1. 2 2 k r Factorial Designs with Replications r replications of 2 k Experiments –2 k r...
-
date post
21-Dec-2015 -
Category
Documents
-
view
229 -
download
2
Transcript of 1. 2 2 k r Factorial Designs with Replications r replications of 2 k Experiments –2 k r...
2
2kr Factorial Designs with Replications
• r replications of 2k Experiments– 2kr observations.– Allows estimation of experimental errors
• Model:y = q0 + qAxA + qBxB +qABxAxB+e
e=Experimental error
3
Computation of Effects• Simply use means of r measurements
I A B AB y Mean y
1111
-11
-11
-1-111
1-1-11
(15, 18, 12)(45, 48, 51)(25, 28, 19)(75, 75, 81)
15482477
16441
8621.5
389.5
205
TotalTotal/4
Effects: q0 = 41, qA = 21.5, qB = 9.5, qAB = 5
4
Estimation of Experimental Errors• Estimated Response:
yi = q0 + qAxAi + qBxBi +qABxAixBi
• Experimental Error = Estimated-Measured
eij = yij - yi
= yij - q0 - qAxAi - qBxBi - qABxAixBi
• Sum of Squared Errors:
SSE =
r
iie
22
1
2
^
^
0,
ji
ije
5
Experimental Errors: Example
Estimated Response:y1 = q0 - qA - qB + qAB = 41 -21.5 -9.5 +5 = 15
Experimental errors:e11 = y11 - y1 = 15 - 15 = 0
^
^
i I A B AB Response Responses Errors41 21.5 9.5 5 yi yi1 yi2 yi3 ei1 ei2 ei3
1 1 -1 -1 1 15 15 18 12 0 3 -32 1 1 -1 -1 48 45 48 51 -3 0 33 1 -1 1 -1 24 25 28 19 1 4 -5
4 1 1 1 1 77 75 75 81 -2 -2 4
Effect Estimated Measured
SSE = 02 + 32 + (-3)2 + (-3)2 + ... + 42 = 102
^
6
Allocation of Variation
Total variation or total sum of squares:SST =
jiij yy
,
2..)(
ijBiAiABBiBAiAij exxqxqxqqy 0
ji
ijABBAji
ij erqrqrqyy,
2222222
,
2 222..)(
SST = SSA + SSB + SSAB + SSE
7
DerivationModel:
ijBiAiABBiBAiAij exxqxqxqqy 0
Since x’s, their products, and all errors add to zero
jiij
jiBiAiAB
jiBiB
jiAiA
jijiij
exxqxq
xqqy
,,,
,,0
,
02
,0
,
2 rqqyjiji
ij
8
Derivation (cont’d)Mean response:
0,
22
1.. qy
ry
jiij
Squaring both sides of the model and ignoring cross product terms:
jiij
jiBiAiAB
jiBiB
jiAiA
jijiij
exxqxq
xqqy
,
2
,
222
,
22
,
22
,
20
,
2
SSY = SS0 + SSA + SSB + SSAB + SSE
9
Derivation (cont’d)Total Variation:
SSESSABSSBSSA
SSSSY
yy
yySST
jijiij
jiij
0
..
,
2
..,
2
,
2
One way to compute SSE:
22220
22 ABBA qqqqrSSYSSE
10
Example: Memory-Cache Study
27204
81757545121815 2222222
SSY
703220172272040
102)55.95.2141(3227204
3005122
1083)5.9(122
5547)5.21(122
20172411220
22222
222
222
222
220
2
SSSSYSST
SSE
rqSSAB
rqSSB
rqSSA
rqSS
AB
B
A
11
Example: Memory-Cache Study(cont’d)
SSA + SSB + SSAB + SSE=5547 + 1083 + 300 + 102= 7032 = SST
Factor A explains 5547/7032 or 78.88%Factor B explains 15.40%Interaction AB explains 4.27%1.45% is unexplained and is attributed to errors.
12
Review: Confidence Interval for the Mean
Problem: How to get a single estimate of the population mean from k sample estimates?Answer: Get probabilistic bounds.
Eg., 2 bounds, C1 & C2 There is a high probability, 1-,
that the mean is in the interval (C1, C2 ):
Pr {C1 C2} = 1 - • Confidence interval (C1, C2 )
• Significance Level• 100 (1-) Confidence Level• 1- Confidence Coefficient.
13
Confidence Interval for the Mean (cont’d)
Note: Confidence Level is traditionally expressed as a percentage (near 100%); whereas, significance level , is expressed as a fraction & is typically near zero; e.g., 0.05 or 0.01.
14
Confidence Interval for the Mean (cont’d)
Example:Given sample with:
mean = x = 3.90SD = s = 0.95n = 32
A 90 % CI for the mean = 3.90 + (1.645)(0.95)/= (3.62, 4.17), used the central limit theorem.
