252y0521s 3/25/05
First take-home problem – based on 9.26 on page E-48
————— 3/25/2005 1:54:04 AM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\252x0502-1.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0502-1.MTW'
Worksheet was saved on Fri Mar 25 2005
Results for: 252x0502-1.MTW
MTB > print c1-c4
Data Display
Row x1 x2 rx1 rx2
1 45.14 9.55 10 2
2 10.11 38.76 2 15
3 29.38 16.65 7 9
4 19.65 19.00 5 12
5 16.25 17.00 4 10
6 29.46 29.01 8 13
7 8.13 12.34 1 6
8 45.63 11.18 11 4
9 24.49 12.15 6 5
10 12.71 14.40 3 7
11 37.04 8.00 9 1
12 16.19 8
13 33.46 14
14 18.37 11
15 9.86 3
Descriptive Statistics: x1, x2
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
x1 11 0 25.27 4.02 13.33 8.13 12.71 24.49 37.04 45.63
x2 15 0 17.73 2.35 9.11 8.00 11.18 16.19 19.00 38.7
MTB > TwoSample c1 c2;
SUBC> Pooled.
Two-Sample T-Test and CI: x1, x2
Two-sample T for x1 vs x2
N Mean StDev SE Mean
x1 11 25.3 13.3 4.0
x2 15 17.73 9.11 2.4
Difference = mu (x1) - mu (x2)
Estimate for difference: 7.54382
95% CI for difference: (-1.52083, 16.60847)
T-Test of difference = 0 (vs not =): T-Value = 1.72 P-Value = 0.099 DF = 24
Both use Pooled StDev = 11.0641
MTB > TwoSample c1 c2.
Two-Sample T-Test and CI: x1, x2
Two-sample T for x1 vs x2
N Mean StDev SE Mean
x1 11 25.3 13.3 4.0
x2 15 17.73 9.11 2.4
Difference = mu (x1) - mu (x2)
Estimate for difference: 7.54382
95% CI for difference: (-2.32775, 17.41539)
T-Test of difference = 0 (vs not =): T-Value = 1.62 P-Value = 0.125 DF = 16
MTB > VarTest c1 c2;
SUBC> Unstacked.
Test for Equal Variances: x1, x2
95% Bonferroni confidence intervals for standard deviations
N Lower StDev Upper
x1 11 8.87606 13.3313 25.6219
x2 15 6.39055 9.1054 15.4332
F-Test (normal distribution)
Test statistic = 2.14, p-value = 0.187
Levene's Test (any continuous distribution)
Test statistic = 2.96, p-value = 0.098
Test for Equal Variances for x1, x2
MTB > Mann-Whitney 95.0 c1 c2;
SUBC> Alternative 0.
Mann-Whitney Test and CI: x1, x2
N Median
x1 11 24.49
x2 15 16.19
Point estimate for ETA1-ETA2 is 6.87
95.1 Percent CI for ETA1-ETA2 is (-2.12,17.31)
W = 178.0
Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.1323
MTB > NormTest c2;
SUBC> KSTest.
Probability Plot of x2
MTB > NormTest c1;
SUBC> KSTest.
Probability Plot of x1
MTB >
First take-home problem – based on 9.26 on page E-48
Lilliefors computations.
————— 4/7/2005 10:01:12 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\notmuch.MTW".
Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My
Documents\Minitab\notmuch.MTW'
Worksheet was saved on Fri Jan 21 2005
Results for: 252x0502-1.MTW
MTB > let c10=c1
MTB > let c11 = c10*c10
MTB > let k10 = sum(c10)
MTB > let k11=sum(c11)
MTB > print k10, k11
Data Display
K10 277.990
K11 8802.55
MTB > let k10 = k10/11
MTB > print k10
Data Display
K10 25.2718
MTB > let k12 = 11*k10*k10
MTB > print k11 k12
Data Display
K11 8802.55
K12 7025.31
MTB > let k11=k12-k11
MTB > print k11
Data Display
K11 -1777.24
MTB > describe c10
Descriptive Statistics: x
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
x 11 0 25.27 4.02 13.33 8.13 12.71 24.49 37.04 45.63
MTB > let k11 = k11/10
MTB > print k11
Data Display
K11 177.724
MTB > let k11 = sqrt(k11)
MTB > print k11
Data Display
K11 13.3313
MTB > Center c10 c13.
MTB > describe c13
Descriptive Statistics: C13
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3
C13 11 0 -1.11022E-16 0.302 1.000 -1.286 -0.942 -0.0586 0.883
Variable Maximum
C13 1.527
MTB > describe c10
Descriptive Statistics: x
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
x 11 0 25.27 4.02 13.33 8.13 12.71 24.49 37.04 45.63
MTB > print k10 k11
Data Display
K10 25.2718
K11 13.3313
MTB > let c12 = c10-k10
MTB > describe c12
Descriptive Statistics: C12
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3
C12 11 0 -1.11022E-15 4.02 13.33 -17.14 -12.56 -0.782 11.77
Variable Maximum
C12 20.36
MTB > let c12 = c10-k10
MTB > print k11
Data Display
K11 13.3313
MTB > let c12 = c12/k11
MTB > let c12 = c12 * 10000
MTB > let c12 = c12/100
MTB > let c14 = c12
MTB > round c12 c14
MTB > let c12 = c14/100
MTB > let c15=c14
MTB > let c16 = c15/11
MTB > CDF c13 c17;
SUBC> Normal 0.0 1.0.
MTB > Sort c10 c11 c12 c17 c10 c11 c12 c17;
SUBC> By c10.
