 # UNIVERSITY OF TORONTO AT SCARBOROUGH Sample Exam STAC52H3

```UNIVERSITY OF TORONTO AT SCARBOROUGH
Sample Exam
STAC52H3
EXPERIMENTAL DESIGNS
Duration - 3 hours
LAST NAME_____________________________________________________
FIRST NAME_____________________________________________________
STUDENT NUMBER___________________________________________
There are 23 pages including this page and t tables. Please check to see if you have all the
pages.
Aids allowed: You are allowed to use the textbook (Design and Analysis of
experiments by Douglas Montgomery.), the class notes and a non-communicating
calculator. No other material will be allowed during the test.
All your work must be presented clearly in order to get credit. Just an answer with
no other work shown will only qualify for zero credit. Show your work and answer in
the space provided, in ink. Pencil may be used, but then any re-grading will NOT be
allowed.
PLEASE CHECK AND MAKE SURE THAT THERE ARE NO MISSING PAGES
IN THIS BOOKLET.
1) Many studies have suggested that there is a link between exercise and healthy bones.
One study examined the effect of jumping on bone density of growing rats. There were
three treatments: a control with no jumping, a low-jump condition (the jump height was
30 cm), and a high jump condition (60 cm). Thirty rats were randomly divided into three
treatment groups (10 in each). After 8 weeks of 10 jumps per day, 5 days per week, the
bone density of the rats ( in mg / cm3 ) was measured. The summary statistics shown
below were obtained from the analysis of the data from this experiment. This question
and the next two questions are based on this study. Assume that the data satisfies the
necessary assumptions.
Note: In this question µc , µl , and µ h denote the means of control, low-jump and high
jump populations respectively.
Summary Statistics:
Treatment
Control
Highjump
Lowjump
N
10
10
10
Mean
601.10
638.70
612.50
StDev
27.36
16.59
19.33
a) [ 5 points] Test whether the mean bone densities are the same for all three treatments.
Test at α = 0.05. State the null and alternative hypothesis and show you work clearly.
b) [2 points] Calculate a 95% confidence interval for µ h .
Sol Grand Total = 10*(601.1+638.7+612.5) = 18523
a) SSTrt = (1010^2)/10+(6387^2)/10+(6125^2)/10 –GT^2/30 = 7433.866667
MSTrt = 7433.866667/(3-1) = 3716.933334
MSE= (27.36^2+16.59^2+19.33^2)/3 = 465.8155333
dfError = 30-3 = 27
F = 3716.933334/465.8155333 = 7.97941045
F~F(2, 27) = 3.35 and so the trt effect is sig at the 5% level.
b)
S=sqrt((27.36^2+16.59^2+19.33^2)/3) = 21.5827601
T(27, 0.025) = 2.056
ME = ts/sqrt(n) = 2.056*21.58/sqrt(10)
CI = 638.70±ME
Page 2 of 17
2) An experiment was conducted to test the effects of five different diets for turkeys. Six
turkeys were randomly assigned to each of five diet groups (labeled 1, 2, …, 5) and were
fed for a period of time. (i.e. a group of 30 turkeys were divided at random into the five
treatment groups, each with 6 turkeys) Their weight gains were measured at the end of
this period. A part of the SAS output used in the analysis of this data set is given below.
In this question µ1 , µ2 ,… , µ5 denote the means of the diet groups 1, 2, …, 5 respectively.
Assume that the data satisfies the necessary assumptions.
The GLM Procedure
t Tests (LSD) for WtGain
NOTE: This test controls the Type I comparisonwise error rate, not the
experimentwise error rate.
Alpha
Error Degrees of Freedom
Error Mean Square
Critical Value of t
Least Significant Difference
0.05
25
0.3154
2.05954
0.6678
Comparisons significant at the 0.05 level are indicated by ***.
