variable (Y). This routine
is set up to compute all the columns and column sums that you needed. However, because of the way the
data was arranged, it has things reversed.
x , sumx 
y , smysq 
x 2 , smxy 
x y , smxsq 
y2 ,
So n = n , sumy 

ybar  x , xbar  y ,
 SSy 
y
2

SSy  SSx   x

2
 nx , Sxy  Sxy 
2

 xy  nx y

and SSx
2
 ny  SST .
Sorry about that, but the alternative was a load of new programming.
15