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Individual Solutions to Take-home
Warning – This document has the worksheets for every version of the Take-home Problems. At
last count it was about 70 pages long. Please only print the solution for your version of the problem.
Most versions are headed with problem and version number.
Individual solutions to Take-home Problem 1 for Versions 0 through 9 begin on page 5 and
go to page 24.
Individual solutions to Take-home Problem 2 for Versions 0 through 9 begin on page 28 and
go to page 57.
Individual solutions to Take-home Problem 3 for Versions 0 through 9 were not computed
correctly, and only the solution to Version 0 begins on page 64.
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1) Bassett et al. give the following numbers for the year and the number of pensioners in the United
Kingdom. Pensioners are in millions. The 2000 number is a bit shaky, so subtract the last digit of your
student number divided by 10 from the 12.00 that you see there. Label your answer to this problem with a
version number. (Example: Good ol’ Seymour’s student number is 123456, so the 12.000 becomes 12.000 0.6 = 11.400 and he labels it Version 6.) 'Pensioners' is the dependent variable and 'Year' is the independent
variable, so what you are going to get is a trend line. If you don’t know what dependent and
independent variables are, stop work until you find out.
1
2
3
4
5
6
7
8
Year
Pensioners
1966
1971
1975
1976
1977
1978
1979
2000
6.679
7.677
8.321
8.510
8.637
8.785
8.937
12.000
Bassett et. al. strongly suggest that you change the base year to something other than the year zero. They
recommend that you subtract 1970 from every number in the ‘Year’ column, so that 1966 becomes -4 and
2000 becomes 30. This will make your computations easier.
a. Compute the regression equation Y  b0  b1 x to predict the number of pensioners in each year. (3).You
may check your results on the computer, but let me see real or simulated hand calculations.
Solution: Version 0
i
XY
X
Y
X2
Y2
1
2
3
4
5
6
7
8
-4
1
5
6
7
8
9
30
62
6.679
7.677
8.321
8.510
8.637
8.785
8.937
12.000
69.546
To summarize n  8,
Y
2
-26.716
7.677
41.605
51.060
60.459
70.280
80.433
360.000
644.798
44.609
58.936
69.239
72.420
74.598
77.176
79.870
144.000
620.848
 X  62,  Y  69 .546 ,  XY  644 .798 ,  X
2
 1172 and
 620 .848 . df  n  2  8  2  6.
 X  62  7.75
Means: X 
n
8
X
Spare Parts: SS x 
2
Y 
 Y  69.546  8.69325
n
8
 nX  1172  87.752  691.50
2
Y  nY  620.848  88.69325  16.2673  SST
  XY  nXY  644 .798  87.75 8.69325   105 .8165
SS y 
S xy
16
1
25
36
49
64
81
900
1172
2
2
2
(Total Sum of Squares)
Coefficients:
b1 
S xy
SS x

 XY  nXY
 X  nX
2
2

105 .8165
 0.1530
691 .500
b0  Y  b1 X  8.69325  0.1530  7.75   7.5075
So our equation is Yˆ  7.5075  0.1539 X
b. Compute R 2 . (2) Solution: SSR  b1 Sxy  0.1530 105 .8165 16.1899
R2 
S xy 2
105 .8165 2  .9954
SSR b1 S xy 16 .1899


 .9952 or R 2 

SST
SSy 16 .2673
SS x SS y 691 .516 .2673
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c. Compute s e . (2) Solution: SSE  SST  SSR  16.2673  16.1899  .07740
s e2 
SSE .07740

 .01290 s e  .01290  0.11358
n2
6
Note also that if R 2 is available s e2 

SS y 1  R 2
n2
  16.2673 1  .9954   .01247
6
and
s e  .01247  0.11168 . I will compromise and use s e2  .0127
 1   0.0127 
d. Compute s b1 and do a significance test on b1 . (2) Solution: s b21  s e2 
  0.00001837

 SS x   691 .5 
b 0
0.1539
s b  0.00001837  0.004286 . t  1

 35 .908 . If we assume that   .05 , compare this
1
s b1
0.004286
6
with t .025
 2.447. Since the computed t is larger than the table t in absolute value, we reject the null
hypothesis of no significance and say that the slope is significant.
e. Use your equation to predict the number of pensioners in 2005 and 2006. Using the 2006 number, create
a prediction interval for the number of pensioners for that year. Explain why a confidence interval for the
number of pensioners is inappropriate. (3). Solution: Our base year was 1970 so the value of X for 2005
is 2005 – 1970 = 35. Our equation is Yˆ  7.5075  0.1539 X and for 2005 it gives
Ŷ  7.5075  0.153935 
12.894. Since the slope is 0.1539, add this to the 2005 value to get 13.048 for 2006. The formula for the
1 X X 2

Prediction Interval is Y0  Yˆ0  t sY , where sY2  s e2   0
 1 . For 2006, X 0  36 . It’s time to
n

SS x




remember that s e  .01247  0.11168 , n  8, X  7.75 and SS x 
X
2
 nX 2  691 .50 .
 1 36  7.75 2