diet
Comparison
5
5
5
5
4
4
4
4
3
3
3
3
2
2
2
2
1
1
1
1
-
4
3
2
1
5
3
2
1
5
4
2
1
5
4
3
1
5
4
3
2
Difference
Between
Means
2.3833
2.4000
3.8833
5.6000
-2.3833
0.0167
1.5000
3.2167
-2.4000
-0.0167
1.4833
3.2000
-3.8833
-1.5000
-1.4833
1.7167
-5.6000
-3.2167
-3.2000
-1.7167
95% Confidence
Limits
1.7155
omitted
3.2155
4.9322
-3.0511
-0.6511
0.8322
2.5489
-omitted
-0.6845
0.8155
2.5322
-4.5511
-2.1678
-2.1511
1.0489
-6.2678
-3.8845
-3.8678
-2.3845
3.0511
***
4.5511
6.2678
-1.7155
0.6845
2.1678
3.8845
***
***
***
0.6511
2.1511
3.8678
-3.2155
-0.8322
-0.8155
2.3845
-4.9322
-2.5489
-2.5322
-1.0489
***
***
***
***
***
***
***
***
***
***
***
***
a) [2 points] Use Fishier’s method to calculate a 95% confidence for the difference
between the means of diet 3 and diet 5.
b) [2 points] Use Tukey’s method to calculate a 95% confidence for the difference
between the means of diet 3 and diet 5.
c) [3 points] Calculate a 95% confidence interval for 2 µ1 − µ 2 − µ3 using Scheffe’s
method.
Page 3 of 17
Sol
a) 2.4+/-0.6678 =( 1.7322, 3.0678)
Descriptive Statistics: Weight Gain (pounds)
Variable
Weight Gain (pou
Group
1
2
3
4
5
N
6
6
6
6
6
Mean
3.783
5.500
6.983
7.000
9.383
StDev
0.527
0.867
0.578
0.358
0.293
For part c) Note that 2µ1 − µ2 − µ3 = ( µ1 − µ2 ) − ( µ3 − µ1 )
And 2 y1 − y2 − y3 = ( y1 − y2 ) − ( y3 − y1 ) = -1.7167 - 3.2000 =
S == sqrt( (5-1)*4.18*0.3154*(4/6+1/6+1/6)) = 2.296407629
F(5-1, 30-5, 0.05) = 4.18
CI = L ±S
-
4.917
3) [ 5points] A researcher conducted a study of two weight reduction programs (program
I and program II) with two diets (diet A and diet B). 16 subjects were randomly into four
groups, each group receiving one of the four treatments (i.e. a program-diet combination)
in a 2 x 2 factorial design with four observations for each treatment. The amount of
weight each subject lost was measured after one month. The table below gives treatment
means and standard deviations for each of the four treatments. The values in parenthesis
are the standard deviations. Assume that the data satisfies the necessary assumptions.
Program A
Program B
Diet I
4.750 (0.957)
5.500 ( 2.380)
Diet II
11.250 (2.986)
13.000 ( 2.160)
Test whether the interaction of program and diet is significant. Test at α = 0.05.
SS_int = (x_bar(I,A)+x_bar(II, B)-x_bar(II, A)- x_bar(I, B) ^2/(1/n1+1/n2+1/n3+1/n4) =
1
with df = (2-1)(2-1) = 1 and so MS(int) = 1
MSE = pooled variance of the 4 treatment variances = the average of the four variances
when the sample sizes are equal = 5.04 and F = MS(int)/MSE = 1 / 5.04 = 0.2 approx
Here is the MINITAB output for ANOVA
Two-way ANOVA: Weight loss versus Exercise Program, Deit
Page 4 of 17
Source
Exercise Program
Deit
Interaction
Error
Total
S = 2.245
DF
1
1
1
12
15
SS
6.25
196.00
1.00
60.50
263.75
R-Sq = 77.06%
MS
6.250
196.000
1.000
5.042
F
1.24
38.88
0.20
P
0.287
0.000
0.664
Question 4 is not in the exam coverage for this year
(
)
4) In assignment 3 we showed that for a BIBD V τˆ i1 − τˆ i2 =
2k 2
σ .
λa
(
)
a) [4 points] Prove that for a randomized complete block design V τˆ i1 − τˆ i2 =
2σ 2
.
b
k
1
≥ and give it a statistical interpretation.
λa b
Note: The expressions above are in the usual notation we discussed in class. You may use
without proof, any results given in class but you should state them all clearly with all the
assumptions under which they are valid.
b) [5 points] Prove that
Sol (nor
a) For RCBD,
τˆ i1 = yi1 . − y.. ⇒ τˆ i1 − τˆ i2 = yi1 . − yi2 .
