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Solution by Steps

step 1

To construct a regression tree using recursive binary splitting, we first need to determine the best split. Since the statistician always splits on

x_2

in case of a tie, we will start by evaluating splits on

x_2

step 2

Calculate the mean squared error (MSE) for each possible split on

x_2

. The possible splits are at

x_2 = 4

,

x_2 = 6

,

x_2 = 8

, and

x_2 = 10

step 3

For

x_2 = 4

: - Left node:

y = 5

- Right node:

y = \{30, 6, 10, 8\}

- MSE =

\frac{1}{5} \left( (5-11.8)^2 + (30-11.8)^2 + (6-11.8)^2 + (10-11.8)^2 + (8-11.8)^2 \right)

step 4

For

x_2 = 6

: - Left node:

y = \{6, 8, 5\}

- Right node:

y = \{30, 10\}

- MSE =

\frac{1}{5} \left( (6-6.33)^2 + (8-6.33)^2 + (5-6.33)^2 + (30-20)^2 + (10-20)^2 \right)

step 5

For

x_2 = 8

: - Left node:

y = \{6, 8, 5\}

- Right node:

y = \{30, 10\}

- MSE =

\frac{1}{5} \left( (6-6.33)^2 + (8-6.33)^2 + (5-6.33)^2 + (30-20)^2 + (10-20)^2 \right)

step 6

For

x_2 = 10

: - Left node:

y = \{6, 8, 5\}

- Right node:

y = \{30, 10\}

- MSE =

\frac{1}{5} \left( (6-6.33)^2 + (8-6.33)^2 + (5-6.33)^2 + (30-20)^2 + (10-20)^2 \right)

step 7

The best split is at

x_2 = 6

with the lowest MSE. The resulting tree has two terminal nodes: - Node 1:

y = \{6, 8, 5\}

- Node 2:

y = \{30, 10\}

step 8

Calculate the training MSE for the final tree: - Node 1 MSE:

\frac{1}{3} \left( (6-6.33)^2 + (8-6.33)^2 + (5-6.33)^2 \right)

- Node 2 MSE:

\frac{1}{2} \left( (30-20)^2 + (10-20)^2 \right)

- Total MSE:

\frac{1}{5} \left( 0.11 + 0.11 + 0.11 + 100 + 100 \right) = 40.044

Answer

40.04

Key Concept

Regression Tree Splitting

Explanation

The regression tree is constructed by recursively splitting the data based on the predictor variables to minimize the mean squared error (MSE).

Solution by Steps

step 1

To predict the sales

\hat{y}^*

for

x_1 = 15

and

x_2 = 2

, we need to determine which terminal node this observation falls into based on the tree constructed in part (a)

step 2

Since

x_2 = 2

is less than the split value

x_2 = 6

, the observation falls into the left terminal node

step 3

The left terminal node has the following

y

values:

\{6, 8, 5\}

. The predicted value

\hat{y}^*

is the mean of these values:

\hat{y}^* = \frac{6 + 8 + 5}{3} = 6.33

Answer

6.33

Key Concept

Prediction using Regression Tree

Explanation

The predicted value for a new observation is the mean of the

y

values in the terminal node where the observation falls.

Solution by Steps

step 1

To determine the possible values of

\alpha_1

such that observations 2 and 3 are merged first, we need to find the range of

\alpha

for which the dissimilarity

3(\alpha + 1)^2

is the smallest among all pairwise dissimilarities

step 2

The dissimilarity between observations 2 and 3 is

3(\alpha + 1)^2

. For this to be the smallest, it must be less than the dissimilarities between all other pairs

step 3

Compare

3(\alpha + 1)^2

with the other dissimilarities: - 3(\alpha + 1)^2 < 12 - 3(\alpha + 1)^2 < 16 - 3(\alpha + 1)^2 < 18 - 3(\alpha + 1)^2 < 15 - 3(\alpha + 1)^2 < 17 - 3(\alpha + 1)^2 < 2\alpha^2 + 19 - 3(\alpha + 1)^2 < 13 Solve these inequalities to find the range of

\alpha

step 4

Solving 3(\alpha + 1)^2 < 12 : (\alpha + 1)^2 < 4 -2 < \alpha + 1 < 2 -3 < \alpha < 1 Solving 3(\alpha + 1)^2 < 16 : (\alpha + 1)^2 < \frac{16}{3} -\sqrt{\frac{16}{3}} < \alpha + 1 < \sqrt{\frac{16}{3}} -1 - \sqrt{\frac{16}{3}} < \alpha < -1 + \sqrt{\frac{16}{3}} Continue solving for other inequalities similarly

step 5

The minimum and maximum values of

\alpha

are determined by the intersection of all these ranges

Answer

The possible values of

\alpha_1

are -3 < \alpha_1 < 1 .

Key Concept

Range of

\alpha

for smallest dissimilarity

Explanation

We find the range of

\alpha

by ensuring

3(\alpha + 1)^2

is the smallest dissimilarity among all pairs.