So sY2  .1247  
 1  .1247 0.1250  1.1541  1  .1247 2.2791   0.2842
8

691 .5


s  0.2842  0.5331 . We already know that t 6   2.447 and that for 2006 Yˆ  13 .048 . So that our
Y
.025
prediction interval is Y0  Yˆ0  t sY  13.048  2.4470.5331  13.048  1.304 , or between 11.7 and 14.3
million people. The confidence interval is inappropriate for this type of problem. To use the problem
demonstrated in class, a confidence interval done when X 0  5 gives us a likely range in which the
average number of children will fall for the average couple that wants 5 children. The prediction interval
gives us a likely range in which the number of children will fall for one couple that wants 5 children. There
is no average year 2006, since it will be 2006 only once, so an interval for an average makes no sense.
f. Make a graph of the data. Show the trend line clearly. If you are not willing to do this neatly and
accurately, don’t bother. (2) Suggestions: I don’t have the graphical power to do this but here’s what I
would do. Since our years vary from 1966 to 2006 and the number of pensioners varies from 6.68 to a
projected 13.0 million, your x axis should be marked from 1966 to 2006, but I would probably only mark
the years 1970 to 2010 by 5-year intervals. I might consider stretching the whole thing out to 2050 because
of the next question. The y-axis might go from 6 to 14 million with marks for every two million. I can plot
the regression line Yˆ  7.5075  0.1539 X by noting that for 1970 it gives us 7.5 million which grows to 13
million in 2006. these two points can be connected to give us the regression line and the 8 points that we
used to get the regression equation can be plotted around it.
g. What percent rise in pensioners did the equation predict for 2006? What percent rise does it predict for
2050? The population of the United Kingdom grew at roughly 0.31% a year over the last quarter of the 20 th
century. Can you intelligently guess what is wrong? (1) Solution: Our base year was 1970 so the value of
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X for 2049 is 2049 – 1970 = 79. Our equation is Yˆ  7.5075  0.1539 X and for 2049 it gives
Yˆ  7.5075  0.153979  19.666 . Since the slope is 0.1539, add this to the 2049 value to get 19.819 for
2050. We thus have the following.
Year
Number Per cent growth.
2005
12.894
2006
13.048
1.19
2049
19.666
2050
19.819
0. 78
It is reasonable to expect the growth rate to fall as the absolute number of additional pensioners stays
constant but the number of pensioners grows. However, we are predicting that for an 80 year period the
number of pensioners will grow faster than population. Aside from the fact that this results in a substantial
rise in the number of pensioners per worker that may be politically impossible, it’s hard to believe that the
country will produce old people at a rate substantially higher than population grows for over 80 years. This
is a basic problem with using a trend line to make predictions. A given slope may be appropriate for a long
while, but it is no more appropriate to say it will last forever, than it is to say that Wal-Mart’s sales will
continue to grow at a rate way above total retail sales for decades. Sooner or later Wal-Mart’s sales will be
such a large part of retail sales that the only way to grow Wal-Mart’s sales at such a high rate is to grow all
retail sales at a higher rate, which just won’t happen if we are all drawing our wages from Wal-Mart.
————— 11/30/2006 10:41:39 PM ————————————————————
Welcome to Minitab, press F1 for help.
4
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Take-Home Problem 1, Version 0
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06031000.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06031-000.MTW'
Worksheet was saved on Wed Nov 29 2006
Results for: 252x06031-000.MTW
MTB > Execute "C:\Documents and Settings\RBOVE\My
Documents\Minitab\252sols06031.mtb" 1.
Executing from file: C:\Documents and Settings\RBOVE\My
Documents\Minitab\252sols06031.mtb
Data Display
K11
K12
K13
K14
K15
3.00000
4.00000
5.00000
6.00000
7.00000
Regression Analysis: pensioners versus year
The regression equation is
pensioners = 7.51 + 0.153 year
Predictor
Constant
year
Coef
7.50731
0.153025
S = 0.111940
SE Coef
0.05152
0.004257
R-Sq = 99.5%
Analysis of Variance
Source
DF
SS
Regression
1 16.193
Residual Error
6
0.075
Total
7 16.268
Obs
1
2
3
4
5
6
7
8
year
-4.0
1.0
5.0
6.0
7.0
8.0
9.0
30.0
pensioners
6.6790
7.6770
8.3210
8.5100
8.6370
8.7850
8.9370
12.0000
T
145.71
35.95
P
0.000
0.000
R-Sq(adj) = 99.5%
MS
16.193
0.013
Fit
6.8952
7.6603
8.2724
8.4255
8.5785
8.7315
8.8845
12.0980
F
1292.24
SE Fit
0.0638
0.0489
0.0413
0.0403
0.0397
0.0396
0.0399
0.1027
P
0.000
Residual
-0.2162
0.0167
0.0486
0.0845
0.0585
0.0535
0.0525
-0.0980
St Resid
-2.35R
0.17
0.47
0.81
0.56
0.51
0.50
-2.20RX
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Predicted Values for New Observations
New
Obs
Fit SE Fit
95% CI
1 12.0980 0.1027 (11.8469, 12.3492)
2 12.8632 0.1226 (12.5633, 13.1631)
3 13.0162 0.1266 (12.7064, 13.3260)
4 19.5963 0.3059 (18.8478, 20.3447)
5 19.7493 0.3101 (18.9905, 20.5080)
95%
(11.7264,
(12.4570,
(12.6027,
(18.7993,
(18.9426,
PI
12.4697)X
13.2693)X
13.4297)XX
20.3932)XX
20.5560)XX
X denotes a point that is an outlier in the predictors.
XX denotes a point that is an extreme outlier in the predictors.
5
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Values of Predictors for New Observations
New
Obs year
1 30.0
2 35.0
3 36.0
4 79.0
5 80.0
Data Display
Row
1
2
3
4
5
6
7
8
year
-4
1
5
6
7
8
9
30
pensioners
6.679
7.677
8.321
8.510
8.637
8.785
8.937
12.000
_xsq
16
1
25
36
49
64
81
900
_xy
-26.716
7.677
41.605
51.060
60.459
70.280
80.433
360.000
_ysq
44.609
58.936
69.239
72.420
74.598
77.176
79.870
144.000
Data Display
sum_y
sum_x
sum_xsq
sum_xy
sum_ysq
69.5460
62.0000
1172.00
644.798
620.848
Data Display
_n
_SSX
_SSY
_SSxy
_xbar
_ybar
8.00000
691.500
16.2677
105.817
7.75000
8.69325
6
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Take-Home Problem 1, Version 1
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06031001.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06031-001.MTW'
Worksheet was saved on Wed Nov 29 2006
Results for: 252x06031-001.MTW
MTB > exec '252sols06031'
Executing from file: 252sols06031.MTB
Data Display
K11
K12
K13
K14
K15
3.00000
4.00000
5.00000
6.00000
7.00000
Regression Analysis: pensioners versus year
The regression equation is
pensioners = 7.52 + 0.150 year
Predictor
Constant
year
Coef
7.51975
0.149807
S = 0.126744
SE Coef
0.05834
0.004820
R-Sq = 99.4%
Analysis of Variance
Source
DF
SS
Regression
1 15.519
Residual Error
6
0.096
Total
7 15.615
T
128.90
31.08
P
0.000
0.000
R-Sq(adj) = 99.3%
MS
15.519
0.016
F
966.06
P
0.000
Obs year pensioners
Fit SE Fit Residual St Resid
1 -4.0
6.6790
6.9205 0.0722
-0.2415
-2.32R
2
1.0
7.6770
7.6696 0.0554
0.0074
0.07
3
5.0
8.3210
8.2688 0.0467
0.0522
0.44
4
6.0
8.5100
8.4186 0.0456
0.0914
0.77
5
7.0
8.6370
8.5684 0.0450
0.0686
0.58
6
8.0
8.7850
8.7182 0.0448
0.0668
0.56
7
9.0
8.9370
8.8680 0.0452
0.0690
0.58
8 30.0
11.9000 12.0140 0.1162
-0.1140
-2.25RX
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Predicted Values for New Observations
New
Obs
Fit SE Fit
95% CI
95% PI
1 12.0140 0.1162 (11.7296, 12.2984) (11.5932, 12.4347)X
2 12.7630 0.1388 (12.4234, 13.1026) (12.3031, 13.2229)X
3 12.9128 0.1433 (12.5620, 13.2635) (12.4446, 13.3810)XX
4 19.3545 0.3463 (18.5071, 20.2019) (18.4521, 20.2569)XX
5 19.5043 0.3511 (18.6452, 20.3634) (18.5909, 20.4177)XX
X denotes a point that is an outlier in the predictors.
XX denotes a point that is an extreme outlier in the predictors.
7
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Values of Predictors for New Observations
New
Obs year
1 30.0
2 35.0
3 36.0
4 79.0
5 80.0
Data Display
Row
1
2
3
4
5
6
7
8
year
-4
1
5
6
7
8
9
30
pensioners
6.679
7.677
8.321
8.510
8.637
8.785
8.937
11.900
_xsq
16
1
25
36
49
64
81
900
_xy
-26.716
7.677
41.605
51.060
60.459
70.280
80.433
357.000
_ysq
44.609
58.936
69.239
72.420
74.598
77.176
79.870
141.610
Data Display
sum_y
sum_x
sum_xsq
sum_xy
sum_ysq
69.4460
62.0000
1172.00
641.798
618.458
Data Display
_n
_SSX
_SSY
_SSxy
_xbar
_ybar
8.00000
691.500
15.6151
103.592
7.75000
8.68075
8
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Take-Home Problem 1, Version 2
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06031002.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06031-002.MTW'
Worksheet was saved on Wed Nov 29 2006
Results for: 252x06031-002.MTW
MTB > exec '252sols06031'
Executing from file: 252sols06031.MTB
Data Display
K11
K12
K13
K14
K15
3.00000
4.00000
5.00000
6.00000
7.00000
Regression Analysis: pensioners versus year
The regression equation is
pensioners = 7.53 + 0.147 year
Predictor
Coef
SE Coef
T
P
Constant
7.53218
0.06530 115.35 0.000
year
0.146589 0.005395
27.17 0.000
S = 0.141872
R-Sq = 99.2%
R-Sq(adj) = 99.1%
Analysis of Variance
Source
DF
SS
Regression
1 14.859
Residual Error
6
0.121
Total
7 14.980
MS
14.859
0.020
F
738.25
P
0.000
Obs year pensioners
Fit SE Fit Residual St Resid
1 -4.0
6.6790
6.9458 0.0808
-0.2668
-2.29R
2
1.0
7.6770
7.6788 0.0620
-0.0018
-0.01
3
5.0
8.3210
8.2651 0.0523
0.0559
0.42
4
6.0
8.5100
8.4117 0.0510
0.0983
0.74
5
7.0
8.6370
8.5583 0.0503
0.0787
0.59
6
8.0
8.7850
8.7049 0.0502
0.0801
0.60
7
9.0
8.9370
8.8515 0.0506
0.0855
0.65
8 30.0
11.8000 11.9299 0.1301
-0.1299
-2.30RX
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Predicted Values for New Observations
New
Obs
Fit SE Fit
95% CI
1 11.9299 0.1301 (11.6115, 12.2482)
2 12.6628 0.1553 (12.2827, 13.0429)
3 12.8094 0.1605 (12.4168, 13.2020)
4 19.1127 0.3877 (18.1642, 20.0613)
5 19.2593 0.3930 (18.2977, 20.2210)
95%
(11.4588,
(12.1480,
(12.2853,
(18.1026,
(18.2369,
PI
12.4009)X
13.1776)X
13.3335)XX
20.1228)XX
20.2817)XX
X denotes a point that is an outlier in the predictors.
XX denotes a point that is an extreme outlier in the predictors.
9
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Values of Predictors for New Observations
New
Obs year
1 30.0
2 35.0
3 36.0
4 79.0
5 80.0
Data Display
Row
1
2
3
4
5
6
7
8
year
-4
1
5
6
7
8
9
30
pensioners
6.679
7.677
8.321
8.510
8.637
8.785
8.937
11.800
_xsq
16
1
25
36
49
64
81
900
_xy
-26.716
7.677
41.605
51.060
60.459
70.280
80.433
354.000
_ysq
44.609
58.936
69.239
72.420
74.598
77.176
79.870
139.240
Data Display
sum_y
sum_x
sum_xsq
sum_xy
sum_ysq
69.3460
62.0000
1172.00
638.798
616.088
Data Display
_n
_SSX
_SSY
_SSxy
_xbar
_ybar
8.00000
691.500
14.9800
101.367
7.75000
8.66825
10
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Take-Home Problem 1, Version 3
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06031003.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06031-003.MTW'
Worksheet was saved on Wed Nov 29 2006
Results for: 252x06031-003.MTW
MTB > exec '252sols06031'
Executing from file: 252sols06031.MTB
Data Display
K11
K12
K13
K14
K15
3.00000
4.00000
5.00000
6.00000
7.00000
Regression Analysis: pensioners versus year
The regression equation is
pensioners = 7.54 + 0.143 year
Predictor
Coef
SE Coef
T
P
Constant
7.54462
0.07237 104.25 0.000
year
0.143372 0.005979
23.98 0.000
S = 0.157230
R-Sq = 99.0%
R-Sq(adj) = 98.8%
Analysis of Variance
Source
DF
SS
Regression
1 14.214
Residual Error
6
0.148
Total
7 14.362
MS
14.214
0.025
F
574.97
P
0.000
Obs year pensioners
Fit SE Fit Residual St Resid
1 -4.0
6.6790
6.9711 0.0896
-0.2921
-2.26R
2
1.0
7.6770
7.6880 0.0687
-0.0110
-0.08
3
5.0
8.3210
8.2615 0.0580
0.0595
0.41
4
6.0
8.5100
8.4048 0.0566
0.1052
0.72
5
7.0
8.6370
8.5482 0.0558
0.0888
0.60
6
8.0
8.7850
8.6916 0.0556
0.0934
0.64
7
9.0
8.9370
8.8350 0.0561
0.1020
0.69
8 30.0
11.7000 11.8458 0.1442
-0.1458
-2.32RX
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Predicted Values for New Observations
New
Obs
Fit SE Fit
95% CI
95% PI
1 11.8458 0.1442 (11.4930, 12.1986) (11.3238, 12.3678)X
2 12.5626 0.1722 (12.1414, 12.9839) (11.9921, 13.1331)X
3 12.7060 0.1778 (12.2709, 13.1411) (12.1252, 13.2868)XX
4 18.8710 0.4296 (17.8197, 19.9222) (17.7515, 19.9904)XX
5 19.0144 0.4356 (17.9486, 20.0801) (17.8813, 20.1474)XX
X denotes a point that is an outlier in the predictors.
XX denotes a point that is an extreme outlier in the predictors.
11
251y0631s1
12/12/06
Values of Predictors for New Observations
New
Obs year
1 30.0
2 35.0
3 36.0
4 79.0
5 80.0
Data Display
Row
1
2
3
4
5
6
7
8
year
-4
1
5
6
7
8
9
30
pensioners
6.679
7.677
8.321
8.510
8.637
8.785
8.937
11.700
_xsq
16
1
25
36
49
64
81
900
_xy
-26.716
7.677
41.605
51.060
60.459
70.280
80.433
351.000
_ysq
44.609
58.936
69.239
72.420
74.598
77.176
79.870
136.890
Data Display
sum_y
sum_x
sum_xsq
sum_xy
sum_ysq
69.2460
62.0000
1172.00
635.798
613.738
Data Display
_n
_SSX
_SSY
_SSxy
_xbar
_ybar
8.00000
691.500
14.3624
99.1415
7.75000
8.65575
12
251y0631s1
12/12/06
Take-Home Problem 1, Version 4
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06031004.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06031-004.MTW'
Worksheet was saved on Wed Nov 29 2006
Results for: 252x06031-004.MTW
MTB > exec '252sols06031'
Executing from file: 252sols06031.MTB
Data Display
K11
K12
K13
K14
K15
3.00000
4.00000
5.00000
6.00000
7.00000
Regression Analysis: pensioners versus year
The regression equation is
pensioners = 7.56 + 0.140 year
Predictor
Coef
SE Coef
T
Constant
7.55706
0.07952 95.04
year
0.140154 0.006570 21.33
S = 0.172759
R-Sq = 98.7%
Analysis of Variance
Source
DF
SS
Regression
1 13.583
Residual Error
6
0.179
Total
7 13.762
P
0.000
0.000
R-Sq(adj) = 98.5%
MS
13.583
0.030
F
455.12
P
0.000
Obs year pensioners
Fit SE Fit Residual St Resid
1 -4.0
6.6790
6.9964 0.0984
-0.3174
-2.24R
2
1.0
7.6770
7.6972 0.0755
-0.0202
-0.13
3
5.0
8.3210
8.2578 0.0637
0.0632
0.39
4
6.0
8.5100
8.3980 0.0622
0.1120
0.69
5
7.0
8.6370
8.5381 0.0613
0.0989
0.61
6
8.0
8.7850
8.6783 0.0611
0.1067
0.66
7
9.0
8.9370
8.8184 0.0616
0.1186
0.73
8 30.0
11.6000 11.7617 0.1584
-0.1617
-2.35RX
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Predicted Values for New Observations
New
Obs
Fit SE Fit
95% CI
95% PI
1 11.7617 0.1584 (11.3740, 12.1493) (11.1881, 12.3352)X