(
)
(
)
( )
( )
⇒ Var τˆ i1 − τˆ i2 = Var yi1 . − yi2 . = Var yi1 . + Var yi2 . =
σ 2 σ 2 2σ 2
+
=
b
b
b
(Note: yi1 . − yi2 . = τi1 − τi2 + εi1 . − εi2 . )
b) From class we have λ =
r ( k − 1)
and ar = bk
a −1
r ( k − 1)
ar ( k − 1)
⇒ λa =
a −1
a −1
ar (k − 1) bk (k − 1)
λa =
=
(∵ ar = bk )
a −1
a −1
bk (a − 1)
≤
= bk
(∵ a ≥ k )
a −1
1 k
⇒ ≤
b λa
λ=
This implies that the estimates of treatment differences in a RCBD have less standard
error than the corresponding estimates from a BIBD. (2 points)
5) This question is question 15.37 p914 Ott.
Page 5 of 17
 A food-processing plant has tested several different formulations of a new breakfast
drink. A panel of six members rated these formulations. Each member of the panel rated
12 different formulations obtained from combining one of the three levels of sweetness,
one of two levels of caloric content, and one of the two colours. Each member of the
panel rated the formulations in a random order.
Identify the experimental design. Give the analysis of variance table showing the sources
of variation and the respective degrees of freedom.
Ans This is a RBD with three factors and panel members are the blocks
SV
Sweetness(S)
df
2
Caloric (Ca )
Colour (Co)
1
1
S x Ca
Ca x Co
S xCo
S x Ca xCo
Blocks
Error
Tot
2
1
2
2
5
55
71
6) An experiment was performed to determine the effects of four different geometrical
shapes of a certain film-type resistor on the current-noise of the resistors. A BIBD was
used because only three resistors could be mounted on one plate. The design layout
below shows the observations on noise measurements from this experiment. Assume that
the data satisfies the necessary assumptions.
Plates
(Blocks)
1
2
3
4
Total
A
1.11
1.70
1.66
4.47
Shapes (Treatments)
B
C
0.95
1.22
1.11
1.52
1.22
1.54
3.55
4.01
D
0.82
0.97
1.18
2.97
Total
2.88
3.89
4.29
3.94
15.00
The sum of squats of all observations = 19.6768
a) [ 10 points] Test whether the treatment effect is significant. (Use α = 0.05)
b) [ 5 points] Calculate a 95% C.I. (Fisher-type) for τˆ1 − τˆ 2 where τˆ1 and τˆ 2 are the
effects of shapes A and B respectively.
Page 6 of 17
Sol
a)
SSTot = 19.6768-(15^2)/12 = 0.9268, dfTot = N-1 = 12 – 1 = 11
Q1 = 4.47-(2.88+3.89+4.29)/3 = 0.7833333333
Q2 = 3.55 – (3.89+4.29+3.94)/3 = -0.49
Q3 = 4.01 – (2.88+4.29+3.94)/3 = 0.3066666667
Q4 = 2.97 – (2.88+3.89+3.94)/3 = -0.6
k=3
a=4
λ=2
k a 2
∑ Qi = (3/8)*( 0.7833333333^2+0.49^2+0.3066666667^2+0.6^2) =
λa i =1
0.4904083333
dfTrt = a – 1 = 4 – 1 = 3
SSBlock = (2.88^2+3.89^2+4.29^2+3.94^2)/3-(15^2)/12 = 0.3680666667
dfBlock = b -1 = 4 – 1 = 3
SSE = SSTot – SSBlock – SSTr(Adj) = 0.0683250003
dfError = dfTot – dfBlock – dfTrt = 11 – 3 – 3 = 5
F= MSTrt(Adj) / MSE = (0.4904083333/3)/(0.0683250003/5) = 11.96263767
The SAS System
1
Here is the full SAS output
03:23 Sunday, December 12, 2010
The GLM Procedure
Class Level Information
^L
Class
Levels
Values
shape
4
A B C D
plate
4
1 2 3 4
12
Number of Observations Used
12
The SAS System
2
03:23 Sunday, December 12, 2010
The GLM Procedure
Page 7 of 17
Dependent Variable: noise
DF
Sum of
Squares
Mean Square
F Value
Pr > F
Model
6
0.85847500
0.14307917
10.47
0.0104
Error
5
0.06832500
0.01366500
11
0.92680000
Source
Corrected Total
R-Square
Coeff Var
Root MSE
noise Mean
0.926279
9.351791
0.116897
1.250000
Source
DF
Type I SS
Mean Square
F Value
Pr > F
plate
shape
3
3
0.36806667
0.49040833
0.12268889
0.16346944
8.98
11.96
0.0186
0.0102
Source
DF
Type III SS
Mean Square
F Value
Pr > F
plate
shape
3
3
0.44700833
0.49040833
0.14900278
0.16346944
10.90
11.96
0.0124
0.0102
b)
τöi =
(
kQi
(λa )
,
)
V τöi − τö j =
2k 2
σ
λa
7) In a large study of a health awareness program, three states (factor A, a random factor)
were randomly selected from all states of the country. Each state selected, independently
devised the health awareness program. From each state (factor A) selected, three cities
(Factor B) were chosen at random from all the cities in the state and five households
within each city were randomly selected to evaluate the effectiveness of the program. All
members of the selected households were interviewed before and after participation in the
program and a composite index was formed for each household measuring the impact of
the health awareness program. The data were analyzed using the model:
yijk = µ + α i + β j ( i ) + εijk i = 1,… , a j = 1,… , b k = 1,… , n ,
where a is the number of levels for factor A, b is the b is the number of levels for factor B
and n is the number of replicates.