MTB > sort c13 c13;
SUBC> by c1.
MTB > let c18 = c16 - c17
MTB > let c18 = abs(c18)
MTB > print c10 c11 c12 c14 c15 c16 c17 c18
Data Display
Row x xsq zrnd O CumO Fo Fe D
1 8.13 66.10 -1.29 1 1 0.09091 0.099251 0.008342
2 10.11 102.21 -1.14 1 2 0.18182 0.127705 0.054114
3 12.71 161.54 -0.94 1 3 0.27273 0.173025 0.099702
4 16.25 264.06 -0.68 1 4 0.36364 0.249286 0.114351
5 19.65 386.12 -0.42 1 5 0.45455 0.336622 0.117924
6 24.49 599.76 -0.06 1 6 0.54545 0.476617 0.068837
7 29.38 863.18 0.31 1 7 0.63636 0.621020 0.015344
8 29.46 867.89 0.31 1 8 0.72727 0.623301 0.103972
9 37.04 1371.96 0.88 1 9 0.81818 0.811314 0.006868
10 45.14 2037.62 1.49 1 10 0.90909 0.931932 0.022842
11 45.63 2082.10 1.53 1 11 1.00000 0.936631 0.063369
Descriptive Statistics: D
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3
D 11 0 0.0614 0.0131 0.0433 0.00687 0.0153 0.0634 0.1040
Variable Maximum
D 0.1179
WilcoxonPaired Problem Lee p612
Row
1 40 62 -22 22 6.0 6.0-
2 35 49 -14 14 5.0 5.0-
3 42 39 3 3 2.5 2.5+
4 30 28 2 2 1.0 1.0+
5 63 66 -3 3 2.5 2.5-
6 36 40 -4 4 4.0 4.0-
Row
1 40 62 -22
2 35 49 -14
3 42 39 3
4 30 28 2
5 55 55 0
6 63 66 -3
7 36 40 -4
Variable Mean SE Mean StDev
7 43.00 4.46 11.80
7 48.43 5.15 13.62
7 -5.43 3.49 9.24
Teaching Problem Bassett p134
MTB > TwoSample c1 c2;
SUBC> Pooled;
SUBC> Alternative -1.
Two-Sample T-Test and CI: x1, x2
Two-sample T for x1 vs x2
N Mean StDev SE Mean
x1 250 68.45 7.96 0.50
x2 150 70.62 7.06 0.58
Difference = mu (x1) - mu (x2)
Estimate for difference: -2.17521
95% upper bound for difference: -0.87525
T-Test of difference = 0 (vs <): T-Value = -2.76 P-Value = 0.003 DF = 398
MTB > TwoSample c1 c2;
SUBC> Alternative -1.
Two-Sample T-Test and CI: x1, x2
Two-sample T for x1 vs x2
N Mean StDev SE Mean
x1 250 68.45 7.96 0.50
x2 150 70.62 7.06 0.58
Difference = mu (x1) - mu (x2)
Estimate for difference: -2.17521
95% upper bound for difference: -0.91322
T-Test of difference = 0 (vs <): T-Value = -2.84 P-Value = 0.002 DF = 343
MTB > VarTest c1 c2;
SUBC> Unstacked.
Test for Equal Variances: x1, x2
F-Test (normal distribution)
Test statistic = 1.27, p-value = 0.107
Marble Problem
One-Sample T: x1
Test of mu = 12 vs < 12
95%
Upper
Variable N Mean StDev SE Mean Bound T P
x1 105 12.0150 0.0498 0.0049 12.0231 3.09 0.999
ChiSquare Problem
————— 3/30/2005 10:22:03 PM ————————————————————
Welcome to Minitab, press F1 for help.
Results for: 252x0502-7.MTW
MTB > WSave "C:\Documents and Settings\rbove\My Documents\Minitab\252x0502-7.MTW";
SUBC> Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0502-7.MTW'
MTB > sum c1
Sum of O
Sum of O = 180
MTB > echo
MTB > Execute "C:\Documents and Settings\rbove\My Documents\Minitab\252chisq.mtb" 1.
Executing from file: C:\Documents and Settings\rbove\My Documents\Minitab\252chisq.mtb
MTB > #252chisq
MTB > Name c1 'O' #O on c1, E in C2
MTB > Name c2 'E'
MTB > let c3='E'-'O'
MTB > let c4=c3*c3
MTB > let c5=c4/'E'
MTB > let c6='O'*'O' #Shortcut method
MTB > let c6=c6/'E'
MTB > print c1-c6
Data Display
Row O E C3 C4 C5 C6
1 42 36 -6 36 1.00000 49.0000
2 33 36 3 9 0.25000 30.2500
3 35 36 1 1 0.02778 34.0278
4 25 36 11 121 3.36111 17.3611
5 45 36 -9 81 2.25000 56.2500
MTB > name k1 'n'
MTB > let k1=sum(c1) #k1=k2=n
MTB > let k2=sum(c2)
MTB > let k3=sum(c3) #Should be zero
MTB > let k5=sum(c5)
MTB > name k5 'chisq'
MTB > let k6=sum(c6)
MTB > name k4 'chisq1'
MTB > let k4=k6-k1
MTB > print k1-k6
Data Display
n 180.000
K2 180.000
K3 0
chisq1 6.88889
chisq 6.88889
K6 186.889
MTB > end
MTB > Save "C:\Documents and Settings\rbove\My Documents\Minitab\252x0502-7.MTW";
SUBC> Replace.
Saving file as: 'C:\Documents and Settings\rbove\My
Documents\Minitab\252x0502-7.MTW'
Existing file replaced.
MTB >
1