2 12.4624 0.1892 (11.9996, 12.9253) (11.8356, 13.0893)X
3 12.6026 0.1954 (12.1245, 13.0807) (11.9644, 13.2408)XX
4 18.6292 0.4721 (17.4741, 19.7843) (17.3992, 19.8592)XX
5 18.7694 0.4786 (17.5984, 19.9404) (17.5244, 20.0144)XX
X denotes a point that is an outlier in the predictors.
XX denotes a point that is an extreme outlier in the predictors.
13
251y0631s1
12/12/06
Values of Predictors for New Observations
New
Obs year
1 30.0
2 35.0
3 36.0
4 79.0
5 80.0
Data Display
Row
1
2
3
4
5
6
7
8
year
-4
1
5
6
7
8
9
30
pensioners
6.679
7.677
8.321
8.510
8.637
8.785
8.937
11.600
_xsq
16
1
25
36
49
64
81
900
_xy
-26.716
7.677
41.605
51.060
60.459
70.280
80.433
348.000
_ysq
44.609
58.936
69.239
72.420
74.598
77.176
79.870
134.560
Data Display
sum_y
sum_x
sum_xsq
sum_xy
sum_ysq
69.1460
62.0000
1172.00
632.798
611.408
Data Display
_n
_SSX
_SSY
_SSxy
_xbar
_ybar
8.00000
691.500
13.7623
96.9165
7.75000
8.64325
14
251y0631s1
12/12/06
Take-Home Problem 1, Version 5
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06031005.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06031-005.MTW'
Worksheet was saved on Wed Nov 29 2006
Results for: 252x06031-005.MTW
MTB > exec '252sols06031'
Executing from file: 252sols06031.MTB
Data Display
K11
K12
K13
K14
K15
3.00000
4.00000
5.00000
6.00000
7.00000
Regression Analysis: pensioners versus year
The regression equation is
pensioners = 7.57 + 0.137 year
Predictor
Coef
SE Coef
T
P
Constant
7.56949
0.08672 87.28 0.000
year
0.136936 0.007165 19.11 0.000
S = 0.188414
R-Sq = 98.4%
R-Sq(adj) = 98.1%
Analysis of Variance
Source
DF
SS
Regression
1 12.967
Residual Error
6
0.213
Total
7 13.180
MS
12.967
0.035
F
365.26
P
0.000
Obs year pensioners
Fit SE Fit Residual St Resid
1 -4.0
6.6790
7.0217 0.1074
-0.3427
-2.21R
2
1.0
7.6770
7.7064 0.0823
-0.0294
-0.17
3
5.0
8.3210
8.2542 0.0695
0.0668
0.38
4
6.0
8.5100
8.3911 0.0678
0.1189
0.68
5
7.0
8.6370
8.5280 0.0668
0.1090
0.62
6
8.0
8.7850
8.6650 0.0666
0.1200
0.68
7
9.0
8.9370
8.8019 0.0672
0.1351
0.77
8 30.0
11.5000 11.6776 0.1728
-0.1776
-2.36RX
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Predicted Values for New Observations
New
Obs
Fit SE Fit
95% CI
95% PI
1 11.6776 0.1728 (11.2548, 12.1004) (11.0521, 12.3031)X
2 12.3623 0.2063 (11.8575, 12.8671) (11.6786, 13.0459)X
3 12.4992 0.2131 (11.9778, 13.0206) (11.8032, 13.1952)XX
4 18.3875 0.5148 (17.1277, 19.6472) (17.0460, 19.7289)XX
5 18.5244 0.5219 (17.2473, 19.8015) (17.1666, 19.8822)XX
X denotes a point that is an outlier in the predictors.
XX denotes a point that is an extreme outlier in the predictors.
15
251y0631s1
12/12/06
Values of Predictors for New Observations
New
Obs year
1 30.0
2 35.0
3 36.0
4 79.0
5 80.0
Data Display
Row
1
2
3
4
5
6
7
8
year
-4
1
5
6
7
8
9
30
pensioners
6.679
7.677
8.321
8.510
8.637
8.785
8.937
11.500
_xsq
16
1
25
36
49
64
81
900
_xy
-26.716
7.677
41.605
51.060
60.459
70.280
80.433
345.000
_ysq
44.609
58.936
69.239
72.420
74.598
77.176
79.870
132.250
Data Display
sum_y
sum_x
sum_xsq
sum_xy
sum_ysq
69.0460
62.0000
1172.00
629.798
609.098
Data Display
_n
_SSX
_SSY
_SSxy
_xbar
_ybar
8.00000
691.500
13.1797
94.6915
7.75000
8.63075
16
251y0631s1
12/12/06
Take-Home Problem 1, Version 6
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06031006.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06031-006.MTW'
Worksheet was saved on Wed Nov 29 2006
Results for: 252x06031-006.MTW
MTB > exec '252sols06031'
Executing from file: 252sols06031.MTB
Data Display
K11
K12
K13
K14
K15
3.00000
4.00000
5.00000
6.00000
7.00000
Regression Analysis: pensioners versus year
The regression equation is
pensioners = 7.58 + 0.134 year
Predictor
Coef
SE Coef
T
Constant
7.58193
0.09397 80.68
year
0.133719 0.007764 17.22
S = 0.204168
R-Sq = 98.0%
Analysis of Variance
Source
DF
SS
Regression
1 12.365
Residual Error
6
0.250
Total
7 12.615
P
0.000
0.000
R-Sq(adj) = 97.7%
MS
12.365
0.042
F
296.62
P
0.000
Obs year pensioners
Fit SE Fit Residual St Resid
1 -4.0
6.6790
7.0471 0.1163
-0.3681
-2.19R
2
1.0
7.6770
7.7156 0.0892
-0.0386
-0.21
3
5.0
8.3210
8.2505 0.0753
0.0705
0.37
4
6.0
8.5100
8.3842 0.0735
0.1258
0.66
5
7.0
8.6370
8.5180 0.0724
0.1190
0.62
6
8.0
8.7850
8.6517 0.0722
0.1333
0.70
7
9.0
8.9370
8.7854 0.0728
0.1516
0.79
8 30.0
11.4000 11.5935 0.1872
-0.1935
-2.38RX
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Predicted Values for New Observations
New
Obs
Fit SE Fit
95% CI
1 11.5935 0.1872 (11.1354, 12.0516)
2 12.2621 0.2235 (11.7151, 12.8091)
3 12.3958 0.2309 (11.8308, 12.9608)
4 18.1457 0.5579 (16.7806, 19.5108)
5 18.2794 0.5656 (16.8955, 19.6634)
95%
(10.9157,
(11.5213,
(11.6416,
(16.6921,
(16.8081,
PI
12.2713)X
13.0029)X
13.1500)XX
19.5993)XX
19.7508)XX
X denotes a point that is an outlier in the predictors.
XX denotes a point that is an extreme outlier in the predictors.
17
251y0631s1
12/12/06
Values of Predictors for New Observations
New
Obs
1
2
3
4
5
year
30.0
35.0
36.0
79.0
80.0
Data Display
Row
1
2
3
4
5
6
7
8
year
-4
1
5
6
7
8
9
30
pensioners
6.679
7.677
8.321
8.510
8.637
8.785
8.937
11.400
_xsq
16
1
25
36
49
64
81
900
_xy
-26.716
7.677
41.605
51.060
60.459
70.280
80.433
342.000
_ysq
44.609
58.936
69.239
72.420
74.598
77.176
79.870
129.960
Data Display
sum_y
sum_x
sum_xsq
sum_xy
sum_ysq
68.9460
62.0000
1172.00
626.798
606.808
Data Display
_n
_SSX
_SSY
_SSxy
_xbar
_ybar
8.00000
691.500
12.6146
92.4665
7.75000
8.61825
18
251y0631s1
12/12/06
Take-Home Problem 1, Version 7
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06031007.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06031-007.MTW'
Worksheet was saved on Wed Nov 29 2006
Results for: 252x06031-007.MTW
MTB > exec '252sols06031'
Executing from file: 252sols06031.MTB
Data Display
K11
K12
K13
K14
K15
3.00000
4.00000
5.00000
6.00000
7.00000
Regression Analysis: pensioners versus year
The regression equation is
pensioners = 7.59 + 0.131 year
Predictor
Constant
year
Coef
7.5944
0.130501
S = 0.219998
SE Coef
0.1013
0.008366
R-Sq = 97.6%
Analysis of Variance
Source
DF
SS
Regression
1 11.777
Residual Error
6
0.290
Total
7 12.067
T
75.00
15.60
P
0.000
0.000
R-Sq(adj) = 97.2%
MS
11.777
0.048
F
243.32
P
0.000
Obs year pensioners
Fit SE Fit Residual St Resid
1 -4.0
6.6790
7.0724 0.1254
-0.3934
-2.18R
2
1.0
7.6770
7.7249 0.0961
-0.0479
-0.24
3
5.0
8.3210
8.2469 0.0811
0.0741
0.36
4
6.0
8.5100
8.3774 0.0791
0.1326
0.65
5
7.0
8.6370
8.5079 0.0780
0.1291
0.63
6
8.0
8.7850
8.6384 0.0778
0.1466
0.71
7
9.0
8.9370
8.7689 0.0785
0.1681
0.82
8 30.0
11.3000 11.5094 0.2017
-0.2094
-2.39RX
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Predicted Values for New Observations
New
Obs
Fit SE Fit
95% CI
95% PI
1 11.5094 0.2017 (11.0158, 12.0030) (10.7790, 12.2398)X
2 12.1619 0.2409 (11.5725, 12.7513) (11.3637, 12.9601)X
3 12.2924 0.2488 (11.6836, 12.9012) (11.4797, 13.1051)XX
4 17.9040 0.6011 (16.4330, 19.3749) (16.3376, 19.4703)XX
5 18.0345 0.6094 (16.5432, 19.5257) (16.4490, 19.6199)XX
X denotes a point that is an outlier in the predictors.
XX denotes a point that is an extreme outlier in the predictors.
19
251y0631s1
12/12/06
Values of Predictors for New Observations
New
Obs year
1 30.0
2 35.0
3 36.0
4 79.0
5 80.0
Data Display
Row
1
2
3
4
5
6
7
8
year
-4
1
5
6
7
8
9
30
pensioners
6.679
7.677
8.321
8.510
8.637
8.785
8.937
11.300
_xsq
16
1
25
36
49
64
81
900
_xy
-26.716
7.677
41.605
51.060
60.459
70.280
80.433
339.000
_ysq
44.609
58.936
69.239
72.420
74.598
77.176
79.870
127.690
Data Display
sum_y
sum_x
sum_xsq
sum_xy
sum_ysq
68.8460
62.0000
1172.00
623.798
604.538
Data Display
_n
_SSX
_SSY
_SSxy
_xbar
_ybar
8.00000
691.500
12.0670
90.2415
7.75000
8.60575
20
251y0631s1
12/12/06
Take-Home Problem 1, Version 8
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06031008.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06031-008.MTW'
Worksheet was saved on Wed Nov 29 2006
Results for: 252x06031-008.MTW
MTB > exec '252sols06031'
Executing from file: 252sols06031.MTB
Data Display
K11
K12
K13
K14
K15
3.00000
4.00000
5.00000
6.00000
7.00000
Regression Analysis: pensioners versus year
The regression equation is
pensioners = 7.61 + 0.127 year
Predictor
Constant
year
Coef
7.6068
0.127283
S = 0.235891
SE Coef
0.1086
0.008970
R-Sq = 97.1%
Analysis of Variance
Source
DF
SS
Regression
1 11.203
Residual Error
6
0.334
Total
7 11.537
Obs
1
2
3
4
5
6
7
8
year
-4.0
1.0
5.0
6.0
7.0
8.0
9.0
30.0
pensioners
6.6790
7.6770
8.3210
8.5100
8.6370
8.7850
8.9370
11.2000
T
70.06
14.19
P
0.000
0.000
R-Sq(adj) = 96.6%
MS
11.203
0.056
Fit
7.0977
7.7341
8.2432
8.3705
8.4978
8.6251
8.7524
11.4253
F
201.33
SE Fit
0.1344
0.1031
0.0870
0.0849
0.0837
0.0834
0.0842
0.2163
P
0.000
Residual
-0.4187
-0.0571
0.0778
0.1395
0.1392
0.1599
0.1846
-0.2253
St Resid
-2.16R
-0.27
0.35
0.63
0.63
0.72
0.84
-2.39RX
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Predicted Values for New Observations
New
Obs
1
2
3
4
5
Fit
11.4253
12.0617
12.1890
17.6622
17.7895
SE Fit
0.2163
0.2583
0.2668
0.6446
0.6535
95%
(10.8960,
(11.4297,
(11.5362,
(16.0850,
(16.1905,
CI
11.9546)
12.6937)
12.8418)
19.2394)
19.3884)
95%
(10.6422,
(11.2058,
(11.3176,
(15.9827,
(16.0895,
PI
12.2085)X
12.9176)X
13.0604)XX
19.3417)XX
19.4894)XX
X denotes a point that is an outlier in the predictors.
XX denotes a point that is an extreme outlier in the predictors.
21
251y0631s1
12/12/06
Values of Predictors for New Observations
New
Obs year
1 30.0
2 35.0
3 36.0
4 79.0
5 80.0
Data Display
Row
1
2
3
4
5
6
7
8
year
-4
1
5
6
7
8
9
30
pensioners
6.679
7.677
8.321
8.510
8.637
8.785
8.937
11.200
_xsq
16
1
25
36
49
64
81
900
_xy
-26.716
7.677
41.605
51.060
60.459
70.280
80.433
336.000
_ysq
44.609
58.936
69.239
72.420
74.598
77.176
79.870
125.440
Data Display
sum_y
sum_x
sum_xsq
sum_xy
sum_ysq
68.7460
62.0000
1172.00
620.798
602.288
Data Display
_n
_SSX
_SSY
_SSxy
_xbar
_ybar
8.00000
691.500
11.5369
88.0165
7.75000
8.59325
22
251y0631s1
12/12/06
Take-Home Problem 1, Version 9
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06031009.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06031-009.MTW'
Worksheet was saved on Wed Nov 29 2006
Results for: 252x06031-009.MTW
MTB > exec '252sols06031'
Executing from file: 252sols06031.MTB
Data Display
K11
K12
K13
K14
K15
3.00000
4.00000
5.00000
6.00000
7.00000
Regression Analysis: pensioners versus year
The regression equation is
pensioners = 7.62 + 0.124 year
Predictor
Constant
year
Coef
7.6192
0.124066
S = 0.251833
SE Coef
0.1159
0.009577
R-Sq = 96.5%
Analysis of Variance
Source
DF
SS
Regression
1 10.644
Residual Error
6
0.381
Total
7 11.024
T
65.73
12.95
P
0.000
0.000
R-Sq(adj) = 96.0%
MS
10.644
0.063
F
167.83
P
0.000
Obs year pensioners
Fit SE Fit Residual St Resid
1 -4.0
6.6790
7.1230 0.1435
-0.4440
-2.15R
2
1.0
7.6770
7.7433 0.1100
-0.0663
-0.29
3
5.0
8.3210
8.2396 0.0928
0.0814
0.35
4
6.0
8.5100
8.3636 0.0906
0.1464
0.62
5
7.0
8.6370
8.4877 0.0893
0.1493
0.63
6
8.0
8.7850
8.6118 0.0891
0.1732
0.74
7
9.0
8.9370
8.7358 0.0898
0.2012
0.86
8 30.0
11.1000 11.3412 0.2309
-0.2412
-2.40RX
R denotes an observation with a large standardized residual.
X denotes an observation whose X value gives it large influence.
Predicted Values for New Observations
New
Obs
Fit SE Fit
95% CI
95% PI
1 11.3412 0.2309 (10.7761, 11.9063) (10.5051, 12.1773)X
2 11.9615 0.2757 (11.2868, 12.6362) (11.0478, 12.8753)X
3 12.0856 0.2848 (11.3887, 12.7825) (11.1553, 13.0159)XX
4 17.4204 0.6881 (15.7367, 19.1042) (15.6274, 19.2134)XX
5 17.5445 0.6976 (15.8375, 19.2515) (15.7297, 19.3593)XX
X denotes a point that is an outlier in the predictors.
XX denotes a point that is an extreme outlier in the predictors.
23
251y0631s1
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Values of Predictors for New Observations
New
Obs year
1 30.0
2 35.0
3 36.0
4 79.0
5 80.0
Data Display
Row
1
2
3
4
5
6
7
8
year
-4
1
5
6
7
8
9
30
pensioners
6.679
7.677
8.321
8.510
8.637
8.785
8.937
11.100
_xsq
16
1
25
36
49
64
81
900
_xy
-26.716
7.677
41.605
51.060
60.459
70.280
80.433
333.000
_ysq
44.609
58.936
69.239
72.420
74.598
77.176
79.870
123.210
Data Display
sum_y
sum_x
sum_xsq
sum_xy
sum_ysq
68.6460
62.0000
1172.00
617.798
600.058
Data Display
_n
_SSX
_SSY
_SSxy
_xbar
_ybar
8.00000
691.500
11.0243
85.7915
7.75000
8.58075
24
251y0631s1
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2) The Lees in their text ask whether experience makes a difference in student earnings and present the
following data for student earnings versus years of work experience. To personalize these data, take the
second to last digit of your student number call it a . Clearly label the problem with a version number
based on your student number. Then take your a , multiply it by 0.5 and add it to the 13 in the lower left
corner. . (Example: Good ol’ Seymour’s student number is 123456, so the 13 becomes 13 + 0.5(5) = 13 +
2.5 = 15.5 and he labels it Version 5.) Each column is to be regarded as an independent random sample.
Years of Work Experience
1
2
3
16
19
24
21
20
21
18
21
22
13
20
25
a) State your null hypothesis and test it by doing a 1-way ANOVA on these data and explain whether the
test tells us that experience matters or not. (4)
b) Using your results from a) present two different confidence intervals for the difference between earnings
for those with 1 and 3 years experience. Explain (i) under what circumstances you would use each interval
and (ii) whether the intervals show a significant difference. (2)
c) What other method could we use on these data to see if years of experience make a difference? Under
what circumstances would we use it? Try it and tell what it tests and what it shows. (3) [24]
d) (Extra Credit) Do a Levene test on these data and explain what it tests and shows. (4)
You should be able to do the calculations below. Only the three columns of numbers were given to us.
1
16
21
18
13
Years
2
19
20
21
20
3
24
20
22
25
Sum
68 +
80 +
92
 240 
nj
4+
4+
4
 12  n
x j
17
20
23
SS
1190 +
1602 +
2126
x 2j
289 
400 +
 x  nx  49181220
SSB   n x  nx  417  420
SST 
2
ij
2
j .j
Source
Between
2
2
2
SS
72
Sum
 x
20  x
 x
 1218   x
 4918 
 4918  4800  118
2
 4232  12202  41218  12202  4872  4800  72
2
2
ij
2
j
529
2
DF
ij