iid
iid
iid
We also assume α i ~ N (0, σ α2 ) , β j (i ) ~ N (0, σβ2 ) , εijk ~ N (0, σε2 ) and α i , β j ( i ) and εijk are
independent.
Page 8 of 17
The ANOVA table constructed from the resulting data set (with some entries deleted) is
shown below: You may assume that the model is appropriate (necessary assumptions are
satisfied) when answering questions related to this study.
(See page 987 Ott)
Analysis of Variance for index
Source
state
city(state)
Error
Total
DF
2
6
36
44
SS
6976.8
167.6
3893.2
11037.6
MS
3488.4
27.9
108.1
F
omitted
omitted
For this design E ( MSE ) = σ ε2 , E ( MSB( A)) = σ ε2 + nσ β2 , and E ( MSA) = σ ε2 + nσ β2 + nbσ α2
in usual notation.
a) [ 3 points] Test the null hypothesis H 0 : σ α2 = 0 against H a : σ α2 > 0 . Use α = 0.05.
Ans n=5, b = 3
b) [ 2 points] Estimate the variance component σ α2
c) [5 points] Calculate the value of the constant c such that


 σβ2  
 σβ2  
MSB( A) 

P 1 + c  2   Fα /2,6,36 <
<  1 + c  2   F1−( α / 2),6,36  = 1 − α where Fp ,6,36 denotes
 
 



MSE
 σε  
 σε  



the value of the inverse cdf of the F distribution with 6 df in the numerator and 36 df in
the denominator evaluated at p. Show your work clearly.


MSB ( A) / (σε2 + nσβ2 )
Sol use P  Fα / 2,6,36 <
<
F
= 1 − α with n=5

1−( α / 2 ),6,36 

MSE / σε2


2
2
 (σε2 + nσβ2 )

MSB( A) (σε + nσβ )
P
Fα /2,6,36 <
<
F1−( α /2 ),6,36  = 1 − α
2
2


σε
MSE
σε




 σβ2  
 σβ2  
MSB( A) 
P  1 + n  2   Fα / 2,6,36 <
< 1 + n  2   F1−( α /2 ),6,36  = 1 − α
 σ 
 



MSE
 ε 
 σε  



i.e c = n= 5
8) A petroleum company was interested in comparing the miles per gallon (MPG)
achieved by four different gasoline blends (1, 2, 3, and 4). Because there can be
considerable variability due to drivers and due to car models, the two extraneous sources
of variability were included as “blocking” variables in a Latin square design. Each of the
four drivers participated in the experiment drove each of the four car models with the
Page 9 of 17
assigned gasoline blend in a Latin square design. Some useful SAS outputs and the SAS
code that produced it are given below. Assume that the data satisfies the necessary
assumptions.
