Multiple Regression
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Solution Summary
The results of multiple regression for Defect are determined. An alternative hypothesis is analyzed.
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See attachment please!
Results of multiple regression for Defect
Summary measures
Multiple R 0.9383
R-Square 0.8803
Adj R-Square 0.8612
StErr of Est 7.2326
ANOVA Table
Source df SS MS F p-value
Explained 4 9621.5292 2405.3823 45.9828 0.0000
Unexplained 25 1307.7628 52.3105
Regression coefficients
Coefficient Std Err t-value p-value
Constant 1.0312 73.8985 0.0140 0.9890
Temperature 17.4189 9.5635 1.8214 0.0805
Density -1.5741 1.7446 -0.9022 0.3755
Rate 0.1184 0.1330 0.8904 0.3817
Morning -0.9186 3.0655 -0.2997 0.7669
Use the output above to answer parts (h) through (l).
h) Returning to the p-value for the indicator variable Morning, what conclusion can you draw?
From the above table, we know that the p-value for the indicator variable Morning is 0.7669 and the corresponding t-value=-0.2997. So, we can roughly know that the indicator variable Morning as a predictor variable has a small impact on the number of defects.
i) Using non-technical language, state and interpret the standard error of the estimate.
From the above table, we know that the standard error of the estimate is 7.2326. Also, we know the regression equation is
Number of defects Y=1.0312+ 17.4189* TemperatureX1 -1.5741* DensityX2
+0.1184 *RateX3-0.9186*MorningX4
So, we can use this equation to predict the number of defects Y*. If we use 95% confidence level, then the real value for the number of defects can be away from Y* by , ie., (Y*-1.96*7.2326, Y*+1.96*7.2326)
j) Is Rate significant? What does Rate's significance or lack thereof imply about ...
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
Recent Feedback
- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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