**Unformatted text preview: **-0.25 0.00 0.25 0.50 b) The mean values are: 3.886, 3.8611, 3.76 (barley, b+l, lupins)
From the ANOVA the estimate of σ can be obtained
Source DF
SS
MS
F
P
C4
2
0.235 0.118 0.72 0.489
Error
76 12.364 0.163
Total
78 12.599
S = 0.4033
R-Sq = 1.87%
R-Sq(adj) = 0.00%
The minimum sample size could be used to calculate the standard error of
a sample mean σˆ X = MS E
=
b 0.163
= 0.081
25 The graph would not show any differences between the diets.
13-24. μ = 55 , τ1 = -5, τ2 = 5, τ3 = -5, τ4 = 5.
n(100)
Φ2 =
= n,
a −1 = 3
4(25)
Various choices for n yield:
n
4
5 Φ2
4
5 a (n − 1) = 4(n − 1) Φ
2
2.24 a(n-1)
12
16 Power=1-β
0.80
0.90 Therefore, n = 5 is needed.
13-25 μ = 188 , τ1 = -13, τ2 = 2, τ3 = -28, τ4 = 12, τ5 = 27.
n(1830)
Φ2 =
= n,
a −1 = 4
a (n − 1) = 5(n − 1)
5(100)
Various choices for n yield:
n
2
3 Φ2
7.32
10.98 Φ
2.7
3.13 a(n-1)
5
10 Power=1-β
0.55
0.95 Therefore, n = 3 is needed.
Section 13-3
13-26 a)
Analysis of Variance for UNIFORMITY
Source
DF
SS
MS
WAFERPOS
3
16.220
5.407
Error
8
5.217
0.652
Total
11
21.437 F
8.29 P
0.008 Reject H0, and conclude that there are significant differences among wafer positions. MSTreatments − MS E 5.407 − 0.652
=
= 1.585
n
3
2
c) σˆ = MS E = 0.652
b) σˆ τ2 = d) Greater variability at wafer position 1. There is some slight curvature in the normal probability plot. 13-19 Residuals Versus WAFERPOS Residuals Versus the Fitted Values (response is UNIFORMI) (response is UNIFORMI) 1 Residual Residual 1 0 -1 0 -1 1 2 3 4 1 2 3 WAFERPOS 4 Fitted Value Normal Probability Plot of the Residuals
(response is uniformi)
2 Normal Score 1 0 -1 -2
-1 0 1 Residual 13-27 a) Analysis of Variance for OUTPUT
Source
DF
SS
MS
LOOM
4
0.3416
0.0854
Error
20
0.2960
0.0148
Total
24
0.6376 F
5.77 P
0.003 Reject H0, there are significant differences among the looms. MSTreatments − MS E 0.0854 − 0.0148
=
= 0.01412
n
5
2
c) σˆ = MS E = 0.0148
b) σˆ τ2 = d) Residuals are acceptable
Residuals Versus the Fitted Values Residuals Versus LOOM (response is OUTPUT) 0.2 0.2 0.1 0.1 Residual Residual (response is OUTPUT) 0.0 0.0 -0.1 -0.1 -0.2 -0.2
1 2 3 4 3.8 5 3.9 4.0 Fitted Value LOOM 13-20 4.1 Normal Probability Plot of the Residuals
(response is OUTPUT)
2 Normal Score 1 0 -1 -2
-0.2 -0.1 0.0 0.1 0.2 Residual 13-28. a) Yes, the different batches of raw material significantly affect mean yield at α = 0.01 because pvalue is small.
Source
Batch
Error
Total DF
5
24
29 SS
56358
58830
115188 MS
11272
2451 F
4.60 P
0.004 b) Variability between batches MS Treatments − MS E 11272 − 2451
=
= 1764.2
n
5
2
c) Variability within batches σˆ = MSE = 2451 σˆτ2 = d) The normal probability plot and the residual plots show that the model assumptions are
reasonable. 13-21 Residual Plots for Yield
Normal Probability Plot of the Residuals Residuals Versus the Fitted Values 99 100
50 Residual Percent 90
50 0
-50 10
1 -100 -50 0
Residual 50 -100 100 100 6 50 4 13-29. -75 -50 -25 0
25
Residual 50 75 1600 0
-50 2
0 1550
Fitted Value Residuals Versus the Order of the Data 8 Residual Frequency Histogram of the Residuals 1500 -100 100 2 4 6 a) Analysis of Variance for BRIGHTNENESS
Source
DF
SS
MS
F
CHEMICAL
3
54.0
18.0
0.75
Error
16
384.0
24.0
Total
19
438.0 8 10 12 14 16 18 20 22 24 26 28 30 Observation Order P
0.538 Do not reject H0, there is no significant difference among the chemical types. 18.0 − 24.0
= −1 . 2
5 b) σˆ τ2 = c) σˆ 2 = 24.0 set equal to 0 d) Variability is smaller in chemical 4. There is some curvature in the normal probability plot.
Residuals Versus CHEMICAL Residuals Versus the Fitted Values (response is BRIGHTNE) (response is BRIGHTNE) 10 5 5 Residual Residual 10 0 0 -5 -5 1 2 3 78 4 79 80 Fitted Value CHEMICAL 13-22 81 82 Normal Probability Plot of the Residuals
(response is BRIGHTNE)
2 Normal Score 1 0 -1 -2
-5 0 5 10 Residual 13-30 a) 2
2
σˆ total
= σˆ position
+ σˆ 2 = 2.237 b) 2
σˆ position
= 0.709
2
σˆ total c) It could be reduced to 0.6522. This is a reduction of approximately 71%. 13-31.
a) Instead of testing the hypothesis that the individual treatment effects are zero, we are testing
whether there is variability in protein content between all diets. H 0 : σ τ2 = 0
H 1 : σ τ2 ≠ 0
b) The statistical model is ⎧i = 1,2,..., a
y = μ + τ i + ε ij ⎨
⎩ j = 1,2,..., n
ε i ~ N (0, σ 2 ) and τ i ~ N (0, σ τ2 )
c)
The last TWO observations were omitted from two diets to generate equal sample sizes with n = 25. ANOVA: Protein versus DietType
Analysis of Variance for Protein
Source
DietType
Error
Total DF
2
72
74 S = 0.405122 SS
0.2689
11.8169
12.0858 MS
0.1345
0.1641 R-Sq = 2.23% F
0.82 P
0.445 R-Sq(adj) = 0.00% σ 2 = MS E = 0.1641
MS tr − MS E 0.1345 − 0.1641
σ τ2 =
=
= −0.001184
n
25 13-23 ...

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- Winter '15
- Olga
- Statistics