Question

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 Cycles

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+γ2∙x2 .

iii. A simple linear regression model with a logarithm transformation on PSA: lny=δ0+δ1∙ln⁡(x) .

1. 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.
2. 2. 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.
   3. 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.
3. 4. Briefly discuss the advantages and disadvantages of each of the three models.

Answer 1

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=2.76822x10-6<0.05 so overall model is significant.

b. (ii) Since p-value corresponding x2=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.