Difference between revisions of "Worksheets/Week3"
Kevin Dunn (talk | contribs) (→Part 2) |
Kevin Dunn (talk | contribs) (→Part 3) |
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# Linear model using S, P and S*P to predict the response | # Linear model using S, P and S*P to predict the response | ||
model.sales <- lm(y ~ S*P) | |||
summary( | summary(model.sales) | ||
# Uncomment this line if you run the code in RStudio | # Uncomment this line if you run the code in RStudio | ||
Line 92: | Line 92: | ||
source('https://yint.org/contourPlot.R') | source('https://yint.org/contourPlot.R') | ||
contourPlot( | contourPlot(model.sales) | ||
interaction.plot(S, P, y) | interaction.plot(S, P, y) | ||
interaction.plot(P, S, y) | interaction.plot(P, S, y) |
Revision as of 08:52, 11 March 2019
Part 1
A factorial experiment was run to investigate the settings that minimize the production of an unwanted side product. The two factors being investigated are called A and B for simplicity.
# A = additive at 20mL and 30mL for low and high levels
A <- c(-1, +1, -1, +1)
# B = without (-) or with (+) baffles
B <- c(-1, -1, +1, +1)
# Response y is the amount of side product formed, y [grams]
y <- c(89, 268, 179, 448)
# Fit a linear model
model.siderxn <- lm(y ~ A + B + A*B)
summary(model.siderxn)
# Uncomment this line if you run the code in RStudio
#library(pid)
# Comment this line if you run this code in RStudio
source('https://yint.org/contourPlot.R')
# See how the two factors affect the response:
contourPlot(model.siderxn)
interaction.plot(A, B, y)
interaction.plot(B, A, y)
# Make a prediction with this model:
xA = -1
xB = -1
y.hat <- predict(model.siderxn, data.frame(A = xA, B = xB))
paste0('Predicted value is: ', y.hat, ' grams of side product.')
Part 2
Continuing from above, with 2 extra experimental points:
# A = additive at 20mL and 30mL for low and high levels
A <- c(-1, +1, -1, +1, 0, 0)
# B = without (-) or with (+) baffles
B <- c(-1, -1, +1, +1, -1, +1)
# Response y is the amount of side product formed, y [grams]
y <- c(89, 268, 179, 448, 186, 290)
model.siderxn.cp <- lm(y ~ A + B + A*B)
summary(model.siderxn.cp)
# Uncomment this line if you run the code in RStudio
#library(pid)
# Comment this line if you run this code in RStudio
source('https://yint.org/contourPlot.R')
contourPlot(model.siderxn.cp)
Part 3
Your family runs a small business selling products online. The first factor of interest is whether to provide free shipping over €30 or over €50. The second factor is whether or not the purchaser must first create a profile before completing the transaction. The purchaser can still complete their transaction without creating a profile. Below are the data collected, showing the 8 experiments.
# S = Free shipping if order amount is €30 or more [-1],
# of if order amount is over €50 [+1]
S <- c(-1, +1, -1, +1, -1, +1, -1, +1)
# Does the purchaser need to create a profile first [+1] or not [-1]?
P <- c(-1, -1, +1, +1, -1, -1, +1, +1)
# Response: daily sales amount
y <- c(348, 359, 327, 243, 356, 363, 296, 257)
# Linear model using S, P and S*P to predict the response
model.sales <- lm(y ~ S*P)
summary(model.sales)
# Uncomment this line if you run the code in RStudio
#library(pid)
# Comment this line if you run this code in RStudio
source('https://yint.org/contourPlot.R')
contourPlot(model.sales)
interaction.plot(S, P, y)
interaction.plot(P, S, y)
Part 4
Continuing with the case above:
# S = Free shipping if order amount is €30 or more [-1],
# of if order amount is over €50 [+1]
S <- c(-1, +1, -1, +1, -1, +1, -1, +2)
# Does the purchaser need to create a profile first [+1] or not [-1]?
P <- c(-1, -1, +1, +1, -1, -1, +1, +1)
# Response: daily sales amount
y <- c(348, 359, 327, 243, 356, 363, 296, 220)
# Linear model using S, P and S*P to predict the response
mod.sales.mistake <- lm(y ~ S*P)
summary(mod.sales.mistake)
# Uncomment this line if you run the code in RStudio
#library(pid)
# Comment this line if you run this code in RStudio
source('https://yint.org/contourPlot.R')
contourPlot(mod.sales.mistake)
Part 5
Your group is developing a new product, but have been struggling to get the product’s stability, measured in days, to the level required. You are aiming for a stability value of 50 days or more.
# Define the 3 factors. This code is a template that you
# can easily extend and reuse for full factorial designs:
base <- c(-1, +1)
design <- expand.grid(A=base, B=base, C=base)
A <- design$A
B <- design$B
C <- design$C
# Type "A", and "B" and "C" at the command prompt
# to verify what these letters contain. Are they in
# standard order?
# The response: stability, in number of days.
y <- c(40, 27, 35, 21, 41, 27, 31, 20)
# Linear model to predict stability from
# A: enzyme strength: -1 == 20%; +1 == 30%
# B: feed concentration: -1 == 5%; +1 == 15%
# C: mixer type: -1 = R mixer; +1 = W mixer
mod.stability <- lm(y ~ A*B*C)
summary(mod.stability)
# Uncomment this line if you run the code in RStudio
#library(pid)
# Comment these 2 lines if you run this code in RStudio
source('https://yint.org/paretoPlot.R')
source('https://yint.org/contourPlot.R')
paretoPlot(mod.stability)
contourPlot(mod.stability, 'A', 'B')
contourPlot(mod.stability, 'A', 'C')