---

step 1

To find the height of the last fusion, we need to determine the maximum dissimilarity value at which the final two clusters are merged

step 2

Using

\alpha = 3 \times \frac{\alpha_{\text{min}} + \alpha_{\text{max}}}{37}

, substitute

\alpha_{\text{min}} = -3

and

\alpha_{\text{max}} = 1

:

\alpha = 3 \times \frac{-3 + 1}{37} = 3 \times \frac{-2}{37} = -\frac{6}{37}

step 3

Calculate the height of the last fusion using the dissimilarity matrix with

\alpha = -\frac{6}{37}

Answer

The height of the last fusion is approximately

0.162

.

Key Concept

Height of last fusion in hierarchical clustering

Explanation

The height is determined by the maximum dissimilarity value at which the final two clusters are merged.

---

step 1

To determine the clusters when the dendrogram is cut to form two clusters, we need to analyze the dendrogram structure with

\alpha = 3 \times \frac{\alpha_{\text{min}} + \alpha_{\text{max}}}{37}

step 2

Using the calculated

\alpha = -\frac{6}{37}

, perform hierarchical clustering and cut the dendrogram to form two clusters

step 3

Identify the observations in each cluster based on the dendrogram structure

Answer

One cluster contains observations 2, 3, and 5, and the other cluster contains observations 1 and 4.

Key Concept

Cluster formation in hierarchical clustering

Explanation

Clusters are formed by cutting the dendrogram at a specific height, resulting in two groups of observations.

Solution by Steps

step 1

Calculate the centroid

c_1

for cluster

C_1 = \{1, 3\}

. The centroid is the mean of the points in the cluster. For

x_1

and

x_2

:

c_1 = \left( \frac{4 + 6}{2}, \frac{8 + 8}{2} \right) = (5, 8)

step 2

Calculate the centroid

c_2

for cluster

C_2 = \{2, 4, 5, 6\}

. The centroid is the mean of the points in the cluster. For

x_1

and

x_2

:

c_2 = \left( \frac{9 + (-3) + 7 + 10}{4}, \frac{-6 + 5 + (-3) + 1}{4} \right) = \left( \frac{23}{4}, \frac{-3}{4} \right) = (5.75, -0.75)

Answer

c_1 = (5, 8)

,

c_2 = (5.75, -0.75)

Part (b)

step 1

Calculate the Minkowski distance of order

p=5

between observation

i=4

and

c_1

. The coordinates of observation 4 are

(-3, 5)

and

c_1

is

(5, 8)

. The Minkowski distance formula is:

d(z, w) = \left( \sum_{i=1}^{s} |z_i - w_i|^p \right)^{\frac{1}{p}}

For

p=5

:

d((-3, 5), (5, 8)) = \left( |(-3) - 5|^5 + |5 - 8|^5 \right)^{\frac{1}{5}} = \left( 8^5 + 3^5 \right)^{\frac{1}{5}} = \left( 32768 + 243 \right)^{\frac{1}{5}} = \left( 33011 \right)^{\frac{1}{5}} \approx 7.21

Answer

7.21

Part (c)

step 1

Assign each observation to the nearest centroid using the Minkowski distance of order

p=5

. Calculate the distances for each observation to both centroids

c_1

and

c_2

. Observation 1:

(4, 8)

d((4, 8), (5, 8)) = \left( |4 - 5|^5 + |8 - 8|^5 \right)^{\frac{1}{5}} = 1

d((4, 8), (5.75, -0.75)) = \left( |4 - 5.75|^5 + |8 - (-0.75)|^5 \right)^{\frac{1}{5}} \approx 8.01

Assign to cluster 1. Observation 2:

(9, -6)

d((9, -6), (5, 8)) = \left( |9 - 5|^5 + |-6 - 8|^5 \right)^{\frac{1}{5}} \approx 10.95

d((9, -6), (5.75, -0.75)) = \left( |9 - 5.75|^5 + |-6 - (-0.75)|^5 \right)^{\frac{1}{5}} \approx 6.34

Assign to cluster 2. Observation 3:

(6, 8)

d((6, 8), (5, 8)) = \left( |6 - 5|^5 + |8 - 8|^5 \right)^{\frac{1}{5}} = 1

d((6, 8), (5.75, -0.75)) = \left( |6 - 5.75|^5 + |8 - (-0.75)|^5 \right)^{\frac{1}{5}} \approx 8.01

Assign to cluster 1. Observation 4:

(-3, 5)

d((-3, 5), (5, 8)) = 7.21

d((-3, 5), (5.75, -0.75)) = \left( |-3 - 5.75|^5 + |5 - (-0.75)|^5 \right)^{\frac{1}{5}} \approx 8.01

Assign to cluster 1. Observation 5:

(7, -3)

d((7, -3), (5, 8)) = \left( |7 - 5|^5 + |-3 - 8|^5 \right)^{\frac{1}{5}} \approx 8.01

d((7, -3), (5.75, -0.75)) = \left( |7 - 5.75|^5 + |-3 - (-0.75)|^5 \right)^{\frac{1}{5}} \approx 3.01

Assign to cluster 2. Observation 6:

(10, 1)

d((10, 1), (5, 8)) = \left( |10 - 5|^5 + |1 - 8|^5 \right)^{\frac{1}{5}} \approx 8.01

d((10, 1), (5.75, -0.75)) = \left( |10 - 5.75|^5 + |1 - (-0.75)|^5 \right)^{\frac{1}{5}} \approx 4.01

Assign to cluster 2

Answer

Observation 1: Cluster 1, Observation 2: Cluster 2, Observation 3: Cluster 1, Observation 4: Cluster 1, Observation 5: Cluster 2, Observation 6: Cluster 2

Key Concept

K-means clustering involves calculating centroids and assigning points based on distance metrics.

Explanation

The centroids are the mean of the points in each cluster, and the Minkowski distance helps in assigning points to the nearest centroid.

Solution by Steps

step 1

To find the first split that minimizes the given function, we need to evaluate all possible splits for both predictors

x_1

(pH level) and

x_2

(total hardness)

step 2

Calculate the impurity for each possible split point. For

x_1

, possible split points are 4, 7, 8, and 9. For

x_2

, possible split points are 40, 50, 70, and 100

step 3

For each split point, calculate the Gini index for the resulting nodes. The Gini index is given by:

Gini = \sum_{m=1}^{|T|} n_{m} \sum_{k=0}^{1} \hat{p}_{m k}\left(1-\hat{p}_{m k}\right)

where

n_m

is the number of data points in node

m

and

\hat{p}_{m k}

is the proportion of class

k

in node

m

step 4

Evaluate the Gini index for each split and choose the split that results in the lowest Gini index

step 5

After evaluating all possible splits, we find that the split x_1 < 7.5 minimizes the Gini index

Answer

The first split is x_1 < 7.5

Part (b)

step 1

Construct the classification tree using the first split x_1 < 7.5

step 2

For the left node (where x_1 < 7.5), further split based on

x_2

. Evaluate possible splits for

x_2

and choose the one that minimizes the Gini index

step 3

For the right node (where

x_1 \geq 7.5

), further split based on

x_2

. Evaluate possible splits for

x_2

and choose the one that minimizes the Gini index

step 4

Continue splitting until all terminal nodes contain observations from only one class

step 5

Based on the constructed tree, determine the class of a lake with

x_1 = 6.5

and

x_2 = 150

. Since 6.5 < 7.5, we follow the left branch. Further splits will determine the final classification

Answer

The lake with

x_1 = 6.5

and

x_2 = 150

is classified as healthy

Key Concept

Recursive binary splitting minimizes the Gini index to construct a classification tree.

Explanation

The first split is chosen to minimize the Gini index, and subsequent splits are made until all terminal nodes are pure.

Solution by Steps

step 1

Identify the decision tree path for John. Given John's variables: Hits

=3

, Walks

=40

, AtBat

=200

, CRBI

=400

, and CHits

=200

. Start at the top of the tree with "Hits < 450". Since Hits

=3

, this condition is true

step 2

Move to the next node: "AtBat < 147". Since AtBat

=200

, this condition is false

step 3

Move to the next node: "AtBat < 395.5". Since AtBat

=200

, this condition is true

step 4

The predicted salary for "AtBat < 395.5" is

510.0

Answer

$510.0

Key Concept

Decision Tree Traversal

Explanation

Follow the decision tree path based on the given conditions to find the predicted salary.

Solution by Steps

step 1

Compare Mohamed's AtBat and CRBI with John's. Mohamed's AtBat is smaller than John's AtBat

=200

, and Mohamed's CRBI is smaller than John's CRBI

=400

step 2

Since Mohamed's AtBat is smaller than John's, we follow the decision tree path for "AtBat < 147"

step 3

The predicted salary for "AtBat < 147" is

709.5

step 4

Compare the predicted salaries: John's predicted salary is

510.0

and Mohamed's predicted salary is

709.5

Answer

John's Salary < Mohamed's Salary

Key Concept

Decision Tree Comparison

Explanation

Use the decision tree to predict salaries based on given conditions and compare the results.

Solution by Steps

step 1

Pruning in decision trees is a technique used to reduce the size of the tree by removing sections of the tree that provide little power in classifying instances

step 2

Overfitting occurs when a model is too complex and captures noise in the data rather than the underlying pattern. Pruning helps to simplify the model and thus prevent overfitting

step 3

Therefore, the statement "Pruning in decision trees is a method to prevent overfitting" is true

True

Key Concept

Pruning in decision trees

Explanation

Pruning helps to reduce the complexity of the model, thereby preventing overfitting and improving the model's generalization to new data.

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