options ls=75;
data a;
infile 'latin1.txt' firstobs=2;
input Row Driver Model Blend MPG;
PROC GLM data=a;
class Driver ;
model MPG= Driver;
PROC GLM data=a;
class Model ;
model MPG= Model;
PROC GLM data=a;
class Blend ;
model MPG= Blend;
PROC GLM data=a;
class Driver Model Blend ;
model MPG= Blend Driver Model;
run;
quit;
The GLM Procedure
Class Level Information
Class
Levels
Driver
^L
4
Values
1 2 3 4
Number of Observations Used
The SAS System
The GLM Procedure
16
16
2
Dependent Variable: MPG
DF
Sum of
Squares
Mean Square
F Value
Pr > F
Model
3
5.8968750
1.9656250
0.03
0.9936
Error
12
869.7025000
72.4752083
Corrected Total
15
875.5993750
Source
DF
Type I SS
Mean Square
F Value
Pr > F
Driver
3
5.89687500
1.96562500
0.03
0.9936
Source
Page 10 of 17
Source
DF
Type III SS
Mean Square
F Value
Pr > F
Driver
3
5.89687500
1.96562500
0.03
0.9936
The GLM Procedure
Class Level Information
Class
Levels
Model
4
Values
1 2 3 4
Number of Observations Used
16
16
^
The GLM Procedure
Dependent Variable: MPG
Source
Model
Error
Corrected Total
Source
Model
Source
Model
DF
Sum of
Squares
Mean Square
F Value
Pr > F
3
736.9118750
245.6372917
21.25
<.0001
12
15
138.6875000
875.5993750
11.5572917
DF
Type I SS
Mean Square
F Value
Pr > F
3
736.9118750
245.6372917
21.25
<.0001
DF
Type III SS
Mean Square
F Value
Pr > F
3
736.9118750
245.6372917
21.25
<.0001
The GLM Procedure
Class Level Information
Class
Levels
Blend
4
Values
1 2 3 4
Number of Observations Used
16
16
The GLM Procedure
Dependent Variable: MPG
Source
DF
Sum of
Squares
Mean Square
F Value
Pr > F
Model
3
108.9818750
36.3272917
0.57
0.6462
Error
12
766.6175000
63.8847917
Page 11 of 17
Corrected Total
Source
Blend
15
DF
3
Source
Blend
875.5993750
Type I SS
108.9818750
Mean Square
36.3272917
F Value
0.57
Pr > F
0.6462
DF
Type III SS
Mean Square
F Value
Pr > F
3
108.9818750
36.3272917
0.57
0.6462
[3 points] Which of the following numbers is closest to the appropriate F-statistic for
testing the equality of means of the four blends? Circle your answer. [Note: This is a
multiple choice question. You do not have to show your work in this question and no part
marks are given for showing work]
A)
B)
C)
D)
E)
7
9
11
13
0.57
Ans B) 9
Note: This is example 15.4 p865 Ott
SS_blend = 109.0 df_blend = 4-1 =3 MS = 109.0/3 =36.3
SSE = SSTot – SS_Driver – SS_Model – SS_Blend = 875.6 – 5.9 – 736.9 – 109.0 = 23.8
Df_error = 15-3-3-3 – 6 and MSE = SSE/df_error = 23.8/6 = 3.97
F= MS_Blend/MSE = 36.3/3.97 = 9.14 =9 (approx)
Here is the full output
The GLM Procedure
Dependent Variable: MPG
Source
Model
Error
Corrected Total
R-Square
DF
Sum of
Squares
Mean Square
F Value
Pr > F
9
6
851.7906250
23.8087500
94.6434028
3.9681250
23.85
0.0005
15
875.5993750
Coeff Var
Root MSE
MPG Mean
Page 12 of 17
0.972809
8.950364
1.992015
22.25625
Source
DF
Type I SS
Mean Square
F Value
Pr > F
Blend
Driver
Model
3
3
3
108.9818750
5.8968750
736.9118750
36.3272917
1.9656250
245.6372917
9.15
0.50
61.90
0.0117
0.6987
<.0001
Source
DF
Type III SS
Mean Square
F Value
Pr > F
Blend
Driver
Model
3
3
3
108.9818750
5.8968750
736.9118750
36.3272917
1.9656250
245.6372917
9.15
0.50
61.90
0.0117
0.6987
<.0001
Page 13 of 17
9) (This is from q13.8 p798 Ott) [ 3 points] An experiment was conducted to compare the
number of major defectives obtained along each of three production lines in which
changes were being instituted. Production was monitored continuously during the period
of changes, and the number of major defectives was recorded per day for each line. The
data are given below:
Production Line
2
54
41
38
33
56
1
34
44
32
36
51
3
75
62
45
10
68
The data showed violations of the one-way ANOVA procedure for comparing the three
lines and so decided to use the Kruskal-Wallace test.
