Stats for Strategy
HOMEWORK 4 Solution
(updated 5/26/2015)
A.
(a) 97.94%
(b) Ronald Fisher
B. Topic 6 Examples
1. (Example 1 Case 2)
(a) P -value = 1 − 0.822021 = 0.177979
(b) 0.396569
(c) P -value > 0.396569
(d) P -value < 0.177979
2. (Example 1 Case 3)
(a)
s21 = 37
s22 = 30.5
s23 = 62.5
s2p = average(37, 30.5, 62.5) = 43.33
(b)
F =
(c)
MSG
500
=
= 11.54
MSE
43.33
The variation between sample means is 11.54 times as large as expected, if the
null hypothesis is true.
(d)
1 − 0.998398 = 0.001602
(e) (1) F = 11.54
(2) P -value = 0.002
(3) Almost! For the Notes Prof. Whitten used additional MINITAB graphing options
to make small changes such as “jiggling” the points horizontally so that they’re not
obscured by the sample means.
1
3. (a)
MSG = 102.8
MSE = 538.9307692
F =
MSG
= 0.190748062
MSE
(b) P -value = 0.828612
(c) F = 0.19, P -value = 0.829
C. Textbook
1. Pretest scores (before reading programs are applied)
(a)
1. µB ( or µ1 ) = mean pretest score of all children taught with Basal
µD ( or µ2 ) = . . . . . . DRTA
µS ( or µ3 ) = . . . . . . Strat
H0 : µ 1 = µ 2 = µ 3
HA : At least one mean diﬀers from the others
2. From MINITAB: F = 1.13 , P -value = 0.329
3. Fail to Reject H0 since P -value = 0.329 > 0.05 = α
4. There is not suﬃcient evidence to show any diﬀerence in mean pretest scores between the three groups of children.
(b) Yes. The diﬀerence in sample means seems small compared to the within-sample variation
in each of the three groups.
Individual Value Plot of Pretest Score vs Group
17.5
Pretest Score
15.0
12.5
10.0
7.5
5.0
Basal
DRTA
Group
2
Strat
(c) Yes. The Tukey CI’s comparing mean scores all contain 0.
Tukey Simultaneous 95% CIs
Differences of Means for Pretest Score
DRTA - Basal
Strat - Basal
Strat - DRTA
-4
-3
-2
-1
0
1
2
If an interval does not contain zero, the corresponding means are significantly different.
µD − µB = (−2.951, 1.406)
(d)
µS − µB = (−3.542, 0.815)
µS − µD = (−2.770, 1.588)
(e) Probably not but it doesn’t hurt to look at Tukey CI’s anyway, just to be sure. The
ANOVA test has already indicated no statistically-signiﬁcant diﬀerences.
(f) Let µ = mean pretest score for all children. A 95% conﬁdence interval for µ is
(9.045, 10.530) points.
Combining the three samples into one large sample makes sense because the F test shows
no real diﬀerences between the three mean scores. In this case µ is the common mean
pretest reading score for all children:
µ = µ1 = µ2 = µ3
(g) Yes.
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2. Reading comprehension scores (after reading programs are applied)
(a)
1. µB = mean test score of all children taught with Basal method
µD = . . . . . . DRTA
µS = . . . . . . Strat
H0 : µB = µD = µS
HA : At least one mean diﬀers from the others
2. From MINITAB: F = 4.48 , P -value = 0.015
3. Reject H0 since P -value = 0.015 < 0.05 = α
4. There is suﬃcient evidence to show that mean test scores diﬀer between teaching
methods.
(b) It’s diﬃcult to judge from the graph whether there are any real diﬀerences in the three
population means. For that reason, we should rely on the P -value.
Individual Value Plot of Comprehensive Reading vs Group
60
Comprehensive Reading
55
50
45
40
35
30
B
D
S
Group
(c) Yes. The F test (ANOVA test) has indicated signiﬁcant diﬀerences in mean reading
scores.
