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In: Statistics and Probability

An experiment is performed to study the fatigue performance of a high strength alloy. The number...

An experiment is performed to study the fatigue performance of a high strength alloy. The number of cycles to crack initiation is measured for twenty specimens over a range of applied pseudo-stress amplitude (PSA) levels. Use the data in the table provided to fit the following three regression models with y = Cycles and x = PSA (note the natural log transform of y for all models):

PSA (x) Cycles (y)

80, 97379

80, 340084

80, 246163

80, 239348

100, 34346

100, 23834

100, 70423

100, 51851

120, 9139

120, 9487

120, 8094

120, 17956

140, 5640

140, 3338

140, 6170

140, 5608

160,    1723

160, 3525

160, 2655

160, 1732

i. A simple linear regression model: lny=β0+β1∙x .

ii. A quadratic polynomial model: lny=γ0+γ1∙x+γ2x2 .

iii. A simple linear regression model with a logarithm transformation on PSA: lny=δ0+δ1ln⁡(x) .

    1. For model (i.), what hypothesis being tested by the ANOVA F-test (in terms of the coefficients of the model)? Interpret the conclusion of this test in the context of the engineering problem.
    1. For model (ii.), what hypothesis being tested by the ANOVA F-test (in terms of the coefficients of the model)? Interpret the conclusion of this test in the context of the engineering problem.
    2. What is the p-value for testing the significance of the quadratic term in model (ii.) (Ho: γ2=0)? Interpret the conclusion of this test in the context of the engineering problem.
    1. Briefly discuss the advantages and disadvantages of each of the three models.

Solutions

Expert Solution

Solution:-

Given that

An experiment is performed to study the fatigue performance of a high strength alloy. The number of cycles to crack initiation is measured for twenty specimens over a range of applied pseudo-stress amplitude (PSA) levels. Use the data in the table provided to fit the following three regression models with y = Cycles and x = PSA (note the natural log transform of y for all models):

a.(i).

SUMMARY OUTPUT

Regression Statistics
Multiple R 0.732833479
R square 0.537044908
Adjusted R square 0.511325181
Standard Error 68732.05708
Observations 20
ANOVA
df SS MS F Significance F
Regression 1 98642240334 9.86E+10 20.88066 0.000237653
Residual 18 85033722059 4.72E+09
Total 19 1.83676E+11
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 356881.15 66991.72252 5.327242 4.6E-05 216136.7636 497625.5364
x -2482.97 543.3746216 -4..56954 0.000238 -3624.557719 -1341.382281

Since p-value of F-test = 0.000237653 < 0.05

so Overall model is significant.

ii.

SUMMARY OUTPUT

Regression Statistics
Multiple R 0.8821044
R square 0.778108173
Adjusted R square 0.752003252
Standard Error 48963.48877
Observations 20
ANOVA
df SS MS F Significance F
Regression 2 1.4292E+11 7.15E+10 29.80695 2.76822E-06
Residual 17 40756194946 2.4E+09
Total 19 1.83676E+11
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 1312922.293 227524.1028 5.770476 2.26E-05 832888.3964 1792956.189
x -19354.28429 3944.850497 -4.90621 0.000133 -27677.19132 -11031.37725
x^2 70.29714286 16.35755352 4.297534 0.000488 35.78572163 104.8085641

Since p-value of F test =

so overall model is significant

b. (iii).

Since p-value corresponding = 0.000488 < 0.05

so is significantly different from zero.

c.

Output of Model (iii)

SUMMARY OUTPUT

Regression Statistics
Multiple R 0.973049464
R square 0.946825259
Adjusted R square 0.943871107
Standard Error 0.398494771
Observations 20
ANOVA
df SS MS F Significance F
Regression 1 50.89583163 50.89583 320.5066 6.46066E-13
Residual 18 2.85836549 0.158798
Total 19 53.75419712
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept 40.67886966 1.733510441 23.46618 6E-15 37.03689936 44.32083995
log (x) -6.513561728 0.363831296 -17.9027 6.46E-13 -7.277942916 -5.74918054

Coefficient of determination for model (i) = 0.537044908

Coefficient of determination for model (ii) = 0.778108173

Coefficient of determination for model (iii) = 0.946825259

Hence model (iii) is the best model since it explains 94.68% of total variation in y and this is maximum among three.

However three models are significant.

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