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Results of multiple regression for Defect

Summary measures
Multiple R 0.9383
R-Square 0.8803
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?

i) Using non-technical language, state and interpret the standard error of the estimate.

j) Is Rate significant? What does Rate's significance or lack thereof imply about controlling the production quality? In particular, should you be slowing down the production rate as Ole has stated?

k) What action would you take to lower the number of defects? Be specific.

l) What is the expected number of defects when the standard deviation in temperature is 1, the density is 25, the rate is 200, and produced by the morning shift?

_____________________________________________________________________________________________________________________

2. You have tested a new system that supposedly reduces variable costs of production. Because the new system involves additional expenditures, you calculated that you would be willing to use the new system only if variable cost would be less than \$6.27 per unit produced. Based on careful data collection and analysis of the new system, you found that the average variable cost under the new system is \$6.05.

Solution Summary

Results of multiple regression for Defect

Summary measures
Multiple R 0.9383
R-Square 0.8803
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?

i) Using non-technical language, state and interpret the standard error of the estimate.

j) Is Rate significant? What does Rate's significance or lack thereof imply about controlling the production quality? In particular, should you be slowing down the production rate as Ole has stated?

k) What action would you take to lower the number of defects? Be specific.

l) What is the expected number of defects when the standard deviation in temperature is 1, the density is 25, the rate is 200, and produced by the morning shift?

__________________________________________________________________________________________________

2. You have tested a new system that supposedly reduces variable costs of production. Because the new system involves additional expenditures, you calculated that you would be willing to use the new system only if variable cost would be less than \$6.27 per unit produced. Based on careful data collection and analysis of the new system, you found that the average variable cost under the new system is \$6.05.

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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!"
• "Thank you"
• "Thank you very much for your valuable time and assistance!"

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