4/24/00 252y0031
ECO252 QBA2
THIRD HOUR EXAM
April 19, 2000
Name
key
Hour of Class Registered (Circle)
MWF TR 10 12 12:30 2:00
(Open this document in 'Page Layout' view!)
I. (10+ points) Do all the following;
1. Hand in your computer printouts for problems 1,2 and 3.(5 points – 3 point penalty for not handing in)
2. Do not do the following unless you handed in at least two outputs.
On the next few pages there are problems very much like the ones you did.
a. An oil price report reveals that the mean national price for gas was $1.46 last week. A random
sample of gas stations (data is "gaspr1") in Southeastern Pennsylvania reveals that the mean price was
$1.49. Using the tests with p-values in Minitab Problem 1 below can we conclude that the price in
Southeastern Pennsylvania is higher than it is nationwide? Do not answer without citing which of the tests
and p-values you used. (2)
b. The regression Problem 2 relates sales of a product (‘sales’) to shelf space ('shsp') in a random
sample of grocery stores. Identify the slope of the equation and explain whether it is significant at the 1%
level.(2) Add a regression line to the graph. (1) Do a 99% confidence interval for the slope (2)
Remember: As I have said innumerable times, your null hypothesis is usually 'no
significance.'
The rule on p-value:
If the p-value is less than the significance level (alpha) reject the null hypothesis; if the pvalue is greater than or equal to the significance level, do not reject the null hypothesis.
Since a p-value is a probability, there is no reason in the universe why you would compare it
with anything but a probability. (The significance level is also a probability.)
See next few pages for the answer to question 2.
7
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Problem 1:
MTB > RETR 'C:\MINITAB\2X0031-3.MTW'.
Retrieving worksheet from file: C:\MINITAB\2X0031-3.MTW
Worksheet was saved on 4/11/2000
MTB > ttest mu=1.46 'gaspr1'
T-Test of the Mean
Test of mu = 1.46000 vs mu not = 1.46000
Variable
gaspr1
N
35
Mean
1.49020
StDev
0.04969
SE Mean
0.00840
T
3.60
P-Value
0.0010
MTB > ttest mu=1.46 'gaspr1';
SUBC> alt=1.
T-Test of the Mean
*****
Test of mu = 1.46000 vs mu > 1.46000
Variable
gaspr1
N
35
Mean
1.49020
StDev
0.04969
SE Mean
0.00840
T
3.60
P-Value
0.0005
T
3.60
P-Value
1.00
MTB > ttest mu=1.46 'gaspr1';
SUBC> alt=-1.
T-Test of the Mean
Test of mu = 1.46000 vs mu < 1.46000
Variable
gaspr1
N
35
Mean
1.49020
StDev
0.04969
SE Mean
0.00840
MTB > Stop.
Solution to 2a: The only part of the printout that is relevant is the starred section which tests
 H 0 :   1.46
. Since p-value is .0005 and this is very small, we reject the null hypothesis at the 5%

