MECH5750 DoE Midterm Exam

MECH5750 DoE Midterm Exam. Fall 2016, due Midnight Sunday October 16^{th}. Name _________________

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Write My Essay For Me**QUIZ C to be answered by student ID last digit 3 or 5. Part II**

**Please show background Calculations. Think of this problem as a combination of treatments and DoE **

- For the experiment to Minimize the yield of popcorn = # of un-popped Kernels (2 factors and one interaction)

Brand Ratio Brand*Ratio Yield (un-popped Kernels)

Test Popcorn of Oil Interaction Replication

__Number (1) (2) (12) 1 2 __ S__Y__2__ Avg. __S__y (____Sy)2__ __V__

1 – – ___ 6 4 52 5 10 100 2

2 + – ___ 5 4 41 4.5 9 81 0.5

3 – + ___ 8 64 8 8 64

4 + + ____ 3 __9 3 3 9__

Total 166 5.13 30 254

- Fill in the blanks for the levels of the interaction, and calculate the correction Factor CF = _______

- Complete the anova treatment Table and determine Experiment significance

Please Show Calculations. Use Assignment 3 problem 3.6 as a guide for the ANOVA. Between = Regression and Within = Error

Effect | DoF | SS | MS | F ratio | Significant
95% |

Between | Yes or No | ||||

Within | |||||

Total |

- Calculate the main
and their interaction.__effects__**Make sure to use the average of each experiment line.**

Calculate 1 = __________ 2 = ________ 12 = _______

- Calculate Pooled estimate of Variance _______ using two methods. Indicate methods (they should be equal).

Method 1 _________________ ; Method 2 __________________ (Hint see assignment 3.4.e solution)

- What levels (+ or -) would be best for lowest un-popped kernels (Yield) values? Hint: one answer is intuitive

With Interaction: Factor 1_____ factor 2 ______ Yield Value______

Without Interaction: Factor 1_____factor 2 ______ Yield Value______

- Based on CI (factors), which factors are significant: 1_______, 2________12__________

- Calculate the average, standard error (s/√n) and 95% confidence interval for the following:

Hint: several CI (+/-) are the same and one answer is intuitive.

** **

- Total Experiment: Average ___
__5.13______+/- ________; - Treatment 1 (Experiment Line 1) only (6, 4)

Experiment line 1 Average _____5.0______+/ ______ (when considering line 1 only and ignoring experiment)

Experiment line 1 Average _____5.0______+/ ________ (when considering total experiment)

- Treatment 3 (Experiment Line 3) only (8)

Experiment line 3 Average _____8.0______+/ ________ (when considering line 3 only and ignoring experiment)

Experiment line 3 Average _____8.0______+/ ________ (when considering total experiment

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