Worksheets/Week4

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) # Uncomment and re-run these lines to understand # the interactions #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) # Uncomment these two lines later to inspect the # contour plots #contourPlot(model.stability, 'A', 'B') #contourPlot(model.stability, 'A', 'C') # Make a prediction with this model: xA = 0 xB = 0 xC = 0 y.hat <- predict(model.stability, data.frame(A = xA, B = xB, C = xC)) paste0('Predicted stability is: ', y.hat, ' days.')