(d) One of the Tukey CI’s shows a signiﬁcant diﬀerence between the Basal and DRTA
methods. A 95% conﬁdence interval for (µD − µB ) is
(1.12, 10.25) points
so we are 95% conﬁdent that using DRTA rather than Basal will result in improved test
scores of between 1.12 and 10.25 points, on average.
Neither of the other two Tukey CI’s shows a statistically signiﬁcant diﬀerence:
• No clear diﬀerence between the Basal and Strat methods
• No clear diﬀerence between the DRTA and Strat methods
(e) With 95% conﬁdence, the company can claim that its DRTA method is superior to Basal
but cannot make a similar claim for Strat.
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3. (Bookstores)
(a)
1. µ1 = mean age of all Bookstore 1 customers (in years)
µ2 = . . . . . . Bookstore 2
µ3 = . . . . . . Bookstore 3
µ4 = . . . . . . Bookstore 4
µ5 = . . . . . . Bookstore 5
H0 : µ1 = µ2 = µ3 = µ4 = µ5
HA : At least one mean diﬀers from the others
2. F = 472.46 , P -value = 0.000
3. Reject H0 since P -value = 0 < 0.05 = α .
4. There is suﬃcient evidence to show that mean customer age diﬀers between bookstores.
(b) There are 7 signiﬁcant comparisons and 3 non-signiﬁcant comparisons.
(c) There are three distinct clusters.
(d) Cluster 1 = Bookstores 1, 4, 5
Cluster 2 = Bookstore 2
Cluster 3 = Bookstore 3
(e) If the clusters are deﬁned as in (d),
Cluster 1 has youngest customers.
Cluster 2 has intermediate customers.
Cluster 3 has oldest customers.
(f) Bookstore customers at the mall tend to be younger than bookstore customers downtown.
(g) Cluster 3 = Bookstore 3 has the oldest customers. A 95% conﬁdence interval for µ3 is
(53.894, 56.906) years.
(h) Cluster 1 = Bookstores 1, 4, 5 has the youngest customers. Let µ represent the (common)
mean age of customers at these three stores, all located at the mall. A 95% conﬁdence
interval for µ is
(18.782, 20.418) years.
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4. (More about clusters)
(a)
• Cluster 1 = Basil and Strat methods
• Cluster 2 = DRTA and Strat methods
(b) The clusters are not distinct since they contain a teaching method (Strat) in common.
(c) Since both clusters contain Strat, neither cluster is shown to be clearly superior to the
other cluster.
5. SE 14.3
(a)
1.
µ0
µ1
µ3
µ5
µ7
=
=
=
=
=
mean Vitamin C content in bread immediately after baking
. . . . . . 1 day after baking
. . . . . . 3 days
. . . . . . 5 days
. . . . . . 7 days
H0 : µ0 = µ1 = µ3 = µ5 = µ7
HA : At least one mean diﬀers from the others
2. F = 367.74 P -value = 0.000
3. Reject H0 since P -value = 0 < 0.05 = α
4. There is suﬃcient evidence to show that mean Vitamin C content in bread changes
over time.
(b) Tukey 95% CI’s for adjacent time points:
• (µ1 − µ0 ) = (−12.05 , −1.45)
• (µ3 − µ1 ) = (−25.46 , −14.86)
• (µ5 − µ3 ) = (−14.68 , −4.08)
• (µ7 − µ5 ) = (−9.39 , 1.20)
There are signiﬁcant diﬀerences among the ﬁrst three pairs of adjacent time points, but
not between the last pair.
Interpret: We are 95% conﬁdent that average Vitamin C content in bread (in mg per
100g of ﬂour) decreases by between
• 1.45 and 12.05 mg immediately after baking to one day after baking.
• 14.86 and 25.46 one day after baking to three days after baking.
• 4.08 and 14.68 three days after baking to ﬁve days after baking.
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(c) Mean loss of Vitamin C in bread begins immediately after baking and continues until
ﬁve days after baking, at which time it appears to level oﬀ between ﬁve and seven days
after baking.
(d) There are 4 distinct clusters of adjacent time points:
Cluster
Cluster
Cluster
Cluster
1
2
3
4
=
=
=
=
Immediately after baking
1 day . . .