 H 1 :   1.46
significance level (or any other level you are likely to use). We thus conclude that the mean Southeastern
PA price is above the mean nationwide price. Testing H 0 :   1.46 is not the same as testing
H 1 :   1.46.
Problem 2:
MTB > RETR 'C:\MINITAB\2X0031-2.MTW'.
Retrieving worksheet from file: C:\MINITAB\2X0031-2.MTW
Worksheet was saved on 4/ 7/2000
MTB > print 'sales''shsp'
Data Display
Row
sales
shsp
1
2
3
4
5
6
7
8
9
10
11
12
1.6
2.2
1.4
1.9
2.4
2.6
2.3
2.7
2.8
2.6
2.9
3.1
5
5
5
10
10
10
15
15
15
20
20
20
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MTB > regress 'sales' on 1 'shsp''resid''pred'
Regression Analysis
The regression equation is
sales = 1.45 + 0.0740 shsp
Predictor
Constant
shsp
Coef
1.4500
0.07400
s = 0.3081
Stdev
0.2178
0.01591
R-sq = 68.4%
t-ratio
6.66
4.65
p
0.000
0.000
R-sq(adj) = 65.2%
Analysis of Variance
SOURCE
Regression
Error
Total
DF
1
10
11
SS
2.0535
0.9490
3.0025
MS
2.0535
0.0949
F
21.64
p
0.000
MTB > plot 'pred'*'shsp'
MTB > plot 'sales'*'shsp'
MTB > plot 'pred'*'shsp' 'sales'*'shsp';
SUBC> symbol;
SUBC> type 3 1;
SUBC> color 8 9;
SUBC> overlay.
MTB > Save 'C:\MINITAB\2X0031-2.MTW';
SUBC>
Replace.
Saving worksheet in file: C:\MINITAB\2X0031-2.MTW
* NOTE * Existing file replaced.
MTB > Stop.
3.0
pred
2.5
2.0
1.5
5
10
15
20
shsp
Answer to 2b: (i)According to the printout, the slope of the regression equation is b1  0.0740 . (Its
H 0 :  1  0
standard deviation is sb1  01591. ) If we assume a confidence level of 1% and test 
, we find
H 1 :  1  0
a p-value of 0.000 which is below .01 (=1%) and reject the null hypothesis. We thus conclude that the slope
is significant. The p-value from the ANOVA has the same effect (ii) Just connect the x's. (iii) From the
10
outline: 1  b1  t sb1  0.07400  3.169 0.01591   0.074  0.050 . Here tn2  t.005
 3.169.
2
2
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Problem 3:
MTB >
Worksheet size: 100000 cells
MTB > RETR 'C:\MINITAB\2X0031-1.MTW'.
Retrieving worksheet from file: C:\MINITAB\2X0031-1.MTW
Worksheet was saved on 4/ 7/2000
MTB > print 'br st''op''mach'
Data Display
Row
br st
op
mach
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
31
32
33
34
35
36
116
116
120
112
109
115
110
111
108
118
115
115
106
103
107
111
114
115
110
111
107
101
104
102
104
103
106
113
116
112
106
108
108
109
112
111
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
4
4
4
4
4
4
4
4
4
1
1
1
2
2
2
3
3
3
1
1
1
2
2
2
3
3
3
1
1
1
2
2
2
3
3
3
1
1
1
2
2
2
3
3
3
MTB > table 'op''mach'
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Tabulated Statistics
ROWS: op
1
2
3
4
ALL
COLUMNS: mach
1
2
3
ALL
3
3
3
3
12
3
3
3
3
12
3
3
3
3
12
9
9
9
9
36
CELL CONTENTS -COUNT
MTB > table 'op''mach';
SUBC> data 'br st' .
Tabulated Statistics
ROWS: op
COLUMNS: mach
1
2
3
1
116.00
116.00
120.00
112.00
109.00
115.00
110.00
111.00
108.00
2
118.00
115.00
115.00
106.00
103.00
107.00
111.00
114.00
115.00
3
110.00
111.00
107.00
101.00
104.00
102.00
104.00
103.00
106.00
4
113.00
116.00
112.00
106.00
108.00
108.00
109.00
112.00
111.00
CELL CONTENTS -br st:DATA
MTB > table 'op''mach';
SUBC> mean 'br st'.
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Tabulated Statistics
ROWS: op
1
2
3
4
ALL
COLUMNS: mach
1
2
3
ALL
117.33
116.00
109.33
113.67
114.08
112.00
105.33
102.33
107.33
106.75
109.67
113.33
104.33
110.67
109.50
113.00
111.56
105.33
110.56
110.11
CELL CONTENTS -br st:MEAN
MTB > twoway 'br st''op''mach'
Two-way Analysis of Variance
Analysis of Variance for br st
Source
DF
SS
MS
op
3
301.11
100.37
mach
2
329.39
164.69
Interaction
6
86.39
14.40
Error
24
90.67
3.78
Total
35
807.56
MTB > Save 'C:\MINITAB\2X0031-1.MTW';
SUBC>
Replace.
Saving worksheet in file: C:\MINITAB\2X0031-1.MTW
* NOTE * Existing file replaced.
MTB > Stop.
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III. Do at least 4 of the following 6 Problems (at least 10 each) (or do sections adding to at least 40 points Anything extra you do helps, and grades wrap around) . Show your work! State H 0 and H1 where
applicable.
1. In Problem 3 in the computer output above we are looking at the effect of operator ('op') and machine
('mach') on the breaking strength ('br st') of a product.
a. Complete the table in the ANOVA by computing all the Fs and finding the relevant Fs for
comparison on your F table. Is there a significant difference between mean breaking strength for the
machines at the 1% significance level? Show what numbers brought you to your conclusion. (4)
b. Test for significant interaction – explain your conclusion. Use a 99% confidence level. (1)
c. Do a 99% confidence interval for the mean of machine 1 (2)
d. Do a 99% confidence interval between the means for machine 2 and 3 that is
(i) Valid when used alone. (1)
(ii) Valid when used with other possible differences between means. (2)
e. I have not asked you any questions about operators. If I was not interested in the effect of
