Answer key to homework due 3/24/05
Chap 12: 4,12, 13
Chap 11: 34, 69
Points: 34 (20 pts); 69 (30 pts); 12.4 (15 pts); 12 (15 pts) 13 (20 pts)
11.34
Two-way ANOVA: Density versus Strength, Time
Source
Strength
Time
Interaction
Error
Total
S = 1.493
Strength
1
2
Time
10
14
DF
1
1
1
12
15
SS
52.5625
1.5625
3.0625
26.7500
83.9375
R-Sq = 68.13%
Mean
2.625
6.250
Mean
4.125
4.750
MS
52.5625
1.5625
3.0625
2.2292
F
23.58
0.70
1.37
P
0.000
0.419
0.264
R-Sq(adj) = 60.16%
Individual 95% CIs For Mean Based on
Pooled StDev
+---------+---------+---------+--------(-------*------)
(-------*------)
+---------+---------+---------+--------1.5
3.0
4.5
6.0
Individual 95% CIs For Mean Based on
Pooled StDev
---+---------+---------+---------+-----(--------------*-------------)
(-------------*--------------)
---+---------+---------+---------+-----3.20
4.00
4.80
5.60
1. Test interaction: Ho: There is no interaction effect; H1: Time and
strength interact;
Since F=1.37 and p=.264, we ACCEPT the Ho and conclude that there is NO
interaction.
2. Effect of developer strength: Ho: Means for different developer
strengths are equal u1=u2; H1: means are not equal, there is a
relationship.
The F-value for this test is 23.58, with p=.000, therefore, we REJECT ho,
the means are NOT equal, there is a relationship.
3. Effect of development time: test Ho: development times have same means,
u1=u2; h1: the means are different and there is a relationship. Since
f=.70 and p-value=.419, we ACCEPT the HO. There is NO relationship, the
means are equal.
4. Draw a graph: They might have a main effects graph for either or both
variables, or they may have an interaction graph. Any receive credit.
These look as follows:
Interaction Plot (data means) for Density
Main Effects Plot (data means) for Density
Strength
6.5
6.0
6
Mean
5.5
Mean of Density
Strength
1
2
7
Time
5.0
5
4
4.5
4.0
3
3.5
2
3.0
10
14
Time
2.5
1
2
10
14
e) We conclude that only developer strength matters. The effect of development time
and the interaction does not affect density. The model explains 68% (r-sq) of the
variation. It appears that strength #2 is significantly higher than #1.
11.69:
Using file Access.mtw
a) Test for equal variances using Levene’s test:
Test for Equal Variances: ReadTime versus FileSize
95% Bonferroni confidence intervals for standard deviations
FileSize
1
2
3
N
8
8
8
Lower
0.100801
0.130136
0.178452
StDev
0.165869
0.214139
0.293644
Upper
0.406234
0.524452
0.719171
Bartlett's Test (normal distribution)
Test statistic = 2.14, p-value = 0.343
Levene's Test (any continuous distribution)
Test statistic = 4.58, p-value = 0.022
The levene’s test has Ho: Groups have equal variances; h1: groups have different
variances; Since p-value=.022, we REJECT ho, the groups have different variances.
This means that TECHNICALLY, we should NOT use ANOVA , however, we will
proceed nevertheless.
b) Do a oneway ANOVA for read times and factor of file size.
One-way ANOVA: ReadTime versus FileSize
Source
FileSize
Error
Total
DF
2
21
23
S = 0.2306
SS
0.2763
1.1172
1.3935
MS
0.1381
0.0532
R-Sq = 19.83%
F
2.60
P
0.098
R-Sq(adj) = 12.19%
Level
1
2
3
N
8
8
8
Mean
2.2438
2.3863
2.5063
StDev
0.1659
0.2141
0.2936
Individual 95% CIs For Mean Based on
Pooled StDev
+---------+---------+---------+--------(---------*----------)
(---------*----------)
(----------*---------)
+---------+---------+---------+--------2.08
2.24
2.40
2.56
Pooled StDev = 0.2306
We test Ho: means are NOT different, H1: means are different.
value=.098, we accept Ho and conclude there is no difference.
Since p-
c) Do a tukey’s test. NOTE: Make this 2 points Extra credit since some may
read problem as only do this if we rejected in part a.
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of FileSize
Individual confidence level = 98.00%
FileSize = 1 subtracted from:
FileSize
2
3
Lower
-0.1478
-0.0278
Center
0.1425
0.2625
Upper
0.4328
0.5528
------+---------+---------+---------+--(-----------*----------)
(----------*-----------)
------+---------+---------+---------+---0.25
0.00
0.25
0.50
FileSize = 2 subtracted from:
FileSize
3
Lower
-0.1703
Center
0.1200
Upper
0.4103
------+---------+---------+---------+--(-----------*----------)
------+---------+---------+---------+---0.25
0.00
0.25
0.50
For all comparisons, zero is in the interval, indicating that there are no true
differences in means.
D) We conclude that read time is not affected by file size.
Part II doing a two way anova using file size and buffer size as factors and
read times as response variable.
