Problem # 2, (Use hand calculator) Thirteen specimens of 90/10 Cu-Ni alloys, eac

Question

Problem # 2, (Use hand calculator) 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 sault seawater at 30 ft/s for 60 days/ The corrosion was measured in weight loss in miligram/square decimeter/day, MDD. The following data were collected. 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 X (Fe) Y(loss in MDD) 127.6,124.0, 110.8, 103.9, 101.5, 130.1, 1 22.0, 92.3, 113.1, 83.7, 128.0,91.4,862 1. Assuming a model, Y- + .x + e, what are the least squares estimates of As and B,? What is the prediction equation? 2. Constructthe analysis of variance table and test the hypothesis H:.-0 with an risk of 0.05 3. What are the confidence limits (a- 0.05) for ? 4, what are the confldence limits (-0.05) for the true mean value of Y when X-3? 5, what are the confidence limits (-0.05) for the difference between the true mean value of Y when X.-3 and the true mean value of Y when X:"-2? 6. Are there any indications that a better model should be tried?

Explanation / Answer

Back-up Theory

Given X = Fe and Y = loss in MDD, the linear regression model is:

Y = 0 + 1X + , where is the error term, which is assumed to be Normally distributed with mean 0 and variance 2.

Let (xi, yi) be a pair of sample observation on (X, Y), i= 1, 2, …., n, where n = sample size.

Then, Mean X = Xbar = (1/n)sum of xi over I = 1, 2, …., n; ……………….(1)

Sxx = sum of (xi – Xbar)2 over i = 1, 2, …., n ………………………………..(2)

Similarly, Mean Y = Ybar =(1/n)sum of yi over i= 1, 2, …., n;…………….(4)

Syy = sum of (yi – Ybar)2 over i = 1, 2, …., n ………………………………………………(5)

Sxy = sum of {(xi – Xbar)(yi – Ybar)} over i = 1, 2, …., n………(7)

Correlation Coefficient of X and Y = rXY= Sxy/sq.rt(Sxx.Syy). …………………………(8)

Least square estimates of 0 and 1 are:

0cap = Ybar – 1capXbar..………………………………………………….(9)

1cap = Sxy/Sxx and aYX = Ybar – bYX.Xbar..…………………….(9)

Prediction Equation (Estimated Regression of Y on X) is given by: Ycap = 0cap + 1capX ….()

Estimate of 2 is given by s2 = (Syy – b2Sxx)/(n - 2).

Standard Error of 1cap is sb, where sb2 = s2/Sxx

Standard Error of 0cap is sa, where sa2 = sb2{(sum of xi2 over i = 1, 2, …., n)/n}

Standard Error of yicap = s[(1/n) + {(xi – Xbar)2/Sxx}]

100(1 - )% Confidence Interval (CI) for 1 = 1cap ± SE(1cap) x tn – 2,/2

100(1 - )% Confidence Interval (CI) for 0 = 0cap ± SE(0cap)xtn – 2,/2

100(1 - )% Confidence Interval (CI) for ycap at x = x0 is                                                                                              (0cap + 1capx0) ± tn – 2,/2 x s[(1/n) + {(x0 – Xbar)2/Sxx}]

Calculations:

Summary of Excel calculations is given below:

n

13

xbar

1

ybar

108.815

Sxx

5.709

Syy

3396.617

Sxy

-137.127

1cap

-24.02

0cap

129.79

s^2

9.3500

sb^2

1.638

sa^2

1.9677

s

3.058

sb

1.28

sa

1.403

Part (a)

1cap

-24.02 ANSWER 1

0cap

129.79 ANSWER2

Prediction Equation: Ycap = 129.79 – 24.02x ANSWER 3

Part (b)

ANOVA

Source

DF

SS

MS

Fcal

Fcrit

Regression

1

3293.766

3293.77

352.27

4.84

Error

11

102.850

9.35

Total

12

3396.617

Since Facl > Fcrit, we conclude that 1 is not zero. ANSWER

n

13

xbar

1

ybar

108.815

Sxx

5.709

Syy

3396.617

Sxy

-137.127

1cap

-24.02

0cap

129.79

s^2

9.3500

sb^2

1.638

sa^2

1.9677

s

3.058

sb

1.28

sa

1.403

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