operators, why would I have done a 2-way analysis of variance instead of a 1-way ANOVA? (1)
f. In a study of the time necessary to repair a VCR, 4 brands (Factor A) , 3 service centers (Factor
B) and two types (standard or deluxe) of product (Factor C) are distinguished. There are six
measurements (replications) per cell. Generate an ANOVA table showing all possible interactions,
using the following data. SSA = 53, SSB = 950, SSC = 38, SSAB = 405, SSAC = 15, SSBC = 20,
SSW = 245, SST = 1800. Using a 5% significance level, which of the differences and interactions
are significant.? (7)
Solution: a. The printout says (with F and F.01 columns added):
Analysis of Variance for br st
Source
DF
SS
MS
F
op
3
301.11
100.37
26.55
mach
2
329.39
164.69
43.57
6
24
35
86.39
90.67
807.56
14.40
3.78
3.81
Interaction
Error
Total
F.01
F 3,24  4.72
F 2,24  5.61
F 6,24  3.67
For the 'mach' line the null hypothesis is 'machine means equal.' We have computed
164 .69
F
 43 .57 . Because this F is larger than F 2,24  5.61 from the F table, we reject the null
3.78
hypothesis and conclude that there is a significant difference between the breaking strengths.
14 .40
 3.81 , and is larger than
b. For the interaction the null hypothesis is 'no interaction'. F 
3.78
F 6,24  3.67 , so we reject the null hypothesis and conclude that there is significant interaction.
c. The table of means in the printout says:
ROWS: op
1
2
3
4
ALL
COLUMNS: mach
1
2
3
ALL
117.33
116.00
109.33
113.67
114.08
112.00
105.33
102.33
107.33
106.75
109.67
113.33
104.33
110.67
109.50
113.00
111.56
105.33
110.56
110.11
CELL CONTENTS -br st:MEAN
4/24/00 252y0031
13
From this we conclude that x1  114 .08 , x2  106 .75 and x3  109 .50 . Note that
P  3, C  3 and R  4.
The outline says:
i. A Single Confidence Interval
If we desire a single interval we use the formula for a Bonferroni Confidence Interval below with m  1 .
ii Scheffé Confidence Interval
2MSW
For column means, use  1   2  x1  x2   C  1FC 1, RC P 1
.
PR
iii. Bonferroni Confidence Interval
2MSW
Use for column means  1   2  x1  x2   t RC P 1
.
2m
PR
You were also told to use the formulas without the second mean if you wanted intervals for one mean. (See
problems.) The ANOVA table tells us that the error mean square term has 24 degrees of freedom so that
24
t RC P1  t.005
 2.797. Under these circumstances, the Bonferroni formula becomes
2
 1  x1   t RC P 1
2
MSW
3.78
 114 .08   2.797
 114 .08  1.57 .
PR
12
d. (i) Using the advice above, the Bonferroni formula becomes  2   3  x2  x3   t RC P 1
2
 106 .75  109 .50   2.797
23.78 
 2.75  2.22
12
2MSW
PR
2,24  5.61 . So the Scheffé interval becomes
(ii) From the ANOVA table FC 1, RC P 1  F.01
 2   3  x2  x3  
C  1FC 1, RCP 1 2MSW  106 .75  109 .50   3  15.61 23.78   2.75  2.66.
PR
12
e. The columns are still not random samples and it is important to eliminate any effects due to the operators
before we look at the machines.
f. b) There are 4  3  2  24 groups with 6 observations in each group, so n  24  6  144 .
‘s’ means ‘significant difference’ ( H 0 rejected), ‘ns’ means ‘no significant difference’ ( H 0 accepted).
Source
SS
DF
MS
F
F.05
Factor A
53
3
Factor B
950
2
475
232.65
Factor C
38
1
38
18.61
Interaction AB
405
6
67.5
33.06
Interaction AC
15
3
5
2.44
Interaction BC
20
2
10
4.89
Interaction ABC
74
6
12.333
6.04
245
1800
120
143
2.04167
Error (Within)
Total
17.667
8.65
F 3,120  2.68 s
F 2,120  3.07 s
F 1,120  3.92 s
F 6,120  2.17 s
F 3,120  2.68 ns
F 2,120  3.07 s
F 6,120  2.17 s
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2. A hospital administrator wishes to compare the distribution of unoccupied beds in three hospitals.
Because she believes that the distributions are quite badly skewed she does not use analysis of variance but
instead ranks the entire sample and uses a test based on these ranks. Data is below. Numbers in boldface
were added in the process of solving the problem.
Day
1
2
3
4
5
6
7
8
9
10
Beds
6
38
3
17
11
30
15
16
25
5
Rank
5
27
2
13
8
21
11
12
17
4
120
Beds
34
28
42
13
40
31
9
32
39
27
Rank
25
19
30
9.5
29
22
7
23
28
18
210.5
Beds
13
35
19
4
29
0
7
33
18
24
Rank
9.5
26
15
3
20
1
6
24
14
16
134.5
Ranks
1 3 2
3 1 2
1 3 2
3 2 1
1 3 2
2 3 1
3 2 1
1 2 3
2 3 1
1 3 2
18 25 17
a. On the basis of the rank sums test the hypothesis that the median number of beds unoccupied differs for
the three hospitals. (5)
b. A statistician claims that her method is inappropriate because she has ignored the fact that each row
corresponds to a specific day, even if the days were chosen randomly. Rerank the data to account for the
fact that it is cross-classified and repeat the analysis. (9)
Solution:
a. H 0 : Columns come from same distribution or medians equal.
Sums of ranks are added above in boldface. To check the ranking, note that the sum of the three rank sums
nn  1 30 31

