In: Statistics and Probability
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 32 parts.
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 |
1.5 |
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 |
PLEASE answer only the STEPWISE part of the question
except utilize stepwise regression to identify a model.]
Answer:
Given That,
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.
(a) 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?
Regression Equation
Pitch = -0.0302 + 0.000029 Temp + 0.002318 SoakTime - 0.00303 SoakPct + 0.00848 DiffTime - 0.00236 DiffPct
Coefficients
Analysis of Variance (ANOVA): provides the analysis of the variance in the model, as the name suggests. Regression statistics: provide numerical information on the variation and how well the model explains the variation for the given data/observations.
From above ANOVA table F-statistic = 158.9236 >Significance F
= 1.25397E-18 so we reject the null hypothesis and conclude that
their is a significant relationship exists between dependent and
independent variables.
(b) Check for multicollinearity. Explain your
findings
.Multicollinearty Assumption
Coefficients
from above table we have VIF that is Variation Inflation Factor which is used detect multicollinearity
If VIF is greater than 10 then we say that their is serious multicollinearity exist we need to remove it.
But for our model all the VIF < 10 so no multicollinearty exist.
c) 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.
Coefficients
this is the output table of t-test which used to check the significance of independent variable in the regreesion model.
From above table
For Temp, SockPct, DiffPct P-value > alpha=0.05 so we accept the null hypothesis and can conclude that these variables are not significant in the model
for SoakTime & DiffTime P-value < alpha=0.05 so we reject the null hypothesis can conclude that these variables are significant in the model
(d) Prepare a normal probability plot of the residuals and check the adequacy of the model
From above normality plot maximum points are lies near to straight line so data follows normality assumption.
NOTE::
I HOPE YOUR HAPPY WITH MY ANSWER....***PLEASE SUPPORT ME WITH YOUR RATING...
***PLEASE GIVE ME "LIKE"...ITS VERY IMPORTANT FOR ME NOW....PLEASE SUPPORT ME ....THANK YOU