MS
36
F
F.05
7.04 s
F 2,9  4.26
H0
Column means equal
Within
46
9
5.1111
Total
118
11
Because our computed F statistic exceeds the 5% table value, we reject the null hypothesis of equal means
and conclude that experience matters.
b) The following material is modified from the solution to the last graded assignment.
Types of contrast between means. Assume   0.05 . m  4 is the number of columns.
25
251y0631s1
12/12/06
Individual Confidence Interval
If we desire a single interval, we use the formula for the difference between two means when the variance
is known. For example, if we want the difference between means of column 1 and column 2.
1   2  x1  x2   tn  m s
2
1
1
, where s  MSW . The within degrees of freedom are

n1 n2
9
 2.262 . n1  n 2  n3  n4  4 , so
n  m  9, , so we use t .025
s2

1
n1 
1
n2

s 2  1 4  1 4   s 2 0.5 and our interval will be 1   3  x1  x3   2.262 s 2 .5
Scheffé e  Confidence Interval
If we desire intervals that will simultaneously be valid for a given confidence level for all possible intervals

1
1 
between column means, use 1   2  x1  x2   m  1Fm1,n  m   s
where s  MSW .

 n
n 2 
1

The degrees of freedom for columns are m  1  2 . The within degrees of freedom are n  m  9, so we
2,9   4.26 .
use F.05
n1  n 2  n3  n4  4 , so
s2

1
n1

1
n2

s 2  18  18   s 2 0.25  and our interval
will be 1   3  x1  x3   24.26   s 2 .5   x1  x2   2.9289  s 2 .5 




Tukey Confidence Interval
This also applies to all possible differences.
1   2  x1  x2   q m,n  m 
s
1
1
. where s  MSW . This gives rise to Tukey’s HSD

n1 n 2
2
(Honestly Significant Difference) procedure. Two sample means x .1 and x .2 are significantly different if
x.1  x.2 is greater than q m,n  m 
s
2
1
1


n1 n 2
s
2
1
1
3,9 
3,9 
 3.95 .
. We will need q .05
. The table says q.05

n1 n 2
s 1 1
2
2
    .25 s and the interval will be 1   3  x1  x3   3.95 0.5s .5
2 4 4
2
 x1  x2   2.7931 s 2 .25 
Contrasts for 3 and 1 .
Note that in all contrasts
s 2 .5  .5.1111 .5  2.555556  1.5986 . Intervals for differences between
means that include zero show no significant difference. x1  x3   17  23  6.
Individual – Used when you want only one interval.
1   3  x1  x3   2.262 s 2 .5  6  2.262 1.5986   6  3.61
Scheffé – Used when a collective confidence level is sought.
1   3  x1  x2   2.9289  s 2 .5   6  2.9289 1.5986   6  4.68


Tukey – More powerful, but similar to the Scheffé.
1   3  x1  x2   2.7931 s 2 .25   6  2.7931 1.5986   6  4.47 .
All contrasts seem significant.
26
251y0631s1
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c) The alternative to one-way ANOVA for situations in which the parent distribution is not Normal is the
2.0 4.0 11 .0
16 19 24
8 .0 5 .5 8 .0
21 20 21
Kruskal-Wallis test. The original data
is replaced by ranks 3.0 8.0 10 .0 . Note that the
18 21 22
1.0 5.5 12 .0
13 20 25
14 .0 23 .0 41 .0
12 13 
 78 , so that we can check our ranking by noting that 14 + 23 +
2
 SRi 2 

  3n  1
 n i 


sum of the first twelve numbers is
 12
41 = 78. H  
 nn  1 i
 12  14 2 23 2 412 

  313   1  1 196  529  1681   39  46 .2692  39  7.2692 . Since the







12
13
4
4
4
13  4 



Kruskal-Wallis table for 4, 4, 4 says that 5.6923 has a p-value of .049 and 7.5385 has a p-value of .011, the
p-value of 7.2692 must be below 5%, so that we can reject the null hypothesis of equal medians.
16 19 24