 465 .
is 120 + 210.5 + 134.5 = 465, and 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  120 2 210 .52 134 .52 

  331  12 1440  4431 .025  1809 .025   93  6.0974 . If we try to







30
31
10
10
10
930



look up this result in the (10,10,10) section of the Kruskal-Wallis table (Table 9) , we find that the problem
is to large for the table. Thus we must use the chi-squared table with 2 degrees of freedom. Since
 .2052   5.9915 reject H 0 .

.
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4/24/00 252y0031
b. H 0 : Columns come from same distribution or medians equal.
Items were ranked within the rows. To check the ranking, note that the sum of the three rank sums is 18 +
cc  1
25 + 17 = 60, and that the sum of the c numbers in a row is
. However, there are r rows, so we
2
rc c  1 10 34
must multiply the expression by r . So we have
SRi 

 60 .
2
2
 12

SRi2   3r c  1
Now compute the Friedman statistic  F2  
 rc c  1 i


 


 12
18 2  25 2  17 2   310 4  3.8 . If we try to find the place on the Friedman Table





10
3
4


(Table 8) for 3 columns and 10 rows, we find that the problem is too big far the table. We thus compare our
Friedman statistic with the  2 distribution, with df  c  1  2 , where c is the number of columns. Since
 2 is not larger than  22   5.9915 , do not reject the null hypothesis.
F
.05
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3. A law enforcement agency has come up with three methods of publicizing burglary-prevention methods
to use during summer vacations. Three communities are selected and the numbers of burglaries are listed
below. You may do this as a 2-way or 1-way ANOVA.
a. In each case state your null hypothesis about the publicity methods and the results (8)
b. Do a confidence interval for the mean for method 3 based on your computations in a. (2)
c. If you did two-way ANOVA explain what else it showed and why this is important if our major
interest is what publicity method to use. (5)
Method
1
15
17
27
Community 1
Community 2
Community 3
Method
2
13
25
23
Method
3
8
13
17
Solution: a) 2-way ANOVA ‘s’ indicates that the null hypothesis is rejected.
Community
Burglaries
Sum
ni
x .1
x .2
x .3
x
1
2
3
Sum
15
17
27
59
13
25
23
+ 61
8
13
17
+ 38
36
55
67
= 158 
nj
3
+3
+3
=9 n
 x
20.3333
12.667
SS
1243
+ 1323
+ 522
17.556  x
3088   xij2
x 2j
386.778
+ 413.444
+ 160.444
= 960.667  x.2j

SSR   n x
SSC 
n j x2j
2
i i.
n
9
3
3
3
9
12.000
18.333
22.333
17.556
458
1083
1547
3088

19.6667
 x  158 . SST 
SS
x i2.
i..
x j
Note that x is not a sum, but is
x i.
 x
2
ij
144.000
336.111
498.778
978.889
 x

