Question

it is a different table so please do the calculation again

Heat treating is often used to carburize metal parts, such as gears. The thickness of the carburized layer is considered a crucial feature of the gear and contributes to the overall reliability of the part. Because of the critical nature of this feature, two different lab tests are performed on each furnace load. One test is run on a sample pin that accompanies each load. The other test is a destructive test, where an actual part is cross-sectioned. This test involves running a carbon analysis on the surface of both the gear pitch (top of the gear tooth) and the gear root (between the gear teeth). Table 12-6 shows the results of the pitch carbon analysis test for

Temp	SoakTime	SoakPct	DiffTime	DiffPct	Pitch
1650	0.58	1.1	0.25	0.9	0.013
1650	0.66	1.1	0.33	0.9	0.016
1650	0.66	1.1	0.33	0.9	0.015
1650	0.66	1.1	0.33	0.95	0.016
1600	0.66	1.15	0.33	1	0.015
1600	0.66	1.15	0.33	1	0.016
1650	1	1.1	0.5	0.8	0.014
1650	1.17	1.1	0.58	0.8	0.021
1650	1.17	1.1	0.58	0.8	0.018
1650	1.17	1.1	0.58	0.8	0.019
1650	1.17	1.1	0.58	0.9	0.021
1650	1.17	1.1	0.58	0.9	0.019
1650	1.17	1.15	0.58	0.9	0.021
1650	1.2	1.15	1.1	0.8	0.025
1650	2	1.15	1	0.8	0.025
1650	2	1.1	1.1	0.8	0.026
1650	2.2	1.1	1.1	0.8	0.024
1650	2.2	1.1	1.1	0.8	0.025
1650	2.2	1.15	1.1	0.8	0.024
1650	2.2	1.1	1.1	0.9	0.025
1650	2.2	1.1	1.1	0.9	0.027
1650	2.2	1.1	1.5	0.9	0.026
1650	3	1.15	1.5	0.8	0.029
1650	3	1.1	1.5	0.7	0.03
1650	3	1.1	1.5	0.75	0.028
1650	3	1.15	1.66	0.85	0.032
1650	3.33	1.1	1.5	0.8	0.033
1700	4	1.1	1.5	0.7	0.039
1650	4	1.1	1.5	0.7	0.04
1650	4	1.15	1.5	0.85	0.035
1700	12.5	1	1.5	0.7	0.056
1700	18.5	1	1.5	0.7	0.068

1. Fit a regression model using all five regressors. Write out the equation. Summarize this analysis by providing the equation, ?², and list the standard errors.    What does the analysis of variance indicate?
2. Check for multicollinearity. Explain your findings.
3. Describe what you notice about the p values for the regressors. Construct a t-test on each regression coefficient. What can you conclude about the variables in this model? Use an alpha = 0.05.
4. Prepare a normal probability plot of the residuals and check the adequacy of the model.
5. Repeat steps (a)-(d) except utilize stepwise regression to identify a model.

Answer 1

(a) The regression equation is:

Pitch = -0.0784 + 0.00004392*Temp + 0.0025*SoakTime + 0.0183*SoakPct + 0.0078*DiffTime - 0.0031*DiffPct

Source	SS	df	MS	F	p-value
Regression	0.00419226	5	0.00083845	162.43	9.54E-19
Residual	0.00013421	26	0.00000516
Total	0.00432647	31
Regression output				confidence interval
variables	coefficients	std. error	t (df=26)	p-value	95% lower	95% upper
Intercept	-0.0784
Temp	0.00004392	0.00003627	1.211	.2369	-0.00003064	0.00011848
SoakTime	0.0025	0.00020822	11.787	6.22E-12	0.0020	0.0029
SoakPct	0.0183	0.0201	0.911	.3707	-0.0231	0.0597
DiffTime	0.0078	0.0013	5.770	4.46E-06	0.0050	0.0106
DiffPct	-0.0031	0.0081	-0.389	.7003	-0.0197	0.0134

The analysis of variance indicates that the model is significant.

(b)

variables	VIF
Intercept
Temp	3.167
SoakTime	3.346
SoakPct	3.186
DiffTime	2.505
DiffPct	2.748

There is no multicollinearity.

(c) The hypothesis being tested is:

H0: β1 = 0

H1: β1 ≠ 0

The p-value is 0.2369.

Since the p-value (0.2369) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Therefore, we cannot conclude that β1 is significant.

The hypothesis being tested is:

H0: β2 = 0

H1: β2 ≠ 0

The p-value is 0.0000.

Since the p-value (0.0000) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that β2 is significant.

The hypothesis being tested is:

H0: β3 = 0

H1: β3 ≠ 0

The p-value is 0.3707.

Since the p-value (0.3707) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Therefore, we cannot conclude that β3 is significant.

The hypothesis being tested is:

H0: β4 = 0

H1: β4 ≠ 0

The p-value is 0.0000.

Since the p-value (0.0000) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that β4 is significant.

The hypothesis being tested is:

H0: β5 = 0

H1: β5 ≠ 0

The p-value is 0.7003.

Since the p-value (0.7003) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Therefore, we cannot conclude that β5 is significant.

(d) The normal probability plot of the residuals is:

The model is adequate.

(e) The regression equation is:

Pitch = 0.0109 + 0.0025*SoakTime + 0.0088*DiffTime

	R²	0.966
	Adjusted R²	0.964
	R	0.983
	Std. Error	0.002
	n	32
	k	2
	Dep. Var.	Pitch
ANOVA table
Source	SS	df	MS	F	p-value
Regression	0.00417950	2	0.00208975	412.35	5.02E-22
Residual	0.00014697	29	0.00000507
Total	0.00432647	31
Regression output				confidence interval
variables	coefficients	std. error	t (df=29)	p-value	95% lower	95% upper
Intercept	0.0109
SoakTime	0.0025	0.00013299	18.516	1.31E-17	0.0022	0.0027
DiffTime	0.0088	0.0010	8.811	1.07E-09	0.0067	0.0108

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