Question

Thirteen specimens of 90/10 Cu-Ni alloys, each with a specific iron were tested in a corrosion-wheel setup. The wheel was rotated in salt seawater at 30 ft/s for 60 days. The corrosion was measured in weight loss in milligram/square decimeter/day, MDD. The following data were collected.

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X(Fe) Yloss in MDD) 127.6, 124.0, 110.8, 103.9, 101.5, 130.1, 122.0,92.3, 113.1,83.7,128.0,91.4,86.2 0.01, 0.48, 0.71, 0.95, 1.19,0.01, 0.48, 1.44,0.71, 1.96, 0.01, 1.44,1.96

1. Assuming a model, Y= β0+β1X+ε, what are the least squares estimates of β0 and β1? What is the prediction equation?
2. Construct an analysis of variance table and test the hypothesis H0: β1=0 with an α risk of 0.05.
3. What are the confidence limits (α=0.05) for β1?
4. What are the confidence limits (α=0.05) for the true mean value of Y when X=3?
5. What are the confidence limits (α=0.05) for the difference between the true mean value of Y when X1=3 and the true mean value of Y when X2=-2?
6. Are there any indications that a better model should be tried?

(Using R is preferable for ANOVA)

Answer 1

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
0.01	127.6	0.7449018	352.8618	-16.2126
0.48	124	0.1545095	230.5725	-5.96872
0.71	110.8	0.0265941	3.938698	-0.32364
0.95	103.9	0.0059172	24.16101	-0.37811
1.19	101.5	0.1004402	53.51485	-2.31841
0.01	130.1	0.7449018	453.0349	-18.3703
0.48	122	0.1545095	173.8341	-5.18257
1.44	92.3	0.3214018	272.7579	-9.36295
0.71	113.1	0.0265941	18.35793	-0.69872
1.96	83.7	1.1814018	630.7825	-27.2985
0.01	128	0.7449018	368.0495	-16.5578
1.44	91.4	0.3214018	303.2956	-9.87318
1.96	86.2	1.1814018	511.4556	-24.5812

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	11.35	1414.6	5.708877	3396.617	-137.127
mean	0.873076923	108.815385	SSxx	SSyy	SSxy

sample size ,   n =   13          
here, x̅ =   0.873076923       ȳ =   108.8153846  
                  
SSxx =    Σ(x-x̅)² =    5.708876923          
SSxy=   Σ(x-x̅)(y-ȳ) =   -137.1266154          
                  
slope ,    ß1 = SSxy/SSxx =   -24.01989345          
                  
intercept,   ß0 = y̅-ß1* x̄ =   129.7865993          
                  
so, regression line is   Ŷ =   129.7866   +   -24.0199   *x

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what are the least squares estimates of β0 and β1? What is the prediction equation?

β0 = 129.7866

and β1 = -24.0199

Ŷ =   129.7866 -24.0199   *x

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Construct an analysis of variance table and test the hypothesis H0: β1=0 with an α risk of 0.05

SSE=   (Sx*Sy - S²xy)/Sx =    102.85
SSR=   S²xy/Sxx =   3293.77

Anova table
variation	SS	df	MS	F-stat	p-value
regression	3293.77	1	3293.77	352.2737	0.0000
error,	102.85	11	9.35
total	3396.62	12

std error ,Se =    √(SSE/(n-2)) =    3.0578
slope hypothesis test               tail=   2
Ho:   ß1=   0          
H1:   ß1╪   0          
n=   13              
alpha=   0.05              
estimated std error of slope =Se(ß1) =                s/√Sxx =    1.2798
                  
t stat =    ß1 /Se(ß1) =        -18.76895613         
                  
p-value =    0.0000              
decision :    p-value<α , reject Ho  

so, slope is not equal to zero

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What are the confidence limits (α=0.05) for β1

confidence interval for slope              
t critical value=   t α/2 =    2.20098516  

estimated std error of slope =Se(ß1) =                s/√Sxx =    1.2798    
margin of error ,E=       tα/2*std error =        2.8167485
lower confidence limit =        ß̂1-E =   -26.83664
upper confidence limit=       ß̂1+E =   -21.2031  
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What are the confidence limits (α=0.05) for the true mean value of Y when X=3?

X Value=   3
Confidence Level=   95%
  
Sample Size , n=   13
Degrees of Freedom,df=n-2 =   11
critical t Value=tα/2 =   2.201
X̅ =Σx/n = 0.873
Σ(x-x̅)² =Sxx   5.709
Standard Error of the Estimate,Se=   3.058 (calculated earlier)
h Statistic = (1/n+(X-X̅)^2/Sxx) =    0.869
Predicted Y (YHat) = Ŷ =   129.7866 -24.0199   *3 = 57.727
  
For Average Y  
margin of error,E=t*Se*√(h stat) =    6.2751
Confidence Lower Limit=Ŷ -E =   51.4519
Confidence Upper Limit=Ŷ +E =   64.0020

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