Which of the following numbers is closest to the value of the Kruskal-Wallace statistic
for comparing the three production lines?
A)
B)
C)
D)
E)
2.0
2.5
3.0
3.5
4.0
Ans B 2.5 The formula on page 191 supp Vukov gives H = 2.66 which is exact the same
in MINITAB output.
2
12
∑
H =
N ( N + 1)
T
i
− 3( N + 1)
n
i
= (12/(15*16))*5*(5.8^2+7.8^2+10.4^2)-3*16=2.66
————— 7/3/2006 2:57:08 AM ————————————————————
Worksheet size: 10000 cells.
Welcome to Minitab, press F1 for help.
Kruskal-Wallis Test: defects versus Line
Kruskal-Wallis Test on defects
Line
1
N
5
Median
36.00
Ave Rank
5.8
Z
-1.35
Page 14 of 17
2
3
Overall
H = 2.66
5
5
15
41.00
62.00
DF = 2
7.8
10.4
8.0
-0.12
1.47
P = 0.264
10) An area in a greenhouse has 10 benches. On each bench, there are 3 pots. Each pot
contains 1 plant. Researchers want to compare the effect of three treatments (A, B, and C)
on the growth of the plants. The researchers randomly assign the three treatments to the
three plants on each bench so that each treatment is represented exactly once on each
bench.
State whether each of the following statements based on this information is true or false.
[ 1 point for each ]
i) This is a completely randomized 1-factor design. [True / False] Ans F
ii) This is a randomized block design.
[True / False] Ans T
iii) This is a two factor factorial design.
[True / False] Ans F
iv) This is an incomplete block design.
[True / False] Ans F
v) The hypothesis of equality of treatment means can be tested using a test statistic that
has an F-distribution with 2 degrees of freedom in the numerator and 18 degrees of
freedom in the denominator.
[True / False] Ans T
And: F(3-1= 2, 29 – 2-9 = 18)
11) This question is based on 15.25 p910 Ott. An experiment was set up to compare the
effect of different soil pH and calcium additives on the trunk diameters of orange trees.
Sulfur, Gypsum and other ingredients were applied to provide pH levels of 4, 5, 6, and 7.
Three levels of calcium supplement (100, 200 and 300 pounds per acre) were also
applied. All factor level combinations were used in the experiment. The trees were
assigned to treatment at random. At the end of a 2-year period, three diameters were
measured at each factor level combination. Some useful outputs from the analysis of this
data set are given below.
ANOVA Table: diameter versus pH, Calcium
Source
pH
Calcium
Interaction
Error
DF
A
B
C
D
SS
4.4608
1.4672
3.2550
1.6267
MS
1.48694
0.73361
0.54250
0.06778
F
21.94
10.82
8.00
P
0.000
0.000
0.000
Page 15 of 17
Normal Scores Plot of Residual
0.50
Residual
0.25
0.00
-0.25
-0.50
-0.75
-2
-1
0
Normal Score
1
2
Interaction Plot (data means) for diameter
C alcium
100
200
300
7.6
Mean
7.2
6.8
6.4
6.0
4.0
5.0
6.0
7.0
pH
State whether each of the following statements based on this information is true or false.
i) The trunk diameter of orange trees is the explanatory variable in this study
[True / False]
Ans F, it is the response variable.
ii) There are two factors of interest in the study.
[True / False]
Page 16 of 17
Ans T, pH, calcium
iii) There are 12 treatments in this study
[True / False]
Ans T
iv) The degrees of freedom for the numerator and the denominator of the F-statistic for
testing the interaction effect are 6 and 24 respectively.
[True / False]
Ans T numerator df = 2 x 3= 6 denominator df = (3 x12 –1) 3 –2- 6 = 24
v) Since the normality of residuals is questionable, we should try a transformation or
some other remedy.
[True / False]
Ans F, normality if good
vi) The effect of calcium on trunk diameter is about the same for all pH levels.
[True / False]
Ans F interaction is sig.
Page 17 of 17
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