2
ij
x
2
i.
 n x  3088  917 .556 2  314 .222 .
2
 n x  3960 .667   917 .556 2  108 .222 . This is SSB in a one way ANOVA.
2
 n x  3978 .889   917 .556 2  162 .888 ( SSW  SST  SSC  SSR  43.111 )
2
Source
SS
DF
MS
F
Rows (Community)
162.888
2
81.444
7.56
Columns(Method)
108.222
2
54.111
5.03
Within (Error)
43.111
4
Total
314.222
8
One way ANOVA (Not blocked by community)
Source
SS
DF
Columns(Method)
108.222
2
F.05
F 2,4  6.94 s
F 2,4  6.94 ns
H0
Row means equal
Column means equal
10.778
( SSW  SST  SSB  19.4447 )
MS
F.05
F
54.111
1.576
F 2,6  5.14 ns
H0
Column means equal
Within (Error)
206.000
6
34.333
Total
314.222
8
Because the computed F is less than the table F , there is no significant difference between the methods.
17
4/24/00 252y0031
b) Using the formulas in the outline for one mean (and k  1 ). In any case s  MSW . For 1-way
ANOVA  3  x3   t n  m  s
2
1
1
34 .333
6
 12 .667   t .025
34 .333
 12 .667   2.447
 12 .67  8.28
n3
3
3
For column means in 2-way ANOVA with P  1 , use  3  x3   t R1C 1
2
MSW
R

 12 .667   t .4025

10 .778
10 .778
 12 .667   2.776
 12 .67  5.26 .
3
3
c) We do find that there is a significant difference between the community means. This could cause us to
believe that there is more difference between the methods than there really is.
18
4/24/00 252y0031
4. a. To test the response of an 800 number, I make 20 attempts to reach the number, continuing to call
until I get through. I hypothesize that the results follow a Poisson distribution with a mean of 2.0. Test this –
data are below. (6)
Number of Unsuccessful
Tries before success.
0
1
2
3
4
5
Observed
Frequency
2
2
4
6
5
1
20
b. A Service station reports the following sales:
12.78
8.89
10.09
10.64
15.98
25.99 (With a sample mean of 14.06 and
a sample standard deviation of 6.35.) Do these data follow a normal distribution? (7)
Solution: a. H 0 :Poisson2.0 Because of the small size of the sample, this is best done by the
Kolmogorov-Smirnov method.
O
D  Fo  Fe
Fo
Fe
x
O
n
0
2
.10
.10
.13534 .03534
1
2
.10
.20
.40601 .20601
2
4
.20
.40
.67668 .27668
3
6
.30
.70
.85713 .15713
4
5
.25
.95
.94735 .00265
5
1
.05
1.00
.98344 .01656
20
1.00
The maximum difference is MaxD  .27668 , which must be checked against the Kolmogorov-Smirnoff
table for n  200 . According to the table, for   .05 , the critical value for n  20 is .294. Since MaxD is
less than .294, accept the null hypothesis.
Of course, most of you wanted to do this problem by the chi-squared method.
x
0
1
2
3
4
5+
Since  .205
O2
E
2 .1353
2.706* 0-1
4
8.120
1.9704
2 .2707
5.414
2
4
5.414
2.9553
4 .2707
5.414
3+
12
6.466 22.2703
6 .1804
3.608*
20
20.000 27.1460
5 .0902
1.804*
20.0000
1 .0527
1.054*
7.1460
20 1.0000
20.000 * Indicates groups with E  5 that had to be merged.
 5.995 is less than our computed  2 , reject the null hypothesis.
O
f
E  fn
x
O
E
19
4/24/00 252y0031
b. H 0 : N  ?, ? H 1 : Not Normal
Because the population mean and standard deviation are unknown and the sample is small, this is a
Lilliefors problem. The x values must be in order From the data we find that x  14 .06 and s  6.35 .
x  x x  14 .06
t

.This is often called z as in a K-S problem and F t  is a cumulative Normal
s
6.35
probability computed just like F z  below.
x
8.89
10 .09
10 .64
12 .78
15 .98
25 .99
t
 0.81
F t  .2090
O
1
O
0.1667
n
Fo 0.1667
D
.0423
 0.63
.2643
1
 0.54
.2946
1
 0.20
.4207
1
0.30
.6179
1
1.88
.9699
1
0.1667
0.3333
.0690
0.1667
0.5000
.2054
0.1667
0.6667
.2460
0.1667
0.8333
.2154
0.1667
1.0000
.0301