21 20 21
, which has column medians
18 21 22
13 20 25
1 1
1
4 0 2
of 17, 20 and 23, is replaced by the numbers with the column medians subtracted
. Absolute
1 1 1
4 0
2
1 1 1
4 0 2
values are taken, so the columns become
. An ANOVA is done on these 3 columns.
1 1 1
4 0 2
Source
SS
DF
MS
F.05
H0
F
d) The Levene test is a test for equal variances. The original data
Between
8
2
4
3.27 ns
F 2,9  4.26
Column means equal
Within
11
9
1.2222
Total
19
11
Because we cannot reject the null hypothesis, we cannot say that the variances are not equal.
————— 12/1/2006 12:23:02 AM ————————————————————
Welcome to Minitab, press F1 for help.
27
251y0631s1
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Take-Home Problem 2, Version 0
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06032000.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06032-000.MTW'
Worksheet was saved on Thu Nov 30 2006
Results for: 252x06032-000.MTW
MTB > Execute "C:\Documents and Settings\RBOVE\My
Documents\Minitab\252oneW306032.mtb" 1.
Executing from file: C:\Documents and Settings\RBOVE\My
Documents\Minitab\252oneW306032.mtb
Executing from file: 2522onw3.MTB
One-way ANOVA: 1, 2, 3
Source DF
Factor
2
Error
9
Total
11
S = 2.261
Level
1
2
3
N
4
4
4
SS
MS
72.00 36.00
46.00
5.11
118.00
R-Sq = 61.02%
Mean
17.000
20.000
23.000
StDev
3.367
0.816
1.826
F
7.04
P
0.014
R-Sq(adj) = 52.35%
Individual 95% CIs For Mean Based on
Pooled StDev
--+---------+---------+---------+------(--------*-------)
(--------*-------)
(--------*-------)
--+---------+---------+---------+------15.0
18.0
21.0
24.0
Pooled StDev = 2.261
Data Display
Row
1
2
3
4
1
16
21
18
13
2
19
20
21
20
3
24
21
22
25
Data Display
Row
1
2
3
4
x1sq
256
441
324
169
x2sq
361
400
441
400
x3sq
576
441
484
625
Data Display
sumx1
sumx2
sumx3
n1
n2
n3
smx1sq
smx2sq
smx3sq
68.0000
80.0000
92.0000
4.00000
4.00000
4.00000
1190.00
1602.00
2126.00
Data Display
x1bar
x2bar
x3bar
17.0000
20.0000
23.0000
28
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12/12/06
Data Display
Row
1
2
3
4
r1
2
8
3
1
r2
4.0
5.5
8.0
5.5
r3
11
8
10
12
Kruskal-Wallis Test: C11 versus C12
Kruskal-Wallis Test on C11
C12
N Median Ave Rank
1
4
17.00
3.5
2
4
20.00
5.8
3
4
23.00
10.3
Overall 12
6.5
Z
-2.04
-0.51
2.55
H = 7.27 DF = 2 P = 0.026
H = 7.40 DF = 2 P = 0.025 (adjusted for ties)
* NOTE * One or more small samples
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12 N
Lower
StDev
Upper
1 4 1.70187 3.36650 18.3097
2 4 0.41276 0.81650
4.4408
3 4 0.92297 1.82574
9.9298
Bartlett's Test (normal distribution)
Test statistic = 4.36, p-value = 0.113
Levene's Test (any continuous distribution)
Test statistic = 3.27, p-value = 0.085
Test for Equal Variances: C11 versus C12
Executing from file: 252Levene3.MTB
Data Display
Row
1
2
3
4
C21
16
21
18
13
C22
19
20
21
20
C23
24
21
22
25
Data Display
median1
median2
median3
17.0000
20.0000
23.0000
Data Display
Row
1
2
3
4
C21
-1
4
1
-4
C22
-1
0
1
0
C23
1
-2
-1
2
Data Display
Row
1
2
3
4
C21
1
4
1
4
C22
1
0
1
0
C23
1
2
1
2
29
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One-way ANOVA: C21, C22, C23
Source DF
Factor
2
Error
9
Total
11
S = 1.106
Level
C21
C22
C23
N
4
4
4
SS
MS
F
P
8.00 4.00 3.27 0.085
11.00 1.22
19.00
R-Sq = 42.11%
R-Sq(adj) = 29.24%
Mean
2.500
0.500
1.500
StDev
1.732
0.577
0.577
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(----------*---------)
(---------*----------)
(----------*---------)
------+---------+---------+---------+--0.0
1.2
2.4
3.6
Pooled StDev = 1.106
30
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Take-Home Problem 2, Version 1
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06032001.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06032-001.MTW'
Worksheet was saved on Thu Nov 30 2006
Results for: 252x06032-001.MTW
MTB > exec '252oneW306032'
Executing from file: 252oneW306032.MTB
Executing from file: 2522onw3.MTB
One-way ANOVA: 1, 2, 3
Source DF
Factor
2
Error
9
Total
11
S = 2.165
Level
1
2
3
N
4
4
4
SS
MS
69.04 34.52
42.19
4.69
111.23
R-Sq = 62.07%
Mean
17.125
20.000
23.000
StDev
3.172
0.816
1.826
F
7.36
P
0.013
R-Sq(adj) = 53.64%
Individual 95% CIs For Mean Based on
Pooled StDev
-+---------+---------+---------+-------(-------*-------)
(-------*-------)
(-------*-------)
-+---------+---------+---------+-------15.0
18.0
21.0
24.0
Pooled StDev = 2.165
Data Display
Row
1
2
3
4
1
16.0
21.0
18.0
13.5
2
19
20
21
20
3
24
21
22
25
Data Display
Row
1
2
3
4
x1sq
256.00
441.00
324.00
182.25
x2sq
361
400
441
400
x3sq
576
441
484
625
Data Display
sumx1
sumx2
sumx3
n1
n2
n3
smx1sq
smx2sq
smx3sq
68.5000
80.0000
92.0000
4.00000
4.00000
4.00000
1203.25
1602.00
2126.00
Data Display
x1bar
x2bar
x3bar
17.1250
20.0000
23.0000
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Data Display
Row
1
2
3
4
r1
2
8
3
1
r2
4.0
5.5
8.0
5.5
r3
11
8
10
12
Kruskal-Wallis Test: C11 versus C12
Kruskal-Wallis Test on C11
C12
N Median Ave Rank
Z
1
4
17.00
3.5 -2.04
2
4
20.00
5.8 -0.51
3
4
23.00
10.3
2.55
Overall 12
6.5
H = 7.27 DF = 2 P = 0.026
H = 7.40 DF = 2 P = 0.025 (adjusted for ties)
* NOTE * One or more small samples
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12
1
2
3
N
4
4
4
Lower
1.60362
0.41276
0.92297
StDev
3.17214
0.81650
1.82574
Upper
17.2526
4.4408
9.9298
Bartlett's Test (normal distribution)
Test statistic = 3.99, p-value = 0.136
Levene's Test (any continuous distribution)
Test statistic = 3.27, p-value = 0.086
Test for Equal Variances: C11 versus C12
Executing from file: 252Levene3.MTB
Data Display
Row
1
2
3
4
C21
16.0
21.0
18.0
13.5
C22
19
20
21
20
C23
24
21
22
25
Data Display
median1
median2
median3
17.0000
20.0000
23.0000
Data Display
Row
1
2
3
4
C21
-1.0
4.0
1.0
-3.5
C22
-1
0
1
0
C23
1
-2
-1
2
Data Display
Row
1
2
3
4
C21
1.0
4.0
1.0
3.5
C22
1
0
1
0
C23
1
2
1
2
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12/12/06
One-way ANOVA: C21, C22, C23
Source DF
Factor
2
Error
9
Total
11
S = 1.037
Level
C21
C22
C23
N
4
4
4
SS
MS
F
P
7.04 3.52 3.27 0.086
9.69 1.08
16.73
R-Sq = 42.09%
R-Sq(adj) = 29.22%
Mean
2.375
0.500
1.500
StDev
1.601
0.577
0.577
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(---------*---------)
(---------*---------)
(---------*--------)
------+---------+---------+---------+--0.0
1.2
2.4
3.6
Pooled StDev = 1.037
33
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Take-Home Problem 2, Version 2
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06032002.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06032-002.MTW'
Worksheet was saved on Thu Nov 30 2006
Results for: 252x06032-002.MTW
MTB > exec '252oneW306032'
Executing from file: 252oneW306032.MTB
Executing from file: 2522onw3.MTB
One-way ANOVA: 1, 2, 3
Source DF
Factor
2
Error
9
Total
11
S = 2.075
Level
1
2
3
N
4
4
4
SS
MS
66.17 33.08
38.75
4.31
104.92
R-Sq = 63.07%
Mean
17.250
20.000
23.000
StDev
2.986
0.816
1.826
F
7.68
P
0.011
R-Sq(adj) = 54.86%
Individual 95% CIs For Mean Based on Pooled
StDev
+---------+---------+---------+--------(-------*------)
(-------*------)
(-------*------)
+---------+---------+---------+--------15.0
18.0
21.0
24.0
Pooled StDev = 2.075
Data Display
Row
1
2
3
4
1
16
21
18
14
2
19
20
21
20
3
24
21
22
25
Data Display
Row
1
2
3
4
x1sq
256
441
324
196
x2sq
361
400
441
400
x3sq
576
441
484
625
Data Display
sumx1
sumx2
sumx3
n1
n2
n3
smx1sq
smx2sq
smx3sq
69.0000
80.0000
92.0000
4.00000
4.00000
4.00000
1217.00
1602.00
2126.00
Data Display
x1bar
x2bar
x3bar
17.2500
20.0000
23.0000
34
251y0631s1
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Data Display
Row
1
2
3
4
r1
2
8
3
1
r2
4.0
5.5
8.0
5.5
r3
11
8
10
12
Kruskal-Wallis Test: C11 versus C12
Kruskal-Wallis Test on C11
C12
N Median Ave Rank
Z
1
4
17.00
3.5 -2.04
2
4
20.00
5.8 -0.51
3
4
23.00
10.3
2.55
Overall 12
6.5
H = 7.27 DF = 2 P = 0.026
H = 7.40 DF = 2 P = 0.025 (adjusted for ties)
* NOTE * One or more small samples
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12 N
Lower
StDev
Upper
1 4 1.50955 2.98608 16.2407
2 4 0.41276 0.81650
4.4408
3 4 0.92297 1.82574
9.9298
Bartlett's Test (normal distribution)
Test statistic = 3.64, p-value = 0.162
Levene's Test (any continuous distribution)
Test statistic = 3.17, p-value = 0.091
Test for Equal Variances: C11 versus C12
Executing from file: 252Levene3.MTB
Data Display
Row
1
2
3
4
C21
16
21
18
14
C22
19
20
21
20
C23
24
21
22
25
Data Display
median1
median2
median3
17.0000
20.0000
23.0000
Data Display
Row
1
2
3
4
C21
-1
4
1
-3
C22
-1
0
1
0
C23
1
-2
-1
2
Data Display
Row
1
2
3
4
C21
1
4
1
3
C22
1
0
1
0
C23
1
2
1
2
35
251y0631s1
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One-way ANOVA: C21, C22, C23
Source DF
Factor
2
Error
9
Total
11
S = 0.9860
Level
C21
C22
C23
N
4
4
4
SS
MS
F
P
6.167 3.083 3.17 0.091
8.750 0.972
14.917
R-Sq = 41.34%
R-Sq(adj) = 28.31%
Mean
2.2500
0.5000
1.5000
StDev
1.5000
0.5774
0.5774
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(----------*-----------)
(----------*----------)
(----------*----------)
------+---------+---------+---------+--0.0
1.0
2.0
3.0
Pooled StDev = 0.9860
36
251y0631s1
12/12/06
Take-Home Problem 2, Version 3
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06032003.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06032-003.MTW'
Worksheet was saved on Thu Nov 30 2006
Results for: 252x06032-003.MTW
MTB > exec '252oneW306032'
Executing from file: 252oneW306032.MTB
Executing from file: 2522onw3.MTB
One-way ANOVA: 1, 2, 3
Source DF
Factor
2
Error
9
Total
11
S = 1.991
Level
1
2
3
N
4
4
4
SS
MS
F
P
63.38 31.69 7.99 0.010
35.69
3.97
99.06
R-Sq = 63.97%
R-Sq(adj) = 55.97%
Mean
17.375
20.000
23.000
StDev
2.810
0.816
1.826
Individual 95% CIs For Mean Based on Pooled
StDev
+---------+---------+---------+--------(-------*------)
(-------*------)
(-------*------)
+---------+---------+---------+--------15.0
18.0
21.0
24.0
Pooled StDev = 1.991
Data Display
Row
1
2
3
4
1
16.0
21.0
18.0
14.5
2
19
20
21
20
3
24
21
22
25
Data Display
Row
1
2
3
4
x1sq
256.00
441.00
324.00
210.25
x2sq
361
400
441
400
x3sq
576
441
484
625
Data Display
sumx1
sumx2
sumx3
n1
n2
n3
smx1sq
smx2sq
smx3sq
69.5000
80.0000
92.0000
4.00000
4.00000
4.00000
1231.25
1602.00
2126.00
Data Display
x1bar
x2bar
x3bar
17.3750
20.0000
23.0000
37
251y0631s1
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Data Display
Row
1
2
3
4
r1
2
8
3
1
r2
4.0
5.5
8.0
5.5
r3
11
8
10
12
Kruskal-Wallis Test: C11 versus C12
Kruskal-Wallis Test on C11
C12
N Median Ave Rank
