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    The Increase in ESS by Applying Ward's Method to Four Clusters

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    Use Ward's method to cluster the four items whose measurements on a single variable X are given in the following table. See attached

    (a) Initially, each item is a cluster and we have the cluster

    Show that ESS=0, as it must

    (b) If we join cluster {1} and {2}, the new cluster {12} jas
    see attached

    (c) Complete the last two algamation steps and construct the dendrogram showing the value of ESS which the mergers take place.

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    https://brainmass.com/statistics/cluster-analysis/increase-ess-applying-ward-method-clusters-594110

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    A. Ward's method is actually just the sum of squared difference. Since in this part, there is only one cluster, ESS1=(2-2)2=0, ESS2=(1-1)2=0, ESS3=(5-5)2=0 and ESS4=(8-8)2=0.

    B. The question already solves for first cluster. So I work out starting from second cluster.
    For second cluster, we join clusters {1} and {3}, then the mean=(2+5)/2=3.5. So ESS2=(2-3.5)2+(5-3.5)2=4.5.
    And ESS associated with the grouping is 4.5+0+0=4.5.
    Hence, the increase in ESS is 4.5-0=4.5. We fill in the 4.5 in the table below.

    For ...

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    The solution gives detailed steps on finding the increase in ESS by applying Ward's Method to four clusters.

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