Two-way ANOVA: ReadTime versus FileSize, Buffer
Source
FileSize
Buffer
Interaction
Error
Total
DF
2
1
2
18
23
SS
0.27630
0.93220
0.05773
0.12723
1.39346
MS
0.138150
0.932204
0.028867
0.007068
F
19.55
131.89
4.08
P
0.000
0.000
0.034
S = 0.08407
R-Sq = 90.87%
R-Sq(adj) = 88.33%
5. test for interaction: Ho: there is NO interaction; h1: there is an
interaction.
For the interaction, f=4.08, p-value=.034, we REJECT the HO and conclude
that there is an interaction.
6. Test for buffer size main effect. Ho: the means for all buffer sizes are
equal; h1: means are not all equal. Since f=131.89, p-value=.000, we
REJECT the Ho and conclude that means are not all equal, buffer size DOES
affect read time.
7. Test for file size main effect. Ho: means for all file sizes are equal;
H1: means are not all equal. Since f=19.55, p-value=.000, we REJECT ho
and conclude that means are not all equal, file size DOES affect read
time.
8. They should graph the main effects AND the interaction plots:
Main Effects Plot (data means) for ReadTime
FileSize
2.6
InteractionPlot (datameans) for ReadTime
2.8
Buffer
FileSize
1
2
3
2.7
2.6
Mean
Mean of ReadTime
2.5
2.4
2.5
2.4
2.3
2.3
2.2
2.2
2.1
1
2
3
1
2
1
2
Buffer
From these graphs it appears that buffer size 2 has higher mean read time
and file size 1 has much lower, while filesize 3 has much higher read time.
The interaction plot seems to show that there is relatively little
difference between the means for different file sizes with buffer #1 , but
the differences between the means are much larger for buffer size two.
(Basically, they should say something about the shape, but actual words are
fairly loose).
9. To have the fastest read time (ie lowest mean) they should use buffer
size 1 and filesize 1. However, the most important factor is buffer size
1.
10. In parts a-d with the Oneway ANOVA using file size we found NO effect.
In parts e-I, with a twoway ANOVA we found significant effects for all
factors, including file size. It appears that buffer size is the most
important factor. Once we remove the variation caused by buffer size
from the unexplained error term, the amount of variation explained by
file size becomes significant, where before the unexplained was so large
that file size was insignificant. We also find that there is an
interaction effect which causes larger variations due to file size when
buffer # 2 is used.
Chapter 12: 4, 12, 13 (note on # 13 don’t have to do part d or give 4 extra
points if they do it).
4.
Test of two proportions using Z.
excel.
They may use minitab, by hand or
Test and CI for Two Proportions
Sample
1
2
X
283
1053
N
1090
2065
Sample p
0.259633
0.509927
Difference = p (1) - p (2)
Estimate for difference: -0.250294
95% CI for difference: (-0.284093, -0.216496)
Test for difference = 0 (vs not = 0): Z = -13.53
a)
b)
c)
P-Value = 0.000
Test if difference: ho: p1-p2=0; h1: p1-p2≠0. Since p-value=.000, we
reject the Ho and conclude that the proportions are different. Sample 2
(ages 8-18) appears to use it more often.
P-value=.0000. This means that there is virtually zero risk of a type
one error by rejecting the null hypothesis.
95% confidence interval for difference (-.28 to -.21)
12. Do same as #4 but using a chi-square test. They may do it by hand, by excel or by
minitab. It should have a contingency table as shown:
yes
283
1053
2-7
8-18
no
807
1012
Using minitab, stat/tables/chi-square test (table in worksheet)
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
1
yes
283
461.57
69.082
no
807
628.43
50.738
Total
1090
2
1053
1012
2065
874.43 1190.57
36.464
26.782
Total
1336
1819
3155
Chi-Sq = 183.066, DF = 1, P-Value = 0.000
a) The null hypothesis: proportions equal h1: proportions not equal
Since p-value=.000, we reject ho and conclude that the proportions are not
equal.
B)
C)
p-value=.000
We get similar answers on both #4 and #12 although we used different
methods. (note: in the chisquare table, we can see that group 1 (2-7 year
olds) was more likely to say no (observed>expected), while older group 2 was
more likely to say yes
13. Test of equal proportions using chisquare.
Chi-Square Test: no particles, particles
Expected counts are printed below observed counts
Chi-Square contributions are printed below expected counts
1
no particles
320
296.89
particles
14
37.11
Total
334
2
1.799
14.393
80
103.11
5.180
36
12.89
41.441
116
Total
400
50
450
Chi-Sq = 62.812, DF = 1, P-Value = 0.000
a) HO: no relationship, proportions same; h1: relation, proportions
different. Since chi-square=62.812, and p-value=.000, we reject Ho and
conclude that there is a relationship.
b) P-value=.000
c) It appears that if there are no particles on the chip, they are more
likely to be in group 1 (good), while if there are particles on the chip,
they are more likely to be in group 2 (bad). We see this by comparing
observed and expected. For particles, we see that fewer than expected
(14 vs. 37.11) are in good and more than expected (36 vs. 12.89) are in
bad.
d) EXTRA CREDIT: The z-test used in #5 and the chisquare results here give
the same result that there is a difference. To get extra credit, must
show results of a z-test as follows:
Test and CI for Two Proportions
Sample
1
2
X
320
14
N
400
50
Sample p
0.800000
0.280000
Difference = p (1) - p (2)
Estimate for difference: 0.52
95% CI for difference: (0.389519, 0.650481)
Test for difference = 0 (vs not = 0): Z = 7.93
P-Value = 0.000