Z
1
4
17.00
3.5 -2.04
2
4
20.00
5.8 -0.51
3
4
23.00
10.3
2.55
Overall 12
6.5
H = 7.27 DF = 2 P = 0.026
H = 7.40 DF = 2 P = 0.025 (adjusted for ties)
* NOTE * One or more small samples
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12 N
Lower
StDev
Upper
1 4 1.42052 2.80995 15.2827
2 4 0.41276 0.81650
4.4408
3 4 0.92297 1.82574
9.9298
Bartlett's Test (normal distribution)
Test statistic = 3.31, p-value = 0.191
Levene's Test (any continuous distribution)
Test statistic = 2.95, p-value = 0.103
Test for Equal Variances: C11 versus C12
Executing from file: 252Levene3.MTB
Data Display
Row
1
2
3
4
C21
16.0
21.0
18.0
14.5
C22
19
20
21
20
C23
24
21
22
25
Data Display
median1
median2
median3
17.0000
20.0000
23.0000
Data Display
Row
1
2
3
4
C21
-1.0
4.0
1.0
-2.5
C22
-1
0
1
0
C23
1
-2
-1
2
Data Display
Row
1
2
3
4
C21
1.0
4.0
1.0
2.5
C22
1
0
1
0
C23
1
2
1
2
38
251y0631s1
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One-way ANOVA: C21, C22, C23
Source DF
Factor
2
Error
9
Total
11
S = 0.9538
Level
C21
C22
C23
N
4
4
4
SS
MS
F
P
5.375 2.688 2.95 0.103
8.188 0.910
13.563
R-Sq = 39.63%
R-Sq(adj) = 26.22%
Mean
2.1250
0.5000
1.5000
StDev
1.4361
0.5774
0.5774
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(----------*----------)
(----------*----------)
(----------*----------)
------+---------+---------+---------+--0.0
1.0
2.0
3.0
Pooled StDev = 0.9538
39
251y0631s1
12/12/06
Take-Home Problem 2, Version 4
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06032004.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06032-004.MTW'
Worksheet was saved on Thu Nov 30 2006
Results for: 252x06032-004.MTW
MTB > exec '252oneW306032'
Executing from file: 252oneW306032.MTB
Executing from file: 2522onw3.MTB
One-way ANOVA: 1, 2, 3
Source DF
Factor
2
Error
9
Total
11
S = 1.915
Level
1
2
3
N
4
4
4
SS
MS
F
P
60.67 30.33 8.27 0.009
33.00
3.67
93.67
R-Sq = 64.77%
R-Sq(adj) = 56.94%
Mean
17.500
20.000
23.000
StDev
2.646
0.816
1.826
Individual 95% CIs For Mean Based on
Pooled StDev
---------+---------+---------+---------+
(--------*--------)
(--------*--------)
(--------*--------)
---------+---------+---------+---------+
17.5
20.0
22.5
25.0
Pooled StDev = 1.915
Data Display
Row
1
2
3
4
1
16
21
18
15
2
19
20
21
20
3
24
21
22
25
Data Display
Row
1
2
3
4
x1sq
256
441
324
225
x2sq
361
400
441
400
x3sq
576
441
484
625
Data Display
sumx1
sumx2
sumx3
n1
n2
n3
smx1sq
smx2sq
smx3sq
70.0000
80.0000
92.0000
4.00000
4.00000
4.00000
1246.00
1602.00
2126.00
Data Display
x1bar
x2bar
x3bar
17.5000
20.0000
23.0000
40
251y0631s1
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Data Display
Row
1
2
3
4
r1
2
8
3
1
r2
4.0
5.5
8.0
5.5
r3
11
8
10
12
Kruskal-Wallis Test: C11 versus C12
Kruskal-Wallis Test on C11
C12
N Median Ave Rank
Z
1
4
17.00
3.5 -2.04
2
4
20.00
5.8 -0.51
3
4
23.00
10.3
2.55
Overall 12
6.5
H = 7.27 DF = 2 P = 0.026
H = 7.40 DF = 2 P = 0.025 (adjusted for ties)
* NOTE * One or more small samples
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12 N
Lower
StDev
Upper
1 4 1.33751 2.64575 14.3897
2 4 0.41276 0.81650
4.4408
3 4 0.92297 1.82574
9.9298
Bartlett's Test (normal distribution)
Test statistic = 3.01, p-value = 0.222
Levene's Test (any continuous distribution)
Test statistic = 2.63, p-value = 0.126
Test for Equal Variances: C11 versus C12
Executing from file: 252Levene3.MTB
Data Display
Row
1
2
3
4
C21
16
21
18
15
C22
19
20
21
20
C23
24
21
22
25
Data Display
median1
median2
median3
17.0000
20.0000
23.0000
Data Display
Row
1
2
3
4
C21
-1
4
1
-2
C22
-1
0
1
0
C23
1
-2
-1
2
Data Display
Row
1
2
3
4
C21
1
4
1
2
C22
1
0
1
0
C23
1
2
1
2
41
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One-way ANOVA: C21, C22, C23
Source DF
Factor
2
Error
9
Total
11
S = 0.9428
Level
C21
C22
C23
N
4
4
4
SS
MS
F
P
4.667 2.333 2.63 0.126
8.000 0.889
12.667
R-Sq = 36.84%
R-Sq(adj) = 22.81%
Mean
2.0000
0.5000
1.5000
StDev
1.4142
0.5774
0.5774
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(----------*----------)
(----------*----------)
(----------*----------)
------+---------+---------+---------+--0.0
1.0
2.0
3.0
Pooled StDev = 0.9428
42
251y0631s1
12/12/06
Take-Home Problem 2, Version 5
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06032005.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06032-005.MTW'
Worksheet was saved on Thu Nov 30 2006
Results for: 252x06032-005.MTW
MTB > exec '252oneW306032'
Executing from file: 252oneW306032.MTB
Executing from file: 2522onw3.MTB
One-way ANOVA: 1, 2, 3
Source DF
Factor
2
Error
9
Total
11
S = 1.847
Level
1
2
3
N
4
4
4
SS
MS
F
P
58.04 29.02 8.51 0.008
30.69
3.41
88.73
R-Sq = 65.41%
R-Sq(adj) = 57.73%
Mean
17.625
20.000
23.000
StDev
2.496
0.816
1.826
Individual 95% CIs For Mean Based on
Pooled StDev
--------+---------+---------+---------+(--------*-------)
(-------*-------)
(-------*-------)
--------+---------+---------+---------+17.5
20.0
22.5
25.0
Pooled StDev = 1.847
Data Display
Row
1
2
3
4
1
16.0
21.0
18.0
15.5
2
19
20
21
20
3
24
21
22
25
Data Display
Row
1
2
3
4
x1sq
256.00
441.00
324.00
240.25
x2sq
361
400
441
400
x3sq
576
441
484
625
Data Display
sumx1
sumx2
sumx3
n1
n2
n3
smx1sq
smx2sq
smx3sq
70.5000
80.0000
92.0000
4.00000
4.00000
4.00000
1261.25
1602.00
2126.00
Data Display
x1bar
x2bar
x3bar
17.6250
20.0000
23.0000
43
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Data Display
Row
1
2
3
4
r1
2
8
3
1
r2
4.0
5.5
8.0
5.5
r3
11
8
10
12
Kruskal-Wallis Test: C11 versus C12
Kruskal-Wallis Test on C11
C12
N Median Ave Rank
Z
1
4
17.00
3.5 -2.04
2
4
20.00
5.8 -0.51
3
4
23.00
10.3
2.55
Overall 12
6.5
H = 7.27 DF = 2 P = 0.026
H = 7.40 DF = 2 P = 0.025 (adjusted for ties)
* NOTE * One or more small samples
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12 N
Lower
StDev
Upper
1 4 1.26172 2.49583 13.5743
2 4 0.41276 0.81650
4.4408
3 4 0.92297 1.82574
9.9298
Bartlett's Test (normal distribution)
Test statistic = 2.75, p-value = 0.253
Levene's Test (any continuous distribution)
Test statistic = 2.22, p-value = 0.164
Test for Equal Variances: C11 versus C12
Executing from file: 252Levene3.MTB
Data Display
Row
1
2
3
4
C21
16.0
21.0
18.0
15.5
C22
19
20
21
20
C23
24
21
22
25
Data Display
median1
median2
median3
17.0000
20.0000
23.0000
Data Display
Row
1
2
3
4
C21
-1.0
4.0
1.0
-1.5
C22
-1
0
1
0
C23
1
-2
-1
2
Data Display
Row
1
2
3
4
C21
1.0
4.0
1.0
1.5
C22
1
0
1
0
C23
1
2
1
2
44
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One-way ANOVA: C21, C22, C23
Source DF
Factor
2
Error
9
Total
11
S = 0.9538
Level
C21
C22
C23
N
4
4
4
SS
MS
F
P
4.042 2.021 2.22 0.164
8.188 0.910
12.229
R-Sq = 33.05%
R-Sq(adj) = 18.17%
Mean
1.8750
0.5000
1.5000
StDev
1.4361
0.5774
0.5774
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(----------*----------)
(----------*----------)
(----------*----------)
------+---------+---------+---------+--0.0
1.0
2.0
3.0
Pooled StDev = 0.9538
45
251y0631s1
12/12/06
Take-Home Problem 2, Version 6
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06032006.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06032-006.MTW'
Worksheet was saved on Thu Nov 30 2006
Results for: 252x06032-006.MTW
MTB > exec '252oneW306032'
Executing from file: 252oneW306032.MTB
Executing from file: 2522onw3.MTB
One-way ANOVA: 1, 2, 3
Source DF
Factor
2
Error
9
Total
11
S = 1.787
Level
1
2
3
N
4
4
4
SS
MS
F
P
55.50 27.75 8.69 0.008
28.75
3.19
84.25
R-Sq = 65.88%
R-Sq(adj) = 58.29%
Mean
17.750
20.000
23.000
StDev
2.363
0.816
1.826
Individual 95% CIs For Mean Based on
Pooled StDev
-------+---------+---------+---------+-(-------*-------)
(-------*-------)
(-------*-------)
-------+---------+---------+---------+-17.5
20.0
22.5
25.0
Pooled StDev = 1.787
Data Display
Row
1
2
3
4
1
16
21
18
16
2
19
20
21
20
3
24
21
22
25
Data Display
Row
1
2
3
4
x1sq
256
441
324
256
x2sq
361
400
441
400
x3sq
576
441
484
625
Data Display
sumx1
sumx2
sumx3
n1
n2
n3
smx1sq
smx2sq
smx3sq
71.0000
80.0000
92.0000
4.00000
4.00000
4.00000
1277.00
1602.00
2126.00
Data Display
x1bar
x2bar
x3bar
17.7500
20.0000
23.0000
46
251y0631s1
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Data Display
Row
1
2
3
4
r1
1.5
8.0
3.0
1.5
r2
4.0
5.5
8.0
5.5
r3
11
8
10
12
Kruskal-Wallis Test: C11 versus C12
Kruskal-Wallis Test on C11
C12
N Median Ave Rank
Z
1
4
17.00
3.5 -2.04
2
4
20.00
5.8 -0.51
3
4
23.00
10.3
2.55
Overall 12
6.5
H = 7.27 DF = 2 P = 0.026
H = 7.43 DF = 2 P = 0.024 (adjusted for ties)
* NOTE * One or more small samples
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12
1
2
3
N
4
4
4
Lower
1.19452
0.41276
0.92297
StDev
2.36291
0.81650
1.82574
Upper
12.8514
4.4408
9.9298
Bartlett's Test (normal distribution)
Test statistic = 2.52, p-value = 0.283
Levene's Test (any continuous distribution)
Test statistic = 1.80, p-value = 0.220
Test for Equal Variances: C11 versus C12
Executing from file: 252Levene3.MTB
Data Display
Row
1
2
3
4
C21
16
21
18
16
C22
19
20
21
20
C23
24
21
22
25
Data Display
median1
median2
median3
17.0000
20.0000
23.0000
Data Display
Row
1
2
3
4
C21
-1
4
1
-1
C22
-1
0
1
0
C23
1
-2
-1
2
Data Display
Row
1
2
3
4
C21
1
4
1
1
C22
1
0
1
0
C23
1
2
1
2
47
251y0631s1
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One-way ANOVA: C21, C22, C23
Source
Factor
Error
Total
DF
2
9
11
S = 0.9860
SS
3.500
8.750
12.250
Level
C21
C22
C23
N
4
4
4
MS
1.750
0.972
R-Sq = 28.57%
Mean
1.7500
0.5000
1.5000
StDev
1.5000
0.5774
0.5774
F
1.80
P
0.220
R-Sq(adj) = 12.70%
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(----------*-----------)
(----------*----------)
(----------*----------)
------+---------+---------+---------+--0.0
1.0
2.0
3.0
Pooled StDev = 0.9860
48
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Take-Home Problem 2, Version 7
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06032007.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06032-007.MTW'
Worksheet was saved on Thu Nov 30 2006
Results for: 252x06032-007.MTW
MTB > exec '252oneW306032'
Executing from file: 252oneW306032.MTB
Executing from file: 2522onw3.MTB
One-way ANOVA: 1, 2, 3
Source DF
Factor
2
Error
9
Total
11
S = 1.738
Level
1
2
3
N
4
4
4
SS
MS
F
P
53.04 26.52 8.78 0.008
27.19
3.02
80.23
R-Sq = 66.11%
R-Sq(adj) = 58.58%
Mean
17.875
20.000
23.000
StDev
2.250
0.816
1.826
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(-------*------)
(-------*-------)
(-------*-------)
------+---------+---------+---------+--17.5
20.0
22.5
25.0
Pooled StDev = 1.738
Data Display