3 days . . .
5 days and 7 days . . .
6. SE 4.5
Vitamin A
(a)
1.
µ0
µ1
µ3
µ5
µ7
=
=
=
=
=
mean Vitamin A content in bread immediately after baking
. . . . . . 1 day after baking
. . . . . . 3 days
. . . . . . 5 days
. . . . . . 7 days
H0 : µ0 = µ1 = µ3 = µ5 = µ7
HA : At least one mean diﬀers from the others
2. F = 12.09
P -value = 0.009
3. Reject H0 since P -value = 0.009 < 0.05 = α
4. There is suﬃcient evidence to show that mean Vitamin A content in bread changes
over time.
(b) Tukey 95% CI’s:
• (µ1 − µ0 ): (−0.354 , 0.134)
• (µ3 − µ1 ): (−0.274 , 0.214)
• (µ5 − µ3 ): (−0.149 , 0.339)
• (µ7 − µ5 ): (−0.584 , −0.096)
There are no signiﬁcant diﬀerences among three pairs but there is a signiﬁcant diﬀerence
between the last pair.
Interpret:
We are 95% conﬁdent that the average Vitamin A content in bread decreases
by between 0.096 and 0.584 mg per 100g of ﬂour from ﬁve days after baking to
seven days after baking.
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(c) There’s no detectable loss in average Vitamin A content in bread until ﬁve days after
baking, but there is a loss between ﬁve and seven days after baking.
(d) There are 2 distinct clusters of time points:
Cluster 1 = Immediately, 1 day, 3 days, 5 days after baking
Cluster 2 = 7 days after baking
Vitamin E
(a)
1.
µ0
µ1
µ3
µ5
µ7
=
=
=
=
=
mean Vitamin E content in bread immediately after baking
. . . . . . 1 day after baking
. . . . . . 3 days
. . . . . . 5 days
. . . . . . 7 days
H0 : µ0 = µ1 = µ3 = µ5 = µ7
HA : At least one mean diﬀers from the others
2. F = 0.69
P -value = 0.630
3. Fail to Reject H0 since P -value = 0.630 > 0.05 = α.
4. There is not suﬃcient evidence to show that mean Vitamin E content in bread
changes over time.
(b) Tukey CI’s:
• (µ1 − µ0 ): (−8.13 , 6.43)
• (µ3 − µ1 ): (−5.88 , 8.68)
• (µ5 − µ3 ): (−6.78 , 7.78)
• (µ7 − µ5 ): (−9.93 , 4.63)
There are no signiﬁcant diﬀerences between any of these pairs of means, which is consistent with the fact that we did not reject the ANOVA null hypothesis!
(c) There is no apparent loss in Vitamin E in bread (on average) over the ﬁrst seven days
after baking.
(d) There is a single cluster of time points:
Cluster 1 = Immediately, 1 day, 3 days, 5 days, 7 days after baking
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(Exercise 7 solution on next page)
D. Answers to Additional Practice
1. (a)
2. (a)
3. (e)
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7. (a) and (b)
Vitamin C, A, E Loss in Bread Over Time
Vitamin C in Bread Over Time
50
40
tn
ten
oC
C 30
ni
m
tai
V 20
10
immediate
Vitamin C:
F = 367.74
one day
three days
P -value = 0.000
five days
seven days
(Also circle Vitamin C clusters on graph.)
Vitamin A in Bread Over Time
3.4
3.3
nte
tn 3.2
oC
A
ni
3.1
am
itV
3.0
2.9
immediate
Vitamin A:
F = 12.09
one day
three days
P -value = 0.009
five days
seven days
(Also circle Vitamin A clusters.)
Vitamin E in Bread Over Time
98
97
tn
et 96
no
C
E 95
in
m
at
iV 94
93
92
immediate
Vitamin E:
F = 0.69
one day
three days
P -value = 0.630
five days
seven days
(Also circle Vitamin E clusters.)
(c) As the F statistics decrease, the P -values increase while the numbers of clusters decrease.
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