Assignment 7 - 2): Suggested answer

[maxschlake]

Task 2)

For one of the latent class clusterings computed in question 1 on the
electionwithna-data produce a heatmap as on slide 289 of the course notes. Note
that this requires a different colour choice for the heatmap "entries" than in the
course notes, as the data now have more than 2 categories (it is part of the exercise
to fi nd out how to do this). Comment on the plots. Do you find the clusters
convincing? Why or why not? Is there evidence against local independence?

Answer

Of course our solutions all depend on the numbers of clusters we chose. I found K=3 to be the most convincing answer (also with the background about the US system with 2 parties and some independent candidates). But: The clusters looked somewhat less convincing in the heatmap.
Regarding local independence, I found this much harder to evaluate, as we have 5 possible responses (instead of the example with 2 responses we had in class). I didn't see any clear evidence that it was violated.

electionwithna_m = data.matrix(electionwithna)

# Order the rows of the matrix based on predclass
cluster_ord = order(electionwithna_lca$predclass)
electionwithna_m_ord = electionwithna_m[cluster_ord, ]
col_vector = c(9,7,5,12,14)

# Create a heatmap
jpeg("EX2_heatmap_lca.jpeg", width = 800, height = 800)
heatmap(
  electionwithna_m_ord[order(electionwithna_lca$predclass),],
  Rowv = NA,
  Colv = NA,
  RowSideColors = palette()[electionwithna_lca$predclass]
  [order(electionwithna_lca$predclass)],
  col = col_vector,
  scale = "none",
)
legend('right', legend=c(1,2,3,4,5), fill=col_vector)
dev.off()

[SebastianVeuskens]

Very nice answer, thanks Max!

To make the plot more comprehensible, I wrote

Colv=as.dendrogram(hclust(daisy(t(electionwithna_m), metric="gower")),method="average"))

This orders the columns and lets the plot separate into different distinctive

areas.