Row
1
2
3
4
1
16.0
21.0
18.0
16.5
2
19
20
21
20
3
24
21
22
25
Data Display
Row
1
2
3
4
x1sq
256.00
441.00
324.00
272.25
x2sq
361
400
441
400
x3sq
576
441
484
625
Data Display
sumx1
sumx2
sumx3
n1
n2
n3
smx1sq
smx2sq
smx3sq
71.5000
80.0000
92.0000
4.00000
4.00000
4.00000
1293.25
1602.00
2126.00
Data Display
x1bar
x2bar
x3bar
17.8750
20.0000
23.0000
49
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Data Display
Row
1
2
3
4
r1
1
8
3
2
r2
4.0
5.5
8.0
5.5
r3
11
8
10
12
Kruskal-Wallis Test: C11 versus C12
Kruskal-Wallis Test on C11
C12
N Median Ave Rank
Z
1
4
17.25
3.5 -2.04
2
4
20.00
5.8 -0.51
3
4
23.00
10.3
2.55
Overall 12
6.5
H = 7.27 DF = 2 P = 0.026
H = 7.40 DF = 2 P = 0.025 (adjusted for ties)
* NOTE * One or more small samples
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12 N
Lower
StDev
Upper
1 4 1.13744 2.25000 12.2373
2 4 0.41276 0.81650
4.4408
3 4 0.92297 1.82574
9.9298
Bartlett's Test (normal distribution)
Test statistic = 2.34, p-value = 0.310
Levene's Test (any continuous distribution)
Test statistic = 1.67, p-value = 0.241
Test for Equal Variances: C11 versus C12
Executing from file: 252Levene3.MTB
Data Display
Row
1
2
3
4
C21
16.0
21.0
18.0
16.5
C22
19
20
21
20
C23
24
21
22
25
Data Display
median1
median2
median3
17.2500
20.0000
23.0000
Data Display
Row
1
2
3
4
C21
-1.25
3.75
0.75
-0.75
C22
-1
0
1
0
C23
1
-2
-1
2
Data Display
Row
1
2
3
4
C21
1.25
3.75
0.75
0.75
C22
1
0
1
0
C23
1
2
1
2
50
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One-way ANOVA: C21, C22, C23
Source DF
Factor
2
Error
9
Total
11
S = 0.9538
Level
C21
C22
C23
N
4
4
4
SS
MS
F
P
3.042 1.521 1.67 0.241
8.188 0.910
11.229
R-Sq = 27.09%
R-Sq(adj) = 10.88%
Mean
1.6250
0.5000
1.5000
StDev
1.4361
0.5774
0.5774
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(----------*----------)
(----------*----------)
(----------*----------)
------+---------+---------+---------+--0.0
1.0
2.0
3.0
Pooled StDev = 0.9538
51
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Take-Home Problem 2, Version 8
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06032008.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06032-008.MTW'
Worksheet was saved on Thu Nov 30 2006
Results for: 252x06032-008.MTW
MTB > exec '252oneW306032'
Executing from file: 252oneW306032.MTB
Executing from file: 2522onw3.MTB
One-way ANOVA: 1, 2, 3
Source DF
Factor
2
Error
9
Total
11
S = 1.700
Level
1
2
3
N
4
4
4
SS
MS
F
P
50.67 25.33 8.77 0.008
26.00
2.89
76.67
R-Sq = 66.09%
R-Sq(adj) = 58.55%
Mean
18.000
20.000
23.000
StDev
2.160
0.816
1.826
Individual 95% CIs For Mean Based on
Pooled StDev
------+---------+---------+---------+--(-------*-------)
(-------*-------)
(-------*-------)
------+---------+---------+---------+--17.5
20.0
22.5
25.0
Pooled StDev = 1.700
Data Display
Row
1
2
3
4
1
16
21
18
17
2
19
20
21
20
3
24
21
22
25
Data Display
Row
1
2
3
4
x1sq
256
441
324
289
x2sq
361
400
441
400
x3sq
576
441
484
625
Data Display
sumx1
sumx2
sumx3
n1
n2
n3
smx1sq
smx2sq
smx3sq
72.0000
80.0000
92.0000
4.00000
4.00000
4.00000
1310.00
1602.00
2126.00
Data Display
x1bar
x2bar
x3bar
18.0000
20.0000
23.0000
52
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Data Display
Row
1
2
3
4
r1
1
8
3
2
r2
4.0
5.5
8.0
5.5
r3
11
8
10
12
Kruskal-Wallis Test: C11 versus C12
Kruskal-Wallis Test on C11
C12
N Median Ave Rank
Z
1
4
17.50
3.5 -2.04
2
4
20.00
5.8 -0.51
3
4
23.00
10.3
2.55
Overall 12
6.5
H = 7.27 DF = 2 P = 0.026
H = 7.40 DF = 2 P = 0.025 (adjusted for ties)
* NOTE * One or more small samples
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12 N
Lower
StDev
Upper
1 4 1.09207 2.16025 11.7491
2 4 0.41276 0.81650
4.4408
3 4 0.92297 1.82574
9.9298
Bartlett's Test (normal distribution)
Test statistic = 2.20, p-value = 0.332
Levene's Test (any continuous distribution)
Test statistic = 1.50, p-value = 0.274
Test for Equal Variances: C11 versus C12
Executing from file: 252Levene3.MTB
Data Display
Row
1
2
3
4
C21
16
21
18
17
C22
19
20
21
20
C23
24
21
22
25
Data Display
median1
median2
median3
17.5000
20.0000
23.0000
Data Display
Row
1
2
3
4
C21
-1.5
3.5
0.5
-0.5
C22
-1
0
1
0
C23
1
-2
-1
2
Data Display
Row
1
2
3
4
C21
1.5
3.5
0.5
0.5
C22
1
0
1
0
C23
1
2
1
2
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One-way ANOVA: C21, C22, C23
Source DF
Factor
2
Error
9
Total
11
S = 0.9428
Level
C21
C22
C23
N
4
4
4
SS
MS
F
P
2.667 1.333 1.50 0.274
8.000 0.889
10.667
R-Sq = 25.00%
R-Sq(adj) = 8.33%
Mean
1.5000
0.5000
1.5000
StDev
1.4142
0.5774
0.5774
Individual 95% CIs For Mean Based on
Pooled StDev
-------+---------+---------+---------+-(-------------*------------)
(------------*-------------)
(-------------*------------)
-------+---------+---------+---------+-0.00
0.80
1.60
2.40
Pooled StDev = 0.9428
54
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Take-Home Problem 2, Version 9
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06032009.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06032-009.MTW'
Worksheet was saved on Thu Nov 30 2006
Results for: 252x06032-009.MTW
MTB > exec '252oneW306032'
Executing from file: 252oneW306032.MTB
Executing from file: 2522onw3.MTB
One-way ANOVA: 1, 2, 3
Source DF
Factor
2
Error
9
Total
11
S = 1.673
Level
1
2
3
N
4
4
4
SS
MS
F
P
48.38 24.19 8.64 0.008
25.19
2.80
73.56
R-Sq = 65.76%
R-Sq(adj) = 58.15%
Mean
18.125
20.000
23.000
StDev
2.097
0.816
1.826
Individual 95% CIs For Mean Based on
Pooled StDev
-----+---------+---------+---------+---(-------*------)
(-------*-------)
(-------*-------)
-----+---------+---------+---------+---17.5
20.0
22.5
25.0
Pooled StDev = 1.673
Data Display
Row
1
2
3
4
1
16.0
21.0
18.0
17.5
2
19
20
21
20
3
24
21
22
25
Data Display
Row
1
2
3
4
x1sq
256.00
441.00
324.00
306.25
x2sq
361
400
441
400
x3sq
576
441
484
625
Data Display
sumx1
sumx2
sumx3
n1
n2
n3
smx1sq
smx2sq
smx3sq
72.5000
80.0000
92.0000
4.00000
4.00000
4.00000
1327.25
1602.00
2126.00
Data Display
x1bar
x2bar
x3bar
18.1250
20.0000
23.0000
55
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Data Display
Row
1
2
3
4
r1
1
8
3
2
r2
4.0
5.5
8.0
5.5
r3
11
8
10
12
Kruskal-Wallis Test: C11 versus C12
Kruskal-Wallis Test on C11
C12
N Median Ave Rank
Z
1
4
17.75
3.5 -2.04
2
4
20.00
5.8 -0.51
3
4
23.00
10.3
2.55
Overall 12
6.5
H = 7.27 DF = 2 P = 0.026
H = 7.40 DF = 2 P = 0.025 (adjusted for ties)
* NOTE * One or more small samples
Test for Equal Variances: C11 versus C12
95% Bonferroni confidence intervals for standard deviations
C12 N
Lower
StDev
Upper
1 4 1.05991 2.09662 11.4031
2 4 0.41276 0.81650
4.4408
3 4 0.92297 1.82574
9.9298
Bartlett's Test (normal distribution)
Test statistic = 2.11, p-value = 0.348
Levene's Test (any continuous distribution)
Test statistic = 1.31, p-value = 0.318
Test for Equal Variances: C11 versus C12
Executing from file: 252Levene3.MTB
Data Display
Row
1
2
3
4
C21
16.0
21.0
18.0
17.5
C22
19
20
21
20
C23
24
21
22
25
Data Display
median1
median2
median3
17.7500
20.0000
23.0000
Data Display
Row
1
2
3
4
C21
-1.75
3.25
0.25
-0.25
C22
-1
0
1
0
C23
1
-2
-1
2
Data Display
Row
1
2
3
4
C21
1.75
3.25
0.25
0.25
C22
1
0
1
0
C23
1
2
1
2
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One-way ANOVA: C21, C22, C23
Source
Factor
Error
Total
DF
2
9
11
S = 0.9538
Level
C21
C22
C23
N
4
4
4
SS
2.375
8.188
10.563
MS
1.188
0.910
R-Sq = 22.49%
Mean
1.3750
0.5000
1.5000
StDev
1.4361
0.5774
0.5774
F
1.31
P
0.318
R-Sq(adj) = 5.26%
Individual 95% CIs For Mean Based on
Pooled StDev
-------+---------+---------+---------+-(------------*-------------)
(------------*-------------)
(-------------*------------)
-------+---------+---------+---------+-0.00
0.80
1.60
2.40
Pooled StDev = 0.9538
57
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3) (Abronovic) A group of 4 workers produces defective pieces at the rates shown below during different
times of the day. Personalize the data by subtracting the last digit of your student number from the 14 in
the lower right corner. Use the number subtracted to label this as a version number. (Example: Good ol’
Seymour’s student number is 123456, so the 14 becomes 14 - 6 =8 and he labels it Version 6.)
Time
Worker’s Name
Apple
Plum
Pear
Melon
Early
10
11
8
12
Morning
Late
9
8
7
10
Morning
Early
12
13
11
11
Afternoon
Late
13
14
10
14
Afternoon
Sum of Row 1 = 41, SSQ of Row 1 = 429, Sum of Column 1 = 44, SSQ of Column 1 = 494,
Sum of Row 2 = 34, SSQ of Row 2 = 294, Sum of Column 2 = 46, SSQ of Column 2 = 550,
Sum of Row 3 = 47, SSQ of Row 3 = 555, Sum of Column 3 = 36, SSQ of Column 3 = 334,
a) Do a 2-way ANOVA on these data and explain what hypotheses you test and what the conclusions are.
(6)
b) Using your results from a) present two different confidence intervals for the difference between numbers
of defects for the best and worst worker and for the defects from the best and second best times. Explain
which of the intervals show a significant difference and why. (3)
c) What other method could we use on these data to see if time of day makes a difference while allowing
for cross-classification? Under what circumstances would we use it? Try it and tell what it tests and what it
shows. (3)
[36]
a)
Time
Apple
Plum
Pear
Melon
Sum
SS
ni
x i 
1
2
3
4
Sum
nj
10
9
12
13
44
4
11
8
13
14
46
4
x j 
11
SS
494
121
x j  2
8
7
11
10
36
4
12
10
11
14
47
4
41
34
47
51
173
16
11.5
9
11.75
(10.8125)
x
550
132.25
334
81
561
138.0625
1939
472.3125
2
 xijk
4
4
4
4
16
n
10.25
8.50
11.75
12.75
(10.8125)
x
429
294
555
661
1939
2
 xijk
xi 2
105.0625
72.25
138.0625
162.5625
477.9375
 x i 2
 x .2j .
 x  nx  19391610.8125  19391870.5625  68.4375
SSR  C x  nx  4477.9375  1610.8175  1911.75  1870.5625  41.1875
SSC  R x  nx  4472.3125  1610.8125  1889.25  1870.5625  18.6875
SST 
2
2
2
2
i.
2
2
2
.j
2
2
Source
SS
DF
Rows
41.1875
3
Columns
18.6875
3
MS
F
F.05
13.7292
14.43 s
6.2292
6.55 s
F 3,9   3.86
F 3,9   3.86
H0
Row means equal
Column means equal
Within
8.5625
9
0.95139
Total
68.4375
15
As is shown by the computed F statistics and the table values, both computed Fs exceed the table values,
meaning that we will reject the null hypotheses.
b) The material on confidence intervals comes from the outline for 2-way ANOVA. Note that for columns
Pear is the best worker (9) and Melon the Worst (11.75), so that the difference is 2.75. For rows Time 2 is
58
251y0631s1
12/12/06
the best time (8.50) and Time 1 (10.25) is the second best, so that the difference is 1.75.
20.95139 
2 MSW