MaxD   .2460
Since the Critical
O  n  6 Value for   .05
is .319 , do not re ject H 0 .
20
4/24/00 252y0031
5. An firm wishes to explain the volume of office sales (in millions of dollars) over a year as a function of
number of customers (in thousands).. It collects data for a random sample of nine offices as follows:
Observation
cust
sales
(The xy column is added here.)
xy
9.1
11.2
101.92
9.3
11.0
102.30
5.1
6.8
34.68
7.4
9.2
68.08
7.9
9.4
74.26
8.9
10.1
89.89
8.7
9.4
81.78
5.1
6.7
34.17
5.3
7.2
38.16
625.24
For your convenience the following values are given:
1
2
3
4
5
6
7
8
9
 x  66.8,  x
2
 521 .48,
 y  81.0,  y
2
 752 .78, n  9.
a. Compute the regression equation Y  b0  b1 x to predict sales. (6)
b. On the basis of your regression, how many millions of dollars of sales do you expect from an office that
has nine thousand customers. (1)
c. Compute R 2 . (4)
Solution:
Spare Parts Computation:
 x  66.8  7.42222
x
n
y
SSxx 
 y  81  9.00
Sxy 
 24 .04
9
Sxy

SSxx
 xy  nxy  24.04  0.93630
 x  nx 25.67556
2
 nx 2  521 .48  97.4222 2
 xy  nx y  625 .24  97.42222 9.00 
SSyy 
a. b1 
2
 25 .67556
9
n
x
2
y
2
 ny  752 .78  99.00 2
2
 23 .78  TSS
b0  y  b1 x  9.00  0.9363 7.4222  2.0506
b. Y  b0  b1 x becomes Yˆ  2.0506  0.9363 x , and Yˆ  2.0506  0.93639  10.4773 is the number of
millions of dollars that we forecast.
RSS 22 .5087
xy  nx y  0.9363 24 .04   22 .5087 R 2 

 0.9465 or
c. RSS  b1 Sxy  b1
TSS
23 .78


 xy  nxy 
Sxy2 
24.04 2


 .9465
SSxxSSyy  x 2  nx 2  y 2  ny 2  25.67556 23.78 
2
R
2
( 0  R 2  1 always!)
21
4/24/00 252y0031
6. Continuing the previous problem. (  .01)
a. Compute s e . (3)
b. Compute s b0 and do a significance test on b0 .(4)
c. Do a confidence interval for b1 (3)
d. Using your SST etc., put together the ANOVA table (6)
Solution:
ESS 1.2713
s e2 

 0.181614 or
a. ESS  TSS  RSS  23.78  22.5087  1.2713
n2
7
s e2 
SSyy  b1 Sxy

n2
 y
  xy  nxy  23.78   0.9363 24.04 

 0.18162 or
 ny 2  b1
1  R TSS  1  R  y

2
2
s e2
2
n2
n2

n2
s e  0.181614  0.4262
b.
 ny 2
2
(
s e2
1
1
x 2 
s b20  s e2  
 s e2  
 n SSxx 
n



or
se2
 y

7
2
 ny
2
  b  x
2
1
2
 nx 2

n2
is always positive!)

sb0  0.40985  0..6402
2

  0.181614  1  7.42222 
 9 25 .67556
x 2  nx 2 


x2

  0.409850


b0  2.0506
H 0:  0  0 H 1 :  0  0
t
b0   00 b0  0 2.0506


 3.203 . Since this is between
s b0
s b0
0.6402
7
 t.n2  t.005
 3.499 , do not reject H 0 and thus conclude that  0 is not significant.
2
c. s b21 
s e2

SSxx
 x
s e2
2
 nx
2


0.18162
 0.007074
25 .67556
sb1  0.007074  0.08411.
So 1  b1  sb1  0.9363 3.4990.08411  0.936  0.294
d. From the previous page and above RSS  22.5087 , TSS  23.7800 and ESS  1.2713 . H 0 is that there
is no relation between Y and X .
Source
SS
DF
MS
F
F.01
Regression
22.5087
1
Error (Within)
Total
1.2713
23.7800
7
8
22.509
123.93
F 1,7   12.25 s
0.1816
Since the table F is less than the computed F, reject H 0 .
22