Worksheets/Week4
Revision as of 19:08, 11 March 2019 by Kevin Dunn (talk | contribs) (Created page with "=== Part 1 === 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 secon...")
Part 1
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],
# or 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 2
Continuing with the case above:
# S = Free shipping if order amount is €30 or more [-1], or if
# order amount is over €50 [+1]. Notice that a mistake was made
# with the last experiment: order minimum for free shipping was €60 [+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
model.sales.mistake <- lm(y ~ S*P)
summary(model.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(model.sales.mistake)
Part 3
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? Look at the "design" variable also.
# The response: stability [units=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
model.stability <- lm(y ~ A*B*C)
summary(model.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(model.stability)
contourPlot(model.stability, 'A', 'B')
contourPlot(model.stability, 'A', 'C')