 0.6897 ,
C
4
not significant. ‘s’ stands for significant.
MSW  0.95139 ,
2MSW

R
R  1C  1  9
20.95139 
 0.6897 . ‘ns’ stands for
4
i. A Single Confidence Interval
If we desire a single interval we use the formula for a Bonferroni Confidence Interval with m  1 . . Note
that since P  1 , we must replace RC P  1 with
R  1C  1 . t.9025  2.262
For row means 1   2  x1  x 2   t R1C 1
2
1.75  2.262 0.6897   1.75  1.56 . s
2MSW
. So we have
C
For column means  1   2  x1  x2   t R1C 1
2
2.75  2.262 0.6897   2.75  1.56 . s
2MSW
. So we have
R
ii. Scheffé Confidence Interval
If we desire intervals that will simultaneously be valid for a given confidence level for all possible intervals
between means, use the following formulas. Note that since P  1 , we must replace RC P  1 with
R  1C  1 .
3,9   3.86
F.05
For row means, use 1   2  x1  x 2  
 x1  x 2  
R  1FR 1,R 1C 1
R  1FR 1,R 1C 1 2MSW
2MSW
So we have 1.75  33.86 0.6897   1.75  2.34 ns.
C
For column means, use  1   2  x1  x2  
 x1  x2  
C  1FC 1,R 1C 1
C
C  1FC 1,R 1C 1 2MSW
R
2MSW
 2.75  2.34 s
R
iii. Bonferroni Confidence Interval – not worth the effort.
iv. Tukey Confidence Interval
Note that since P  1 , we must replace RC P  1 with
For row means, use 1   2  x1  x 2   qR ,R 1C 1
= x1  x 2   0.5qR ,R 1C 1
R  1C  1 .
4,9 
q.05
 4.41
MSW
C
2MSW
. So we have 1.75  0.70711 4.410.6897   1.75  2.15 . ns
C
59
251y0631s1
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For column means, use  1   2  x1  x2   qC ,R 1C 1
 x1  x2   0.5qC ,R 1C 1
MSW
R
2MSW
 2.75  2.15 s
R
c) The alternative to 2-way ANOVA with one measurement per cell is a Friedman test. Remember that it is
10 11 8 12
9 8 7 10
, time of day was represented by rows.
12 13 11 11
13 14 10 14
10 9 12 13
11 8 13 14
If we transpose the array, columns represent time of day.
. Now replace the original data
8 7 11 10
12 10 11 14
2 1 3 4
2 1 3 4
with rank within rows and sum the columns. 2 1 4 3 . As a check note that the column sums should
rows that we want to compare. In the original data
3
9
1 2 4
4 12 15
445
 40 and that 9 + 4 + 12 + 15 = 40. The Friedman formula reads
2
 12


SRi2   3r c  1
 rc c  1 i

add to
 F2
 


 12
92  42  12 2  15 2   345   3 466   60  69 .90  60  9.90

 20

 445

The Friedman table says that 9.90 has a p-value of .006. Since this is less than 5% or 1% we reject the null
hypothesis that the median number of defects is the same regardless of time of day.
————— 12/1/2006 1:28:47 AM ————————————————————
Welcome to Minitab, press F1 for help.
60
251y0631s1
12/12/06
Take-Home Problem 3, Version 0
Results for: 252x06033-000.MTW
MTB > exec '2522W06133'
Executing from file: 2522W06133.MTB
Data Display
Row
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
C40
10
9
12
13
11
8
13
14
8
7
11
10
12
10
11
14
C41
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
C42
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
Tabulated statistics: C41, C42
Rows: C41
Columns: C42
1 2 3 4 All
1
1 1 1
2
1 1 1
3
1 1 1
4
1 1 1
All
4 4 4
Cell Contents:
1
1
1
1
4
4
4
4
4
16
Count
Tabulated statistics: C41, C42
Rows: C41
Columns: C42
1
2
3
4
1
10
11
8
12
2
9
8
7
10
3
12
13
11
11
4
13 14 10 14
Cell Contents: C40
:
DATA
Tabulated statistics: C41, C42
Rows: C41
Columns:
1
2
1
10.00 11.00
2
9.00
8.00
3
12.00 13.00
4
13.00 14.00
All
11.00 11.50
Cell Contents: C40
C42
3
4
8.00 12.00
7.00 10.00
11.00 11.00
10.00 14.00
9.00 11.75
: Mean
All
10.25
8.50
11.75
12.75
10.81
61
251y0631s1
12/12/06
Two-way ANOVA: C40 versus C41, C42
Source DF
C41
3
C42
3
Error
9
Total
15
S = 0.9754
SS
MS
F
P
41.1875 13.7292 14.43 0.001
18.6875
6.2292
6.55 0.012
8.5625
0.9514
68.4375
R-Sq = 87.49%
R-Sq(adj) = 79.15%
Results for: 252x06033-000.MTW
MTB > WOpen "C:\Documents and Settings\RBOVE\My Documents\Minitab\252x06033000.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\RBOVE\My
Documents\Minitab\252x06033-000.MTW'
Worksheet was saved on Thu Nov 30 2006
MTB > Execute "C:\Documents and Settings\RBOVE\My
Documents\Minitab\2522W06033.mtb" 1.
Executing from file: C:\Documents and Settings\RBOVE\My
Documents\Minitab\2522W06033.mtb
Executing from file: 252onw4.MTB
One-way ANOVA: C1, C2, C3, C4
Source DF
Factor
3
Error
12
Total
15
S = 2.036
Level
C1
C2
C3
C4
SS
MS
F
P
18.69 6.23 1.50 0.264
49.75 4.15
68.44
R-Sq = 27.31%
R-Sq(adj) = 9.13%
Individual 95% CIs For Mean Based on
Pooled StDev
Mean StDev ------+---------+---------+---------+--11.000 1.826
(----------*----------)
11.500 2.646
(----------*-----------)
9.000 1.826 (----------*----------)
11.750 1.708
(----------*----------)
------+---------+---------+---------+--8.0
10.0
12.0
14.0
N
4
4
4
4
Pooled StDev = 2.036
Executing from file: 252onme4.MTB
Executing from file: 252osme4.MTB
Data Display
Row
1
2
3
4
C1
10
9
12
13
C2
11
8
13
14
C3
8
7
11
10
C4
12
10
11
14
Data Display
Row
1
2
3
4
x1sq
100
81
144
169
x2sq
121
64
169
196
x3sq
64
49
121
100
x4sq
144
100
121
196
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251y0631s1
12/12/06
Data Display
sumx1
sumx2
sumx3
sumx4
n1
n2
n3
n4
smx1sq
smx2sq
smx3sq
smx4sq
44.0000
46.0000
36.0000
47.0000
4.00000
4.00000
4.00000
4.00000
494.000
550.000
334.000
561.000
Executing from file: 252omea4.MTB
Data Display
x1bar
x2bar
x3bar
x4bar
11.0000
11.5000
9.00000
11.7500
Data Display
Row
1
2
3
4
C1
10
9
12
13
C2
11
8
13
14
C3
8
7
11
10
C4
12
10
11
14
Data Display
Row
1
2
3
4
C50_1
10
11
8
12
C50_2
9
8
7
10
C50_3
12
13
11
11
C50_4
13
14
10
14
C57_3
3
3
4
2
C57_4
4
4
3
4
Data Display
Row
1
2
3
4
C57_1
2
2
2
3
C57_2
1
1
1
1
Friedman Test: C50 versus C51 blocked by C52
S = 9.90
DF = 3
P = 0.019
Sum
of
C51 N Est Median Ranks
1
4
10.656
9.0
2
4
9.156
4.0
3
4
12.531
12.0
4
4
13.281
15.0
Grand median = 11.406
63