Note: A 90 % CI => We can state with 90 % confidence that the population mean is between 3.62 & 4.17. The chance of error in this statement is 10 %
32
15
Testing for a Zero Mean
Difference in processor times of two different implementations of the same algorithms was measured on 7 similar workloads. The differences are:
{1.5, 2.6, -1.8, 1.3, -0.5, 1.7, 2.4}Can we say with 99 % confidence that one implementation is superior to the other
16
Testing for a Zero Mean (cont’d)Sample size = n = 7mean = x = 1.03sample variance = s2 = 2.57sample deviation = s = 1.60CI = 1.03 t x 1.60/ = 1.03 0.605t 100 (1- ) = 99, = 0.01, 1- /2 = 0.995
From Table, the t value at six degrees of freedom is:
t[0.995; 6] = 3.707 & the 99% CI = (-1.21, 3.27).
Since the CI includes zero, we can not say with 99%
confidence that the mean difference is significantly
different from Zero.
7
17
Type I & Type II ErrorsIn testing a NULL, hypothesis, the level of significance is the probability of rejecting a true hypothesis.
HYPOTHESIS
Actually True Actually False
Correct Error (Type II)
Correct Error (Type I)
To Accept
To Reject
DECISION
Note: The letters & denote the probability related to these errors
18
Since q0 = Linear combination of normal variables=> q0 is normal with variance
Variance of errors:
Effects are random variables.Errors ~ N(0,σe) => y ~ N( y.., σe)
Confidence Intervals For Effects
)2/( 22 re
MSEr
SSEe
rs
ijije
)1(2)1(2
12
22
2
19
Confidence Intervals For Effects (cont’d)
Denominator = 22(r - 1) = # of independent terms in SSE=> SSE has 22(r - 1) degrees of freedom.Estimated variance of q0 :
Similarly,
Confidence intervals (CI) for the effects:
CI does not include a zero => significant
)2/( 222
0rss eq
r
Sqqq
e
ABBAsss
22
iqri stq)1(2;2/1 2
20
Example
For Memory-cache study:Standard deviation of errors:
57.375.128102
)1(22 r
SSEes
Standard deviation of effects:
03.112/57.32/ 2 rss eqi
For 90% Confidence : 86.18,95.0 t
21
Example (cont’d)Confidence intervals:
)91.6,08.3(
)41.11,58.7(
)41.23,58.19(
)91.42,08.39(
92.1)03.1)(86.1(
0
AB
B
A
ii
q
q
q
q
No zero crossing=> All effects are significant.
22
Confidence Intervals for Contrasts
Contrast Linear combination with coefficients = 0
Variance of hiqi:
r
hss ie
qh ii 2
222
2
For 100 ( 1 - ) % confidence interval,use .)1(2;2/1 2 r
t
23
u = qA + qB -2qAB
Coefficients = 0,1,1, and -2 => ContrastMean u = 21.5 + 9.5 - 2 x 5 = 21
Variance
Example: Memory-cache study
375.632
62
22
eu
ss
Standard deviation 52.2375.6 ust[0.95;8] = 1.8690% Confidence interval for u :
52.286.121 utsu (16.31, 25.69)
24
Mean response y :y = q0 + qA xA + qB xB + qABxA xB
The standard deviation of the mean of m response:
CI for Predicted Response
^^
21
11ˆ mney effm
ss
neff = Effective deg of freedom
=Total number of runs
1 + Sum of DFs of params used in y
5
22 r
25
CI for Predicted Response (cont’d)
21
225
ˆ:)(rey ssm
100 ( 1 - ) % confidence interval:
myrsty ˆ)1(2;2/1 2ˆ
A single run (m = 1) : 21
21
12
5ˆ
rey ss
Population mean
26
Example: Memory-cache Study
For xA = -1 and xB = -1:•A single confirmation experiment:y1 = q0 - qA - qB + qAB
= 41 - 21.5 - 9.5 + 5 = 15Standard deviation of the prediction:
^
25.4157.31 125
25
ˆ2
1
21
rey ss
Using t[0.95;8]=1.86, the 90% confidence interval is:
)91.22,09.8(25.486.115
27
Example: Memory-cache Study (cont’d)
•Mean response for 5 experiments in future:
80.257.3 51
1251
25
ˆ2
1
21
mrey ss
The 90% confidence interval is:
)29.20,79.9(80.286.115 •Mean response for a large number of experiments in future:
30.257.3 125
25
ˆ2
1
2 rey ss
The 90% confidence interval is:
)28.19,72.10(30.286.115
28
Example: Memory-cache Study (cont’d)
•Current mean response: Not for future.(Use the formula for contrasts):
06.212475.12
2ˆ 2
22
r
hs
yies
90% confidence interval:
)83.18,17.11(06.286.115 Notice: Confidence intervals become narrower.